diff --git a/.github/workflows/pretext-cli.yml b/.github/workflows/pretext-cli.yml index 4e97ede3..bc62df49 100644 --- a/.github/workflows/pretext-cli.yml +++ b/.github/workflows/pretext-cli.yml @@ -1,7 +1,4 @@ -# -# (delete the above line to manage this file manually) - -name: PreTeXt-CLI Actions +name: PreTeXt-CLI Actions (TBIL Customization) on: # Runs on pull requests pull_request: @@ -24,8 +21,17 @@ jobs: - name: install deps run: pip install -r requirements.txt + - name: quick build instructor versions + if: github.ref != format('refs/heads/{0}', github.event.repository.default_branch) + run: | + pretext build precal-web-instructor -q + pretext build calc-web-instructor -q + pretext build la-web-instructor -q + - name: build deploy targets + if: github.ref == format('refs/heads/{0}', github.event.repository.default_branch) || github.event_name == 'workflow_dispatch' run: pretext build --deploys + - name: stage deployment run: pretext deploy --stage-only diff --git a/.gitignore b/.gitignore index 728b01fc..66464048 100644 --- a/.gitignore +++ b/.gitignore @@ -1,4 +1,4 @@ -# + # (delete the above line to manage this file manually) # Boilerplate list of files in a PreTeXt project for git to ignore @@ -97,7 +97,3 @@ GitHub.sublime-settings # Don't track codechat config (will be generated automatically) codechat_config.yaml - -# Don't track deprecated workflows -.github/workflows/deploy.yml -.github/workflows/test-build.yml diff --git a/calculus/assets/3_4_PA1.pdf b/calculus/assets/3_4_PA1.pdf new file mode 100644 index 00000000..8f84a28e Binary files /dev/null and b/calculus/assets/3_4_PA1.pdf differ diff --git a/calculus/assets/3_4_PA1.svg b/calculus/assets/3_4_PA1.svg new file mode 100644 index 00000000..5afc674f --- /dev/null +++ b/calculus/assets/3_4_PA1.svg @@ -0,0 +1,13 @@ + + + + + + + + + + + + + diff --git a/calculus/assets/4_1_Act1.pdf b/calculus/assets/4_1_Act1.pdf new file mode 100644 index 00000000..5a26fbfa Binary files /dev/null and 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mode 100644 index 00000000..f6085f7c --- /dev/null +++ b/calculus/source/01-LT/01.ptx @@ -0,0 +1,323 @@ + + +
+ Limits graphically (LT1) + + + + + Activities + + +

+ In the graph of a function is given, but something is wrong. The graphic card failed and one portion did not render properly. We can’t see what is happening in the neighborhood of x=2. +

+
+ + A graph of a function that has not been rendered properly. +
+ + + +

+ Imagine moving along the graph toward the missing portion from the left, so that you are climbing up and to the right toward the obscured area of the graph. What y-value are you approaching? +

+
    +

  1. 0.5

    2. +

  2. 1

    3. +

  3. 1.5

    4. +

  4. 2

    5. +

  5. 2.5

    6. +
+ + +

+ Think of the same process, but this time from the right. You're falling down and to the left this time as you come close to the missing portion. What y-value are you approaching? +

+
    +

  1. 0.5

    2. +

  2. 1

    3. +

  3. 1.5

    4. +

  4. 2

    5. +

  5. 2.5

    6. +
+ + + + +

+ In the graphic card is working again and we can see more clearly what is happening in the neighborhood of x=2. +

+
+ + A graph of a function that has rendered properly +
+ + + +

+ What is the value of f(2)? +

+ + +

+ What is the y-value that is approached as we move toward x = 2 from the left? +

+
    +

  1. 0.5

    2. +

  2. 1

    3. +

  3. 1.5

    4. +

  4. 2

    5. +

  5. 2.5

    6. +
+ + +

+ What is the y-value that is approached as we move toward x = 2 from the right? +

+
    +

  1. 0.5

    2. +

  2. 1

    3. +

  3. 1.5

    4. +

  4. 2

    5. +

  5. 2.5

    6. +
+ + + + +

+ When studying functions in algebra, we often focused on the value of a function given a specific x-value. For instance, finding f(2) for some function f(x). In calculus, and here in and , we have instead been exploring what is happening as we approach a certain value on a graph. This concept in mathematics is known as finding a limit. +

+ + + + +

Based on and , write your first draft of the definition of a limit. What is important to include? (You can use concepts of limits from your daily life to motivate or define what a limit is.) +

+ + + + + +

+ Given a function f, + a fixed input x = a, and a real number L, + we say that f has limit L as x approaches a, + limitdefinition + and write + + \lim_{x \to a} f(x) = L + + provided that we can make f(x) as close to L as we like by taking x sufficiently close + (but not equal) + to a. + If we cannot make f(x) as close to a single value as we would like as x approaches a, + then we say that f does not have a limit as x approaches a. +

+ + + + + + + +

+ +

+
+ + A piecewise-defined function +
+ + + +

+ What is the limit as x approaches 0 in ? +

+
    +

  1. The limit is 1

    2. +

  2. The limit is -1

    3. +

  3. The limit is 0

    4. +

  4. The limit is not defined

    5. +
+ + + + + +

+ We say that f has limit L_1 as x approaches a from the left and write + + \lim_{x \to a^-} f(x) = L_1 + + provided that we can make the value of f(x) as close to L_1 as we like by taking x sufficiently close to a while always having x \lt a. + We call L_1 the left-hand limit of f as x approaches a. + Similarly, we say L_2 is the right-hand limit of f as x approaches a and write + + \lim_{x \to a^+} f(x) = L_2 + + provided that we can make the value of f(x) as close to L_2 as we like by taking x sufficiently close to a while always having x \gt a. +

+ + + + + +

+ Refer again to from . +

+ + + + +

+ Which of the following best matches the definition of right and left limits? (Note that DNE is short for "does not exist.") +

+
    +

  1. The left limit is -1. The right limit is 1.

    2. +

  2. The left limit is 1. The right limit is -1.

    3. +

  3. The left limit DNE. The right limit is 1.

    4. +

  4. The left limit is -1. The right limit DNE.

    5. +

  5. The left limit DNE. The right limit DNE.

    6. +
+ + +

+ What do you think the overall limit equals? +

+
    +

  1. The limit is 1

    2. +

  2. The limit is -1

    3. +

  3. The limit is 0

    4. +

  4. The limit is not defined

    5. +
+ + + + + +

+ Consider the following graph: +

+
+ + Another piecewise-defined function +
+ + + + +

+ Find \displaystyle \lim_{x \to -3^-} f(x) and \displaystyle \lim_{x \to -3^+} f(x). +

+ + +

+ Find \displaystyle \lim_{x \to -1^-} f(x) and \displaystyle \lim_{x \to -1^+} f(x). +

+ + +

+ Find \displaystyle \lim_{x \to 2^-} f(x) and \displaystyle \lim_{x \to 2^+} f(x). +

+ + +

+ Find \displaystyle \lim_{x \to 4^-} f(x) and \displaystyle \lim_{x \to 4^+} f(x). +

+ + +

+ For which x-values does the overall limit exist? Select all. If the limit exists, find it. If it does not, explain why. +

+
    +
  1. -3
    2. +
  2. -1
    3. +
  3. 2
    4. +
  4. 4
    5. +
+ + + + + +

Sketch the graph of a function f(x) that meets all of the following criteria. Be sure to scale your axes and label any important features of your graph.

+
    + +

  1. \displaystyle \lim_{x\to 5^-} f(x) is finite, but \displaystyle \lim_{x\to 5^+} f(x) is infinite.

    2. + + + + +

  2. \displaystyle \lim_{x\to -3} f(x)=-4, but f(-3)=0.

    3. + + + + +

  3. \displaystyle \lim_{x\to -1^-} f(x)=-1 but \displaystyle \lim_{x\to -1^+} f(x)\neq-1.

    4. +
+ + + + + +

Suppose that:

+
    +

  • for an interval around x=a, we have that f(x) \leq g(x) \leq h(x) ;

    • +

  • the limit as x approaches a of f(x) is equal to the same value L as the limit of h(x), so \displaystyle\lim_{x\to a} f(x) = L =\lim_{x\to a} h(x) .

    • +
+

Then, the limit of g(x) as x \to a is also L, so \displaystyle\lim_{x\to a} g(x) = L .

+ + + +

In this activity we will explore a mathematical theorem, the Squeeze Theorem .

+ + + +

The part of the theorem that starts with “Suppose…” forms the assumptions of the theorem, while the part of the theorem that starts with “Then…” is the conclusion of the theorem. What are the assumptions of the Squeeze Theorem? What is the conclusion? +

+ + + + +

The assumptions of the Squeeze Theorem can be restated informally as “the function g is squeezed between the functions f and h around a.” Explain in your own words how the two assumptions result into a “squeezing effect.” +

+ + + + +

Let’s see an example of the application of this theorem. First examine the following picture. Explain why, from the picture, it seems that both assumptions of the theorem hold.

+
+ + A pictorial example of the Squeeze Theorem. +
+ + + + +

Match the functions f(x), g(x), h(x) in the picture to the functions \cos(x), 1, \frac{\sin(x)}{x}.

+ + + + +

Using trigonometry, one can show algebraically that \cos(x) \leq \frac{\sin(x)}{x} \leq 1 for x values close to zero. Moreover, \displaystyle\lim_{x \to 0} \cos(x) = \cos(0)=1 (we say that cosine is a continuous function). Use these facts and the Squeeze Theorem, to find the limit \displaystyle\lim_{x\to 0} \frac{\sin(x)}{x} .

+ + + + + + + Videos +
+
+ + +
+ diff --git a/calculus/source/01-LT/02.ptx b/calculus/source/01-LT/02.ptx new file mode 100644 index 00000000..c2f9ced7 --- /dev/null +++ b/calculus/source/01-LT/02.ptx @@ -0,0 +1,480 @@ + + +
+ Limits numerically (LT2) + + + + + Activities + + + + + + + x + + + 6.9 + + + 6.99 + + + 6.999 + + + 7 + + + 7.001 + + + 7.01 + + + 7.1 + + + + + f(x) + + + 0.1695 + + + 0.1699 + + + 0.1667 + + + \unknown + + + 0.1667 + + + 0.1664 + + + 0.1639 + + + +
+

+ Based on the values of , what is the best approximation for \displaystyle\lim_{x\to 7} f(x)? +

    +

  1. the limit is approximately 7

    2. +

  2. the limit is approximately 0.17

    3. +

  3. the limit is approximately 0.16

    4. +

  4. the limit is approximately 0.1667

    5. +

  5. the limit is approximately 6.9999

    6. +
+

+ + + +

+ Notice that the value we obtained in is only an approximation, based on the trends that we have seen within the table. +

+ + + + + + + + x + + + 1.25 + + + 1.5 + + + 1.75 + + + 2 + + + 2.25 + + + 2.5 + + + 2.75 + + + + + f(x) + + + -0.7606 + + + -0.13 + + + 0.4881 + + + \unknown + + + 1.3119 + + + 1.33 + + + 0.9606 + + + +
+

+In 's we found an approximation to the limit of the function as x tends to 2. +Now let us say you are also given a table of numerical values +() for the function. Given this new information which of the choices below best describes the limit of +the function as x tends to 2? +

+
    +

  1. There is not enough information because we do not know the value of the function at x = 2.

    2. +

  2. The limit can be approximated to be 1 because the data in the table and the graph show that from the left and the right the function approaches 1 as x goes to 2.

    3. +

  3. The limit can be approximated to be 1 because the values appear to approach 1 and the graph appears to approach 1, but we should zoom in on the graph to be sure.

    4. +

  4. The limit cannot be approximated because the function might not exist at x = 2.

    5. +
+ + + + + + + + x + + + 0.9 + + + 0.99 + + + 0.999 + + + 1 + + + 1.001 + + + 1.01 + + + 1.1 + + + + + f(x) + + + -0.4 + + + -0.49 + + + -0.499 + + + \unknown + + + 0.499 + + + 0.49 + + + 0.4 + + + +
+

Based on , what information can be inferred about \displaystyle \lim_{x\to 1^-}f(x), \displaystyle \lim_{x\to 1^+}f(x), and \displaystyle \lim_{x\to 1}f(x)? +

+
    +

  1. \displaystyle \lim_{x\to 1^-}f(x) = -0.5,\,\, \lim_{x\rightarrow 1^+}f(x) = 0.5, and \displaystyle\lim_{x\rightarrow 1}f(x) = 0

    2. +

  2. \displaystyle \lim_{x\to 1^-}f(x) = -0.5,\,\, \lim_{x\rightarrow 1^+}f(x) = 0.5, and \displaystyle\lim_{x\rightarrow 1}f(x) does not exist

    3. +

  3. \displaystyle \lim_{x\rightarrow 1^-}f(x) = 0.5,\,\, \lim_{x\rightarrow 1^+}f(x) = -0.5, and \displaystyle\lim_{x\rightarrow 1}f(x) does not exist

    4. +

  4. \displaystyle \lim_{x\rightarrow 1^-}f(x) = 0.5,\,\, \lim_{x\rightarrow 1^+}f(x) = -0.5, and \displaystyle\lim_{x\rightarrow 1}f(x) = 0

    5. +
+ + + + + +

Consider the following function f(x)= 3x^3 + 2x^2- 5x+20.

+ + +

Of the following options, at which values given would you evaluate f(x) to best determine \displaystyle \lim_{x \to 2}f(x) numerically?

+ +
    +

  1. 1.9, 1.99, 2.0, 2.01, 2.1

    2. +

  2. 1.98, 1.99, 2.0, 2.01, 2.02

    3. +

  3. 1.8, 1.9, 2.0, 2.1, 2.2

    4. +

  4. 1.0, 1.5, 2.0, 2.5, 3.0

    5. +
+ + + + +

Use the values that you chose in part (a) to calculate an approximation for \displaystyle\lim_{x \to 2}f(x).

+ + +

Which value best describes the limit that you obtained in part (b)?

+ +
    +

  1. The approximate value is 41.25

    2. +

  2. The approximate value is 41.5

    3. +

  3. The approximate value is 41.75

    4. +

  4. The approximate value is 42

    5. +
+ + + + + + + +

In is the graph for f(x) = \sin\left(\frac{1}{x}\right). Several values for f(x) in the neighborhood of x = 0 are approximated in .

+
+ + Graph of f(x) = \sin(1/x). +
+ + + + + x + + + -0.1 + + + -0.01 + + + -0.001 + + + 0 + + + 0.001 + + + 0.01 + + + 0.1 + + + + + f(x) + + + 0.54402 + + + 0.50637 + + + -0.82688 + + + \unknown + + + 0.82688 + + + -0.50637 + + + -0.54402 + + + +
+ +

Based on the graph and table what is the best explanation for the limit as x tends to zero?

+
    +

  1. The limit does not exist because the left and right limits have opposite values.

    2. +

  2. The limit does not exist because we do not have enough information to answer the question.

    3. +

  3. The limit does not exist because the function is oscillating between -1 and 1.

    4. +

  4. The limit does not exist because you are dividing by zero when x = 0 for f(x).

    5. + +
+ + +

Would your conclusion that resulted from change if the function was f(x) = \cos(1/x) or f(x) = \tan(1/x)?

+ + + + + +

+Use technology to complete the following table of values. +f(x)=\frac{ x^{2} - x - 12 }{ x^{2} + 16 \, x + 39 } + +\begin{array}{c|ccccccc} +x & -3.1& -3.01& -3.001& -3& -2.999& -2.99& -2.9 \\\hline +f(x) & & & & & & & \\ +\end{array} + +

+

+Then explain how to use it to make an educated guess as to the value of +the limit \displaystyle \lim_{ x\to -3 } \frac{ x^{2} - x - 12 }{ x^{2} + 16 \, x + 39 } +

+ + + + + + + +

In this activity you will study the velocity of Usain Bolt in his Beijing 100 meters + dash. + He completed 100 meters in 9.69 seconds for an overall average speed of 100/9.69 = 10.32 meters per second (about 23 miles per hour). + But this is the average velocity on the whole interval. How fast was he at different instances? What was his maximum velocity? + Let's explore this. The table shows his split times recorded every 10 meters.

+ + + + + t\, \text{(seconds)} + + + 1.85 + + + 2.87 + + + 3.78 + + + 4.65 + + + 5.5 + + + 6.32 + + + 7.14 + + + 7.96 + + + 8.79 + + + 9.69 + + + + + d \, \text{(meters)} + + + 10 + + + 20 + + + 30 + + + 40 + + + 50 + + + 60 + + + 70 + + + 80 + + + 90 + + + 100 + + + +
+ +

What was the average velocity on the first 50 meters? On the second 50 meters?

+ +

What was the average velocity between 30 and 50 meters? Between 50 and 70 meters?

+ +

What was the average velocity between 40 and 50 meters? Between 50 and 60 meters?

+ +

What is your best estimate for the Usain's velocity at the instant when he passed the 50 meters mark? This is your estimate for the instantaneous velocity.

+ +

Using the table of values, explain why 50 meters is NOT the best guess for when the instantaneous velocity was the largest. What other point would be more reasonable?

+ + + + + + + Videos +
+
+ + +
diff --git a/calculus/source/01-LT/03.ptx b/calculus/source/01-LT/03.ptx new file mode 100644 index 00000000..5c9dc1c7 --- /dev/null +++ b/calculus/source/01-LT/03.ptx @@ -0,0 +1,407 @@ + + +
+ Limits analytically (LT3) + + + + + Activities + + + +

Recall that in we used numerical methods and table of values to find the limit of a relatively simple degree three polynomial at a point. This was inefficient, “there’s gotta be a better way!”

+ + + + +

Given f(x)=3x^2-\frac{1}{2}x+4, evaluate f(2) and approximate \displaystyle \lim_{x\to2}f(x) numerically (or graphically). What do you think is more likely?

+
    +
  1. \displaystyle \lim_{x \to 2}f(x)=f(2)
    2. +
  2. \displaystyle \lim_{x \to 2}f(x) \approx f(2)
    3. +
  3. \displaystyle \lim_{x \to 2}f(x) \neq f(2)
    4. +
+ + + + +

The table below gives values of a few different functions.

+ + + + + x + + + 6.99 + + + 6.999 + + + 7.001 + + + 7.01 + + + + + f(x) + + + 13.99 + + + 13.999 + + + 14.001 + + + 14.01 + + + + + g(x) + + + 22.97 + + + 22.997 + + + 23.003 + + + 23.03 + + + + + 3f(x) + + + 41.97 + + + 41.997 + + + 42.003 + + + 42.03 + + + + + f(x)+g(x) + + + 36.96 + + + 36.996 + + + 37.004 + + + 37.04 + + + + + f(x)g(x) + + + 321.350 + + + 321.935 + + + 322.065 + + + 322.650 + + + +
+

Using the table above, which of the following is least likely to be true?

 +
    +
  1. \displaystyle \lim_{x\to 7}f(x)= 14 and \displaystyle \lim_{x\to 7}g(x)= 23
    2. +
  2. \displaystyle \lim_{x \to 7}3f(x) = 3 \lim_{x\to 7}f(x)
    3. +
  3. \displaystyle \lim_{x\to 7}\left( f(x)+g(x) \right) = \lim_{x\to 7}f(x) + \lim_{x\to 7}g(x)
    4. +
  4. \displaystyle \lim_{x\to 7}\left( f(x)g(x) \right) =f(7) \left( \lim_{x\to 7}g(x) \right)
    5. +
+ + +

In we observed that limits seem to be "well-behaved" when combined with standard operations on functions. The next theorems, known as Limit Laws, tell us how limits interact with combinations of functions.

+ + Limit Laws, I + +

+ Let f and g be functions defined on an open interval I containing the real number c satisfying + + \lim_{x \to c} f(x) = L \, \text{and} \, \lim_{x \to c} g(x) = K, + + for L and K some real numbers. Then we have the following limits. +

    +
  1. Constant Law: \displaystyle \lim_{x \to c} b = b, for b any constant real number;
    2. +
  2. Identity Law: \displaystyle \lim_{x \to c} x = c ;
    3. +
  3. Sum/Difference Law: \displaystyle \lim_{x \to c} (f(x) \pm g(x)) = L \pm K;
    4. +
  4. Scalar Multiple Law: \displaystyle \lim_{x \to c} b \cdot f(x) = bL, for b any constant real number;
    5. +
  5. Product Law: \displaystyle \lim_{x \to c} f(x) \cdot g(x) = LK;
    6. +
  6. Quotient Law: if K \ne 0, then \displaystyle \lim_{x \to c} f(x) / g(x) = L/K.
    7. +
+

+ + + + + +

If \displaystyle \lim_{x\to 2} f(x) = 2 and \displaystyle \lim_{x\to 2} g(x) = -3, which of the following statements are true? Select all that apply!

+
    +
  1. \displaystyle \lim_{x\to 2} (f(x) \cdot g(x)) = -6
    2. +
  2. \displaystyle \lim_{x\to 2} (f(x) + g(x)) = -1
    3. +
  3. \displaystyle \lim_{x\to 2} (f(x) - g(x)) = -2
    4. +
  4. \displaystyle \lim_{x\to 2} (f(x)/g(x)) = -2/3
    5. +
+ + + + + + Limit Laws, II + +

Let f and g be functions defined on an open interval I containing c satisfying + + \lim_{x \to c} f(x) = L, \, \, \, \, \lim_{x \to L} g(x) = K, \text{and} \, g(L)=K. + Then we have the following limits as well. +

+
    +
  1. Power Law: \displaystyle \lim_{x \to c} f(x)^n = L^n, for n a positive integer;
    2. +
  2. Root Law: \displaystyle \lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{L}, for n a positive integer;
    3. +
  3. Composition Law: \displaystyle \lim_{x \to c} g(f(x)) = K.
    4. +
+ + + + + +

Below you are given the graphs of two functions. Compute the limits below (if possible). +

+ +
+ + + \begin{tikzpicture}[scale=1] +\begin{axis}[ %human, + axis lines=middle, + grid=major, + xmin=-2, xmax=8.5, + ymin=-3, ymax=3, + xtick={-2,-1,...,8}, + ytick={-4,-3,...,4}, + tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel=$y$, + xlabel=$x$, + ] + \addplot[domain=-2:1, blue, ultra thick] {(1/3)*(x)-1/3}; + \addplot[domain=1:5, blue, ultra thick] {-0.5*x +0.5}; + \addplot[domain=5:7, blue, ultra thick] {x-5}; + \addplot[mark=none, color=blue, nodes near coords={$f(x)$}] coordinates {(2,1)}; +\addplot [only marks, blue] table { +5 0 \\ +1 2 \\ +8 2 \\ +}; + + \addplot[only marks, color=blue, mark=*, fill=white] coordinates{(1,0) }; + + \addplot[only marks, color=blue, mark=*, fill=white] coordinates {(5,-2) }; + + \addplot[only marks, color=blue, mark=*, fill=white] coordinates {(7,2) }; + + +\end{axis} +\end{tikzpicture} + + + The graph of f(x). +
+
+ + +\begin{tikzpicture}[scale=1] + +\begin{axis}[ axis lines=middle, %human, + grid=major, grid,xtick={-3,-2,...,3}, ytick={-2,-1,...,4}, ymin=-2.2,xmin=-3.5,ymax=4.5,xmax=3.5, ylabel={}, xlabel={$x$}] +\addplot[domain=-3:0, color=blue, samples=100]{0.5*x}; +\addplot[domain=0:2, color=blue, samples=100]{2^x}; +\addplot[domain=2:4, color=blue, samples=50]{0.5*x+3}; + + \addplot[only marks, color=blue, mark=*, fill=blue] coordinates{(0,1) }; + + \addplot[only marks, color=blue, mark=*, fill=white] coordinates {(0,0) }; + + + \addplot[mark=none, color=blue, nodes near coords={$g(x)$}] coordinates {(2,1)}; +\end{axis} +\end{tikzpicture} + + + The graph of g(x). +
+ + +

\displaystyle \lim_{x \to 1} f(x) + g(x) .

+ +

\displaystyle \lim_{x \to 5^+} 3f(x) .

+ +

\displaystyle \lim_{x \to 0^+ } f(x)g(x) .

+ +

(Challenge) \displaystyle \lim_{x \to 1} g(x) / f(x) .

+ +

(Challenge) \displaystyle \lim_{x \to 0^+} f(g(x)) .

+ + + + + +

Given p(x) = -3x^2 - 5x+7, which of the following limit laws would use to determine \displaystyle \lim_{x\to 2} p(x)? Choose all that apply.

+
    +
  1. Sums/Difference Law
    2. +
  2. Scalar Multiple Law
    3. +
  3. Product Law
    4. +
  4. Identity Law
    5. +
  5. Power Law
    6. +
  6. Constant Law
    7. +
+ + + + + + + Limits of Polynomials + +

If p(x) is a polynomial and c is a real number, then \displaystyle \lim_{x \to c} p(x) = p(c). This is also known as the Direct Substitution PropertyDirect Substitution Property for polynomials. +

+ + + + + +

Given p(x)=-3x^2-5x+7 and q(x)=x^4-x^2+3, which of the following describes the most efficient way to determine \displaystyle \lim_{x \to -1} \frac{p(x)}{q(x)}? +

+
    +
  1. Sums/difference, scalar multiple, and product laws
    2. +
  2. and the quotient law
    3. +
  3. Power, sums/difference, scalar multiple, and constant laws
    4. +
  4. Quotient and root law
    5. + +
+ + + + + Limits of Rational Functions + +

If p(x) and q(x) are polynomials, c is a real number, and q(c) \neq 0 then \displaystyle \lim_{x \to c} \frac{p(x)}{q(x)}=\frac{p(c)}{q(c)}.

+ + + + +

Consider taking the limit of a rational function \frac{p(x)}{q(x)} as x \to c. If q(c)=0, is it possible for \displaystyle\lim_{x \to c}\frac{p(x)}{q(x)} to equal a number?

+
    +
  1. No, because \frac{p(x)}{q(x)} is not defined at x=c since q(c)=0.
    2. +
  2. Yes, because if you graph f(x)=\frac{x^2-1}{x-1}, the value f(1) is not defined, but the graph shows that the limit of f(x) does exist as x \to 1.
    3. +
  3. No, because if you graph g(x)=\frac{x^2+1}{x-1}, the value g(1) is not defined and the graph shows that the limit of \displaystyle\lim_{x \to c}g(x) does not exist.
    4. +
  4. Yes, because we can use .
    5. +
+ + + + + + +

Let f(x) = 2x and g(x) = x, which of the following statements is true?

+
    +
  1. \displaystyle \lim_{x\to 0} (f(x)/g(x)) = 0
    2. +
  2. \displaystyle \lim_{x\to 0} (f(x)/g(x)) = 2
    3. +
  3. \displaystyle \lim_{x\to 0} (f(x)/g(x)) cannot be determined
    4. +
  4. \displaystyle \lim_{x\to 0} (f(x)/g(x)) does not exist
    5. +
+ + + + +

When we compute the limit of a ratio where both the numerator and denominator have limit equal to zero, we have to compute the value of a \frac{0}{0} indeterminate form. The value of an indeteminate form can be any real number or even infinity or not existent, we just do not know yet! We can usually determine the value of an indeterminate form using some algebraic manipulations of the expression given.

+ + +

A function f(x) has a holehole at x=c if f(c) does not exist but \displaystyle \lim_{x \to c} f(x) does exist and is equal to a real number.

+ +

The function f(x)=\frac{x^2-1}{x-1} has a hole at x=1 because f(1) is not defined but + \lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2 , + so the limit exists and is equal to a real number. Notice that \displaystyle \lim_{x \to 1} \frac{x^2-1}{x-1} is also an example of a limit giving an indeterminate form \frac{0}{0} which we could then compute using an algebraic manipulation of the function given. +

+ + + +

+ Determine the following limits and explain your reasoning. +

+ + \lim_{x\to-6 } \frac{ x^{2} - 6 \, x + 5 }{ x^{2} - 3 \, x - 18 } + \lim_{x\to-1 } \frac{ x^{2} - 1 }{ x^{2} + 3 \, x + 2 } + \lim_{x\to5 } \frac{ x - 5 }{ \sqrt{x + 31} - 6 } + + + + + + + + +

In activity you studied the velocity of Usain Bolt in his Beijing 100 meters dash. We will now study this situation analytically. To make our computations simpler, we will approximate that he could run 100 meters in 10 seconds and we will consider the model d=f(t)=t^2, where d is the distance in meters and t is the time in seconds.

+ + +

The average velocity is the ratio distance covered over time elapsed. If we consider the interval that starts at t=a and has width h, written [a,a+h], the average velocity on this interval is \displaystyle \frac{f(a+h)-f(a)}{(a+h) - a} = \frac{f(a+h)-f(a)}{h}. The instantaneous velocity at time t=a is given by:

+ \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}. + +

Compute the average velocity on the interval [5,6]. We think of this interval as [5,5+h] for the value of h=1.

+ +

Compute the average velocity starting at 5 seconds, but now with h=0.5 seconds.

+ +

We want to study the instantaneous velocity at a=5 seconds. Find an expression for the average velocity on the interval [5,5+h], where h is an unspecified value.

+ +

Expand your expression. When h \neq 0, you can simplify it!

+ +

Recall that the instantaneous velocity is the limit of your expression as h\to 0. Find the instantaneous velocity given by this model at t=5 seconds.

+ +

The model d=f(t)=t^2 does not really capture the real-world situation. Think of at least one reason why this model does not fit the scenario of Usain Bolt's 100 meters dash.

+ + + + + + Videos +
+
+ + +
diff --git a/calculus/source/01-LT/04.ptx b/calculus/source/01-LT/04.ptx new file mode 100644 index 00000000..b790f3c8 --- /dev/null +++ b/calculus/source/01-LT/04.ptx @@ -0,0 +1,432 @@ + + +
+ Continuity (LT4) + + + + + Activities + + +

A continuous function is one whose values change smoothly, with no jumps or gaps in the graph. We'll explore the idea first, and arrive at a mathematical definition soon.

+ + + + +

+ Which of the following scenarios best describes a continuous function? +

+
    +
  1. +

    + The age of a person reported in years +

    +
    2. +
  2. +

    + The price of postage for a parcel depending on its weight +

    +
    3. +
  3. +

    + The volume of water in a tank that is gradually filled over time +

    +
    4. +
  4. +

    + The number of likes on my latest TikTok depending on the time since I posted it +

    +
    5. +
+ + + + + + +

+ How would you use the language of limits to clarify the definition of continuity? +

+ + + + +

+ A function f defined on -4 \lt x \lt 4 has the graph pictured below. Use the graph to answer each of the following questions. +

+
+ + I need help with alt text for this that doesn't give the answers away. + +
+ + +

+ For each of the values a = -3, -2, -1, 0, 1, 2, 3, determine whether the limit \displaystyle\lim_{x \to a} f(x) exists. If the limit does not exist, be ready to explain why not. +

+ + +

+ For each of the values of a where the limit of f exists, determine the value of f(a) at each such point. +

+ + +

+ For each such a value, is f(a) equal to \displaystyle\lim_{x \to a} f(x)? +

+ + +

+ Use your understanding of continuity to determine whether f is continuous at each value of a. +

+ + +

+ Any revisions would you want to make to your definition of continuity that you arrived at toward the end of ? +

+ + + + + +

+ A function f is continuouscontinuous function at x = a provided that +

    +
  1. +

    + f has a limit as x \to a +

    +
    2. +
  2. +

    + f is defined at x = a (equivalently, a is in the domain of f), and +

    +
    3. +
  3. +

    + \displaystyle\lim_{x \to a} f(x) = f(a). +

    +
    4. +
+

+ + + + + +

+ Suppose that some function h(x) is continuous at x = -3. Use to decide which of the following quantities are equal to each other. +

+
    +
  1. +

    + \displaystyle\lim_{x \to -3^+} h(x) +

    +
    2. +
  2. +

    + \displaystyle\lim_{x \to -3^-} h(x) +

    +
    3. +
  3. +

    + \displaystyle\lim_{x \to -3} h(x) +

    +
    4. +
  4. +

    + h(-3) +

    +
    5. +
+ + + + + +

+ Consider the function f whose graph is pictured below (it's the same graph from ). In the + questions below, consider the values a = -3, -2, -1, 0, 1, 2, 3. +

+
+ + (for accessibility) + +
+ + +

+ For which values of a do we have \displaystyle\lim_{x \to a^-} f(x) \ne \lim_{x \to a^+} f(x)? +

+ + +

+ For which values of a is f(a) not defined? +

+ + +

+ For which values of a does f have a limit at a, yet f(a) \ne \lim_{x \to a} f(x)? +

+ + +

+ For which values of a does f fail to be continuous? Give a complete list of intervals on which f is continuous. +

+ + + + + +

+ Which condition is stronger, meaning it implies the other? +

+
    +
  1. +

    + f has a limit at x = a +

    +
    2. +
  2. +

    + f is continuous at x = a +

    +
    3. +
+ + + + + +

+ Previously, you have used graphs, tables, and formulas to answer questions about limits. Which of those are suitable for answering questions about continuity? +

+
    +
  1. +

    + Graphs only +

    +
    2. +
  2. +

    + Formulas only +

    +
    3. +
  3. +

    + Graphs and formulas only +

    +
    4. +
  4. +

    + Tables and formulas only +

    +
    5. +
+ + + + + +

+ Consider the function f whose graph is pictured below. +

+
+ + + \begin{tikzpicture}[scale=1] +\begin{axis}[ %human, + axis lines=middle, + grid=major, + xmin=-2, xmax=8.5, + ymin=-3, ymax=3, + xtick={-2,-1,...,8}, + ytick={-4,-3,...,4}, + tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel=$y$, + xlabel=$x$, + ] + \addplot[domain=-2:1, blue, ultra thick] {(1/3)*(x)-1/3}; + \addplot[domain=1:5, blue, ultra thick] {-0.5*x +0.5}; + \addplot[domain=5:7, blue, ultra thick] {x-5}; + \addplot[mark=none, color=blue, nodes near coords={$f(x)$}] coordinates {(2,1)}; +\addplot [only marks, blue] table { +5 0 \\ +1 2 \\ +8 2 \\ +}; + + \addplot[only marks, color=blue, mark=*, fill=white] coordinates{(1,0) }; + + \addplot[only marks, color=blue, mark=*, fill=white] coordinates {(5,-2) }; + + \addplot[only marks, color=blue, mark=*, fill=white] coordinates {(7,2) }; + + +\end{axis} +\end{tikzpicture} + + + The graph of f(x). +
+

+ Give a list of x-values where f(x) is not continuous. Be prepared to defend your answer based on . +

+ + + + + +

+ When \displaystyle\lim_{x \to a} f(x) exists but is not equal to f(a), we say that f has a removable discontinuityremovable discontinuity at x = a. This is because if f(a) were redefined to be equal to \displaystyle\lim_{x \to a} f(x), the redefined function would be continuous at x = a, thus removing the discontinuity. +

+

+ When the left and right limit exist separately, but are not equal, the discontinuity is not removable and is called a jump discontinuityjump discontinuity. +

+ + + + + + + + +

Determine the value of b to make h(x) continuous at x=5.

+

+ + h(x) = \begin{cases} + b - x, & x < 5 \\ + -x^{2} + 6 \, x - 6, & x \geq 5 + \end{cases} + +

+ + +

+Classify the type of discontinuity present at x=-6 for +the function f(x).

+

+ + f(x) = \begin{cases} + -8 \, x - 46, & x < -6 \\ + 6, & x = -6 \\ + 4 \, x + 30, & x > -6 \\ + \end{cases} + +

+ + + + + + + + + +

+ If f and g are continuous at x = a and c is a real number, then the functions f + g, f - g, cf, and fg are also continuous at x = a. Moreover, f/g is continuous at x = a provided that g(a) \ne 0. +

+ + + + + +

+ Answer the questions below about piecewise functions. It may be helpful to look at some graphs. +

+ + +

+ Which values of c, if any, could make the following function continuous on the real line? + g(x) = \begin{cases} + x + c \amp x \leq 2 \\ + x^2 \amp x \gt 2 + \end{cases} + +

+ + +

+ Which values of c, if any, could make the following function continuous on the real line? + h(x) = \begin{cases} + 4 \amp x \leq c \\ + x^2 \amp x \gt c + \end{cases} + +

+ + +

+ Which values of c, if any, could make the following function continuous on the real line? + k(x) = \begin{cases} + x \amp x \leq c \\ + x^2 \amp x \gt c + \end{cases} + +

+ + + + +Intermediate Value TheoremIntermediate Value Theorem + +

Suppose that:

+
    +

  • the function f is continuous on the interval [a,b] ;

    • +

  • you pick a value N such that f(a)\leq N \leq f(b) or f(b)\leq N \leq f(a).

    • +
+

Then, there is some input c in the interval [a,b] such that f(c)=N.

+ + + + +

In this activity we will explore a mathematical theorem, the Intermediate Value Theorem.

+ + + +

To get an idea for the theorem, draw a continuous function f(x) on the interval [0,10] such that f(0)=8 and f(10)=2. Find an input c where f(c)=5. +

+ + + + +

Now try to draw a graph similar to the previous one, but that does not have any input corresponding to the output 5. Then, find where your graph violates these conditions: f(x) is continuous on [0,10], f(0)=8, and f(10)=2. +

+ + + + +

The part of the theorem that starts with “Suppose…” forms the assumptions of the theorem, while the part of the theorem that starts with “Then…” is the conclusion of the theorem. What are the assumptions of the Intermediate Value Theorem? What is the conclusion? +

 + + +

Apply the Intermediate Value Theorem to show that the function f(x) = x^3 +x -3 has a zero (so crosses the x-axis) at some point between x=-1 and x=2. (Hint: What interval of x values is being considered here? What is N? Why is N between f(a) and f(b)?)

+ + + + + + + Videos +
+
+ + +
diff --git a/calculus/source/01-LT/05.ptx b/calculus/source/01-LT/05.ptx new file mode 100644 index 00000000..95e5f594 --- /dev/null +++ b/calculus/source/01-LT/05.ptx @@ -0,0 +1,507 @@ + + +
+ Limits with infinite inputs (LT5) + + + + + Activities + + + + + +

+ Consider the graph of the polynomial function f(x) = x^3 . We want to think about what the long term behavior of this function might be. Which of the following best describes its behavior? +

+
+ + + + The graph of x^3. +
+ + +
    +

  1. As x gets larger, the function x^3 gets smaller and smaller.

    2. +

  2. As x gets more and more negative, the function x^3 gets more and more negative.

    3. +

  3. As x gets more and more positive, the function x^3 gets more and more negative.

    4. +

  4. As x gets smaller, the function x^3 gets smaller and smaller.

    5. +
+ + + +

We say that the limit as x tends to negative infinity of x^3 is negative infinity and that the limit as x tends to positive infinity of x^3 is positive infinity. In symbols, we write

+ + \lim_{x \to + \infty} x^3 = + \infty \, , \, \lim_{x \to - \infty} x^3 = - \infty. + + + + +

+ Consider the graph of the rational function f(x) = 1/ x^3 . We want to think about what the long term behavior of this function might be. Which of the following best describes its behavior? +

+
+ + + + The graph of 1/x^3. +
+ + + +
    +

  1. As x tends to positive infinity, the function 1/x^3 tends to positive infinity

    2. +

  2. As x tends to negative infinity, the function 1/x^3 tends to 0

    3. +

  3. As x tends to positive infinity, the function 1/x^3 tends to negative infinity

    4. +

  4. As x tends to 0, the function 1/x^3 tends to 0

    5. +
+ + + + + +

A function has a horizontal asymptotehorizontal asymptote at y=b when

+ + \lim_{x\to +\infty} f(x) = b + +

or

+ + \lim_{x\to -\infty} f(x) = b + +

This means that we can make the output of f(x) as close as we want to b, as long as we take x a large enough positive number (x \to \infty) or a large enough negative number (x \to -\infty).

+ + + + + +

We say that the function 1/x^3 has horizontal asymptote y=0 because the limit as x tends to positive infinity of 1/x^3 is 0. Alternatively, we could also justify it by saying that the limit as x goes to negative infinity is 0.

+ + + + + +

+ Which of the following functions have horizontal asymptotes? Select all! + +

+ +
+ + + + \begin{tikzpicture}[scale=0.6] + \begin{axis}[ + axis x line=middle, + axis y line=middle, + x label style={at={(axis description cs:1,-0.2)}}, + y label style={at={(axis description cs:0.3,1)}}, + ylabel=$y$, + xlabel=$x$ + ] + \addplot[samples=500, domain=-5:5, blue, ultra thick] {e^x}; + \end{axis} + \end{tikzpicture} + + + A +
+
+ + + \begin{tikzpicture}[scale=0.6] + \begin{axis}[ + axis x line=middle, + axis y line=middle, + x label style={at={(axis description cs:1,-0.2)}}, + y label style={at={(axis description cs:0.3,1)}}, + ylabel=$y$, + xlabel=$x$ + ] + \addplot[samples=500, domain=0.01:5, blue, ultra thick] {ln(x)}; + \end{axis} + \end{tikzpicture} + + + B +
+
+ + + \begin{tikzpicture}[scale=0.6] + \begin{axis}[ + axis x line=middle, + axis y line=middle, + x label style={at={(axis description cs:1,-0.2)}}, + y label style={at={(axis description cs:0.3,1)}}, + ylabel=$y$, + xlabel=$x$ + ] + \addplot[domain=-5:-0.1, blue, ultra thick] {-1/x}; + \addplot[domain=0.1:5, blue, ultra thick] {-1/x}; + \end{axis} + \end{tikzpicture} + + + C +
+ + +
+ + + \begin{tikzpicture}[scale=0.6] + \begin{axis}[ + axis x line=middle, + axis y line=middle, + x label style={at={(axis description cs:1,-0.2)}}, + y label style={at={(axis description cs:0.3,1)}}, + ylabel=$y$, + xlabel=$x$ + ] + \addplot[domain=-5:0.9, blue, ultra thick] { (5*x+1)/(x-1)}; + \addplot[domain=1.1:5, blue, ultra thick] { (5*x+1)/(x-1)}; + \end{axis} + \end{tikzpicture} + + + D +
+
+ + + \begin{tikzpicture}[scale=0.6] + \begin{axis}[ + axis x line=middle, + axis y line=middle, + x label style={at={(axis description cs:1,-0.2)}}, + y label style={at={(axis description cs:0.3,1)}}, + ylabel=$y$, + xlabel=$x$ + ] + \addplot[domain=-2:-0.1, blue, ultra thick] { (e^x+1)/(e^x-1)}; + \addplot[domain=0.1:2, blue, ultra thick] { (e^x+1)/(e^x-1)}; + \end{axis} + \end{tikzpicture} + + + E +
+
+ + + \begin{tikzpicture}[scale=0.6] + \begin{axis}[ + axis x line=middle, + axis y line=middle, + x label style={at={(axis description cs:1,-0.2)}}, + y label style={at={(axis description cs:0.3,1)}}, + ylabel=$y$, + xlabel=$x$ + ] + \addplot[samples=500, domain=-5:0.9, blue, ultra thick] { (x^2+1)/(x-1)}; + \addplot[samples=500, domain=1.1:5, blue, ultra thick] { (x^2+1)/(x-1)}; + \end{axis} + \end{tikzpicture} + + + F +
+ + + + + + +

+ Recall that a rational function is a ratio of two polynomials. For any given rational function, what are all the possible behaviors as x tends to + or - infinity? +

+ + +
    +

  1. The only possible limit is 0

    2. +

  2. The only possible limits are 0 or \pm \infty +

    3. +

  3. The only possible limits are 0, 1 or \pm \infty +

    4. +

  4. The only possible limits are any constant number or \pm \infty

    5. +
+ + + + + + In this activity we will examine functions whose limit as x approaches positive and negative infinity is a nonzero constant. + +

Graph the following functions and consider their limits as x approaches positive and negative infinity. Which function(s) have a limit that is nonzero and constant? Find each of these limits. +

+ + +
    +

  1. \displaystyle f(x)= \frac{x^3-x+3}{2x^3-6x+1}

    2. +

  2. \displaystyle f(x)= \frac{x^2-3}{5x^3-2x^2+5}

    3. +

  3. \displaystyle f(x)= \frac{x^4-3x-2}{3x^3-5x+1}

    4. +

  4. \displaystyle f(x)= \frac{10x^5-3x+2}{5x^5-3x^2+1}

    5. +

  5. \displaystyle f(x)= \frac{-8x^2-5x+1}{2x^2-2x+3}

    6. +
+

Conjecture a rule for how to determine that a rational function has a nonzero constant limit as x approaches positive and negative infinity. Test your rule by creating a rational function whose limit as x \to \infty equals 3 and then check it graphically.

+ + + + +

What about when the limit is not a nonzero constant? How do we recognize those? In this activity you will first conjecture the general behavior of rational functions and then test your conjectures.

+ +

+ Consider a rational function r(x) = \frac{p(x)}{q(x)}. Looking at the numerator p(x) and the denominator q(x), when does the function r(x) have limit equal to 0 as x \to \infty? +

+
    + +

  1. When the ratio of the leading terms is a constant.

    2. +

  2. When the degree of the numerator is greater than the degree of the denominator. +

    3. +

  3. When the degree of the numerator is less than the degree of the denominator. +

    4. +

  4. When the degree of the numerator is equal to the degree of the denominator. +

    5. +
+ +

+ Consider a rational function r(x) = \frac{p(x)}{q(x)}. Looking at the numerator p(x) and the denominator q(x), when does the function r(x) have limit approaching infinity as x \to \infty? +

+
    + +

  1. When the ratio of the leading terms is a constant.

    2. +

  2. When the degree of the numerator is greater than the degree of the denominator. +

    3. +

  3. When the degree of the numerator is less than the degree of the denominator. +

    4. +

  4. When the degree of the numerator is equal to the degree of the denominator. +

    5. +
+ +

Conjecture a rule for the each of the previous two parts of the activity. Test your rules by creating a rational function whose limit as x \to \infty equals 0 and another whose limit as x \to \infty is infinite. Then check them graphically.

 + + + + + +

+Explain how to find +the value of each limit. +

+ +

+\lim_{x\to-\infty } -\frac{6 \, {x^4} + 7 \, {x^3} - 7}{6 \, x - {x^4} + 9} +\,\text{ and }\, +\lim_{x\to+\infty } -\frac{6 \, {x^4} + 7 \, {x^3} - 7}{6 \, x - {x^4} + 9} +

+

+\lim_{x\to-\infty } -\frac{7 \, {x^4} - 5 \, {x^3} + 8}{3 \, {\left(2 \, {x^5} + 3 \, {x^2} - 3\right)}} +\,\text{ and }\, +\lim_{x\to+\infty } -\frac{7 \, {x^4} - 5 \, {x^3} + 8}{3 \, {\left(2 \, {x^5} + 3 \, {x^2} - 3\right)}} +

+

+\lim_{x\to-\infty } \frac{3 \, {x^6} + {x^3} - 8}{7 \, x - 6 \, {x^5} + 7} +\,\text{ and }\, +\lim_{x\to+\infty } \frac{3 \, {x^6} + {x^3} - 8}{7 \, x - 6 \, {x^5} + 7} +

+ + + + + + + +

+What is your best guess for the limit as x goes to + \infty of the function graphed below? +

+
+ + + \begin{tikzpicture} +\begin{axis}[ + axis lines=middle, + % grid=major, + % xmin=-3.14, + % xmax=9.42, + ymin=-1.5, + ymax=1.5, + % xtick={-3.14,0,...,9.42}, + % ytick={-1,-0.75,...,1}, + ] + \addplot[samples=500,domain=-4*pi:4*pi, blue, thick]{sin(deg(x))}; +\end{axis} +\end{tikzpicture} + + + A mysterious periodic function. +
+ + +
    +

  1. The limit is 0

    2. +

  2. The limit is 1 +

    3. +

  3. The limit is -1 +

    4. +

  4. The limit is +\infty

    5. +

  5. The limit DNE

    6. +
+ + + + + For a periodic function, a function whose outputs repeat periodically, there is not one distinguished long term behavior, so the limit DNE. Notice that this is different from the limit being +infinity in which case the outputs have a clear behavior: they are getting larger and larger. Unfortunately, you will find in many cases that the notation DNE is used both for a limit equal to infinity and a limit that does not tend to one distinguished value. Beware! + + + +

Compute the following limits.

+ + +

\displaystyle \lim_{x\to -\infty} \frac{x^3-x+83}{1}

+

\displaystyle \lim_{x\to -\infty} \frac{1}{x^3-x+83}

+

\displaystyle \lim_{x\to +\infty} \frac{x+3}{2-x}

 +

\displaystyle \lim_{x\to -\infty} \frac{\pi-3x}{\pi x-3}

 +

(Challenge) \displaystyle \lim_{x\to + \infty} \frac{3e^x+2}{2e^x+3}

 +

(Challenge) \displaystyle \lim_{x\to - \infty} \frac{3e^x+2}{2e^x+3}

 + + + + + + + +

The graph below represents the function f(x) = \displaystyle\frac{2(x+3)(x+1)}{x^2-2x-3}.

+
+ + +\begin{tikzpicture}[scale=1] +\begin{axis}[ + axis lines=middle, + grid=major, + xmin=-10, xmax=10, + ymin=-5, ymax=8, + xtick={-10,-8,...,10}, + ytick={-4,-2,...,10}, + tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel=$y$, + xlabel=$x$, + ] + \addplot[domain=-10:-1.1, blue, ultra thick] {(2*x^2+8*x+6)/(x^2-2*x-3)}; + \addplot[domain=-0.9:2.9, blue, ultra thick] {(2*x^2+8*x+6)/(x^2-2*x-3)}; + \addplot[domain=3.1:10, blue, ultra thick] {(2*x^2+8*x+6)/(x^2-2*x-3)}; + \addplot[mark=none, color=blue, nodes near coords={$f(x)$}] coordinates {(2,1)}; + \draw[fill=white, draw=blue](axis cs:-1,-1) circle(1mm); + + +\end{axis} +\end{tikzpicture} + + + + The graph of f(x) +
+ + + +

Find the horizontal asymptote of f(x) . First, guess it from the graph. Then, prove that your guess is right using algebra.

+ + + + +

Use limit notation to describe the behavior of f(x) at its horizontal asymptotes.

+ + + + + +

Come up with the formula of a rational function that has horizontal asymptote y=3 .

 + + +

What do you think is happening around x=3? We will come back to this in the next section!

+ + + + + + + +

An exponential function P(t) = a \, b^t exhibiting exponential decay will have the long term behavior P(t) \to 0 as t \to \infty. If we shift the graph up by c units, we obtain the new function Q(t) = a \, b^t + c, with the long term behavior \displaystyle\lim_{t \to \infty} Q(t) = c. A cooling object can be represented by the exponential decay model Q(t) = a \, b^t + c.

+ + + + +

In this activity you will explore an exponential model for a cooling object. +

+

Consider a cup of coffee initially at 100 degrees Fahrenheit. The said cup of coffee was forgotten this morning on the kitchen counter where the thermostat is set at 72 degrees Fahrenheit. From previous observations, we can assume that a cup of coffee looses 10 percent of its temperature each minute.

+ + + +

In the long run, what temperature do you expect the coffee to tend to? Write your observation with limit notation. +

+ + + + +

In the model Q(t) = a \, b^t +c, your previous answer gives you the value of one of the parameters in this model. Which one? +

+ + + + +

From the information given, we notice that the cup of coffee has decay rate of 10% or r = -0.1. When an exponential model has decay rate r, its exponential base b has value b=1+r. Use this to find the value of b for the exponential model described in this scenario.

+ + + + +

Assume that the initial temperature corresponds to input t=0. Use the data about the initial temperature to find the value of the parameter a in the model Q(t) = a \, b^t + c.

+ + + + +

You should have found that this scenario has exponential model Q(t) = 28 \, (0.9)^t + 72. If you go back to drink the cup of coffee 30 minutes after it was left on the counter, what temperature will the coffee have reached?

+ + + + + + + + + Videos +
+
+ + +
diff --git a/calculus/source/01-LT/06.ptx b/calculus/source/01-LT/06.ptx new file mode 100644 index 00000000..b641f7bd --- /dev/null +++ b/calculus/source/01-LT/06.ptx @@ -0,0 +1,326 @@ + + +
+ Limits with infinite outputs (LT6) + + + + + Activities + + + + +

+ Consider the graph in . +

+
+ + + + The graph of 1/x^2. +
+ + + +

+Which of the following best describes the limit as x approaches zero in the graph? +

+
    +

  1. The limit is 0

    2. +

  2. The limit is positive infinity

    3. +

  3. The limit does not exist

    4. +

  4. This limit is negative infinity

    5. +
+ + +

+ Which of the following best describes the relationship between the line x=0 and the graph of the function? +

+
    +

  1. The line x=0 is a horizontal asymptote for the function

    2. +

  2. The function is not continuous at the point x=0

    3. +

  3. The function is moving away from the line x=0

    4. +

  4. The function is getting closer and closer to the line x=0

    5. +

  5. The function has a jump in outputs around x=0

    6. +
+ + + + + +

A function has a vertical asymptotevertical asymptote at x=a when

+ + \lim_{x\to a} f(x) = + \infty + +

or

+ + \lim_{x\to a} f(x) = - \infty + +

The limit being equal to positive infinity means that we can make the output of f(x) as large a positive number as we want as long as we are sufficiently close to x=a. Similarly, the limit being equal to negative infinity means that we can make the output of f(x) as large a negative number as we want as long as we are sufficiently close to x=a.

+ + + + + + +

+ Select all of the following graphs which illustrate functions with vertical asymptotes. + +

+
+ + + + Choices for vertical asymptotes +
+ + + + + +

If x=a is a vertical asymptote for the function f(x) , the function f(x) is not defined at x=a. As f(a) does not exist, the function is NOT continuous at x=a. Moreover, the function's output tends to plus or minus infinity and so the limit is not equal to a number.

+ + + + + +

+ Notice that as x goes to 0, the value of x^2 goes to 0 but the value of 1/x^2 goes to infinity. What is the best explanation for this behavior? +

+ +
    +

  1. When dividing by an increasingly small number we get an increasing big number +

    2. +

  2. When dividing by an increasingly large number we get an increasing small number +

    3. +

  3. A rational function always has a vertical asymptote +

    4. +

  4. A rational function always has a horizontal asymptote +

    5. +
+ + +

Informally, we say that the limit of "\frac{1}{0}" is infinite. Notice that this could be either positive or negative infinity, depending on how whether the outputs are becoming more and more positive or more and more negative as we approach zero.

 + + + +

+Consider the rational function f(x) = \frac{2}{x-3} . Which of the following options best describes the limits as x approaches 3 from the right and from the left? +

+
    +

  1. As x \to 3^+, the limit DNE, but as x \to 3^- the limit is -\infty. +

    2. +

  2. As x \to 3^+, the limit is +\infty, but as x \to 3^- the limit is -\infty. +

    3. +

  3. As x \to 3^+, the limit is +\infty, but as x \to 3^- the limit is +\infty. +

    4. +

  4. As x \to 3^+, the limit is -\infty, but as x \to 3^- the limit is -\infty. +

    5. +

  5. As x \to 3^+, the limit DNE and as x \to 3^- the limit DNE. +

    6. +
+ + + + +

When considering a ratio of functions f(x)/g(x) , the inputs a where g(a)=0 are not in the domain of the ratio. If g(a)=0 but f(a) is not equal to 0, then x=a is a vertical asymptote.

+ + + + + +

+Consider the function f(x)=\frac{x^2-1}{x-1}. The line x=1 is NOT a vertical asymptote for f(x). Why? +

+
    +

  1. When x is not equal to 1, we can simplify the fraction to x-1, so the limit is 1. +

    2. +

  2. When x is not equal to 1, we can simplify the fraction to x+1, so the limit is 2. +

    3. +

  3. The function is always equal to x+1. +

    4. +

  4. The function is always equal to x-1. +

    5. +
+ + + +

Recall the definition of a hole from . In we have a hole at x=1.

+ + + +

Find all the vertical asymptotes of the following rational functions.

+ + + +

y = \frac{3x-4}{7x+1} +

+ + + + +

y= \frac{x^2+10x+24}{x^2-2x+1} +

+ + + + +

y= \frac{(x^2-4)(x^2+1)}{x^6}

+ + + + +

y= \frac{2x+1}{2x^2+8x-10}

+ + + + + + + +

+Explain and demonstrate how to find the value of each limit. +

+ +

+ \lim_{x\to-3^- } \frac{{\left(x + 4\right)}^{2} {\left(x - 2\right)}}{{\left(x + 3\right)} {\left(x - 5\right)}} +

+ + +

+ \lim_{x\to-3^+ } \frac{{\left(x + 4\right)}^{2} {\left(x - 2\right)}}{{\left(x + 3\right)} {\left(x - 5\right)}} +

+ + +

+ \lim_{x\to-3 } \frac{{\left(x + 4\right)}^{2} {\left(x - 2\right)}}{{\left(x + 3\right)} {\left(x - 5\right)}} +

+ + + + + + + + +

The graph below represents the function f(x) = \displaystyle\frac{(x+2)(x+4)}{x^2+3x-4}.

+
+ + + + The graph of f(x) +
+ + + +

Explain the behavior of f(x) at x=-4 .

+ + + + +

Find the vertical asymptote(s) of f(x). First, guess it from the graph. Then, prove that your guess is right using algebra.

+ + + + +

Find the horizontal asymptote(s) of f(x). First, guess it from the graph. Then, prove that your guess is right using algebra.

+ + + + +

Use limit notation to describe the behavior of f(x) at its asymptotes.

+ + + + + + +

+ Consider the following rational function. + r(x) = \frac{ 5 \, {\left(x - 3\right)} {\left(x - 6\right)}^{3} }{ 6 \, {\left(x + 2\right)}^{3} {\left(x - 3\right)} } +

+ +

Explain how to find the horizontal asymptote(s) of r(x), if there are any. Then express your findings using limit notation.

 +

Explain how to find the hole(s) of r(x), if there are any. Then express your findings using limit notation.

 +

Explain how to find the vertical asymptote(s) of r(x), if there are any. Then express your findings using limit notation.

 +

Draw a rough sketch of r(x) that showcases all the limits that you have found above.

+ + + + + + +

You want to draw a function with all these properties.

+
    +

  • \displaystyle \lim_{x\to 3} f(x)=5

    • +

  • f(3)=0

    • +

  • \displaystyle \lim_{x\to 0^{-}} f(x) = -\infty

    • +

  • \displaystyle \lim_{x\to 0^{+}} f(x) = 0

    • +

  • \displaystyle \lim_{x\to +\infty} f(x) = 2

    • +
+ Before you start drawing, consider the following guiding questions. + + + +

At which x values will the limit not exist? +

+ + + + +

What are the asymptotes of this function? +

+ + + + +

At which x values will the function be discontinuous?

+ + + + +

Draw the graph of one function with all the properties above. Make sure that your graph is a function! You only need to draw a graph, writing a formula would be very challenging!

+ + + + + + + Videos +
+
+ + +
+ diff --git a/calculus/source/01-LT/main.ptx b/calculus/source/01-LT/main.ptx new file mode 100644 index 00000000..4b2b5de1 --- /dev/null +++ b/calculus/source/01-LT/main.ptx @@ -0,0 +1,13 @@ + + + + Limits (LT) + + + + + + + + + diff --git a/calculus/source/01-LT/outcomes/01.ptx b/calculus/source/01-LT/outcomes/01.ptx new file mode 100644 index 00000000..70593df9 --- /dev/null +++ b/calculus/source/01-LT/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Find limits from the graph of a function. +

\ No newline at end of file diff --git a/calculus/source/01-LT/outcomes/02.ptx b/calculus/source/01-LT/outcomes/02.ptx new file mode 100644 index 00000000..9c0b7e56 --- /dev/null +++ b/calculus/source/01-LT/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Infer the value of a limit based on nearby values of the function. +

\ No newline at end of file diff --git a/calculus/source/01-LT/outcomes/03.ptx b/calculus/source/01-LT/outcomes/03.ptx new file mode 100644 index 00000000..06320cb2 --- /dev/null +++ b/calculus/source/01-LT/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Compute limits of functions given algebraically, using proper limit properties. +

\ No newline at end of file diff --git a/calculus/source/01-LT/outcomes/04.ptx b/calculus/source/01-LT/outcomes/04.ptx new file mode 100644 index 00000000..6e2bc159 --- /dev/null +++ b/calculus/source/01-LT/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Determine where a function is and is not continuous. +

\ No newline at end of file diff --git a/calculus/source/01-LT/outcomes/05.ptx b/calculus/source/01-LT/outcomes/05.ptx new file mode 100644 index 00000000..5d03b8b7 --- /dev/null +++ b/calculus/source/01-LT/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+Determine limits of functions at infinity. +

\ No newline at end of file diff --git a/calculus/source/01-LT/outcomes/06.ptx b/calculus/source/01-LT/outcomes/06.ptx new file mode 100644 index 00000000..2e251330 --- /dev/null +++ b/calculus/source/01-LT/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+Determine limits of functions approaching vertical asymptotes. +

\ No newline at end of file diff --git a/calculus/source/01-LT/outcomes/main.ptx b/calculus/source/01-LT/outcomes/main.ptx new file mode 100644 index 00000000..faaed135 --- /dev/null +++ b/calculus/source/01-LT/outcomes/main.ptx @@ -0,0 +1,31 @@ + +> + +

+ How do we measure close-by values? +

+

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
+ \ No newline at end of file diff --git a/calculus/source/01-LT/readiness.ptx b/calculus/source/01-LT/readiness.ptx new file mode 100644 index 00000000..a2a124b2 --- /dev/null +++ b/calculus/source/01-LT/readiness.ptx @@ -0,0 +1,106 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Use function notation and evaluate functions

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    • +

      Practice:

      +
        +

      • Evaluate functions

        • +

      • Evaluate functions from their graphs

        • +

      • Function notation word problems

        • +
      +
      • +
    +
    2. + +
  2. +

    Find the domain of a function +

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    • +

      Practice: Determine the domain of functions

      +
      • +
    +
    3. +
  3. +

    Determine vertical asymptotes, horizontal asymptotes, and holes (removable discontinuities) of rational functions

    +
      +
    • +

      Review:

      +
        +

      • Discontinuities of rational functions

        • +

      • Finding horizonal asymptotes

        • +
      +
      • +
    • +

      Practice:

      +
        +

      • Rational functions: zeros, asymptotes, and undefined points

        • +

      • Finding horizonal asymptotes

        • + +
      +
      • +
    +
    4. +
  4. +

    Perform basic operations with polynomials

    +
      +
    • +

      Review:

      +
        +

      • Adding and subtracting polynomials

        • +

      • Multiplying polynomials

        • + +
      +
      • +
    • +

      Practice:

      +
        +

      • Add polynomials

        • +

      • Subtract polynomials

        • +

      • Multiply binomials by polynomials

        • +
      +
      • +
    +
    5. + +
  5. +

    Factor quadratic expressions

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    • +

      Practice: Factoring quadratics intro

      + +
      • +
    +
    6. + +
  6. +

    Represent intervals using number lines, inequalities, and interval notation +

    +
      +
    • +

      Review: Varsity Tutor

      +
      • +
    +
    7. + +
+ + \ No newline at end of file diff --git a/calculus/source/01-LT/tikz/LT5-infinity-intuition1.tex b/calculus/source/01-LT/tikz/LT5-infinity-intuition1.tex new file mode 100644 index 00000000..fcbd459c --- /dev/null +++ b/calculus/source/01-LT/tikz/LT5-infinity-intuition1.tex @@ -0,0 +1,14 @@ +\begin{tikzpicture}[scale=0.5] + \begin{axis}[ + axis lines=middle, + % grid=major, + % xmin=-3.14, + % xmax=9.42, + % ymin=-1.5, + % ymax=1.5, + % xtick={-3.14,0,...,9.42}, + % ytick={-1,-0.75,...,1}, + ] + \addplot[samples=500,domain=-4:4, blue, thick]{x^3}; + \end{axis} + \end{tikzpicture} \ No newline at end of file diff --git a/calculus/source/01-LT/tikz/LT5-infinity-intuition2.tex b/calculus/source/01-LT/tikz/LT5-infinity-intuition2.tex new file mode 100644 index 00000000..34fc6df7 --- /dev/null +++ b/calculus/source/01-LT/tikz/LT5-infinity-intuition2.tex @@ -0,0 +1,16 @@ + +\begin{tikzpicture}[scale=0.5] + \begin{axis}[ + axis lines=middle, + % grid=major, + % xmin=-3.14, + % xmax=9.42, + % ymin=-1.5, + % ymax=1.5, + % xtick={-3.14,0,...,9.42}, + % ytick={-1,-0.75,...,1}, + ] + \addplot[samples=500,domain=-2:-0.1, blue, thick]{1/x^3}; + \addplot[samples=500,domain=0.1:2, blue, thick]{1/x^3}; + \end{axis} + \end{tikzpicture} \ No newline at end of file diff --git a/calculus/source/01-LT/tikz/asymptote-vert.tex b/calculus/source/01-LT/tikz/asymptote-vert.tex new file mode 100644 index 00000000..743d0c3f --- /dev/null +++ b/calculus/source/01-LT/tikz/asymptote-vert.tex @@ -0,0 +1,55 @@ + +\begin{tikzpicture}[scale=0.5] + \begin{scope}[shift={(0,0)}]\node[black,circle] at (0,5) {A}; + \begin{axis}[ + axis x line=middle, + axis y line=middle, + ] + \addplot[samples=500, domain=-5:5, blue, ultra thick] {e^x}; + \end{axis} + \end{scope} + \begin{scope}[shift={(8,0)}]\node[black,circle] at (6,0) {B}; + \begin{axis}[ + axis x line=middle, + axis y line=middle, + ] + \addplot[samples=500, domain=0.01:5, blue, ultra thick] {ln(x)}; + \end{axis} + \end{scope} + \begin{scope}[shift={(16,0)}]\node[black,circle] at (0,0) {C}; + \begin{axis}[ + axis x line=middle, + axis y line=middle, + ] + \addplot[domain=-5:-0.1, blue, ultra thick] {-1/x}; + \addplot[domain=0.1:5, blue, ultra thick] {-1/x}; + \end{axis} + \end{scope} + \begin{scope}[shift={(0,-8)}]\node[black,circle] at (0,0) {D}; + \begin{axis}[ + axis x line=middle, + axis y line=middle, + ] + \addplot[domain=-5:0.9, blue, ultra thick] { (5*x+1)/(x-1)}; + \addplot[domain=1.1:5, blue, ultra thick] { (5*x+1)/(x-1)}; + \end{axis} + \end{scope} + \begin{scope}[shift={(8,-8)}]\node[black,circle] at (0,0) {E}; + \begin{axis}[ + axis x line=middle, + axis y line=middle, + ] + \addplot[domain=-2:-0.1, blue, ultra thick] { (e^x+1)/(e^x-1)}; + \addplot[domain=0.1:2, blue, ultra thick] { (e^x+1)/(e^x-1)}; + \end{axis} + \end{scope} + \begin{scope}[shift={(16,-8)}]\node[black,circle] at (0,0) {F}; + \begin{axis}[ + axis x line=middle, + axis y line=middle, + ] + \addplot[samples=500, domain=-5:0.9, blue, ultra thick] { (x^2+1)/(x-1)}; + \addplot[samples=500, domain=1.1:5, blue, ultra thick] { (x^2+1)/(x-1)}; + \end{axis} + \end{scope} +\end{tikzpicture} \ No newline at end of file diff --git a/calculus/source/01-LT/tikz/rational-practice2.tex b/calculus/source/01-LT/tikz/rational-practice2.tex new file mode 100644 index 00000000..806964f5 --- /dev/null +++ b/calculus/source/01-LT/tikz/rational-practice2.tex @@ -0,0 +1,19 @@ +\begin{tikzpicture}[scale=1] + \begin{axis}[ + axis lines=middle, + grid=major, + xmin=-6, xmax=8, + ymin=-5, ymax=8, + xtick={-6,-5,...,8}, + ytick={-5,-4,...,8}, + tick style={thick}, + ylabel=$y$, + xlabel=$x$, + ] + \addplot[domain=-6:-4.1, blue, thick] {(x^2+6*x+8)/(x^2+3*x-4)}; + \addplot[domain=-3.9:0.99, blue, thick] {(x^2+6*x+8)/(x^2+3*x-4)}; + \addplot[domain=1.1:8, blue, thick] {(x^2+6*x+8)/(x^2+3*x-4)}; + \addplot[mark=none, color=blue, nodes near coords={$f(x)$}] coordinates {(2,1)}; + \draw[fill=white, draw=blue](axis cs:-4,0.4) circle(1mm); + \end{axis} + \end{tikzpicture} \ No newline at end of file diff --git a/calculus/source/01-LT/tikz/vert-asymp.tex b/calculus/source/01-LT/tikz/vert-asymp.tex new file mode 100644 index 00000000..b81c4bc3 --- /dev/null +++ b/calculus/source/01-LT/tikz/vert-asymp.tex @@ -0,0 +1,17 @@ +\begin{tikzpicture}[scale=0.5] + \begin{scope}[shift={(0,0)}] + \begin{axis}[ + axis lines=middle, + % grid=major, + % xmin=-3.14, + % xmax=9.42, + % ymin=-1.5, + % ymax=1.5, + % xtick={-3.14,0,...,9.42}, + % ytick={-1,-0.75,...,1}, + ] + \addplot[samples=500,domain=-2:-0.1, blue, thick]{1/x^2}; + \addplot[samples=500,domain=0.1:2, blue, thick]{1/x^2}; + \end{axis} + \end{scope} +\end{tikzpicture} \ No newline at end of file diff --git a/calculus/source/02-DF/01.ptx b/calculus/source/02-DF/01.ptx new file mode 100644 index 00000000..9e0574a9 --- /dev/null +++ b/calculus/source/02-DF/01.ptx @@ -0,0 +1,496 @@ + + +
+ Derivatives graphically and numerically (DF1) + + + + + Activities + + + + +

In this activity you will study the velocity of a ball falling under gravity. The height of the ball (in feet) is given by the formula f(t) = 64-16(t-1)^2, where t is measured in seconds. We want to study the velocity at the instant t=2, so we will look at smaller and smaller intervals around t=2. For your convenience, below you will find a table of values for f(t). Recall that the average velocity is given by the change in height over the change in time. +

+ + + + + t\, \text{(seconds)} + + + 1 + + + 1.5 + + + 1.75 + + + 2 + + + 2.25 + + + 2.5 + + + 3 + + + + + f(t) \, \text{(feet)} + + + 64 + + + 60 + + + 55 + + + 48 + + + 39 + + + 28 + + + 0 + + + + +
+ +

To start we will look at an interval of length one before t=2 and after t=2, so we consider the intervals [1,2] and [2,3]. What was the average velocity on the interval [1,2]? What about on the interval [2,3]?

+ +

Now let's consider smaller intervals of length 0.5. What was the average velocity on the interval [1.5,2]? What about on the interval [2,2.5]?

+ + +

What was the average velocity on the interval [1.75,2]? What about on the interval [2,2.25]?

+ + +

If we wanted to approximate the velocity at the instant t=2, what would be your best estimate for this instantaneous velocity?

+ + + +

If we want to study the velocity at the instant t=2, it is helpful to study the average velocity on small intervals around t=2. If we consider the interval [2,2+h], where h is the width of the interval, the average velocity is given by the difference quotient

+ \frac{f(2+h)-f(2)}{(2+h)-2} = \frac{f(2+h)-f(2)}{h}. + + +

We want to be able to consider intervals before and after t=2. A positive value of h will give an interval after t=2. For example, the interval [2,3] for h=1. A negative value of h will give an interval before t=2. For example, the interval [1,2] corresponds h=-1. In the formula above, it looks like the interval would be [2,1], but the standard notation in an interval is to write the smallest number first. This does not change the difference quotient because

+ \frac{f(2+h)-f(2)}{(2+h)-2} = \frac{f(2)-f(2+h)}{2-(2+h)}. + + + + +

+ Consider the height of the ball falling under gravity as in .

+ +

What was the average velocity on the interval [2,2+h] for h=1 and h=-1?

+ +

What was the average velocity on the interval [2,2+h] for h=0.5 and h=-0.5?

+ +

What was the average velocity on the interval [2,2+h] for h=0.25 and h=-0.25?

+ + +

What is your best estimate for the limiting value of these velocities as h\to 0? Notice that this is your estimate for the instantaneous velocity at t=2!

+ + + + +

The instanteous velocity at t=a is the limit as h \to 0 of the difference quotient \frac{f(a+h)-f(a)}{h}. In the activity above the instantaneous velocity at t=2 is given by the limit +

+ + v(2) = \lim_{h\to 0} \frac{f(2+h)-f(2)}{h} + + + +

+ The slope of the secant line to f(x) through the points x=a and x=b is given by the difference quotient +

+ \frac{f(b)-f(a)}{b-a}. + + + + +

In this activity you will study the slope of a graph at a point. The graph of the function g(x) is given below. For your convenience, below you will find a table of values for g(x). +

+
+ + +\begin{tikzpicture}[scale=1] +\begin{axis}[ + axis lines=middle, + grid=major, + xmin=0, xmax=6, + ymin=-2, ymax=5, + xtick={0,0.5,...,6}, + ytick={-2,-1.5,...,7}, + tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel=$y$, + xlabel=$x$, + ] + \addplot[domain=0:6, blue, ultra thick] {0.5*x^2-2}; +\end{axis} +\end{tikzpicture} + + + + The graph of g(x) +
+ + + + + x + + + 1 + + + 1.5 + + + 2 + + + 2.5 + + + 3 + + + + + g(x) + + + -1.5 + + + -0.875 + + + 0 + + + 1.125 + + + 2.5 + + + +
+ +

What is the slope of the line through (1,g(1)) and (2,g(2))? Draw this line on the graph of g(x).

+ +

What is the slope of the line through (1.5,g(1.5)) and (2,g(2))? Draw this line on the graph of g(x).

+ +

Draw the line tangent to g(x) at x=2. What would be your best estimate for the slope of this tangent line?

+ +

Notice that the slope of the tangent line at x=2 is positive. What feature of the graph of f(x) around x=2 do you think causes the tangent line to have positive slope?

+
    +

  1. The function f(x) is concave up

    2. +

  2. The function f(x) is increasing

    3. +

  3. The function f(x) is concave down

    4. +

  4. The function f(x) is decreasing

    5. +
+ + + + +

+ The slope of the secant line to f(x) through the points x=a and x=b is given by the difference quotient + \frac{f(b)-f(a)}{b-a}. As the point x=b gets closer to x=a, the slope of the secant line tends to the slope of the tangent line. In symbols, the slope at x=a is given by the limit

+ \lim_{x\to a} \frac{f(x)-f(a)}{x-a}. + +

+ Letting b=a+h, we can also say that the slope of the tangent line at x=a is given by the limit

+ \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}. + + + +

The derivative of f(x) at x=a, denoted f'(a), is given by +

+ f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}. + + +

In and you studied a ball falling under gravity and estimated the instantaneous velocity as a limiting value of average velocities on smaller and smaller intervals. Drawing the corresponding secant lines, we see how the secant lines approximate better the tangent line, showing graphically what we previouly saw numericaly. Here is a Desmos animation showing the secant lines approaching the tangent line .

+ + + +

Suppose that the function f(x) gives the position of an object at time x. Which of the following quantities are the same? Select all that apply!

+
    +

  1. The value of the derivative of f(x) at x=a

    2. +

  2. The slope of the tangent line to f(x) at x=a

    3. +

  3. The instantaneous velocity of the object at x=a

    4. +

  4. The difference quotient \frac{f(a+h)-f(a)}{h}

    5. +

  5. The limit \displaystyle\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}

    6. +
+ + +

We can use the difference quotient \frac{f(a+h)-f(a)}{h} for small values of h to estimate f'(a), the value of the derivative at x=a.

+ + +

Suppose that you know that the function g(x) has values g(-0.5)=7, g(0)=4, and g(0.5)=2. What is your best estimate for g'(0)?

+
    +

  1. g'(0) \approx -3

    2. +

  2. g'(0) \approx -2

    3. +

  3. g'(0) \approx -6

    4. +

  4. g'(0) \approx -4

    5. +

  5. g'(0) \approx -5

    6. +
+ + + +

Suppose that you know that the function f(x) has value f(1)=3 and has derivative at x=1 given by f'(1)=2. Which of the following scenarios is most likely?

+
    +

  1. f(2)=3 because the function is constant

    2. +

  2. f(2)=2 because the derivative is constant

    3. +

  3. f(2) \approx 1 because the function's output decreases by about 2 units for each increase by 1 unit in the input

    4. +

  4. f(2) \approx 5 because the function's output increases by about 2 units for each increase by 1 unit in the input

    5. +
+ + +

We can use the derivative at x=a to estimate the increase/decrease of the function f(x) close to x=a. A positive derivative at x=a suggests that the output values are increasing around x=a approximately at a rate given by the value of the derivative. A negative derivative at x=a suggests that the output values are decreasing around x=a approximately at a rate given by the value of the derivative.

 + + + +

In this activity you will study the abolute value function f(x)=|x|. The absolute value function is a piecewise defined function which outputs x when x is positive (or zero) and outputs -x when x is negative. So the absolute value always outputs a number which is positive (or zero). Here is the graph of this function.

+
+ + +\begin{tikzpicture}[scale=1] +\begin{axis}[ + axis lines=middle, + grid=major, +% xmin=-10, xmax=10, + % ymin=-5, ymax=8, + xtick={-4,-3,...,4}, + ytick={0,1,...,4}, + tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel=$y$, + xlabel=$x$, + ] + \addplot[domain=-4:4, blue, thick] {abs(x)}; +\end{axis} +\end{tikzpicture} + + + + The graph of |x| +
+ +

What do you think is the slope of the function for any x value smaller than zero?

+
    +

  1. 0

    2. +

  2. 1

    3. +

  3. -1

    4. +

  4. DNE

    5. +
+ +

What do you think is the slope of the function for any x value greater than zero?

+
    +

  1. 0

    2. +

  2. 1

    3. +

  3. -1

    4. +

  4. DNE

    5. +
+ +

What do you think is the slope of the function at zero?

+
    +

  1. 0

    2. +

  2. 1

    3. +

  3. -1

    4. +

  4. DNE

    5. +
+ + + +

Because the derivative at a point is defined in terms of a limit, the quantity f'(a) might not exist! In that case we say that f(x) is not differentiable at x=a. This might happen when the slope on the left of the point is different from the slope on the right, like in the case of the absolute value function. We call this behavior a corner in the graph.

+ + + +

Consider the graph of function h(x).

+
+ + + \begin{tikzpicture}[scale=1] +\begin{axis}[ + axis lines=middle, + grid=major, + xtick={-2,-1,...,7}, + ytick={-2,-1,...,4}, + ymin=-1.2, ymax=1.2, + tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel=$y$, + xlabel=$x$, + ] + \addplot[domain=-2:1, blue, ultra thick] {(1/3)*(x)-1/3}; + \addplot[domain=1:5, blue, ultra thick] {0.0625*(x-5)^2-1}; + \addplot[domain=5:7, blue, ultra thick] {(x-6)^2 }; +\addplot [only marks, blue] table { +5 1 \\ +1 1 \\ +}; +\draw[fill=white, draw=blue](axis cs:1,0) circle(1mm); +\draw[fill=white, draw=blue](axis cs:5,-1) circle(1mm); +\end{axis} +\end{tikzpicture} + + + + The graph of h(x). +
+ +

For which of the following points a is h'(a) positive? Select all that apply!

+
    +

  1. -1

    2. +

  2. 1

    3. +

  3. 2

    4. +

  4. 5

    5. +

  5. 6

    6. +
+ +

For which of the following points a is h'(a) negative? Select all that apply!

+
    +

  1. -1

    2. +

  2. 1

    3. +

  3. 2

    4. +

  4. 5

    5. +

  5. 6

    6. +
+ +

For which of the following points a is h'(a) zero? Select all that apply!

+
    +

  1. -1

    2. +

  2. 1

    3. +

  3. 2

    4. +

  4. 5

    5. +

  5. 6

    6. +
+ +

For which of the following points a the quantity h'(a) does NOT exist? Select all that apply!

+
    +

  1. -1

    2. +

  2. 1

    3. +

  3. 2

    4. +

  4. 5

    5. +

  5. 6

    6. +
+ + + + + + + + +

+Sketch the graph of a function f(x) +that satisfies the following criteria. +(You do not need to define the function algebraically.) +

+
    +
  • +

    +Defined and continuous on the interval [-5,5]. +

    +
    • +
  • +

    +f'(x) does not exist at x=0 +

    +
    • +
  • +

    +\displaystyle\lim_{h\to 0} \frac{f( 2 +h)-f( 2 )}{h}< 0 +

    +
    • +
  • +

    +The slope tangent to the graph of f(x) at x=3 is zero +

    +
    • +
  • +

    +The rate of change of f(x) when x=-1 is positive +

    +
    • +
+ + + + + +

You are given the graph of the function f(x).

+
+ + +\begin{tikzpicture}[scale=1] +\begin{axis}[ + axis lines=middle, + grid=major, +% xmin=-10, xmax=10, + % ymin=-5, ymax=8, + xtick={-1,-0.5,...,3}, + ytick={-7,-6,...,1}, + tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel=$y$, + xlabel=$x$, + ] + \addplot[domain=-1:3, blue, thick] {-x^2+2}; +\end{axis} +\end{tikzpicture} + + + + The graph of f(x) +
+ +

Using the graph, estimate the slope of the tangent line at x=2. Make sure you can carefully describe your process for obtaining this estimate!

+ +

If you call your approximation for the slope m, which one of the following expression gives you the equation of the tangent line at x=2?

+
    +

  1. y - 2 = m (x-2)

    2. +

  2. y + 2 = m (x-2)

    3. +

  3. y - 2 = m (x+2)

    4. +

  4. y + 2 = m (x+2)

    5. +
+ +

Find the equation of the tangent line at x=2.

+ + + + + Videos +
+
+ + +
\ No newline at end of file diff --git a/calculus/source/02-DF/02.ptx b/calculus/source/02-DF/02.ptx new file mode 100644 index 00000000..1e257250 --- /dev/null +++ b/calculus/source/02-DF/02.ptx @@ -0,0 +1,289 @@ + + +
+ Derivatives analytically (DF2) + + + + + Activities + + +

Recall that f'(a), the derivative of f(x) at x=a, was defined as the limit as h \to 0 of the difference quotient on the interval [a,a+h] as in . If f'(a) exists, then say that f(x) is differentiable at a. If for some open interval (a,b), we have that f'(x) exists for every point x in (a,b), then we say that f(x) is differentiable on (a,b).

+ + + + +

For the function f(x)=x-x^2 use the limit definition of the derivative at a point to compute f'(2).

+
    +
  1. f'(2)= \displaystyle \lim_{h\to 0} \frac{(2+h)-(2+h)^2-2+4}{h}=-3
    2. +
  2. The limit f'(2)= \displaystyle \lim_{h\to 0} \frac{(2+h)-(2+h)^2-2}{h} simplifies algebraically to \displaystyle \lim_{h\to 0} \frac{-3h - h^2}{h} which does not exist, thus f'(2) is not defined.
    3. +
  3. The limit f'(2)= \displaystyle \lim_{h\to 0} \frac{(2+h)-(2+h)^2-2}{h} simplifies algebraically to \displaystyle \lim_{h\to 0} \frac{h -h^2}{h} which does not exist, thus f'(2) is not defined.
    4. +
  4. f'(2) = \displaystyle \lim_{h\to 0} \frac{(2+h)-(2^2+h^2)-2+4}{h}= 1
    5. +
+ + + + +

Consider the function f(x)=3-2x. Which of the following best summarizes the average rates of changes of on f on the intervals [1,4], [3,7], and [5, 5+h]?

+
    +
  1. The average rate of change on the intervals [1,4] and [3,7] are equal to the slope of f(x), but the average rate of change of f cannot be determined on [5,5+h] without a specific value of h.
    2. +
  2. The average rate of change on the intervals [1,4], [3,7], and [5, 5+h] are all different values.
    3. +
  3. The average rate of change on the intervals [1,4], [3,7], and [5,5+h] are all equal to -2.
    4. +
+ + + + +

Can you find f'(\pi) when f(x)=3-2x without doing any computations?

+
    +
  1. No, because we cannot compute the value f(\pi).
    2. +
  2. No, because we cannot compute the average rate of change on the interval [\pi, \pi+h].
    3. +
  3. Yes, f'(\pi)=3 because the intercept of the tangent line at any point is equal to the constant intercept of f(x).
    4. +
  4. Yes, f'(\pi)=-2 because the slope of the tangent line at any point is equal to the constant slope of f(x).
    5. +
+ + + + +

Let f(x) be function that is differentiable on an open interval (a,b). The derivative function of f(x), denoted f'(x), is given by the limit +

+ f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}. +

At any particular input x=a, the derivative function outputs f'(a), the value of derivative at the point x=a.

+ + +

To specify the indendent variable of our function, we say that f'(x) is the derivative of f(x) with respect to x. For the derivative function of y=f(x) we also use the notation:

+ f'(x)=y'(x)=\frac{dy}{dx}=\frac{df}{dx}. +

The last type of notation is known as differential (or Leibniz) notation for the derivative.

 + + +

Notice that our notation for the derivative function is based on the name that we assign to the function along with our choice of notation for indendent and dependent variables. For example, if we have a differentiable function y=v(t), the derivative function of v(t) with respect to t can be written as v'(t)=y'(t)=\frac{dy}{dt}=\frac{dv}{dt}.

+ + + +

In this activity you will consider f(x)=-x^2+4 and compute its derivative function f'(x) using the limit definition of the derivative function . +

+ +

What expression do you get when you simplify the difference quotient

+ \frac{f(x+h)-f(x)}{h}= \frac{(-(x+h)^2+4)-(-x^2+4)}{h}? + +
    +
  1. \frac{x^2 + h^2 +4 -x^2 -4}{h} = \frac{h^2}{h}
    2. +
  2. \frac{-x^2 - h^2 +4 +x^2 -4}{h} = \frac{-h^2}{h}
    3. +
  3. \frac{-x^2 -2xh - h^2 +4 +x^2 -4}{h} = \frac{-2xh - h^2}{h}
    4. +
  4. \frac{x^2 + 2xh + h^2 +4 - x^2 -4}{h} = \frac{2xh + h^2}{h}
    5. +
+

After taking the limit as h \to 0, which of the following is your result for the derivative function f'(x)?

+
    +
  1. f'(x)=x
    2. +
  2. f'(x)=-x
    3. +
  3. f'(x)=2x
    4. +
  4. f'(x)=-2x
    5. +
+ + + + + +

Using the limit definition of the derivative, find f'(x) for f(x)=-x^2+2x-4. Which of the following is an accurate expression for f'(x)?

+
    +
  1. f'(x)=2x +2
    2. +
  2. f'(x)=-2x
    3. +
  3. f'(x)=-2x+2
    4. +
  4. f'(x)=-2x -2
    5. +
+ + + + +

Using the limit definition of the derivative, you want to find f'(x) for f(x)=\frac{1}{x}. We will do this by first simplifying the difference quotient and then taking the limit as h\to 0.

+

What expression do you get when you simplify the difference quotient

+ \frac{f(x+h)-f(x)}{h}= \frac{\frac{1}{x+h}-\frac{1}{x}}{h}? +
    +
  1. \frac{\frac{1}{x+h}}{h} = \frac{1}{(x+h)h}
    2. +
  2. \frac{\frac{h}{x+h}}{h} = \frac{h}{h(x+h)}
    3. +
  3. \frac{\frac{x-(x+h)}{(x+h)x}}{h} = \frac{-h}{h(x+h)x}
    4. +
  4. \frac{\frac{x-(x+h)}{(x+h)x}}{h} = \frac{-h^2}{(x+h)x}
    5. +
  5. \frac{\frac{h}{(x+h)x}}{h} = \frac{h}{h(x+h)x}
    6. +
+

After taking the limit as h \to 0, which of the following is your result for the derivative function f'(x)?

+
    +
  1. f'(x)=0
    2. +
  2. f'(x)=1/x
    3. +
  3. f'(x)=-1/x
    4. +
  4. f'(x)=1/x^2
    5. +
  5. f'(x)=-1/x^2
    6. +
+ + + + + +

+ Find f'(x) using the limit definition of the derivative. Then evaluate at x=8. +

+ + f(x) = x^{2} - 5 \, x - 5 + + + + + +

Once we have computed the first derivative f'(x), the second derivative of f(x) is the first derivative of f'(x) or + f''(x) = \lim_{h\to 0} \frac{f'(x+h)-f'(x)}{h}. +

+ + +

Consider the function f(x)=-x^2+2x-4. Earlier you saw that f'(x)=-2x+2. What is the second derivative of f(x)?

+
    +
  1. f''(x)=2
    2. +
  2. f''(x)=-2
    3. +
  3. f''(x)=2x
    4. +
  4. f''(x)=-2x
    5. +
+ + + +

The first derivative encodes information about the rate of change of the original function. In particular, +

    +
  • If f'\gt 0 , then f is increasing;
    • +
  • If f'\lt 0 , then f is decreasing;
    • +
  • If f' = 0 , then f has a horizontal tangent line (and it might have a max or min or it might just be changing pace).
    • +
+

+

+The second derivative is the derivative of the derivative. It encodes information about the rate of change of the rate of change of the original function. In particular, +

    +
  • If f'' \gt 0 , then f' is increasing;
    • +
  • If f'' \lt 0 , then f' is decreasing;
    • +
  • If f'' = 0 , then f' has a horizontal tangent line (and it might have a max or min or it might just be changing pace).
    • +

 + + +

Consider the function f(x)=-x^2+2x-4. Earlier you saw that f'(x)=-2x+2 and f''(x)=-2. What does this tell you about the graph of f(x) for x \gt 1?

+
    +
  1. The graph is increasing and concave up
    2. +
  2. The graph is increasing and concave down
    3. +
  3. The graph is decreasing and concave up
    4. +
  4. The graph is decreasing and concave down
    5. +
+ + + +

We have two ways to compute analytically the derivative at a point. For example, to compute f'(1), the derivative of f(x) at x=1, we have two methods

+
    +
  1. We can directly compute f'(1) by finding the difference quotient on the interval [1,1+h] and then taking the limit as h\to 0.
    2. +
  2. We can first find the derivative function f'(x) by computing the difference quotient on the interval [x,x+h], then taking the limit as h\to 0, and finally evaluating the expression for f'(x) at the input x=1.
    3. +
+

The latter approach is more convenient when you want to consider the value of the derivative function at multiple points!

+ + + +

Consider the function f(x) = \frac{1}{x^2}. You will find f'(1) in two ways!

+

Using the limit definition of the derivative at a point, compute the difference quotient on the interval [1,1+h] and then take the limit as h\to 0. What do you get?

+
    +

  1. -1

    2. +

  2. 1

    3. +

  3. 2

    4. +

  4. -2

    5. +
+

Now, using the limit definition of the derivative function, find f'(x). Which of the following is your result for the derivative function f'(x)?

+
    +
  1. f'(x)=-1/x^3
    2. +
  2. f'(x)=1/x^3
    3. +
  3. f'(x)=-2/x^3
    4. +
  4. f'(x)=2/x^3
    5. +
+ +

Make sure that your answers match! So if you plug in x=1 in f'(x), you should get the same number you got when you computed f'(1).

+ + + + + +

In this activity you will study (again!) the velocity of a ball falling under gravity. A ball is tossed vertically in the air from a window. The height of the ball (in feet) is given by the formula + f(t) = 64-16(t-1)^2, where t is the seconds after the ball is launched. Recall that in , you used numerical methods to approxmiate the instantaneous velocity of f(t) to calculate v(2)!

+

Using the limit definition of the derivative function, find the velocity function v(t)=f'(t).

+

Using the velocity function v(t), what is v'(1), the instantaneous velocity at t=1?

+
    +

  1. -32 feet per second

    2. +

  2. 32 feet per second

    3. +

  3. 0 feet per second

    4. +

  4. -16 feet per second

    5. +

  5. 16 feet per second

    6. +
+ +

What behavior would explain your finding?

+
    +

  1. After 1 second the ball is falling at a speed of 32 meters per second.

    2. +

  2. After 1 second the ball is moving upwards at a speed of 32 meters per second.

    3. +

  3. After 1 second the ball reaches its highest point and it stops for an instant.

    4. +

  4. After 1 second the ball is falling at a speed of 16 meters per second.

    5. +

  5. After 1 second the ball is moving upwards at a speed of 16 meters per second.

    6. +
+ + + + +

A function can only be differentiable at x=a if it is also continuous at x=a. But not all continuous functions are differentiable: when we have a corner in the graph of a the function, the function is continuous at the corner point, but it is not differentiable at that point!

+ + + +

+ In , we said that a function is not differentiable when the limit that defines it does not exist. In this activity we will study differentiability analytically.

+ +

+ Consider the following continuous function

+ g(x) = \begin{cases} + x + 2 \amp x \leq 2 \\ + x^2 \amp x \gt 2 + \end{cases} + +

Consider the interval [2,2+h]. When h \lt 0 , the interval falls under the first definition of g(x) and the derivative is always equal to 1. What is the derivative function for x values greater than 2? Show that at x=2 the value of this derivative is not equal to 1 and so g(x) is not differentiable at x=2. +

+ + +

+ Consider the following discontinuous function

+ g(x) = \begin{cases} + x + 2 \amp x \leq 2 \\ + x \amp x \gt 2 + \end{cases} + +

On both sides of x=2 it seems that the slope is the same, but this function is still not differentiable at x=2. Notice that g(2)=4. When h \gt 0 , the interval [2,2+h]falls under the second definition of g(x), but g(2) is always fixed at 4. Compute the difference quotient \frac{g(2+h)-g(2)}{h} assuming that h \gt 0 and notice that this does not simplify as expected! Moreover, if you take the limit as h\to 0, you will get infinity and not the expected slope of 1! +

+ + +

+ Consider the following function

+ g(x) = \begin{cases} + ax + 2 \amp x \leq 2 \\ + bx^2 \amp x \gt 2 + \end{cases} + +

where a,b are some nonzero parameters you will find. Find an equation in a,b that needs to be true if we want the function to be continuous at x=2. Also, find an equation in a,b that needs to be true if we want the function to be differentiable at x=2. Solve the system of two linear equations... you should find that a=-2 and b=-1/2 are the only values that make the function differentiable (and continuous!). +

+ + + + + + Videos +
+
+ + +
\ No newline at end of file diff --git a/calculus/source/02-DF/03.ptx b/calculus/source/02-DF/03.ptx new file mode 100644 index 00000000..6e3faf9f --- /dev/null +++ b/calculus/source/02-DF/03.ptx @@ -0,0 +1,292 @@ + + +
+ Elementary derivative rules (DF3) + + + + + Activities + + We know how to find the derivative function using the limit definition of the derivative. From the activities in the previous section, we have seen that this process gets cumbersome when the functions are more complicated. In this section we will discuss shortcuts to calculate derivatives, known as differentiation rules. + + + +

+ In this activity we will try to deduce a rule for finding the derivative of a power function. Note, a power function is a function of the form f(x) = x^{n} where n is any real number. +

+ + +

Using the limit definition of the derivative, what is f'(x) for the power function f(x) = x?

+
    +

  1. -1

    2. +

  2. 1

    3. +

  3. 0

    4. +

  4. Does not exist

    5. +
+ + + + + +

Using the limit definition of the derivative, what is f'(x) for the power function f(x) = x^{2}?

+
    +

  1. 0

    2. +

  2. -2x

    3. +

  3. 2x

    4. +

  4. 2x+1

    5. +
+ + + + + +

Using the limit definition of the derivative, what is f'(x) for the power function f(x) = x^{3}?

+
    +

  1. 3x^2

    2. +

  2. -3x^2

    3. +

  3. 3x^2-3x

    4. +

  4. -3x^2+3x

    5. +
+ + + + + +

WITHOUT using the limit definition of the derivative, what is your best guess for f'(x) when f(x) = x^{4}? (See if you can find a pattern from the first three tasks of this activity.)

+
    +

  1. 3x^2

    2. +

  2. 3x^3

    3. +

  3. 4x^2

    4. +

  4. 4x^3

    5. +
+ + + + + + The Power Rule

The derivative of the power functionf(x) = x^n, for any real number n, is +

+ f'(x) = n \, x^{n-1}. + + + +

+ We have been using f'(x), read f prime, to denote a derivative of the function f(x). There are other ways to denote the derivative of y=f(x): y' or \frac{df}{dx}, pronounced dee-f dee-x. If you want to take the derivative of f'(x), y', or \frac{df}{dx} to get the second derivative of f(x), the notation is f''(x), y'', or \frac{d^2f}{dx^2}.

+ + + + +

Using , which of the following statement(s) are true? For those statements that are wrong, give the correct derivative.

+
    +
  1. The derivative of y = x^{10} is y' = 10x^{11}.
    2. +
  2. The derivative of y = x^{-8} is y' = -8x^{-9}.
    3. +
  3. The derivative of y = x^{100} is y' = 100x^{99}.
    4. +
  4. The derivative of y = x^{-17} is y' = -17x^{-16}.
    5. +
+ + + + The Derivative of a Constant Function

If f(x) = c, for some constant real number c, then f'(x) = 0. +

+ + + +

Using , which of the following statement(s) are true? Note: Pay attention to the independent variable (the input) of the function. +

    +
  1. The derivative of y(x) = 10 is y'(x) = 9.
    2. +
  2. The derivative of y(t) = x is y'(t) = 0.
    3. +
  3. The derivative of y(a) = x^2 is y'(a) = 2x.
    4. +
  4. The derivative of y(x) = -5 is y'(x) = -4.
    5. +
+

+ + + + The Scalar Multiple Rule

If c is a real number and f(x) is a differentiable function, then using the Scalar Multiple Law for limits, we have that + \frac{d}{dx}\left[cf(x)\right] = c\frac{d}{dx}\left[f(x)\right] +

+ + + +

What is the derivative of the function y(x) = 12x^{2/3}? +

    +
  1. y'(x) = 8x^{5/3}.
    2. +
  2. y'(x) = 18x^{-1/3}.
    3. +
  3. y'(x) = 8x^{-1/3}.
    4. +
  4. y'(x) = 18x^{5/3}.
    5. +
+

+ + + + The Sum/Difference Rule

If f(x) and g(x) are both differentiable, then using the Sum/Differences Law for limits, we have that + \frac{d}{dx}\left[f(x) \pm g(x)\right] = \frac{d}{dx}\left[f(x)\right]\pm\frac{d}{dx}\left[g(x)\right] +

+ + + +

What are the first and second derivatives for the arbitrary quadratic function given by f(x) = ax^2 + bx + c, where a,\,b,\,c are any real numbers?

+
    +
  1. f'(x) = 2ax + bx + c, \, f''(x)=2a +b.
    2. +
  2. f'(x) = 2x + 1, \, f''(x)=2.
    3. +
  3. f'(x) = 2ax + b , \, f''(x)=2a.
    4. +
  4. f'(x) = ax + b, \, f''(x)=a.
    5. +
+ + + +

We can look at power functions with fractional exponents like f(x)= x^{\frac{1}{4}}=\sqrt[4]{x} or with negative exponents like g(x)= x^{-4} = \frac{1}{x^4}. What is the derivative of these two functions?

 +
    +
  1. f'(x) = \frac{1}{4 \sqrt[4]{x^3}}, \, g'(x) = \frac{-4}{x^3}.
    2. +
  2. f'(x) = \frac{1}{4} \sqrt[4]{x^3}, \, g'(x) = \frac{-4}{x^5}.
    3. +
  3. f'(x) = \frac{1}{4} \sqrt[4]{x^3}, \, g'(x) = \frac{-4}{x^3}.
    4. +
  4. f'(x) = \frac{1}{4 \sqrt[4]{x^3}}, \, g'(x) = \frac{-4}{x^5}.
    5. +
+ + + + The Derivative of an Exponential Function +

+The derivative of the exponential functionf(x) = b^x, for any positive number b, is +f'(x) =\ln(b) \, b^x=\log(b)\,b^x=\log_e(b)b^x. +

(In this book, we use both \ln and \log to denote the natural logarithm with base e. +While \log is sometimes used to denote the common logarithm base 10, we prefer to write +\log_{10} in that case.)

+

+ + +

A special case of is when b = e, where e is the base of the natural logarithm function. In this case let f(x) = e^x. Then + f'(x) =\ln(e) \, e^{x} = e^{x}. + So f(x)=e^x is a special function for which f'(x)=f(x).

+ + + +

The first derivative of the function g(x) = x^e + e^{x} is given by g'(x) = ex^{e-1} + e^{x}. What is the second derivative of g(x)?

+
    +
  1. g''(x) = x^{e} + e^{x}.
    2. +
  2. g''(x) = e(e-1)x^{e-2} + e^x.
    3. +
  3. g''(x) = ex^{e-1} + e^x.
    4. +
  4. g''(x) = e^x.
    5. +
+ + + + + The Derivative of the Sine and Cosine Functions

If f(x) = \sin(x), then f'(x) = \cos(x). If f(x) = \cos(x), then f'(x) = -\sin(x).

+ + + +

The derivative of f(x) = 7\sin(x) + 2e^x + 3x^{1/3} - 2 is,

+
    +
  1. f'(x) = 7\cos(x) + 2e^{x} + x^{-2/3} - 2x.
    2. +
  2. f'(x) = 7\cos(x) + 2e^{x} + -2x^{-2/3} - 2.
    3. +
  3. f'(x) = -7\sin(x) + e^{x} + x^{-2/3}.
    4. +
  4. f'(x) = -7\cos(x) + 2e^{x}\ln(x) + x^{-2/3}.
    5. +
  5. f'(x) = 7\cos(x) + 2e^x + x^{-2/3}.
    6. +
+ + + + + + The Derivative of the Natural Log Function

If f(x) = \ln(x), then f'(x) = \frac{1}{x}.

+ + + +

Which of the following statements is NOT true?

+
    +
  1. The derivative of y = 2\ln(x) is y' = \frac{2}{x}.
    2. +
  2. The derivative of y = \frac{\ln(x)}{2} is y' = \frac{1}{2x}.
    3. +
  3. The derivative of y = \frac{2}{3}\ln(x) is y' = \frac{3}{2x}.
    4. +
  4. The derivative of y = \ln(x^{2}) is y' = \frac{2}{x}.
    5. +
+ + + + + +

+ Demonstrate and explain how to find the derivative of the following functions. + Be sure to explicitly denote which derivative rules (scalar multiple, sum/difference, etc.) you are using in your work. +

+ + +

+g(x) = 2 \, \cos\left(x\right) - 3 \, e^x +

+ + +

+h(w) = \sqrt[5]{w^7} + \frac{6}{w^{5}} +

+ + +

+f(t) = -4 \, t^{5} + 5 \, t^{3} + t - 8 +

+ + + + + + +

+ Suppose that the temperature (in degrees Fahrenheit) of a cup of coffee, t minutes after forgetting it on a bench outside, is given by the function + f(t) = 40 \, (0.5)^t + 50 + Find f(1) and f'(1) and try to interpret your result in the context of this problem.

+ + + +

In this activity you will use our first derivative rules to study the slope of tangent lines.

+

The graph of y=x^3-9x^2-16x+1 has a slope of 5 at two points. Find the coordinates of these points.

+

+Find the equations of the two lines tangent to the parabola y=(x-2)^2 which pass through the origin. You will want to think about slope in two ways: as the derivative at x=a and the rise over the run in a linear function through the origin and the point (a, f(a)). Use a graph to check your work and sketch the tangent lines on your graph. +

+ + + + +

+Find the values of the parameters a,b,c for the quadratic polynomial + q(x) = ax^2 +bx +c that best approximates the graph of f(x)=e^x at x=0. This means choosing a,b,c such that

+
    +
  • q(0) = f(0)
    • +
  • q'(0) = f'(0)
    • +
  • q''(0) = f''(0)
    • +
+

Hint: find the values of f(0),f'(0),f''(0). The values of q(0),q'(0),q''(0) at zero will involve some parameters. You can solve for these parameters using the equations above.

+ + + + + + Videos +
+
+ +
\ No newline at end of file diff --git a/calculus/source/02-DF/04.ptx b/calculus/source/02-DF/04.ptx new file mode 100644 index 00000000..ef1684b3 --- /dev/null +++ b/calculus/source/02-DF/04.ptx @@ -0,0 +1,400 @@ + + +
+ The product and quotient rules (DF4) + + + + + Activities + + + +

+ Let f and g be the functions defined by + f(t) = 2t^2 \, , \, g(t) = t^3 + 4t. +

+ +

+ Find f'(t) and g'(t). +

+ +

+ Let P(t) = 2t^2 \, (t^3 + 4t) and observe that P(t) = f(t) \cdot g(t). + Rewrite the formula for P by distributing the 2t^2 term. + Then, + compute P'(t) using the power, sum, and scalar multiple rules. +

+ + + +

+ True or false: P'(t) = f'(t) \cdot g'(t). +

+ + + + + Product Rule +

+ + If f and g are differentiable functions, + then their product P(x) = f(x) \cdot g(x) is also a differentiable function, and + + P'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x) + . +

+ + + + + + + +

+ The product rule is a powerful tool, but sometimes it isn’t necessary; a more elementary rule may suffice. For which of the following functions can you find the derivative without using the product rule? Select all that apply. + +

+ + + + + +
    +
  1. f(x)=e^x \sin x
    2. +
  2. f(x)=\sqrt{x}(x^3+3x-3)
    3. +
  3. f(x)=(4)(x^5)
    4. +
  4. f(x)=x \ln x
    5. + +
+ + + + +

+ Find the derivative of the following functions using the product rule. +

+ + f(x)=(x^2+3x)\sin x + f(x)=e^x \cos x + f(x)=x^2\ln x + + + + +

+ Let f and g be the functions defined by + f(t) = 2t^2 \, , \, g(t) = t^3 + 4t. +

+ +

+ Determine f'(t) and g'(t). (You found these previously in .) +

+ + + + +

+ Let Q(t) = \frac{t^3 + 4t}{2t^2} and observe that Q(t) = \frac{g(t)}{f(t)}. + Rewrite the formula for Q by dividing each term in the numerator by the denominator and use rules of exponents to write Q as a sum of scalar multiples of power functions. + Then, + compute Q'(t) using the sum and scalar multiple rules. +

+ + + +

+ True or false: Q'(t) = \frac{g'(t)}{f'(t)}. +

+ + + + + Quotient Rule +

+ + If f and g are differentiable functions, + then their quotient Q(x) = \frac{f(x)}{g(x)} is also a differentiable function for all x where g(x) \ne 0 and + + Q'(x) = \frac{ f'(x) \cdot g(x) - f(x) \cdot g'(x)}{g(x)^2} + . +

+ + + + + +

+ Just like with the product rule, there are times when we can find the derivative of a quotient using elementary rules rather than the quotient rule. For which of the following functions can you find the derivative without using the quotient rule? Select all that apply. + +

+ + + + + +
    +
  1. f(x) = \frac{6}{x^3}
    2. +
  2. f(x) = \frac{2}{\ln x}
    3. +
  3. f(x) = \frac{e^x}{\sin x}
    4. +
  4. f(x) = \frac{x^3+3x}{x}
    5. + +
+ + + + +

+ Find the derivative of the following functions using the quotient rule (or, if applicable, an elementary rule). + +

+ + f(x) = \frac{6}{x^3} + f(x) = \frac{2}{\ln x} + f(x) = \frac{e^x}{\sin x} + f(x) = \frac{x^3+3x}{x} + + + + +

+ Demonstrate and explain how to find the derivative of the following functions. + Be sure to explicitly denote which derivative rules (product, quotient, sum and difference, etc.) you are using in your work. +

+ + f(w)= -\frac{3 \, w^{2} + 5 \, w - 2}{\sin\left(w\right)} + g(t)= \frac{t^{2} + 6 \, t + 1}{t^{2}} + h(t)= -2 \, {\left(t^{2} + 3 \, t + 3\right)} \cos\left(t\right) + + + + + +

+ We have found the derivatives of \sin x and \cos x, + but what about the other trigonometric functions? It turns out that the quotient rule along + with some trig identities can help us! (See + Khan Academy + for a reminder of trig identities.) +

+ + + +

+ Consider the function f(x) = \tan x , and remember that + \tan x = \frac{\sin x}{\cos x}. + + +

+ +

What is the domain of f?

 +

Use the quotient rule to show that one expression for f'(x) is + + f'(x) = \frac{(\cos x)(\cos x) + (\sin x)(\sin x)}{(\cos x)^2} + .

 +

Which trig identity might be useful here to simplify this expression? + How can this identity be used to find a simpler form for f'(x)?

 +

Recall that \sec x = \frac{1}{\cos x}. + How can we express f'(x) in terms of the secant function?

 +

For what values of x is f'(x) defined? + How does this domain compare to the domain of f?

 + + + + + +

Let g(x) = \cot x , and recall that + \cot x = \frac{\cos x }{\sin x }. + + +

+ +

What is the domain of g(x)?

 +

Use the quotient rule to develop a formula for g'(x) that is expressed completely in terms of \sin x and \cos x. +

 +

Use other relationships among trigonometric functions to write g'(x) only in terms of the cosecant function.

 +

What is the domain of g'(x)? + How does this domain compare to the domain of g'(x)?

 + + + + + +

Let h(x) = \sec x , and recall that + \sec x = \frac{1}{\cos x }. + + +

+ +

What is the domain of h(x)?

 +

Use the quotient rule to develop a formula for h'(x) that is expressed completely in terms of \sin x and \cos x. +

 +

Use other relationships among trigonometric functions to write h'(x) only in terms of the the tangent and secant functions.

 +

What is the domain of h'(x)? + How does this domain compare to the domain of h'(x)?

 + + + + + +

Let p(x) = \csc x , and recall that + \csc x = \frac{1}{\sin x }. + + +

+ +

What is the domain of p(x)?

 +

Use the quotient rule to develop a formula for p'(x) that is expressed completely in terms of \sin x and \cos x. +

 +

Use other relationships among trigonometric functions to write h'(x) only in terms of the the cotangent and cosecant functions.

 +

What is the domain of p'(x)? + How does this domain compare to the domain of p'(x)?

 + + + +

We can now summarize the derivatives of all six trigonometric functions.

+
    +
  • \frac{d}{dx} \sin x = \cos x
    • +
  • \frac{d}{dx} \cos x = -\sin x
    • +
  • \frac{d}{dx} \tan x = (\sec x)^2
    • +
  • \frac{d}{dx} \cot x = -(\csc x)^2
    • +
  • \frac{d}{dx} \sec x = \sec x \tan x
    • +
  • \frac{d}{dx} \csc x = -\csc x \cot x
    • +
+ + + + + + +

+ Consider the functions + f(x)=3 \, \cos\left(x\right), \, \, g(x)=x^{2} + 3 \, e^{x} + and the function h(x) for which a table of values is given. + +\begin{array}{c|ccc} +x & -1& 0& 2 \\\hline +h(x) & -4& -1& 3 \\\hline +h'(x) & 0& -1& 1 \\ +\end{array} + +In answering the following questions, be sure to explicitly denote which derivative rules (product, quotient, sum/difference, etc.) you are using in your work. +

+ + +

+ Find the derivative of f(x)\cdot g(x). +

+ + +

+ Find the derivative of \displaystyle \frac{f(x)}{g(x)}. +

+ + +

+ Find the value of the derivative of f(x) \cdot h(x) at x=-1. +

+ + +

+ Find the value of the derivative of \displaystyle \frac{g(x)}{h(x)} at x=0. +

+ + +

+ Consider the function + r(x) = 3 \, \cos\left(x\right) \cdot x . + Find r'(x), r''(x), r'''(x), and r^{(4)}(x) so the first, second, third, and fourth derivative of r(x). What pattern do you notice? What do you expect the twelfth derivative of r(x) to be? +

+ + + + + + +

Differentiate y = \displaystyle \frac{e^x}{x}, y = \displaystyle \frac{e^x}{x^2}, y = \displaystyle \frac{e^x}{x^3}. Simplify your answers as much as possible. +

+

What do you expect the derivative of y = \displaystyle \frac{e^x}{x^n} to be? Prove your guess! +

+

What do your answers above tell you above the shape of the graph of y = \displaystyle \frac{e^x}{x^n}? Study how the sign of the numerator and the denominator change in the first derivative to determine when the behavior changes! +

+ + + + + +

The quantity q of skateboards sold depends on the selling price p of a skateboard, so we write q=f(p). You are given that + f(140) = 15000, \, \, \, f'(140)= -100 +

+ +

+ What does the data provided tell you about the sales of skateboards? +

+ +

+ The total revenue, R, earned by the sale of skateboards is given by R=q \cdot p = f(p) \cdot p. Explain why. +

+ +

+ Find the derivative of the revenue when p=140, so find the value of + \frac{dR}{dp}\Big|_{p=140}.

+ +

+ What is the sign of the quantity above? What do you think would happen to the revenue if the price was changed from $140 to $141?

+ + + + +

+ Let f(v) be the gas consumption in liters per kilometer (l/km) of a car going at velocity v kilometers per hour (km/hr). So if the car is going at velocity v, then f(v) tells you how many liters of gas the car uses to go one kilometer. You are given the following data + f(50)=0.04, \, \, \, f'(50)=0.0004 +

+

Let g(v) be the distance (in kilometers) that the same car covers per liter of gas at velocity v. What are the units of the output of g(v)? Use these units to infer how to write g(v) in terms of f(v), then find g(50) and g'(50). +

+

Let h(v) be the gas consumption over time, so the liters of gas consumed per hour by the same car going at velocity v. What are the units of the output of h(v)? Use these units to infer how to write h(v) in terms of f(v), then find h(50) and h'(50). +

+

How would you explain the practical meaning of your findings to a driver who knows no calculus? +

+ + + + + Videos +
+
+ + +
\ No newline at end of file diff --git a/calculus/source/02-DF/05.ptx b/calculus/source/02-DF/05.ptx new file mode 100644 index 00000000..22d7b50a --- /dev/null +++ b/calculus/source/02-DF/05.ptx @@ -0,0 +1,308 @@ + + +
+ The chain rule (DF5) + + + + + Activities + +

When we consider the consider the composition f \circ g of the function f with the function g, we mean the composite function f(g(x)), where the function g is applied first and then f is applied to the output of g. We also call f the outside function whilst g is the inside function.

+ + + +

Consider the function f(x) = -x^2+5 and g(x)=2x-1. Which of the following is a formula for f(g(x))?

+
    +

  1. -4x^2 +4x +4

    2. +

  2. 4x^2 -4x +5

    3. +

  3. -2x^2 +9

    4. +

  4. -2x^2 +4

    5. +
+ + +

One of the options above is a formula for g(f(x)). Which one?

+ + + + +

Consider the composite function f(g(x))= \sqrt{e^x}. Which function is the outside function f(x) and which one is the inside function g(x)?

+
    +

  1. f(x)=x^2 \, , \, g(x)=e^x

    2. +

  2. f(x)=\sqrt{x} \, , \, g(x)=e^x

    3. +

  3. f(x)=e^x \, , \, g(x)=\sqrt{x}

    4. +

  4. f(x)=e^x \, , \, g(x)=x^2

    5. +
+ + +

Using properties of exponents, we can rewrite the original function as \displaystyle e^{\frac{x}{2}}. Using this new expression, what is your new inside function and your new outside function?

 + +

Consider the function \displaystyle e^{\sqrt{x}}. In this case, what are the inside and outside functions?

+ + +

In this activity we will build the intuition for the chain rule using a real-world scenario and differential notation for derivatives. Consider the following scenario.

+

My neighborhood is being invaded! The squirrel population grows based on acorn availability, at a rate of 2 squirrels per bushel of acorns. Acorn availability grows at a rate of 100 bushels of acorns per week. How fast is the squirrel population growing per week?

+ + +

The scenario gives you information regarding the rate of growth of s(a), the squirrel population as a function of acorn availability (measured in bushels). What is the current value of \frac{ds}{da}?

+
    +

  1. 2

    2. +

  2. 100

    3. +

  3. 200

    4. +

  4. 50

    5. +
+ + +

The scenario gives you information regarding the rate of growth of a(t), the acorn availability as a function of time (measured in weeks). What is the current value of \frac{da}{dt}?

+
    +

  1. 2

    2. +

  2. 100

    3. +

  3. 200

    4. +

  4. 50

    5. +
+ + +

Given all the information provided, what is your best guess for the value of \frac{ds}{dt}, the rate at which the squirrel population is growing per week?

+
    +

  1. 2

    2. +

  2. 100

    3. +

  3. 200

    4. +

  4. 50

    5. +
+ + +

Given your answers above, what is the relationship between \frac{ds}{da}, \frac{da}{dt}, \frac{ds}{dt}?

+ + + + + + Chain Rule +

+ When looking at the composite function f(g(x)), we have that + \frac{d}{dx}\Big(f( g(x) )\Big)= f'(g(x)) \cdot g'(x). + Using differential notation, if we consider the composite function (v \circ u)(x), we have that + \frac{dv}{dx}= \frac{dv}{du} \cdot \frac{du}{dx} . + This is known as the chain rulechain rule. +

+ + +

It is important to consider the input of a function when taking the derivative! In fact, f'(g(x)) and f'(x) are different functions... So computing \frac{dv}{dx} gives a different result than computing \frac{dv}{du} .

+ + + +

Consider the function f(x) = -x^2+5 and g(x)=2x-1. Notice that f(g(x))= -4x^2 +4x +4 . Which of the following is the derivative function of the composite function f(g(x))?

+
    +

  1. -8x +4

    2. +

  2. -4x

    3. +

  3. -2x

    4. +

  4. 2

    5. +
+ + +

One of the options above is a formula for f'(x) \cdot g'(x) . Which one? Notice that this is not the same as the derivative of f(g(x))!

+ + + +

Consider the composite function \displaystyle h(x) =\sqrt{e^x} = e^{\frac{x}{2}}. For each of the two expressions, find the derivative using the chain rule. Which of the following expressions are equal to h'(x)? Select all!

+
    +

  1. \displaystyle \frac{1}{2} \left( e^x \right)^{\frac{-1}{2}} \cdot e^x

    2. +

  2. \displaystyle \frac{1}{2} \left( e^x \right)^{\frac{3}{2}} \cdot e^x

    3. +

  3. \displaystyle \frac{1}{2} e^{\frac{-x}{2}}

    4. +

  4. \displaystyle e^{\frac{x}{2}} \cdot \frac{1}{2}

    5. +

  5. \displaystyle \frac{1}{2} \sqrt{e^x}

    6. +

  6. \displaystyle\sqrt{e^x} \cdot e^x

    7. +
+ + + +

Below you are given the graphs of two functions: a(x) and b(x). Use the graphs to compute vaules of composite functions and of their derivatives, when possible (there are points where the derivative of these functions is not defined!). Notice that to compute the derivative at a point, you first want to find the derivative as a function of x and then plug in the input you want to study.

+
+ + +\begin{tikzpicture} +\begin{axis}[ + axis lines=middle, + xmin=-5,xmax=5,ymin=-3,ymax=3, +% xtick={-6.28, -4.71,...,6.28}, +% ytick={-4, -3,...,4}, +% xticklabels={$-2\pi$, $-3\pi/2$, $-\pi$ , $-\pi/2$, 0 , $\pi/2$, $\pi$, $3\pi/2$, $-2\pi$}, + xlabel={$x$}, + ylabel={$y$} + ] + \addplot[domain=-4:-0.1,blue, ultra thick,samples=500] {-2} ; + \addplot[domain=0:4,blue,ultra thick, samples=500] {2} ; + \addplot [only marks, blue] coordinates {(0,2)}; +\addplot [only marks, blue, mark=o] coordinates {(0,-2)}; + \addplot[mark=none, color=blue, nodes near coords={$a(x)$}] coordinates {(-4,2)}; +\end{axis} +\end{tikzpicture} +\begin{tikzpicture} +\begin{axis}[ + axis lines=middle, + xmin=-5,xmax=5,ymin=-3,ymax=3, +% xtick={-6.28, -4.71,...,6.28}, +% ytick={-4, -3,...,4}, +% xticklabels={$-2\pi$, $-3\pi/2$, $-\pi$ , $-\pi/2$, 0 , $\pi/2$, $\pi$, $3\pi/2$, $-2\pi$}, + xlabel={$x$}, + ylabel={$y$} + ] + \addplot[domain=-4:0,blue, ultra thick, samples=500] {2+x} ; + \addplot[domain=0:4,blue, ultra thick, samples=500] {2-x} ; + \addplot[mark=none, color=blue, nodes near coords={$b(x)$}] coordinates {(4,2)}; +\end{axis} +\end{tikzpicture} + + + The graphs of a(x) and b(x) +
+ + +

Notice that the derivative of a \circ b is given by a'(b(x)) \cdot b'(x), so the derivative of a \circ b at x= 4 is given by the quantity a'(b(4)) \cdot b'(4) = a'(-2) \cdot b'(4), because b(4)=-2. Using the graphs to compute slopes, what is the derivative of a \circ b at x= 4?

+
    +

  1. 0

    2. +

  2. -1

    3. +

  3. 1

    4. +

  4. -2

    5. +

  5. 2

    6. +

  6. The derivative does not exist at this point.

    7. +
+ + +

Which of the following values is the derivative of a \circ b at x=2 ?

+
    +

  1. 0

    2. +

  2. -1

    3. +

  3. 1

    4. +

  4. -2

    5. +

  5. 2

    6. +

  6. The derivative does not exist at this point.

    7. +
+ + +

Which of the following values is the derivative of b \circ a (different order!) at x=-2 ?

+
    +

  1. 0

    2. +

  2. -1

    3. +

  3. 1

    4. +

  4. -2

    5. +

  5. 2

    6. +

  6. The derivative does not exist at this point.

    7. +
+ + + + + +

In this activity you will study the derivative of \cos^n(x) for different powers n.

+ +

Consider the function \cos^2(x) = \left( \cos(x) \right)^2. Combining power and chain rule, what do you get if you differentiate \cos^2(x)?

+
    +

  1. -\cos^2(x)\sin(x)

    2. +

  2. -\cos^2(x)\sin(x)

    3. +

  3. 2\cos(x)\sin(x)

    4. +

  4. - 2\cos(x)\sin(x)

    5. +
+ +

Consider the function \cos^3(x). Find its derivative.

+

Consider the function \cos^n(x), for n any number. Find the general formula for its derivative.

+ + + + + +

In this activity you will study the derivative of b^{\cos(x)} for different bases b.

+ +

Consider the function e^{\cos(x)} . Combining exponential and chain rule, what do you get if you differentiate e^{\cos(x)}?

+
    +

  1. e^{\cos(x)}

    2. +

  2. - e^{\cos(x)} \sin(x)

    3. +

  3. e^{-\sin(x)}

    4. +

  4. e^{\cos(x)}\sin(x)

    5. +
+ +

Consider the function 2^{\cos(x)}. Find its derivative.

+

Consider the function b^{\cos(x)}, for b any positive number. Find the general formula for its derivative.

+ + +

Remember that exponential and power functions obey very different differentiation rules. This behavior continues when we consider composite function. The composite power function f(x)^3 has derivative

+ 3 [f(x)]^2 \cdot f'(x) +

but the composite exponential function 3^{f(x)} has derivative

+ \ln(3) \, 3^{f(x)} \cdot f'(x) + + + + +

+ Demonstrate and explain how to find the derivative of the following functions. + Be sure to explicitly denote which derivative rules (chain, product, quotient, sum/difference, etc.) you are using in your work. +

+
    +
  1. f(x)= -{\left(4 \, x - 3 \, e^{x} + 4\right)}^{3}
    2. +
  2. k(w)= 9 \, \cos\left(w^{\frac{7}{5}}\right)
    3. +
  3. h(y)= -3 \, \sin\left(-5 \, y^{2} + 2 \, y - 5\right)
    4. +
  4. g(t)= 9 \, \cos\left(t\right)^{\frac{7}{5}}
    5. +
+ + +
    +
  1. f' (x)= 3 \, {\left(4 \, x - 3 \, e^{x} + 4\right)}^{2} {\left(3 \, e^{x} - 4\right)}
    2. +
  2. k' (w)= -\frac{63}{5} \, w^{\frac{2}{5}} \sin\left(w^{\frac{7}{5}}\right)
    3. +
  3. h' (y)= 6 \, {\left(5 \, y - 1\right)} \cos\left(-5 \, y^{2} + 2 \, y - 5\right)
    4. +
  4. g' (t)= -\frac{63}{5} \, \cos\left(t\right)^{\frac{2}{5}} \sin\left(t\right)
    5. +
+ + + + +

Notice that + \displaystyle\left(\frac{f(x)}{g(x)}\right) = +\left( f(x)\cdot g(x)^{-1}\right) + Use this observation, the chain rule, the product rule, and the power rule (plus some fraction algebra) to deduce the quotient rule in a new way! +

+ + + +

Remember my neighborhood squirrel invasion? The squirrel population grows based on acorn availability, at a rate of 2 squirrels per bushel of acorns. Acorn availability grows at a rate of 100 bushels of acorns per week. Considering this information as pertaining to the moment t=0, you are given the following possible model for the squirrel:

+ s(a(t))= 2a(t)+10 += 2\left(50\sin(2t)+60\right)+10. + + +

Check that the model satisfies the data \frac{ds}{da}=2 and \frac{da}{dt}\big|_{t=0} = 100

+ + + +

Find the derivative function \frac{ds}{dt} and check that \frac{ds}{dt}|_{t=0} = 200.

+ + +

According to this model, what is the maximum and minimum squirrel population? What is the fastest rate of increase and decrease of the squirrel population? When will these extremal scenarions occur?

+ + + + +

Suppose that a fish population at t months is approximated by + P(t) = 100 \cdot 4^{0.05 t} +

+ + +

Find P(10) and use units to explain what this value tells us about the population. +

+ + +

Find P'(10) and use units to explain what this value tells us about the population. (If you want to avoid using a calculator, you can use the approximation \ln(4) = 1.4.)

+ + + + + Videos +
+
+ + +
\ No newline at end of file diff --git a/calculus/source/02-DF/06.ptx b/calculus/source/02-DF/06.ptx new file mode 100644 index 00000000..de7c2042 --- /dev/null +++ b/calculus/source/02-DF/06.ptx @@ -0,0 +1,247 @@ + + +
+ Differentiation strategy (DF6) + + + + + Activities + + + +

Consider the functions defined below:

+ + f(x)=\sin((x^2+3x)\cos(2x)) + g(x)=\sin(x^2+3x)\cos(2x) + +

+ What do you notice that is similar about these two functions? +

+

+ What do you notice that is different about these two functions? +

+

+ Imagine that you are sorting functions into different categories based on how you would differentiate them. In what category (or categories) might these functions fall? +

+ + + + +

To take a derivative, we need to examine how the function is built and then proceed accordingly. Below are some questions you might ask yourself as you take the derivative of a function, especially one where multiple rules might need to be used: +

    +
  1. How is this function built algebraically? What kind of function is this? What is the big picture?
    2. +
  2. Where do you start?
    3. +
  3. Is there an easier or more convenient way to write the function?
    4. +
  4. Are there products or quotients involved?
    5. +
  5. Is this function a composition of two (or more) elementary functions? If so, what are the outside and inside functions?
    6. +
  6. What derivative rules will be needed along the way?
    7. +
+

+ + + + + + + +

Consider the function f(x)=x^3\sqrt{3-8x^2}.

+ + +

+ You will need multiple derivative rules to find f'(x). Which rule would need to be applied first? In other words, what is the big picture here? +

    +
  1. Chain rule
    2. +
  2. Power rule
    3. +
  3. Product rule
    4. +
  4. Quotient rule
    5. +
  5. Sum/difference rule
    6. +
+

+ +

+ What other rules would be needed along the way? Select all that apply. +

    +
  1. Chain rule
    2. +
  2. Power rule
    3. +
  3. Product rule
    4. +
  4. Quotient rule
    5. +
  5. Sum/difference rule
    6. +
+

+ +

Write an outline of the steps needed if you were asked to take the derivative of f(x).

+ + + + + + +

Consider the function f(x)= \left(\frac{\ln x}{(3x-4)^3} \right)^5.

+ + +

+ You will need multiple derivative rules to find f'(x). Which rule would need to be applied first? In other words, what is the big picture here? +

    +
  1. Chain rule
    2. +
  2. Power rule
    3. +
  3. Product rule
    4. +
  4. Quotient rule
    5. +
  5. Sum/difference rule
    6. +
+

+ +

+ What other rules would be needed along the way? Select all that apply. +

    +
  1. Chain rule
    2. +
  2. Power rule
    3. +
  3. Product rule
    4. +
  4. Quotient rule
    5. +
  5. Sum/difference rule
    6. +
+

+ +

Write an outline of the steps needed if you were asked to take the derivative of f(x).

+ + + + + + +

Consider the function f(x)= \sin(\cos(\tan(2x^3-1))).

+ + +

+ You will need multiple derivative rules to find f'(x). Which rule would need to be applied first? In other words, what is the big picture here? +

    +
  1. Chain rule
    2. +
  2. Power rule
    3. +
  3. Product rule
    4. +
  4. Quotient rule
    5. +
  5. Sum/difference rule
    6. +
+

+ +

+ What other rules would be needed along the way? Select all that apply. +

    +
  1. Chain rule
    2. +
  2. Power rule
    3. +
  3. Product rule
    4. +
  4. Quotient rule
    5. +
  5. Sum/difference rule
    6. +
+

+ +

Write an outline of the steps needed if you were asked to take the derivative of f(x).

+ + + + + +

Consider the function f(x)= \frac{x^2 e^x}{2x^3-5x+\sqrt{x}}.

+ + +

+ You will need multiple derivative rules to find f'(x). Which rule would need to be applied first? In other words, what is the big picture here? +

    +
  1. Chain rule
    2. +
  2. Power rule
    3. +
  3. Product rule
    4. +
  4. Quotient rule
    5. +
  5. Sum/difference rule
    6. +
+

+ +

+ What other rules would be needed along the way? Select all that apply. +

    +
  1. Chain rule
    2. +
  2. Power rule
    3. +
  3. Product rule
    4. +
  4. Quotient rule
    5. +
  5. Sum/difference rule
    6. +
+

+ +

Write an outline of the steps needed if you were asked to take the derivative of f(x).

+ + + + + + +

Find the derivative of the following functions. For each, include an explanation of the steps involved that references the algebraic structure of the function.

+ + + f(x) = e^{5x}(x^2+7^x)^3 + f(x) = \left( \frac{3x + 1}{2x^{6} - 1} \right)^{ 5 } + f(x) = \sqrt{\cos\left(2 \, x^{2} + x\right)} + f(x) = \tan(xe^x) + + + + + + +

+Demonstrate and explain how to find the derivative of the following functions. +Be sure to explicitly denote which derivative rules (constant multiple, sum/difference, etc.) +you are using in your work. +

+ + + +

+ f(y) = \sqrt{\cos\left(6 \, y^{4} - 6 \, y\right)} +

+ + + + + +

+ g(t) = \left( \frac{5 \, t^{3} + 2}{4 \, t^{4} - 3} \right)^{ 4 } +

+ + + + + +

+ h(x) = -{\left(5 \, x^{4} - 7 \, x^{3}\right)}^{5} x^{\frac{1}{4}} +

+ + + + + + + + Videos +
+
+ + +
\ No newline at end of file diff --git a/calculus/source/02-DF/07.ptx b/calculus/source/02-DF/07.ptx new file mode 100644 index 00000000..8dcd4b65 --- /dev/null +++ b/calculus/source/02-DF/07.ptx @@ -0,0 +1,155 @@ + + +
+ Differentiating implicitly defined functions (DF7) + + + + + Activities + + Many of the equations that has been discussed so far fall under the category of an explicit equation. An explicit equation is one in which the relationship between x and y is given explicitly, such as y = f(x). In this section we will examine when the relationship between x and y is given implicity. An implicit equation looks like f(x,y) = g(x,y) where both sides of the equation may depend on both x and y. + + + + Note that if we are taking the derivative of f(x) with respect to x, then + \frac{d}{dx}(f(x)) = f'(x). + However, if we are taking the derivative of g(y(x)) with respect to x, then + \frac{d}{dx}(g(y)) = g'(y) \cdot \frac{dy}{dx}. + + + + +

+ For this activity we want to find the equation of a tangent line for a circle with radius 5 centered at the origin, x^2+y^2 = 25, at the point (-3,-4). +

+ + + +

The derivative with respect to x for the equation of the circle is given by which expression.

+
    +

  1. 2x + 2y\frac{dy}{dx} = 25

    2. +

  2. 2x + y\frac{dy}{dx} = 0

    3. +

  3. 2x + 2y\frac{dy}{dx} = 0

    4. +

  4. 2x + 2\frac{dy}{dx} = 25

    5. +
+ + + + +

Solving for \frac{dy}{dx} gives?

+
    +

  1. \frac{dy}{dx} = \frac{25-2x}{2y}

    2. +

  2. \frac{dy}{dx} = -\frac{2x}{y}

    3. +

  3. \frac{dy}{dx} = -\frac{x}{y}

    4. +

  4. \frac{dy}{dx} = \frac{25-2x}{2}

    5. +
+ + + +

Plug the point (-3,-4) into the expression found above for the derivative to get the slope of the tangent line.

+ + + + +

Use the value for the slope of the tangent line to obtain the equation of the tangent line.

+ + + + + + +

The curve given in is an example of an astroid. The equation of this astroid is x^{2/3} + y^{2/3} = 3^{2/3}. What is the derivative with respect x for this astroid? (Solve for \frac{dy}{dx}).

+ +
+ + Graph of x^{2/3} + y^{2/3} = 3^{2/3}. +
+
    +

  1. \frac{dy}{dx} = \frac{x^{-1/3}}{y^{-1/3}}

    2. +

  2. \frac{dy}{dx} = \frac{y^{-1/3}}{x^{-1/3}}

    3. +

  3. \frac{dy}{dx} = \frac{3^{-1/3}-x^{-1/3}}{y^{-1/3}}

    4. +

  4. \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}}

    5. +
+ + + + +

An example of a lemniscate is given in . The equation of this lemniscate is (x^{2} + y^{2})^2 = x^2 - y^2. What is the derivative with respect x for this lemniscate? (Solve for \frac{dy}{dx}).

+ +
+ + Graph of (x^{2} + y^{2})^2 = x^2 - y^2. +
+
    +

  1. \frac{dy}{dx} = \frac{x(1-2(x^2+y^2))}{y+2(x^2+y^2)}

    2. +

  2. \frac{dy}{dx} = \frac{x(1-2(x^2+y^2))}{y(1+2(x^2+y^2))}

    3. +

  3. \frac{dy}{dx} = \frac{y(1+2(x^2+y^2))}{x(1-2(x^2+y^2))}

    4. +

  4. \frac{dy}{dx} = \frac{y+2(x^2+y^2)}{x(1-2(x^2+y^2))}

    5. +
+ + + + +

+Explain how to use implicit differentiation to find +\frac{dy}{dx} for each of the following equations. +

+ +

-5 \, x^{5} - 5 \, \cos\left(y\right) = 3 \, y^{4} + 2

 +

-5 \, y e^{x} + 5 \, \sin\left(x\right) = 0

 + + + + + + +

To take the derivative of some explicit equations you might need to make it an implicit equation. For this activity we will find the derivative of y = x^x. Make the equation an implicit equation by taking natural logarithm of both sides, this gives \ln(y) = x\ln(x). Knowing this, what is \frac{dy}{dx}? This process to find a derivative is known as logarithmic differentiation.

+ +
    +

  1. \frac{dy}{dx} = x^x(\ln(x) + 1)

    2. +

  2. \frac{dy}{dx} = \frac{(\ln(x)+1)}{x^x}

    3. +

  3. \frac{dy}{dx} = x^x(\ln(x) + x)

    4. +

  4. \frac{dy}{dx} = \frac{(\ln(x)+x)}{x^x}

    5. +
+ + + + + Valerie is building a square chicken coop with side length x. Because she needs a separate place for the rooster, she needs to put fence around the square and also along the diagonal line shown. The fence costs $20 per linear meter, and she has a budget of $900. +
+ + A diagram of the chicken coop. +
+ +

Which of the following equations gives the relationship between x and y? Make sure you can explain why!

+
    +

  1. 20x + \frac{80x}{\cos(y)}=900

    2. +

  2. 80x + \frac{20x}{\cos(y)}=900

    3. +

  3. 80x + \frac{20x}{\sin(y)}=900

    4. +

  4. 20x + \frac{80x}{\sin(y)}=900

    5. +
+ +

If Valerie builds the coop with y=\pi/3 (and wants to use her whole budget), find what side length x she needs to use.

 +

Find the slope of the curve at this point and interpret what this value tells Valerie.

 + + + + + Videos +
+
+ + +
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+ Differentiating inverse functions (DF8) + + + + + Activities + + + + +

Let f^{-1} be the inverse function of f. The relationship between a function and its inverse can be expressed with the identity +

+f(f^{-1}(x))=x. + + + +

In this activity you will use implicit differentiation and the inverse function identity in to find the derivative of y = \ln(x).

 +

Suppose that y=\ln(x). Then we have that

+ e^y = x . +

Using implicit implicit differentiation, what do you get?

+
    +

  1. \frac{dy}{dx} = \frac{x}{y}

    2. +

  2. \frac{dy}{dx} = \frac{1}{e^x}

    3. +

  3. \frac{dy}{dx} = \frac{x}{e^y}

    4. +

  4. \frac{dy}{dx} = \frac{1}{e^y}

    5. +
+ +

Notice that we started with the relationship e^y=x. Use this to simplify \frac{dy}{dx}. You should get that when y=\ln(x) we have that \frac{dy}{dx} = \frac{1}{x}... as expected!

 + + + +

In this activity we will try to find a general formula for the derivative of the inverse function. Let g be the inverse function of f. We have also used the notation f^{-1} before, but for the purpose of this problem, let us use g to avoid too many exponents. We can express the relationship g is the inverse of f with the equation from

+f(g(x))=x. +

+ Looking at the equation f(g(x))=x, what is the derivative with respect to x of the right hand side of the equation? +

+
    +

  1. x

    2. +

  2. 1

    3. +

  3. 0

    4. +

  4. x^2

    5. +
+ +

+ Looking at the equation f(g(x))=x, what is the derivative with respect to x of the left hand side of the equation? +

+
    +

  1. f'(g(x))

    2. +

  2. f'(g'(x))

    3. +

  3. f(g(x))\, g'(x)

    4. +

  4. f'(g(x))\, g'(x)

    5. +
+ +

+ Setting the two sides of the equation equal after differentiating, we can solve for g'(x). What do you get? +

+
    +

  1. g'(x) = \frac{x}{f(g(x))}

    2. +

  2. g'(x) = \frac{x}{f'(g(x))}

    3. +

  3. g'(x) = \frac{1}{f(g(x))}

    4. +

  4. g'(x) = \frac{1}{f'(g(x))}

    5. +
+ + + +

In the above activity you should have found that +the derivative of g = f^{-1}, the inverse function of f, is given by +

+(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}. +

Notice that because of the chain rule, the derivative of f has to be evaluated at \boldsymbol{f^{-1}(x)}

+ + + +

In this problem you will apply the general formula for the derivative of the inverse function to find the values of some derivatives graphically.

+
+ + + \begin{tikzpicture}[scale=1] +\begin{axis}[ + axis lines=middle, + grid=major, + xmin=0, xmax=80, + ymin=0, ymax=25, + xtick={0,10,...,80}, + ytick={0,4,...,24}, + % yticklabels={0, 0.01, 0.02, 0.03, 0.04, 0.05}, +% tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel={}, + xlabel=$x$, + ] + \addplot[domain=0:80, blue, ultra thick] {(0.6*(x)^3)/(3*(60^2))} node [pos=0.8,above left, ultra thick] {$\boldsymbol{f(x)}$}; + \end{axis} +\end{tikzpicture} +\begin{tikzpicture}[scale=1] +\begin{axis}[ + axis lines=middle, + grid=major, + xmin=0, xmax=80, + ymin=0, ymax=1.2, + xtick={0,10,...,80}, + ytick={0,0.2,...,1.2}, + % yticklabels={0, 0.0002, 0.0004, 0.0006, 0.0008, 0.001}, +% tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel={}, + xlabel=$x$, + ] + \addplot[domain=0:80, blue, ultra thick] {0.6*(x)^2/((60^2))} node [pos=0.8,above left, ultra thick] {$\boldsymbol{f'(x)}$}; + \end{axis} +\end{tikzpicture} + + + The graphs of f(x) and f'(x). +
+ +

The derivative of the inverse function at x=12 given by (f^{-1})' (12) = \frac{1}{f'(f^{-1}(12))}. Using the graphs, what is your best approximation for this quantity?

+
    +

  1. (f^{-1})' (12)\approx \frac{1}{0.2} = 5

    2. +

  2. (f^{-1})' (12) \approx \frac{1}{0.6} \approx 1.67

    3. +

  3. (f^{-1})' (12) \approx \frac{1}{0.4} = 2.5

    4. +

  4. (f^{-1})' (12) \approx \frac{1}{0.1} = 10

    5. +
+ +

What is your best estimate for (f^{-1})' (6) ?

+
    +

  1. (f^{-1})' (6) \approx \frac{1}{0.2} = 5

    2. +

  2. (f^{-1})' (6) \approx \frac{1}{0.6} \approx 1.67

    3. +

  3. (f^{-1})' (6) \approx \frac{1}{0.4} = 2.5

    4. +

  4. (f^{-1})' (6) \approx \frac{1}{0.1} = 10

    5. +
+ + + + +

Use the general formula for the derivative of the inverse function from to find...

 +

The derivative of the inverse function of f(x) = e^x... This should match the result of !

 +

The derivative of the inverse function of f(x) = \frac{1}{x}... This should match a derative that you have seen before! See if you can explain why.

 + + + +

We can only invert the function y=\sin(x) on the restricted domain [-\pi/2,\pi/2] (Why?). On this domain we define arcsine by the condition

+ x = \sin^{-1}(y) \quad \text{ when } \quad y=\sin(x). + + + +

In this activity you will study the arcsine function.

 +

Consider the values of y=\sin(x) given in the table below for an angle x between -\pi/2 and \pi/2. Fill in the corresponding values for the inverse function arcsine x = \sin^{-1}(y). In other words, you need to provide the angle in [-\pi/2, \pi/2] whose sine value is given. You can use the unit circle to help you remember which angles yield the given values of sine. The first entry is provided: a sine value of -1 corresponds to the angle -\pi/2. + + + + + y = \sin(x) + + + -1 + + + -\sqrt{3}/2 + + + -1/2 + + + 0 + + + 1/2 + + + \sqrt{3}/2 + + + 1 + + + + + x = \sin^{-1}(y) + + + -\pi/2 + + + +
+

 +

From the graph of y=\sin(x) and your table above, graph the arcsine function y=\sin^{-1}(x)

+
+ + +\begin{tikzpicture} +\begin{axis}[ + axis lines=middle, + xmin=-2,xmax=2,ymin=-1,ymax=1, + xtick={-1.57, -1.04,...,1.99}, + ytick={-1, -0.75,...,1}, + xticklabels={$-\pi/2$, $-\pi/3$, $-\pi/6$ , 0 , $\pi/6$, $\pi/3$, $\pi/2$, $2\pi/3$}, + xlabel={$x$}, + ylabel={$y$} + ] + \addplot[domain=-pi/2:pi/2,blue, thick, samples=500] {sin(deg(x))} node [pos=0.65,below right] {$f(x)=\sin(x)$}; + % \addplot[domain=-2*pi:2*pi,blue] {tan(deg(x))}; + % \addplot[domain=pi/2:3*pi/2,blue] {tan(deg(x))}; +\end{axis} +\end{tikzpicture} +\begin{tikzpicture}[scale=1] +\begin{axis}[ + axis lines=middle, + xmin=-1,xmax=1,ymin=-2,ymax=2, + ytick={-1.57, -1.04 ,..., 1.99}, + xtick={-1, -0.75,...,1}, + yticklabels={$-\pi/2$, $-\pi/3$, $-\pi/6$ , 0 , $\pi/6$, $\pi/3$, $\pi/2$, $2\pi/3$}, + xlabel={$x$}, + ylabel={$y$} + ] +% \addplot[domain=-2*pi:2*pi,blue, samples=500] {sin(deg(x))} ; + % \addplot[domain=-2*pi:2*pi,blue] {tan(deg(x))}; + % \addplot[domain=pi/2:3*pi/2,blue] {tan(deg(x))}; + \addplot [only marks, blue] coordinates {(-1,-1.57)}; +\end{axis} +\end{tikzpicture} + + + The graphs of \sin(x) and one point on \sin^{-1}(x). +
+ +

Let's now work with the function arccosine. Again, we need to restrict the domain of cosine to be able to invert the function (Why?). The convention is to restrict cosine to the domain [0,\pi] in order to define arccosine. Given this restriction, what are the domain and range of arccosine? Create a table of values and graph the function arccosine.

 +

Let's now work with the function arctangent. Again, we need to restrict the domain of tangent to be able to invert the function (Why?). The convention is to restrict tangent to the domain (-\pi/2,\pi/2) in order to define arctangent. Given this restriction, what are the domain and range of arctangent? Create a table of values and graph the function arctangent.

 + + + +

In this activity you will find a formula for the derivative of arctangent.

 +

Differentiate the implicit equation \tan(y) = x, what do you get for \frac{dy}{dx}?

+
    +

  1. \frac{dy}{dx} = \frac{x}{\tan(y)}

    2. +

  2. \frac{dy}{dx} = \frac{1}{\tan(y)}

    3. +

  3. \frac{dy}{dx} = \frac{x}{\sec^2(y)}

    4. +

  4. \frac{dy}{dx} = \frac{1}{\sec^2(y)}

    5. +
+ + +

For what function y=g(x) have you found the derivative \frac{dy}{dx}?

+ +

We want to rewrite \frac{dy}{dx} only in terms of x. Notice that

+ \tan^2(y) = \frac{\sin^2(y)}{\cos^2(y)} = \frac{1 - \cos^2(y)}{\cos^2(y)}. +

Multiplying out by the denominator, isolating, and solving for \cos^2(y), we get that

+
    +

  1. \cos^2(y) = \frac{\tan^2(y)}{\cos^2(y)}

    2. +

  2. \cos^2(y) = \frac{1}{\tan^2(y) + 1 }

    3. +

  3. \cos^2(y) = \frac{1- \cos^2(y)}{\tan^2(y)}

    4. +

  4. \cos^2(y) = \frac{1}{\tan^2(y) -1 }

    5. +
+ +

Finally, rewrite \frac{dy}{dx} as \frac{dy}{dx} = \cos^2(y) and use the fact that \tan(y)=x to get a nice formula for the derivative of the arctangent function of x.

 + + + + +

+ Consider the functions y = \tan^{-1}(x). Using your algebra above, you should have found that

+\frac{d}{dx}\Big(\tan^{-1}(x)\Big)=\frac{1}{1+x^2}. +

In a similar fashion, one can find that

+ \frac{d}{dx}\Big(\sin^{-1}(x)\Big)=\frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx}\Big(\cos^{-1}(x)\Big)= - \frac{1}{\sqrt{1-x^2}}. + + + + + + +

+ Demonstrate and explain how to find the derivative of the following functions. + Be sure to explicitly denote which derivative rules (product, quotient, sum and difference, etc.) you are using in your work. +

+
    +
  1. k(t)= \frac{\arctan\left(-4 \, t\right)}{\ln\left(-4 \, t\right)}
    2. +
  2. j(u)= -5 \, \arcsin\left(u\right) \log\left(u^{6} + 2\right)
    3. +
  3. n(x)= \ln\left(-\arcsin\left(x\right) + 4 \, \arctan\left(x\right)\right)
    4. +
+ + +
    +
  1. k' (t)= -\frac{\arctan\left(-4 \, t\right)}{t \log\left(-4 \, t\right)^{2}} - \frac{4}{{\left(16 \, t^{2} + 1\right)} \log\left(-4 \, t\right)}
    2. +
  2. j' (u)= -\frac{30 \, u^{5} \arcsin\left(u\right)}{u^{6} + 2} - \frac{5 \, \log\left(u^{6} + 2\right)}{\sqrt{-u^{2} + 1}}
    3. +
  3. n' (x)= \frac{\frac{1}{\sqrt{-x^{2} + 1}} - \frac{4}{x^{2} + 1}}{\arcsin\left(x\right) - 4 \, \arctan\left(x\right)}
    4. +
+ + + + +

Find the equation of the tangent line to y=\tan^{-1}(x) at x=0. Draw the function and the tangent on a graphing calculator to check your work!

 +

Find the equation of the tangent line to y=\sin^{-1}(x) at x=0.5. Draw the function and the tangent on a graphing calculator to check your work!

 +

Find the equation of the tangent line to y=\cos^{-1}(x) at x=-0.5. Draw the function and the tangent on a graphing calculator to check your work!

+ + + +

Let y=f(v) be the gas consumption (in ml/km) of a car at velocity v (in km/hr). We use the notation: ml for milliliters, km for kilometers, and hr for hours. Also consider the function g(y), where v=g(y) is the function that gives the velocity v (in km/hr) when the gas consumption is y (in ml/km). +You are given the graphs of f(v), f'(v) below.

+
+ + + \begin{tikzpicture}[scale=0.8] +\begin{axis}[ + axis lines=middle, + grid=major, + xmin=0, xmax=80, + ymin=0, ymax=25, + xtick={0,10,...,80}, + ytick={0,2,...,24}, + % yticklabels={0, 0.01, 0.02, 0.03, 0.04, 0.05}, +% tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel={}, + xlabel=$v$, + ] + \addplot[domain=0:80, blue, ultra thick] {(0.6*(x)^3)/(3*(60^2))} node [pos=0.8,above left, ultra thick] {$\boldsymbol{f(v)}$}; + \end{axis} +\end{tikzpicture} +\begin{tikzpicture}[scale=0.8] +\begin{axis}[ + axis lines=middle, + grid=major, + xmin=0, xmax=80, + ymin=0, ymax=1.2, + xtick={0,10,...,80}, + ytick={0,0.2,...,1.2}, + % yticklabels={0, 0.0002, 0.0004, 0.0006, 0.0008, 0.001}, +% tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel={}, + xlabel=$v$, + ] + \addplot[domain=0:80, blue, ultra thick] {0.6*(x)^2/((60^2))} node [pos=0.8,above left, ultra thick] {$\boldsymbol{f'(v)}$}; + \end{axis} +\end{tikzpicture} + + + + The graphs of f(v), f'(v). +
+ +

+ Estimate f^{-1}(6). What does this value mean in the context of the problem? +

+

+ Using your answer from part (a), estimate the derivative of the inverse function of f(x) at x=6 i.e., compute (f^{-1})'(6). +

+

+ What is the relationship between the functions f and g? +

+

+ Use the relationship between the functions f and g to estimate g(12) and g'(12). What do these values mean in the context of the problem? +

+ + + + + + Videos +
+
+ + +
\ No newline at end of file diff --git a/calculus/source/02-DF/main.ptx b/calculus/source/02-DF/main.ptx new file mode 100644 index 00000000..5284fa45 --- /dev/null +++ b/calculus/source/02-DF/main.ptx @@ -0,0 +1,15 @@ + + + + Derivatives (DF) + + + + + + + + + + + \ No newline at end of file diff --git a/calculus/source/02-DF/outcomes/01.ptx b/calculus/source/02-DF/outcomes/01.ptx new file mode 100644 index 00000000..44293553 --- /dev/null +++ b/calculus/source/02-DF/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Estimate the value of a derivative using difference quotients, and draw corresponding secant and tangent lines on the graph of a function. +

\ No newline at end of file diff --git a/calculus/source/02-DF/outcomes/02.ptx b/calculus/source/02-DF/outcomes/02.ptx new file mode 100644 index 00000000..15619a8b --- /dev/null +++ b/calculus/source/02-DF/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Find derivatives using the definition of derivative as a limit. +

\ No newline at end of file diff --git a/calculus/source/02-DF/outcomes/03.ptx b/calculus/source/02-DF/outcomes/03.ptx new file mode 100644 index 00000000..a1f97682 --- /dev/null +++ b/calculus/source/02-DF/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+ Compute basic derivatives using algebraic rules. +

diff --git a/calculus/source/02-DF/outcomes/04.ptx b/calculus/source/02-DF/outcomes/04.ptx new file mode 100644 index 00000000..262b6e81 --- /dev/null +++ b/calculus/source/02-DF/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+ Compute derivatives using the Product and Quotient Rules. +

diff --git a/calculus/source/02-DF/outcomes/05.ptx b/calculus/source/02-DF/outcomes/05.ptx new file mode 100644 index 00000000..2ec2f1bc --- /dev/null +++ b/calculus/source/02-DF/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+ Compute derivatives using the Chain Rule. +

diff --git a/calculus/source/02-DF/outcomes/06.ptx b/calculus/source/02-DF/outcomes/06.ptx new file mode 100644 index 00000000..7157c776 --- /dev/null +++ b/calculus/source/02-DF/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+ Compute derivatives using a combination of algebraic derivative rules. +

diff --git a/calculus/source/02-DF/outcomes/07.ptx b/calculus/source/02-DF/outcomes/07.ptx new file mode 100644 index 00000000..28baa1c0 --- /dev/null +++ b/calculus/source/02-DF/outcomes/07.ptx @@ -0,0 +1,4 @@ + +

+ Compute derivatives of implicitly-defined functions. +

diff --git a/calculus/source/02-DF/outcomes/08.ptx b/calculus/source/02-DF/outcomes/08.ptx new file mode 100644 index 00000000..69d1d8e2 --- /dev/null +++ b/calculus/source/02-DF/outcomes/08.ptx @@ -0,0 +1,4 @@ + +

+ Compute derivatives of inverse functions. +

diff --git a/calculus/source/02-DF/outcomes/main.ptx b/calculus/source/02-DF/outcomes/main.ptx new file mode 100644 index 00000000..15f680f4 --- /dev/null +++ b/calculus/source/02-DF/outcomes/main.ptx @@ -0,0 +1,37 @@ + + + +

+ How can we measure the instantaneous rate of change of a function? +

+

+ By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
  7. + +
    8. +
  8. + +
    9. +
+ \ No newline at end of file diff --git a/calculus/source/02-DF/readiness.ptx b/calculus/source/02-DF/readiness.ptx new file mode 100644 index 00000000..505112a1 --- /dev/null +++ b/calculus/source/02-DF/readiness.ptx @@ -0,0 +1,30 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Write equations of lines using slope-intercept and/or point-slope form (Math is Fun)

    +
    2. +
  2. +

    Find the average rate of change of a function over some interval (Khan Academy)

    +
    3. +
  3. +

    Evaluate functions at variable expressions (Purple Math)

    +
    4. +
  4. +

    Use the laws of exponents to rewrite a given expression (Khan Academy (1) and Khan Academy (2) )

    +
    5. +
  5. +

    Compose and decompose functions (Khan Academy )

    +
    6. +
  6. +

    Recall special trig values on the unit circle (Khan Academy (1) and Khan Academy (2) )

    +
    7. +
+ + \ No newline at end of file diff --git a/calculus/source/03-AD/01.ptx b/calculus/source/03-AD/01.ptx new file mode 100644 index 00000000..0e61a81a --- /dev/null +++ b/calculus/source/03-AD/01.ptx @@ -0,0 +1,260 @@ + + +
+ Tangents, motion, and marginals (AD1) + + + + + Activities + + +

+The tangent line of a function f(x) at x=a is the linear function L(x) +L(x) = f'(a)(x-a) + f(a). +Notice that this is the linear function with slope f'(a) and passing through (a,f(a)) in point-slope form. +

 + + + +

For the following functions, find the required tangent line.

+ +

Find the tangent line to f(x) = \ln(x) at x=1

+
    +

  1. L(x) = x

    2. +

  2. L(x) = x+1

    3. +

  3. L(x)= x - 1

    4. +

  4. L(x)= -x + 1

    5. +
+ +

Find the tangent line to f(x) = e^x at x=0

+
    +

  1. L(x) = x

    2. +

  2. L(x) = x+1

    3. +

  3. L(x)= x - 1

    4. +

  4. L(x)= -x + 1

    5. +
+ + + + + +

+ Let f(x) = -2 \, x^{4} + 4 \, x^{2} - x + 5. Find an equation of the line tangent to the graph at the point (-2, -9). +

+ + +

+ If a particle has position function s = f(t), where t is measured in seconds and s is measured in meters, then the derivative of the position function tells us how the position is changing over time, so f'(t) gives us the (instantenous) velocity in meters per second. Also, the derivative of the velocity gives us the change in velocity over time, so so f''(t) gives us the (instantenous) acceleration in meters per second squared. Summarizing, +

    +

  • v(t) = f'(t) is the velocity of the particle in m/s.

    • +

  • a(t) = f''(t) is the acceleration of the particle in m/s^2.

    • +
+ +

+ + + +

A particle moves on a vertical line so that its y coordinate at time t is

+ y = t^{3}-9t^{2}+24t + 3 +

for t\geq 0. Here t is measured in seconds and y is measured in feet.

+ +

Find the velocity and acceleration functions.

 +

Sketch graphs of the position, velocity and acceleration functions for 0 \leq t \leq 5.

 +

When is the particle moving upward and when is it moving downward?

 +

When is the particle's velocity increasing?

 +

Find the total distance that the particle travels in the time interval 0 \leq t \leq 5. Careful: the total distance is not the same as the displacement (the change in position)! Compute how much the particle moves up and add it to how much the particle moves down.

+ + + + +

+Suppose the position of an object in miles +is modeled by the following function: +s(t)=-t^{3} - 3 \, t^{2} - 5 \, t + 8. +

+

+Explain and demonstrate how to find the object's position, + +velocity, + + and acceleration at 2 +seconds. Use appropriate units for each. +

+ + + + + + + +

In some cases, we want to also consider the speed of a particle, which is the absolute value of the velocity. In symbols |v(t)|= |f'(t)| is the speed of the particle. A particle is speeding up when the speed is increasing.

+ + + +

+ Consider the speed of a particle. What is the behavior of the speed in relation to velocity and acceleration?

+ +
    +

  1. The speed is always positive and it is increasing when the velocity and the acceleration have the same sign.

    2. +

  2. The speed is positive when the velocity is positive and negative when the velocity is negative.

    3. +

  3. The speed is positive when the acceleration is positive and negative when the acceleration is negative.

    4. +

  4. The speed is always positive and it is increasing when the velocity and the acceleration have opposite signs.

    5. + +
+ + +

In a parametric motion on a curve C given by x=f(t) and y=g(t) we have that

+
    +

  • \frac{dx}{dt}=f'(t) is the rate of change of f(t), one component of the slope (or velocity)

    • +

  • \frac{dy}{dt}=g'(t) is the rate of change of g(t), one component of the slope (or velocity)

    • +

  • \frac{dy}{dx} is the actual slope (or velocity) of the object and by the chain rule \frac{dy}{dx} = \frac{g'(t)}{f'(t)}

    • +
+ + + +

An airplane is cruising at a fixed height and traveling in a pattern described by the parametric equations

+ x = 4 t , \quad y = -t^4 + 4t - 1 , +

where x, y have units of miles, and t is in hours.

+ +

Find the slope of the curve. +

+ +

What is the slope of the curve at (0,-1). +

+ +

Write the equation of the tangent line to the curve at (0,-1). +

+ + +

+ If C (x) is the cost of producing x items and R(x) is the revenue from selling x items, then P(x)= R(x) - C(x) is the profit. We can study their derivatives, the marginals +

    +
  • C'(x) is the marginal cost, the rate of change of the cost per unit change in production;
    • +
  • R'(x) is the marginal revenue, the rate of change of the revenue per unit change in sales;
    • +
  • P'(x)= R'(x) - C'(x) is the marginal profit, the rate of change of the profit per unit change in sales (assuming we are selling all the items produced).
    • +
+

+ + +

The manager of a computer shop has to decide how many computers to store in the back of the shop. If she stores a large number, she has to pay +extra in storage costs. If she stores only a small number, she will have to reorder +more often, which will involve additional handling costs. She has found that if she +stores x computers, the storage and handling costs will +be C dollars, where +C(x) = 10x^{3}-900x^{2}+16000x+210000 +

 + +

What is the fixed cost of the computer shop, the cost when no computers are in storage? In practical terms this may account for rent and utilities expenses. +

+ +

Find the marginal cost +

+ +

Now suppose that x computers give revenue R(x) = 1000x. What is the marginal revenue? What is the real world interpretation of your finding? +

+ +

Find a formula for the profit function P(x) and find the marginal profit using the marginal revenue and the marginal cost (assuming the number of items produced and sold is equal and given by x). +

+ + + + +

+A gizmo is sold for \$63 per item. Suppose that the number of items produced is equal to the number of items sold and that +the cost (in dollars) of producing x gizmos is given by the +following function: +C(x)=4 \, x^{3} + 10 \, x^{2} + 7 \, x + 4. +

+

+ +Explain and demonstrate how to find the marginal revenue, the marginal cost, and the marginal profit +in this situation. +

+ + + + + + + + + +

A cooling object has temperature modelled by + y= a e^{-kt} +c , + where a,c,k are positive constants determined by the local conditions. +

+ + + + + + +

Consider a cup of coffee initially at 100^\circF. The said cup of coffee was forgotten this morning in my living room where the thermostat is set at 72^\circF. I also observed that when I initially prepared the coffee, the temperature was decreasing at a rate of 3.8 degrees per minute. +

+ +

In the long run, what temperature do you expect the coffee to tend to? Use this information in the model y=ae^{-kt}+c to determine the value of c. +

+ +

Using the initial temperature of the coffee and your value of c, find the value of a in the model y=ae^{-kt}t+c. +

+ +

The scenario also gives you information about the value of the rate of change at t=0. Use this additional information to determine the model y=ae^{-kt}t+c completely. +

+ +

You should find that the temperature model for this coffee cup is y= 72 + 38 e^{-0.1t}. Explain how the values of each parameter connects to the information given. +

+ + + + + Videos +
+
+ + +
\ No newline at end of file diff --git a/calculus/source/03-AD/02.ptx b/calculus/source/03-AD/02.ptx new file mode 100644 index 00000000..69733271 --- /dev/null +++ b/calculus/source/03-AD/02.ptx @@ -0,0 +1,280 @@ + + +
+ Linear approximation (AD2) + + + + + Activities + + + + +

+The linear approximation (or tangent line approximation or linearization) of a function f(x) at x=a is the tangent line L(x) at x=a. In formulas, L(x) is the linear function +L(x) = f'(a)(x-a) + f(a). +Notice that this is obtained by writing the tangent line to f(x) at (a,f(a)) in point-slope form and calling the resulting linear function L(x). The linear approximation L(x) is a linear function that looks like f(x) when we zoom in near x=a. +

 + + + + +

Without using a calculator, we will use calculus to approximate \ln(1.1).

+ + + +

Find the equation of the tangent line to \ln(x) at x=1. This will be your linear approximation L(x). What do you get for L(x)?

+
    +

  1. L(x) = x

    2. +

  2. L(x) = x+1

    3. +

  3. L(x)= x - 1

    4. +

  4. L(x)= -x + 1

    5. +
+ +

As 1.1 is close to 1, we can use L(1.1) to approximate \ln(1.1). What approximation do you get?

+
    +

  1. \ln(1.1) \approx 1.1

    2. +

  2. \ln(1.1) \approx 2.1

    3. +

  3. \ln(1.1) \approx 0.1

    4. +

  4. \ln(1.1) \approx -0.1

    5. +
+ +

Sketch the tangent line L(x) on the same plane as the graph of \ln(x). What do you notice? +

+
+ + +\begin{minipage}{\textwidth} +\begin{center} + \begin{tikzpicture}[scale=0.9] +\begin{axis}[ + axis lines=middle, + grid=major, + xmin=0, xmax=5, + ymin=-2, ymax=2, + % xtick={0,10,...,80}, + % ytick={0,10,...,50}, + % yticklabels={0, 0.01, 0.02, 0.03, 0.04, 0.05}, + tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel=$y$, + xlabel=$x$, + ] + \addplot[domain=0.1:5, blue, ultra thick] {ln(x)} node [pos=0.8,above left , ultra thick] {$\boldsymbol{\ln(x)}$}; +% \addplot [only marks, blue] table { +% 40 20 +% }; +% \addplot[only marks, color=blue, nodes near coords={$(40,20)$}] coordinates {(30,20)}; +% \addplot[domain=20:70, black, thick] {(x-40)+20} node [pos=0.6, below right, ultra thick] {$\boldsymbol{L(x)}$}; +% \addplot [only marks, black] table { +% 20 0 +% }; +% \addplot[only marks, color=black, nodes near coords={$(20,0)$}] coordinates {(30,0)}; + \end{axis} +\end{tikzpicture} +\end{center} + \end{minipage} + + + The graph of \ln(x) +
+ + + + +

+ Using the equation of the tangent line to the graph of \ln(x) at x=1 and the shape of this graph, you can show that for all values of x, we have that +\ln(x) \leq x-1.

+ +

+Compute the second derivative of \ln(x). What do you notice about the sign of the second derivative of \ln(x)? What does this tell you about the shape of the graph? +

 +

+Conclude that because the graph of \ln(x) has a certain shape, the graph will bend below the tangent line and so that \ln(x) will always be smaller than the tangent line approximation L(x)= x -1 . +

 + + + + +

In this activity you will approximate power functions near x=1.

 +

+Find the tangent line approximation to x^2 at x=1.

+
    +

  1. L(x) = 2x

    2. +

  2. L(x) = 2x+1

    3. +

  3. L(x)= 2x - 1

    4. +

  4. L(x)= -2x + 1

    5. +
+ +

Show that for any constant k, the tangent line approximation to x^k at x=1 is L(x) = k(x -1) + 1.

+

Someone claims that the square root of 1.1 is about 1.05. Use the linear approximation to check this estimate. Do you think this estimate is about right? Why or why not?

+

Is the actual value \sqrt{1.1} above or below 1.05? What feature of the graph of \sqrt{x} makes this an over or under estimate?

+ + + +

+If a function f(x) is concave up around x=a, then the function is turning upwards from its tangent line. So when we use a linear approximation, the value of the approximation will be below the actual value of the function and the approximation is an underestimate. If a function f(x) is concave down around x=a, then the function is turning downwards from its tangent line. So when we use a linear approximation, the value of the approximation will be above the actual value of the function and the approximation is an overestimate. +

 + + +

Suppose f has a continuous positive second derivative and \Delta x is a small increment in x (like h in the limit definition of the derivative). + Which one is larger... +

+ f(1 + \Delta x ) \quad \text{or} \quad f'(1) \Delta x + f(1) \quad? + + + + + +

+A certain function p(x) satisfies +p(7) = 49 and p'(7) = 8. +

    +
  1. +

    +Explain how to find the local linearization L(x) +of p(x) at 7. +

    +
    2. +
  2. +

    +Explain how to estimate the value of p(6.951). +

    +
    3. +
  3. +

    +Suppose that p'(7)=0 and you know that +p''(x) < 0 for x < 7. +Explain how to determine if your estimate of p(6.951) +is too large or too small. +

    +
    4. +
  4. +

    +Suppose that p''(x) > 0 for x > 7. +Use this fact and the additional information above to sketch an accurate +graph of y=p(x) near x=7. +

    +
    5. +
+

+ + +
    +

  1. L(x)=8 \, x - 7

    2. +

  2. p(6.951) \approx 48.6064

    3. +

  3. The estimate is too large.

    4. +
+ + + + + +

+ Let's find the quadratic polynomial

+ q(x) = ax^2 +bx +c +

where a,b,c are parameters to be determined so that q(x) best approximates the graph of f(x)=\ln(x) at x=1. +

+ +

We want to choose a,b,c such that our quadratic polynomial resembles f(x) at x=1. First thing, we want f(1)=q(1). What equation in a,b,c does this condition give you?

+
    +

  1. a+b+c = 1

    2. +

  2. a+b+c = 0

    3. +

  3. c = 0

    4. +

  4. c = 1

    5. +
+ +

We also want f'(1)=q'(1). What equation in a,b,c does this condition give you?

 +

Finally, we want f''(1)=q''(1). What equation in a,b,c does this condition give you?

 +

Find a solution to this system of linear equations! Your answer will give you values of a,b,c that can be used to draw a quadratic approximating the natural logarithm. You can check your answer on Desmos

+ + + + +

A linear approximation L(x) to f(x) at x=a is a linear function with

+ L(a) = f(a), \quad L'(a) = f'(a). +

A quadratic approximation Q(x) to f(x) at x=a is a quadratic function with

+ Q(a) = f(a), \quad Q'(a) = f'(a), \quad Q''(a) = f''(a) . + + +

Find the linear approximation L(x) of \cos(x) at x=0. Then find the quadratic approximation Q(x) of \cos(x) at x=0. Graph both and compare the two approximations!

+ + + + +

+Suppose the function p(x) satisfies +p(-2) = 5, p'(-2) = 1, +and p''(x) < 0 for x + values nearby -2.

 + + +

+Explain and demonstrate +how to find the linearization L(x) +of p(x) at x =-2. +

+ + +

+Explain and demonstrate +how to estimate the value of p(-2.03) using +this linearization. +

+ + +

+Explain why your estimate of +p(-2.03) +is greater than or less than the actual value. +

+ + +

+Sketch a possible graph of p(x) and its linearization L(x) nearby x =-2 +to illustrate your findings. +

+ + + + + + + + + Videos +
+
+ + +
\ No newline at end of file diff --git a/calculus/source/03-AD/03.ptx b/calculus/source/03-AD/03.ptx new file mode 100644 index 00000000..3c524520 --- /dev/null +++ b/calculus/source/03-AD/03.ptx @@ -0,0 +1,347 @@ + + +
+ Related rates (AD3) + + + + + Activities + + +

+ So far + we have been interested in the instantaneous rate at which one variable, + say y, changes with respect to another, say x, + leading us to compute and interpret \frac{dy}{dx}. + We also have situations where several variables change together and often each quantity is a function of time, + represented by the variable t. + Knowing how the quantities are related, + we will determine how their rates of change with respect to time are related. +

+ + +

In a sense, the chain rule is our first example of related rates: +recall that when y is a function of x, which in turn is a function of t, we are considering the composite function y(x(t)), and we learned that by the chain rule

+\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t} +

Notice that the chain rule gives a relationship between three rates: \frac{d y}{d t} , \frac{d y}{d x}, \frac{d x}{d t}.

+ + + +

Remember the squirrels taking over my neighborhood? The population s grows based on acorn availability a, at a rate of 2 squirrels per bushel. The acorn availability a is currently growing at a rate of 100 bushels per week. What is \frac{ds}{dt} in this situation?

+
    +

  1. 2

    2. +

  2. 100

    3. +

  3. 200

    4. +

  4. Not enough information

    5. +
+ + + +

+ In a more serious example, + suppose that air is being pumped into a spherical balloon so that its volume increases at a constant rate of 20 cubic inches per second. + Since the balloon's volume and radius are related, + by knowing how fast the volume is changing, + we ought to be able to discover how fast the radius is changing. + Can we determine how fast is the radius of the balloon increasing when the balloon's diameter is 12 inches? +

+ + + +

+ A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. + How fast is the radius of the balloon changing at the instant the balloon's diameter is 12 inches? + Is the radius changing more rapidly when d = 12 or when d = 16? + Why? + + + +

+ Draw several spheres with different radii, + and observe that as volume changes, the radius, + diameter, and surface area of the balloon also change. +

+ + + +

+ Recall that the volume of a sphere of radius r is V = \frac{4}{3} \pi r^3. + Note as well that in the setting of this problem, + both V and r are changing with time t. + Hence both V and r may be viewed as implicit functions of t, + with respective derivatives + \frac{dV}{dt} and \frac{dr}{dt}. + Differentiate both sides of the equation + V = \frac{4}{3} \pi r^3 with respect to t + (using the chain rule on the right) + to find a formula for + \frac{dV}{dt} that depends on both r and \frac{dr}{dt}. +

+ + + +

+ At this point in the problem, + by differentiating we have + related the rates + of change of V and r. + Recall that we are given in the problem that the balloon is being inflated at a constant rate + of 20 cubic inches per second. + Is this rate the value of + \frac{dr}{dt} or \frac{dV}{dt}? + Why? +

+ + + +

+ From part (c), we know the value of + \frac{dV}{dt} at every value of t. + Next, + observe that when the diameter of the balloon is 12, + we know the value of the radius. + In the equation \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}, + substitute these values for the relevant quantities and solve for the remaining unknown quantity, + which is \frac{dr}{dt}. + How fast is the radius changing at the instant d = 12? +

+ + + +

+ How is the situation different when d = 16? + When is the radius changing more rapidly, + when d = 12 or when d = 16? +

+ + +

+ + +

+ In problems where two or more quantities are related to one another, + like in the case that all of the variables involved are functions of time + t, + we are interested in finding out how their rates of change are related; + we call these related rates problems. + related rates + Once we have an equation establishing the relationship among the variables, + we differentiate the equation, usually implicitly with respect to time, to find connections among the rates of change. +

+ + + A guide to solving related rated problems +

+

    +
  1. Picture it! Draw a diagram to represent the situation.
    2. +
  2. What do we know? Make a list of all quantities you are given in the problem, choosing clearly defined variable names for them. If a quantity is changing (a rate), then it should be labeled as a derivative.
    3. +
  3. What do we want to know? Make a list of all quantities to be determined. Again, choose clearly defined variable names.
    4. +
  4. How are the variables related to each other? Find an equation that relates the variables whose rates of change are known to those variables whose rates of change are to be found.
    5. +
  5. How are the rates related? Differentiate implicitly with respect to time. This will give an equation that relates the rates together.
    6. +
  6. Time to evaluate! Evaluate the derivatives and variables at the information relevant to the instant at which a certain rate of change is sought.
    7. +

+ + + Volume formulas +

+

    +
  • A sphere of radius r has volume V= \frac{4}{3} \pi r^3
    • +
  • A vertical cylinder of radius r and height h has volume V= \pi r^2 h
    • +
  • A cone of radius r and height h has volume V= \frac{\pi}{3} r^2 h
    • +

+ + + + +

+ A vertical cylindrical water tank has a radius of 1 meter. If water is pumped out at a rate of 3 cubic meters per minute, at what rate will the water level drop? +

+ + + +

+ Draw a figure to represent the situation. Introduce variables that measure the radius of the water's surface, the water's depth in the tank, and the volume of the water. Label your diagram. +

+ +

What information about rates of changes does the problem give you?

+ +

+ Recall that the volume of a cylinder of radius r and height h is V = \pi r^2 h. What is the related rates equation in the context of the vertical cylindrical tank? What derivative rules did you use to find this equation? +

+
    +
  1. \frac{dV}{dt}= \pi 2 r \frac{dh}{dt}
    2. +
  2. \frac{dV}{dt}= \pi r^2 \frac{dh}{dt}
    3. +
  3. \frac{dV}{dt}= \pi \frac{dr}{dt} h
    4. +
  4. \frac{dV}{dt}= \pi 2r \frac{dr}{dt} h + \pi r^2 \frac{dh}{dt}
    5. +
  5. \frac{dV}{dt}= \pi 2r h + \pi r^2
    6. +
+ + + +

+ Which variable(s) have a constant value in this situation? Why? +

+
    +

  1. The variable measuring the radius of the water's surface

    2. +

  2. The variable measuring the depth of the water

    3. +

  3. The variable measuring the volume of the water

    4. + +
+ + +

+ Which variable(s) have a constant rate of change in this situation? Why? +

+
    +

  1. The variable measuring the radius of the water's surface

    2. +

  2. The variable measuring the depth of the water

    3. +

  3. The variable measuring the volume of the water

    4. +
+ + +

+ Using your finding above, find at what rate the water level is dropping. +

+ + + +

+ If the full tank contains 12 cubic meters of water, how long does it take to empty the tank? +

+ + +

+ Confirm your finding in the previous part by finding the initial water level for 12 cubic centimeters of water and determine how long it takes for the water level to reach 0. +

+ + + + + + + + +

+ A water tank has the shape of an inverted circular cone (the cone points downwards) + with a base of radius 6 feet and a depth of 8 feet. + Suppose that water is being pumped into the tank at a constant instantaneous rate of 4 cubic feet per minute. +

+ + +

+ Draw a picture of the conical tank, + including a sketch of the water level at a point in time when the tank is not yet full. + Introduce variables that measure the radius of the water's surface and the water's depth in the tank, + and label them on your figure. +

+ + +

+ Say that r is the radius and h the depth of the water at a given time, + t. Notice that at any point of time there is a fixed proportion between the depth and the radius of the volume of water, forced by the shape of the tank. What proportional equation relates the radius and height of the water, and why? +

+ + +

+ Determine an equation that gives the volume of water in the tank as a function of only the depth h of the water (so eliminate the radius from the volume equation using the previous part). +

+ + +

+ Through differentiation, + find an equation that relates the instantaneous rate of change of water volume with respect to time to the instantaneous rate of change of water depth at time t. +

+ + + +

+ Find the instantaneous rate at which the water level is rising when the water in the tank is 3 feet deep. +

+ + + +

+ When is the water rising most rapidly? +

+
    +

  1. h = 3

    2. +

  2. h = 4

    3. +

  3. h = 5

    4. +

  4. The water level rises at a constant rate

    5. +
+ + + + + + + +

Recall that in a right triangle with sides a,b and hypotenuse c we have the relationship

+ a^2 + b^2 = c^2, +

also known in the western world as the Pythagorean theorem (even though this result was well know well before his time by other civilizations).

+ + + + +

Notice that by differentiating the equation above with respect to t we get a relationship between + a,b,c, \frac{da}{dt}, \frac{db}{dt}, \frac{dc}{dt}. Find this related rates equation. +

+ + + + + +

A rectangle has one side of 8 cm. How fast is the diagonal of the rectangle changing at the instant when the other + side is 6 cm and increasing at a rate of 3 cm per minute? +

+ + + + + +

A 10 m ladder leans against a vertical wall and the bottom of the ladder slides away at a rate of 0.5 m/sec. + When is the top of the ladder sliding the fastest down the wall? +

+
    +

  1. When the bottom of the ladder is 4 meters from the wall.

    2. +

  2. When the bottom of the ladder is 8 meters from the wall.

    3. +

  3. The top of the ladder is sliding down at a constant rate.

    4. +
+ + + + + + + +

+ Suppose a car was 75 miles east of a town, traveling west at 75 mph. A second car was 120 miles north of the same town, traveling south at 70 mph. At this exact moment, how fast is the distance between the cars changing? +

+ + + + + + + Videos +
+
+
+
+ + +
+ diff --git a/calculus/source/03-AD/04.ptx b/calculus/source/03-AD/04.ptx new file mode 100644 index 00000000..33ce9736 --- /dev/null +++ b/calculus/source/03-AD/04.ptx @@ -0,0 +1,278 @@ + + +
+ Extreme values (AD4) + + + + + Activities + +

+ In many different settings, + we are interested in knowing where a function achieves its least and greatest values. + These can be important in applicationssay to identify a point at which maximum profit or minimum cost occursor in theory to characterize the behavior of a function or a family of related functions. +

+ + + +

+ Consider the familiar example of a parabolic function such as s(t) = -16t^2 + 32t + 48. This function + represents the height of an object tossed vertically straight up: its maximum value occurs at the vertex of the parabola and represents the greatest height the object reaches. This maximum value is an especially important point on the graph and we can notice that the function changes from increasing to decreasing at this point. +

+ +
+ + +\begin{tikzpicture}[scale=1] +\begin{axis}[ + samples = 900, + axis lines=middle, + grid=major, + % xmin=0, xmax=4, + ymin=0, ymax=70, + % xtick={0,0.5,...,4}, + % ytick={-2,-1,...,10}, + % tick style={thick}, +% x label style={at={(axis description cs:1,0.7)}}, +% y label style={at={(axis description cs:0.4,1)}}, + ylabel=$s(t)$, + xlabel=$t$, + ] + \addplot[domain=0:3, blue, thick] {-16*x^2+32*x+48}; +\end{axis} +\end{tikzpicture} + + + + The graph of s(t) = -16t^2 + 32t + 48 +
+ + + + +

+ We say that f(x) has a global maximum at x=c provided that f(c)\geq f(x) for all x + in the domain of the function. + We also say that f(c) is a global maximum value for the function. + On the other hand, we say that f(x) has a global minimum at x=c provided that f(c)\leq f(x) for all x + in the domain of the function. + We also say that f(c) is a global minimum value for the function. + The global maxima and minima are also known as the global extrema (or extreme values or absolute extrema) of the function. +

+ + + + + + + + +

According to , which of the following statements best describes the global extrema of the function in ?

+
    +

  1. The global maximum is t = 1, because this is where the function goes from increasing to decreasing.

    2. +

  2. The global maximum is s(1) = 64, because s(t)\leq 64 for every other input t.

    3. +

  3. The graph has two global minima at the endpoints because the endpoints must be global extrema.

    4. +

  4. The graph has no global minimum.

    5. +
+ + + + +

+ There can be some issues when trying to determine the global mimimum and maximum values of a function only using its graph. The Extreme Value Theorem will guarantee the existence of global extrema on a closed interval. Then we will see how to use derivatives to find algebraically the extrema of a function. +

+ + + + Extreme Value Theorem + +

If f is continuous on a closed interval [a,b], then f has both a global maximum and a global minimum on the interval.

+ + + + + + +

For each of the following figures, decide where the global extrema are located.

+ + +
+ + +
+ +
+ + +
+
+ + +
+ +
+ + +
+ + + + + +

The Extreme Value Theorem (EVT) guarantees a global maximum and a global minumum for which of the following?

+
    +
  1. f(x)=\dfrac{x^{2}}{x^{2}-4x-5} on [-5,0].
    2. +
  2. f(x)=\dfrac{x^{2}}{x^{2}-4x-5} on [0,4].
    3. +
  3. f(x)=\dfrac{x^{2}}{x^{2}-4x-5} on [4,6].
    4. +
  4. f(x)=\dfrac{x^{2}}{x^{2}-4x-5} on [6,10].
    5. +
+ + + + +

For the following activity, draw a sketch of a function that has the following properties.

 + +

The function is continuous and has an global minimum but no global maximum.

+ + +

The function is continuous and has an global maximum but no global minimum.

+ + + + +

+ We say that x=c is a critical point (or critical number) of f(x) if x=c is in the domain of f(x) and either f'(c) = 0 or f'(c) does not exist. +

 + + + + + + +

Which of the following are critical numbers for f(x) = \frac{1}{3}x^3 - 2x + 2?

+
    +
  1. x = \sqrt{2} and x = -\sqrt{2}.
    2. +
  2. x = \sqrt{2}.
    3. +
  3. x = 2 and x = 0.
    4. +
  4. x = 2.
    5. +
+ + + + + + +The Closed Interval Method +

The following is a way of finding the global extrema of a continuous function f on a closed interval [a,b].

+
    +
  1. Make a list of all critical points of f in (a,b). (Do not include any critical points outside of the interval).
    2. +
  2. Add the endpoints a and b to the list.
    3. +
  3. Evaluate f at all points on your list.
    4. +
  4. The smallest output occurs at the global minimum. The largest output occurs at the global maximum.
    5. +
+ + + + + + + + +

What are the global extrema for f(x) = 3x^4 - 4x^3 on [-1,2]. +

+
    +
  1. Global maximum is when x = 0 and global minimum when x = 1.
    2. +
  2. Global maximum is when x = 2 and global minimum when x = -1.
    3. +
  3. Global maximum is when x = 2 and global minimum when x = 1.
    4. +
  4. Global maximum is when x = 0 and global minimum when x = -1.
    5. +
+ + + + + +

What are the global extrema for f(x) = x\sqrt{4-x} on [-2,4]. +

+
    +
  1. Global maximum is when x = -2 and global minimum when x = \frac{8}{3}.
    2. +
  2. Global maximum is when x = 4 and global minimum when x = \frac{8}{3}.
    3. +
  3. Global maximum is when x = \frac{8}{3} and global minimum when x = -2.
    4. +
  4. Global maximum is when x = 4 and global minimum when x = -2.
    5. +
+ + + + + +

+Explain how to find the global minimum and global maximum values of the +function f(x)=-2 \, x^{3} + 18 \, x^{2} + 42 \, x + 33 on the interval [-2,2]. +

+ + + + + + + +

In this problem you will consider the function g(x). + g(x) = \left\{ \begin{array}{ll} +x^3-3x & x \lt 0\\ +x^2 -4x +2 & x\geq 0 +\end{array} +\right. + +

 +

What can you say about the point x=0?

 + +

In addition to x=0, find the other two critical points. What are the critical points of g(x)? +

+
    +
  1. x=0, \, x=1, \, x= 2
    2. + +
  2. x=0, \, x=-1, \, x= 2
    3. + +
  3. x=0, \, x=-1, \, x=-2
    4. + +
  4. x=0, \, x=1, \, x= -2
    5. +
+ +

Can you use the Closed Interval Method on [-4,-1]? If you can, find the global max and min. If you can't, explain why.

 +

Can you use the Closed Interval Method on [1,4]? If you can, find the global max and min. If you can't, explain why.

 +

Can you use the Closed Interval Method on [-1,1]? If you can, find the global max and min. If you can't, explain why.

 + + + + + + Videos +
+
+ + +
diff --git a/calculus/source/03-AD/05.ptx b/calculus/source/03-AD/05.ptx new file mode 100644 index 00000000..10bf179a --- /dev/null +++ b/calculus/source/03-AD/05.ptx @@ -0,0 +1,260 @@ + + +
+ Derivative tests (AD5) + + + + + Activities + + + +

+ We say that f(x) has a local maximum at x=c provided that f(c)\geq f(x) for all x near c. We also say that f(c) is a local maximum value for the function. On the other hand, we say that f(x) has a local minimum at x=c provided that f(c)\leq f(x) for all x near c. We also say that f(c) is a local minimum value for the function. The local maxima and minima are also known as the local extrema (or relative extrema) of the function. +

+ + + + +

+To find the extreme values of a function we can consider all its local extrema (local maxima and minima) and study them to find which one(s) give the largest and smallest values on the function. But how do you find the local/relative extrema? We will see that we can detect local extrema by computing the first derivative and finding the critical points of the function. By finding the critical points, we will produce a list of candidates for the extrema of the function. +

 + + + +

We have encountered several terms recently, so we should make sure that we understand how they are related. Which of the following statements are true?

 +
    +

  1. In a closed interval an endpoint is always a local extrema but it might or might not be a global extrema.

    2. +

  2. In a closed interval an endpoint is always a global extrema.

    3. +

  3. A critical point is always a local extrema but it might or might not be a global extrema.

    4. +

  4. A local extrema only occurs where the first derivative is equal to zero.

    5. +

  5. A local extrema always occurs at a critical point.

    6. +

  6. A local extrema might occur at a critical point or at an endpoint of a closed interval.

    7. +
+ + + + + +

+ Sketch the graph of a continuous function that is increasing on (-\infty, -2), constant on the interval (3,5), and decreasing on the interval (-2,3). +

+ + + +

How would you describe the derivative of the function on each interval?

+
    +

  1. For x < -2 we have f'(x) < 0, then f'(x) < 0 on the interval (-2,3), and on the interval (3,5) we have f'(x) > 0.

    2. +

  2. For x < -2 we have f'(x) > 0, then f'(x) < 0 on the interval (-2,3), and on the interval (3,5) we have f'(x) is undefined.

    3. +

  3. For x < -2 we have f'(x) > 0, then f'(x) < 0 on the interval (-2,3), and on the interval (3,5) we have f'(x)=0 .

    4. +

  4. For x < -2 we have f'(x) < 0, then f'(x) < 0 on the interval (-2,3), and on the interval (3,5) we have f'(x) is constant.

    5. +
+ + + + + + +

+ Look back at the graph you made for . +

+

Which of the following best describes what is occurring when graph changes behavior?

+
    +

  1. There is a critical point.

    2. +

  2. There is a local maximum or minimum.

    3. +

  3. The derivative is undefined.

    4. +

  4. The derivative is equal to zero.

    5. +
+ + + + +

Critical points detect changes in the behavior of a function. We will use critical points as "break points" in studying the behavior of a function. To understand what happens at the critical points we use the Derivative Tests.

 + + The First Derivative Test + +

+ Suppose that x=c is a critical point of f(x) and that f(x) is continuous at x=c. +

+ +

+

    +
  1. If f'(x) > 0 before x=c and f'(x) < 0 after x=c, then a local maximum occurs at x=c.
    2. + +
  2. If f'(x) < 0 before x=c and f'(x) > 0 after x=c, then a local minimum occurs at x=c.
    3. +
  3. If f'(x) is the same sign on both sides of x=c, then neither a local maximum nor a local minimum occur when x=c.
    4. +

+ + + + +

Let f(x)=x^4-4x^3+4x^2

+ + +

Find all critical points of f(x). Draw them on the same number line.

+ + +

What intervals have been created by subdividing the number line at the critical points?

+ + +

Pick an x-value that lies in each interval. Determine whether f'(x) is positive or negative at each point.

+ + +

On which intervals is f(x) increasing? On which intervals is f(x) decreasing?

+ + +

List all local extrema.

+ + + + + The Second Derivative Test + +

+ Suppose that x=c is a critical point of f(x) and that f''(x) is continuous at x=c. +

+ +

+

    +
  1. If f'(c) = 0 and f''(c) < 0 , then a local maximum occurs when x=c.
    2. +
  2. If f'(c) = 0 and f''(c) > 0 , then a local minimum occurs when x=c.
    3. +
  3. If f'(c) = 0 and f''(c) = 0 , then this test is inconclusive.
    4. +

+

In case of an inconclusive Second Derivative Test, we need to go back to the First Derivative Test to classify the critical point.

+ + + + +

Consider the function f(x)=-x^{3} + 3 \, x + 4.

 + +

Find the open intervals where f(x) is increasing or decreasing. +

+ +

Find the local extrema of f(x). +

+ + + + + + + Dealing with discontinuities + +

Our previous activity dealt with a function that was continuous for all real numbers. Because of that, we could trust our chart to point out local extrema. Let's now consider what might happen if a function has any discontinuities.

+ + + + +

Draw a function that is increasing on the left of x=1, discontinuous at x=1, such that f(1)=\displaystyle \lim_{x \to 1^+}f(x), and decreasing to the right of x=1. Does the derivative of f(x) exist at x=1? Does your graph have a local maximum or minimum at x=1?

+ + + + +

Let f(x)=\frac{x}{(x-2)^2}.

+ + +

Note that f(x) is not defined for x=2. But the function may be increasing on one side of x=2 and decreasing on the other! So we include x=2 on your number line.

 + +

Find all critical points of f(x). Plot them and any discontinuities for f(x) on the same number line.

+ + +

What intervals have been created by subdividing the number line at the critical points and at the discontinuities?

+ + +

Pick an x-value that lies in each interval. Determine whether f'(x) is positive or negative each point.

+ + +

On which intervals is f(x) increasing? On which intervals is f(x) decreasing?

+ + +

List all local maxima and local minima.

+ + + + + +

For each of the following functions, find the intervals on which f(x) is increasing or decreasing. Then identify any local extrema using either the First or Second Derivative Test.

+ + +

f(x)=x^3+3x^2+3x+1

+ + +

f(x)=\frac{1}{2}x+\cos x on (0,2\pi)

+ + +

f(x)=(x^2-9)^{2/3}

+ + +

f(x)= \ln(2x-1). (Hint: think about the domain of this one before you get started!)

+ + +

f(x)=\frac{x^2}{x^2-4}

+ + + + + + The Mean Value Theorem +

If f is continuous and differentiable on the closed interval [a,b], then there is some point c in the interval where f'(c) is equal to the slope of the secant line through the points (a,f(a)) and (b,f(b)). In symbols, for some c in (a,b) we have that

+f'(c) = \frac{f(b)-f(a)}{b-a} + + + +

Suppose f is continuous and differentiable on [a,b] and also suppose that f(a)=f(b). What is the average rate of change of f(x) on [a,b]? What does the MVT (Mean Value Theorem) tell you?

 +

Use part (a) to show with the MVT that f(x) = (x-1)^2 + 3 has a critical point on [0,2] .

+ + + + + Videos +
+
+ + +
+ + + + + + + + + + + + + + + + + diff --git a/calculus/source/03-AD/06.ptx b/calculus/source/03-AD/06.ptx new file mode 100644 index 00000000..9b48f756 --- /dev/null +++ b/calculus/source/03-AD/06.ptx @@ -0,0 +1,466 @@ + + +
+ Concavity and inflection (AD6) + + + + + Activities + +

In addition to asking whether + a function is increasing or decreasing, + it is also natural to inquire how + a function is increasing or decreasing. + describes three basic behaviors that an increasing function can demonstrate on an interval, + as pictured in

+ + + + +

+ Sketch a sequence of tangent lines at various points to each of the following curves in . +

+ +
+ Three increasing functions + +
+ + + +

Look at the curve pictured on the left of . How would you describe the slopes of the tangent lines as you move from left to right?

+
    +

  1. The slopes of the tangent lines decrease as you move from left to right.

    2. +

  2. The slopes of the tangent lines remain constant as you move from left to right.

    3. +

  3. The slopes of the tangent lines increase as you move from left to right.

    4. +
+ + + + +

Look at the curve pictured in the middle of . How would you describe the slopes of the tangent lines as you move from left to right?

+
    +

  1. The slopes of the tangent lines decrease as you move from left to right.

    2. +

  2. The slopes of the tangent lines remain constant as you move from left to right.

    3. +

  3. The slopes of the tangent lines increase as you move from left to right.

    4. +
+ + + + +

Look at the curve pictured on the right of . How would you describe the slopes of the tangent lines as you move from left to right?

+
    +

  1. The slopes of the tangent lines decrease as you move from left to right.

    2. +

  2. The slopes of the tangent lines remain constant as you move from left to right.

    3. +

  3. The slopes of the tangent lines increase as you move from left to right.

    4. +
+ + + + + + + +

On the leftmost curve in , + as we move from left to right, + the slopes of the tangent lines will increase. + Therefore, the rate of change of the pictured function is increasing, + and this explains why we say this function is + increasing at an increasing rate. +

+ + + +

We must be extra careful with our language when dealing with negative numbers. + For example, it can be tempting to say that + -100 is bigger than -2. + But we must remember that greater than + describes how numbers lie on a number line: -100 is less than -2 becomes it comes earlier on the number line. + It might be helpful to say that + -100 is "more negative" than -2. +

+ + + + +

+ Sketch a sequence of tangent lines at various points to each of the following curves in . +

+
+ From left to right, three functions that are all decreasing. + +
+ + + + +

Look at the curve pictured on the left in . How would you describe the slopes of the tangent lines as you move from left to right?

+
    +

  1. The slopes of the tangent lines decrease as you move from left to right.

    2. +

  2. The slopes of the tangent lines remain constant as you move from left to right.

    3. +

  3. The slopes of the tangent lines increase as you move from left to right.

    4. +
+ + + + +

Look at the curve pictured in the middle in . How would you describe the slopes of the tangent lines as you move from left to right?

+
    +

  1. The slopes of the tangent lines decrease as you move from left to right.

    2. +

  2. The slopes of the tangent lines remain constant as you move from left to right.

    3. +

  3. The slopes of the tangent lines increase as you move from left to right.

    4. +
+ + + + +

Look at the curve pictured on the right in . How would you describe the slopes of the tangent lines as you move from left to right?

+
    +

  1. The slopes of the tangent lines decrease as you move from left to right.

    2. +

  2. The slopes of the tangent lines remain constant as you move from left to right.

    3. +

  3. The slopes of the tangent lines increase as you move from left to right.

    4. +
+ + + + + + +

Recall the terminology of concavity: when a curve bends upward, we say its shape is concave up. When a curve bends downwards, we say its shape is concave down.

 + + + + +

Look at in . Which curve is concave up? Which one is concave down? Why? Try to explain using the graph!

+ +
+ Two concavity, which is which? + +
+ + + + + +

+ Let f be a differentiable function on some interval (a,b). + Then f is concave up + concave up + on (a,b) if and only if f' is increasing on (a,b); + f is concave down + concave down + on (a,b) if and only if f' is decreasing on (a,b). +

+ + + + + +

+ Look at how the slopes of the tangent lines change from left to right for each of the two graphs in +

+ + + +

Look at the curve pictured on the left in . How would you describe the slopes of the tangent lines as you move from left to right?

+
    +

  1. The slopes of the tangent lines decrease as you move from left to right.

    2. +

  2. The slopes of the tangent lines increase as you move from left to right.

    3. +

  3. The slopes of the tangent lines go from increasing to decreasing as you move from right to left.

    4. +

  4. The slopes of the tangent lines go from decreasing to increasing as you move from right to left.

    5. +
+ + + + +

Which of the following statements is true about the function on the left in ?

+
    +

  1. f'(x) > 0 on the entire interval shown.

    2. +

  2. f'(x) < 0 on the entire interval shown.

    3. +

  3. f''(x) > 0 on the entire interval shown.

    4. +

  4. f''(x) < 0 on the entire interval shown.

    5. +
+ + + + +

Look at the curve pictured on the right in . How would you describe the slopes of the tangent lines as you move from left to right?

+
    +

  1. The slopes of the tangent lines decrease as you move from left to right.

    2. +

  2. The slopes of the tangent lines increase as you move from left to right.

    3. +

  3. The slopes of the tangent lines go from increasing to decreasing as you move from right to left.

    4. +

  4. The slopes of the tangent lines go from decreasing to increasing as you move from right to left.

    5. +
+ + + + +

Which of the following statements is true about the function on the right in ?

+
    +

  1. f'(x) > 0 on the entire interval shown.

    2. +

  2. f'(x) < 0 on the entire interval shown.

    3. +

  3. f''(x) > 0 on the entire interval shown.

    4. +

  4. f''(x) < 0 on the entire interval shown.

    5. +
+ + + + + Test for Concavity + +

+ Suppose that f(x) is twice differentiable on some interval (a,b). If f'' > 0 on (a,b), then f is concave up on (a,b). If f'' < 0 on (a,b), then f is concave down on (a,b). +

+ + + + +

In the previous section, we saw in how to use critical points of the function and the sign of the first derivative to identify intervals of increase/decrease of a function. The next activity uses the critical points of the first derivative function and the sign of the second derivative (accordingly to ) to identify where the original function is concave up/down.

+ + + +

+ Let f(x)=x^4-54x^2. +

+ + +

Find all the zeros of f''(x).

+ + +

What intervals have been created by subdividing the number line at zeros of f''(x)?

+ + +

Pick an x-value that lies in each interval. Determine whether f''(x) is positive or negative at each point.

+ + +

On which intervals is f'(x) increasing? On which intervals is f'(x) decreasing?

+ + +

List all the intervals where f(x) is concave up and all the intervals where f(x) is concave down.

+ + + + + + + +

+ If x=c is a point where f''(x) changes sign, then the concavity of graph of f(x) changes at this point and we call x=c an inflection point of f(x). +

+ + + + +

+ Use the results from to identify all of the inflection points of f(x)=x^4-4x^3+4x^2. +

+ + + + + + +

+For each of the following functions, describe the open intervals +where it is concave up or concave down, and any inflection points. +

+ + +

f(x)=-\frac{1}{4} \, x^{5} - \frac{5}{2} \, x^{4} - \frac{15}{2} \, x^{3}

+ + +

f(x)=\frac{3}{20} \, x^{5} + x^{4} - \frac{5}{2} \, x^{3}

+ + + + + + + +

+ Consider the following table. The values of the first and second derivatives of f(x) are given on the domain [0,7]. The function f(x) does not suddenly change behavior between the points given, so the table gives you enough information to completely determine where f(x) is increasing, decreasing, concave up, and concave down. +

+

+ +\begin{array}{c|cccccccc} +x + & 0 + & 1 + & 2 + & 3 + & 4 + & 5 + & 6 + & 7 + \\\hline +f'(x) + & 2 + & 0 + & -2 + & 0 + & 2 + & 1 + & 0 + & -1 + + \\\hline +f''(x) + & -2 + & -1 + & 0 + & 1 + & 0 + & -1 + & 0 + & 3 + \\ +\end{array} + +

+ + +

List all the critical points of f(x) that you can find using the table above.

+ + +

Use the First Derivative Test to classify the critical numbers (decide if they are a max or min). Write full sentence stating the conclusion of the test for each critical number.

+ + +

On which interval(s) is f(x) increasing? On which interval(s) is f(x) decreasing? List all the critical points of f(x) that you can find using the table above.

+ + +

There is one critical number for which the Second Derivative Test is inconclusive. Which one? You can still determine if it is a max or min using the First Derivative Test!

+ + +

List all the critical points of f'(x) that you can find using the table above.

+ + +

On which intervals is f(x) concave up? On which intervals is f(x) concave down?

+ + +

List all the inflection points of f(x) that you can find using the table above.

+ + + + + + Videos +

See .

+ + +
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+ Graphing with derivatives (AD7) + + + + + Activities + +

In and we learned how the first and second derivatives give us information about the graph of a function. Specifically, we can determine the intervals where a function is increasing, decreasing, concave up, or concave down as well as any local extrema or inflection points. Now we will put that information together to sketch the graph of a function.

 + + + +

+ Which of the following features best describe the curve graphed below?

+ + +
+ + +
+ + + +
    +

  1. Increasing and concave up

    2. +

  2. Increasing and concave down

    3. +

  3. Decreasing and concave up

    4. +

  4. Decreasing and concave down

    5. +
+ + + + + + + +

+ Which of the following features best describe the curve graphed below?

+ + +
+ + +
+ + + +
    +

  1. f'>0 and f''>0

    2. +

  2. f'>0 and f''<0

    3. +

  3. f'<0 and f''>0

    4. +

  4. f'<0 and f''<0

    5. +
+ + + +

For each of the other three answer choices, sketch a curve that matches that description.

+ + + + + + + + +

+ For each prompt that follows, + sketch a possible graph of a function on the interval + -3 \lt x \lt 3 that satisfies the stated properties.

+ + + + +

+ A function f(x) that is increasing on -3 \lt x \lt 3, + concave up on -3 \lt x \lt 0, + and concave down on 0 \lt x \lt 3. +

+ + + +

+ A function g(x) that is increasing on -3 \lt x \lt 3, + concave down on -3 \lt x \lt 0, + and concave up on 0 \lt x \lt 3. +

+ +

+ A function h(x) thatis decreasing on -3 \lt x \lt 3, + concave up on -3 \lt x \lt -1, + neither concave up nor concave down on -1 \lt x \lt 1, + and concave down on 1 \lt x \lt 3. +

+ +

+ A function p(x) that is decreasing and concave down on + -3 \lt x \lt 0 and is increasing and concave down on 0 \lt x \lt 3. +

+ + + + + + +

To draw an accurate sketch, we must keep in mind additional characteristics of a function, such as the domain and the horizontal and vertical asymptotes (when they exist). The next problem includes those aspects in addition to increasing, decreasing, and concavity features. +

+ + + + + +

+The following chart describes the values of f(x) and its +first and second derivatives at or between a few given values of x, +where \nexists denotes that f(x) does not exist at that value +of x. +

+

+ +\begin{array}{c|cccccccccccccccccccc} +x + & & -8 + & & -6 + & & -3 + & & 0 + & & 2 + & & 5 + & & 8 + & & 11 + & & 13 + \\\hline +f(x) + & & 3 + & & 5 + & & \nexists + & & -5 + & & \nexists + & & 4 + & & \nexists + & & -5 + & & -3 + \\ +f'(x) & + + & & + + & & + - & & + - & & + - & & + - & & + + & & + + & & + + & & + + + \\ +f''(x) & + + & & + - & & + - & & + + & & + - & & + + & & + + & & + - & & + - & & + - + \\ +\end{array} + +

+

+Assume that f(x) has vertical asymptotes at each +x-value where f(x) does not exist, that +\displaystyle \lim_{x\to-\infty}f(x)= 1, and that +\displaystyle \lim_{x\to\infty}f(x)= -1. +

+ +

+List all the asymptotes of f(x) and mark them on the graph. +

+ +

+Does f(x) have any local maxima or local minima? If so, at what point(s)? +

+

+Does f(x) have any inflection points? If so, at what point(s)? +

+ + +

+Use the information provided to sketch a reasonable graph of f(x). Watch changes in behavior due to changes in the sign of each derivative. +

+ + + + + + + + A guide to curve sketching +

+

    +
  1. Identify the domain of the function.
    2. +
  2. Identify any vertical or horizontal asymptotes, if they exist.
    3. +
  3. Find f'(x). Then use it to determine the intervals where the function is increasing and the intervals where the function is decreasing. State any local extrema.
    4. +
  4. Find f''(x). Then use it to determine the intervals where the function is concave up and the intervals where the function is concave down. State any inflection points.
    5. + +
  5. Put everything together and draw sketch.
    6. +

+ + + + + + + Sketch the graph of each of the following functions using the guide to curve sketching found in + f(x)=x^4-4x^3+10 + f(x)=\frac{x^2-4}{x^2-9} +

f(x)=x+2\cos x on the interval [0,2\pi]

 + f(x)=\frac{x^2+x-2}{x+3} + f(x)=\frac{x}{\sqrt{x^2+2}} + f(x)=x^6+\frac{12}{5}x^5-12x^4+10 + + + + Videos +

No video is available for this learning outcome.

+ +
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+ Applied optimization (AD8) + + + + + Activities + + The box + +

+ Help your company design an open box (no lid) with maximum volume given the following constraints: +

    +
  • The box must be made from the following material: an 8 by 8 inches piece of cardboard.
    • +
  • To create the box, you are asked to cut out a square from each corner of the 8 by 8 inches piece of cardboard and to fold up the flaps to create the sides.
    • +
+

+ + +

Draw a diagram illustrating how the box is created.

+ + +

Explain why the volume of the box is a function of the side length x of the cutout squares.

+ + +

Express the volume of the box V as a function of the length of the cuts x.

+ + +

What is a realistic domain of the function V(x)?

+ + +

What cut length x maximizes the volume of the box?

+ + + + A guide for optimization problems +

+

    +
  1. Draw a diagram and introduce variables. +
    2. +
  2. Determine a function of a single variable that models the quantity to be optimized. +
    3. +
  3. Decide the domain on which to consider the function being optimized. +
    4. +
  4. Use calculus to identify the global maximum and/or minimum of the quantity being optimized. +
    5. +
  5. Conclusion: what are the optimal points and what optimal values do we obtain at these points? +
    6. +

+ + + + + + + +

+ According to U.S.postal regulations, + the girth plus the length of a parcel sent by mail may not exceed 108 inches, + where the girth is the perimeter of the smallest end. + What is the largest possible volume of a rectangular parcel with a square end that can be sent by mail? + What are the dimensions of the package of largest volume? +

+ + + + + +

+ Let x represent the length of one side of the square end and y the length of the longer side. + Label these quantities appropriately on the image shown in Figure. +

+ +
+ A rectangular parcel with a square end. + +
+ + + + +

+ What is the quantity to be optimized in this problem? + +

+
    +

  1. maximize volume (call this V)

    2. +

  2. maximize the girth plus length (call this P)

    3. +

  3. minimize volume (call this V)

    4. +

  4. minimize the girth plus length (call this P)

    5. +
+ +

Which formula below represents the quantity you want to optimize in terms of x and y?

+
    +

  1. V=x^2y

    2. +

  2. V=xy^2

    3. +

  3. P=2x+y

    4. +

  4. P=4x+y

    5. +
+ + + +

+ The problem statement tells us that the parcel's girth plus length (P) may not exceed 108 inches. + In order to maximize volume, + we assume that we will actually need the girth plus length P to equal 108 inches. + What equation does this constraint give us involving x and y? +

+
    +

  1. 108=4x+y

    2. +

  2. 108 = 2x +y

    3. +

  3. 108 = x^2 +y

    4. +

  4. 108 = xy^2

    5. +
+ +

The equation above gives the relationship between x and y. For ease of notation, solve this equation for y as a function on x and then find a formula for the volume of the parcel as a function of the single variable x. What is the formula for V(x)?

+
    +

  1. V(x) = x^2 (108-4x)

    2. +

  2. V(x) = x(108-4x)^2

    3. +

  3. V(x) =x ^2 (108-2x)

    4. +

  4. V(x) =x (108-2x)^2

    5. +
+ + + +

Over what domain should we consider this function? To answer this question, notice that the problem gives us the constraint that P (girth plus length) is 108 inches. This constraint produces intervals of possible values for x and y.

+
    +

  1. 0 \leq x \leq 108

    2. +

  2. 0 \leq y \leq 108

    3. +

  3. 0 \leq x \leq 27

    4. +

  4. 0 \leq y \leq 27

    5. +
+ + +

+ Use calculus to find the global maximum of the volume of the parcel on the domain you just determined. Justify that you have found the global maximum using either the Closed Interval Method, the First Derivative Test, or the Second Derivative Test! +

+ + + + +

Notice that a critical point might or might not be an global maximum or minimum, so just finding the critical points is not enough to answer an optimization problem. Moreover, some of the critical points might be outside of the domain imposed by the context and thus they cannot be feasible optimal points.

+ + + + Revenue = Number of tickets <m>\times</m> Price of ticket + +

+ Waterford movie theater currently charges $8 for a ticket. At this price, the theater sells 200 tickets daily. The general manager wonders if they can generate more revenue by increasing the price of a tickets. A survey shows that they will lose 20 customers for every dollar increase in the ticket price. +

+ + +

If the price of a movie ticket is increased by d dollars, write a formula for the price P in terms of d.

+ + +

If the price of a ticket is increased by one dollar, how many many customers will the theater lose?

+ + +

Write a formula for the number of tickets sold T as a function of a price increase of d dollars.

+ + +

Consider the new price of a ticket P(d) and the new number of tickets sold T(d). Write a formula for the +revenue earned by ticket sales R(d) as a function of a price increase of d dollars.

+ + +

What is a realistic domain for the function R(d)?

+ + +

What increase in price d should the general manager choose to maximize the revenue? What price would a movie ticket cost then and what would the revenue be at that price?

+ + +

Suppose now that the cost of running the business when the price is increased by d dollars is given by +C(d) = 10d^3 −40d^2 +40d+600. +If the manager decides that they will definitely increase the price, what price increase d maximizes the +profit? (Recall that Profit = Revenue - Cost).

+ + + + + + Modeling given a geometric shape + +

+ The city council is planning to construct a new sports ground in the shape of a rectangle with semicircular ends. A running track 400 meters long is to go around the perimeter. +

+ + +

What choice of dimensions will make the rectangular area in the center as large as possible?

+ + +

What should the dimensions so the total area enclosed by the running track is maximized?

+ + + + + Modeling in algebraic situations + +

Find the coordinates of the point on the curve y=\sqrt{x} closest to the point (1,0).

+ + +

The sum of two positive numbers is 48. What is the smallest possible value of the sum of their squares?

+ + + + + + + + + + +

+Suppose that if a widget is priced at \$176, then +you are able to sell 672 units each day. According to +a survey of customers, increasing this price by \$1 will result +in losing 4 daily sales; decreasing by \$1 will gain 4 daily sales. +Your manager asks you how to adjust the price of a widget to +maximize the revenue (widgets sold times price). Write an explanation +of what this change in price should be and why. +

+ + + + + + + + + + + Videos +
+
+
+
+ +
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+ Limits and Derivatives (AD9) + + + + + Activities + + +

When we compute a limit algebraically, we often encounter the indeterminate form

+\frac{0}{0} +

but this means that limit can equal any number, infinity, or it might not exist. When we encounter an indeterminate form, we just do not know (yet) what the value of the limit is.

+ + +

We can compute limits that give indeterminate forms via algebraic manipulations. Consider

+ \lim_{x\to 1} \frac{4x - 4}{x^2 -1} . + +

Verify that this limit gives an indeterminate form of the type \frac{0}{0}. +

+ +

As you are computing a limit, you can cancel common factors. After you simplify the fraction, what is the limit? +

+
    +

  1. 4

    2. +

  2. 2

    3. +

  3. \frac{1}{2}

    4. +

  4. The limit does not exist.

    5. +
+ + + + + +

Consider the limits

+ \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} = f'(a). + \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = f'(a). +

Notice that these limits give indeterminate forms of the type \frac{0}{0}. However, these limits are equal to f'(a), the derivative of f(x) at x=a. If you can compute f'(a), then you have computed the value of the limit!

+ + +

Use the limit definition of the derivative to compute the following limits. Each limit is f'(a), the derivative of some function f(x) at some point x=a. You need to determine the function and the point to find the value of the limit: f'(a).

+ + +

Notice that \displaystyle \lim_{x\to 0} \frac{e^{2+x}-e^2}{x} is the derivative of e^x at x=2 (where x was used for h). Given this observation, what is this limit equal to? +

+
    +

  1. 2

    2. +

  2. e

    3. +

  3. e^2

    4. +

  4. The limit does not exist.

    5. +
+ + +

Consider \displaystyle \lim_{x\to 0} \frac{\ln(1+x)}{x}. This limit is also the limit definition of some derivative at some point. What is the value of this limit? +

+
    +

  1. 1

    2. +

  2. 0

    3. +

  3. \ln(2)

    4. +

  4. The limit does not exist.

    5. +
+ + + +

Compute the following limits using the limit defintion of the derivative at a point.

 +

\displaystyle \lim_{x\to 0} \frac{\sin(x)}{x}

 +

\displaystyle \lim_{x\to 0} \frac{\tan(x)}{x}

 +

\displaystyle \lim_{x\to 0} \frac{\cos(\frac{\pi}{3}+x) - \frac{1}{2} }{x}

 + + + +

When we compute a limit algebraically, we might encounter the indeterminate form

+\frac{\infty}{\infty} +

but this means that limit can equal any number, infinity, or it might not exist. When we encounter an indeterminate form, we just do not know (yet) what the value of the limit is.

+ + +

We can compute limits that give indeterminate forms via algebraic manipulations. Consider

+ \lim_{x\to +\infty} \frac{2x^2 + 1}{x^2 -1} . + +

Verify that this limit gives an indeterminate form of the type \frac{\infty}{\infty}. +

+ +

You can manipulate this fraction algebraically by dividing numerator and denominator by x^2. Then, notice that \pm \frac{1}{x^2} \to 0 as x \to \infty . Given these observations, what is the given limit equal to? +

+
    +

  1. 2

    2. +

  2. 1

    3. +

  3. \frac{1}{2}

    4. +

  4. The limit does not exist.

    5. +
+ + + + + + L' Hôpital's Rule

If the functions f(x), g(x) are both differentiable around x = a and for the limit of \frac{f(x)}{g(x)} as x \to a (or x \to \pm \infty) we have one of the indeterminate forms \frac{0}{0} or \frac{\infty}{\infty}, then

+ + \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)} +

provided that the limit exists!

+ + + + + +

Look back at some limits that gave you an indeterminate form. Can you use L'Hôpital's Rule to find the limit? + If using the L'Hôpital's Rule is appropriate, then try to compute the limit this way. It should give you the same result.

+ + +

In , when we started to study limits, we encountered the Squeeze Theorem and computed the limit \displaystyle \lim_{x\to 0} \frac{\sin(x)}{x} using this theorem. Let's find new ways to compute this limit.

+ + +

Thinking about x as the length of an interval h, this limit is actually equal to the value of some derivative, so f'(a) = \displaystyle \lim_{h\to 0} \frac{\sin(h)}{h}. What function f(x) and what point x=a would lead to this limit? Use these to find f'(a), the value of this limit (in a new way!).

+ + +

Verify, one more time, that this limit is indeed an indeterminate form. Then use L'Hôpital's Rule to find this limit (again, in another way!).

+ + + + +

+ For the following limits, check if they give an indeterminate form. If they do, try to use L'Hôpital's Rule. Does it help? It may or may not, or you may just need to use the rule repeatedly. Either way, try to compute the value of the following limits.

+ + +

\displaystyle \lim_{x\to 0} \frac{\sin(x)}{x}

+ + +

\displaystyle \lim_{x\to 0} \frac{e^x-1}{x}

+ + +

\displaystyle \lim_{x\to \infty} \frac{3x^2+3}{x^2+ 2x}

+ + +

\displaystyle \lim_{x\to 0^{+}} \frac{\ln(x)}{-x}

+ + +

\displaystyle \lim_{x\to 0^{+}} \frac{\ln(x)}{1/x}

+ + +

\displaystyle \lim_{x\to 0} \frac{\sin^2(3x)}{5x^3 - 3x^2}

+ + + + + +

+For each limit, explain if L'Hôpital's Rule may be applied. +If it can, explain how to use this rule to find the limit. +

+ + +

+ +\lim_{x\to \infty } \frac{ -8 \, x + 3 \, e^{x} }{ 7 \, x - 3 \, e^{x} } + +

+ + +

+ +\lim_{x\to 0 } \frac{ 6 \, \cos\left(8 \, x\right) }{ 4 \, x - 7 } + +

+ + +

+ +\lim_{x\to 0 } \frac{ -9 \, \cos\left(3 \, x\right) + 9 }{ -3 \, x } + +

+ + +

+ +\lim_{x\to 4 } \frac{ x^{2} - x - 12 }{ x^{2} - 13 \, x + 36 } + +

+ + + + + + + +

+ There are situations in which using L'Hôpital's Rule does not help and you do need some algebra skills! Consider the function r(x)=\frac{x}{\sqrt{x^2 +2}} and suppose that we want to find the limits as x tends to \pm \infty.

+ + +

Check that the limit as x \to + \infty gives an indeterminate form \frac{\infty}{\infty}. Then try to use L'Hôpital's Rule... what happens? What if you use it again?

+ + +

We need to use algebra to handle this limit. Informally, we would like to cancel the highest powers at the numerator and denominator. Look at the denominator, \sqrt{x^2 +2}. We want to factor out an x^2 under the square root. What do you get?

+
    +

  1. \sqrt{ x^2\left(1 +\frac{2}{x}\right) }

    2. +

  2. \sqrt{ x^2\left(1 +\frac{2}{x^2}\right) }

    3. +

  3. \sqrt{ x^2\left(1 + x \right)}

    4. +

  4. \sqrt{ x^2\left(1 + x^2 \right)}

    5. +
+ + +

Now we need to be careful when computing \sqrt{x^2} as \sqrt{x^2}=|x|. The absolute value function |x| equals +x when we have a positive input and -x when we have a negative output. So we have the two limits. + \lim_{x \to +\infty} \frac{x}{|x|\sqrt{(1 +\frac{2}{x^2}) }} + \lim_{x \to -\infty} \frac{x}{|x|\sqrt{(1 +\frac{2}{x^2}) }} + Thinking about what happens to the absolute values as you go towards positive or negative infinity, find the values of these two limits... The two limits have different values!

+ + + + + Videos +
+
+ +
+ + diff --git a/calculus/source/03-AD/main.ptx b/calculus/source/03-AD/main.ptx new file mode 100644 index 00000000..37b677aa --- /dev/null +++ b/calculus/source/03-AD/main.ptx @@ -0,0 +1,16 @@ + + + + Applications of Derivatives (AD) + + + + + + + + + + + + \ No newline at end of file diff --git a/calculus/source/03-AD/outcomes/01.ptx b/calculus/source/03-AD/outcomes/01.ptx new file mode 100644 index 00000000..312ae09e --- /dev/null +++ b/calculus/source/03-AD/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Use derivatives to answer questions about rates of change and equations of tangents. +

\ No newline at end of file diff --git a/calculus/source/03-AD/outcomes/02.ptx b/calculus/source/03-AD/outcomes/02.ptx new file mode 100644 index 00000000..a10f4c42 --- /dev/null +++ b/calculus/source/03-AD/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Use tangent lines to approximate functions. +

\ No newline at end of file diff --git a/calculus/source/03-AD/outcomes/03.ptx b/calculus/source/03-AD/outcomes/03.ptx new file mode 100644 index 00000000..352fa7a5 --- /dev/null +++ b/calculus/source/03-AD/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+ Model and analyze scenarios using related rates. +

diff --git a/calculus/source/03-AD/outcomes/04.ptx b/calculus/source/03-AD/outcomes/04.ptx new file mode 100644 index 00000000..f9204adb --- /dev/null +++ b/calculus/source/03-AD/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+ Use the Extreme Value Theorem to find the global maximum and minimum values of a continuous function on a closed interval. +

diff --git a/calculus/source/03-AD/outcomes/05.ptx b/calculus/source/03-AD/outcomes/05.ptx new file mode 100644 index 00000000..ae58bf2f --- /dev/null +++ b/calculus/source/03-AD/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+ Determine where a differentiable function is increasing and decreasing and classify the critical points as local extrema. +

diff --git a/calculus/source/03-AD/outcomes/06.ptx b/calculus/source/03-AD/outcomes/06.ptx new file mode 100644 index 00000000..9e8601fe --- /dev/null +++ b/calculus/source/03-AD/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+ Determine the intervals of concavity of a twice differentiable function and find all of its points of inflection. +

diff --git a/calculus/source/03-AD/outcomes/07.ptx b/calculus/source/03-AD/outcomes/07.ptx new file mode 100644 index 00000000..1427c36e --- /dev/null +++ b/calculus/source/03-AD/outcomes/07.ptx @@ -0,0 +1,4 @@ + +

+ Sketch the graph of a differentiable function whose derivatives satisfy given criteria. +

diff --git a/calculus/source/03-AD/outcomes/08.ptx b/calculus/source/03-AD/outcomes/08.ptx new file mode 100644 index 00000000..7ac29fed --- /dev/null +++ b/calculus/source/03-AD/outcomes/08.ptx @@ -0,0 +1,4 @@ + +

+ Apply optimization techniques to solve various problems. +

diff --git a/calculus/source/03-AD/outcomes/09.ptx b/calculus/source/03-AD/outcomes/09.ptx new file mode 100644 index 00000000..e67cc368 --- /dev/null +++ b/calculus/source/03-AD/outcomes/09.ptx @@ -0,0 +1,4 @@ + +

+ Compute the values of indeterminate limits using L'Hôpital's Rule. +

diff --git a/calculus/source/03-AD/outcomes/main.ptx b/calculus/source/03-AD/outcomes/main.ptx new file mode 100644 index 00000000..f054745f --- /dev/null +++ b/calculus/source/03-AD/outcomes/main.ptx @@ -0,0 +1,40 @@ + +> + +

+ How can we use derivatives to solve application questions? +

+

+ By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
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    5. +
  5. + +
    6. +
  6. + +
    7. +
  7. + +
    8. +
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    9. +
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    10. +
+ \ No newline at end of file diff --git a/calculus/source/03-AD/readiness.ptx b/calculus/source/03-AD/readiness.ptx new file mode 100644 index 00000000..a676338b --- /dev/null +++ b/calculus/source/03-AD/readiness.ptx @@ -0,0 +1,33 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Find the derivative of a function using methods from , , , and .

    +
    2. +

  2. Find the second derivative of a function. (Khan Academy)

    +
    3. +
  3. +

    Determine whether values are in the domain of a function (Khan Academy)

    +
    4. +
  4. +

    Find local and global extrema given the graph of a function (Khan Academy)

    +
    5. +
  5. +

    Find the intervals where a graph is increasing or decreasing (Khan Academy)

    +
    6. +
  6. +

    Determine the zeros of a polynomial (Khan Academy)

    +
    7. +
  7. +

    Modeling with multiple variables (Khan Academy)

    +
    8. + +
+ + \ No newline at end of file diff --git a/calculus/source/04-IN/01.ptx b/calculus/source/04-IN/01.ptx new file mode 100644 index 00000000..43fb3be9 --- /dev/null +++ b/calculus/source/04-IN/01.ptx @@ -0,0 +1,131 @@ + + +
+ Geometry of definite integrals (IN1) + + + + + Activities + + +

The definite integral for a positive function f(x) \geq 0 between the points x=a and x=b is the area between the function and the x-axis. We denote this quantity as \displaystyle \int_a^b f(x) \, dx

+ + +

+ For some functions which have known geometric shapes (like pieces of lines or circles) we can already compute these area exactly and we will do so in this section. But for most functions we do not know quite yet how to compute these areas. In the next section, we will see that because we can compute the areas of rectangles quite easily, we can always try to approximate a shape with rectangles, even if this could be a very coarse approximation. +

+ + + +

+ Consider the linear function f(x)=2x. Sketch a graph of this function. Consider the area between the x-axis and the function on the interval [0,1]. What is \int_0^1 f(x) \, dx?

+
    +

  1. 1

    2. +

  2. 2

    3. +

  3. 3

    4. +

  4. 4

    5. +
+ + + + +

+ Consider the linear function f(x)=4x. What is \int_0^1 f(x) \, dx?

+
    +

  1. 1

    2. +

  2. 2

    3. +

  3. 3

    4. +

  4. 4

    5. +
+ + + +

+ Consider the linear function f(x)=2x +2. Notice that on the interval [0,1], the shape formed between the graph and the x-axis is a trapezoid. What is \int_0^1 f(x) \, dx?

+
    +

  1. 1

    2. +

  2. 2

    3. +

  3. 3

    4. +

  4. 4

    5. +
+ + + + +

+ Consider the function f(x)=\sqrt{4-x^2}. Notice that on the domain [-2,2], the shape formed between the graph and the x-axis is a semicircle. What is \int_{-2}^2 f(x) \, dx?

+
    +

  1. \pi

    2. +

  2. 2\pi

    3. +

  3. 3\pi

    4. +

  4. 4\pi

    5. +
+ + + +

+ + If a function f(x) \leq 0 on [a,b], then we define the integral between a and b to be

+ \int_a^b f(x) \, dx \, = \, (-1) \times \text{area between the graph of f(x) and the x axis on the interval [a,b].} +

So the definite integral for a negative function is the "negative" of the area between the graph and the x-axis.

+ + + + + +

+ Explain how to use geometric formulas for area to compute the following definite + integrals. For each part, sketch the function to support your explanation. +

    +
  1. \int_{ 1 }^{ 6 }\left(-3 \, x + 6\right)dx
    2. +
  2. \int_{ 2 }^{ 6 }\left(-3 \, x + 6\right)dx
    3. +
  3. \int_{ 1 }^{ 5 }\left(-\sqrt{-{\left(x - 1\right)}^{2} + 16}\right)dx
    4. + +
+

+ + + + + + + + + + + +

+ The graph of g(t) and the areas A_1, A_2, A_3 are given below.

+
+ +
+ +

Find \int_{3}^{3} g(t) \, dt

 +

Find \int_{3}^{6} g(t) \, dt

 +

Find \int_{0}^{10} g(t) \, dt

 +

Suppose that g(t) gives the velocity in fps at time t (in seconds) of a particle moving in the vertical direction. A positive velocity indicates that the particle is moving up, a negative velocity indicates that the particle is moving down. If the particle started at a height of 3ft, at what height would it been after 3 seconds? After 6 seconds? After 10 seconds? At what time does the particle reach the highest point in this time interval?

 + + + + + + Videos +
+
+ +
\ No newline at end of file diff --git a/calculus/source/04-IN/02.ptx b/calculus/source/04-IN/02.ptx new file mode 100644 index 00000000..c73e9459 --- /dev/null +++ b/calculus/source/04-IN/02.ptx @@ -0,0 +1,198 @@ + + +
+ Approximating definite integrals (IN2) + + + + + Activities + + + + +

+ Suppose that a person is taking a walk along a long straight path and walks at a constant rate of 3 miles per hour. +

+ + + +

+ On the left-hand axes provided in Figure, + sketch a labeled graph of the velocity function v(t) = 3. +

+ +
+ At left, + axes for plotting y = v(t); + at right, for plotting + y = s(t). + +
+ +

+ Note that while the scale on the two sets of axes is the same, + the units on the right-hand axes differ from those on the left. + The right-hand axes will be used in question (d). +

+ + + +

+ How far did the person travel during the two hours? + How is this distance related to the area of a certain region under the graph of y = v(t)? +

+ + + +

+ Find an algebraic formula, s(t), + for the position of the person at time t, + assuming that s(0) = 0. + Explain your thinking. +

+ + + +

+ On the right-hand axes provided in , + sketch a labeled graph of the position function y = s(t). +

+ + + +

+ For what values of t is the position function s increasing? + Explain why this is the case using relevant information about the velocity function v. +

+ + + + + +

+ Suppose that a person is walking in such a way that her velocity varies slightly according to the information given in Table + and graph given in Figure. +

+ + + + + Velocity data for the person walking. + + + t + v(t) + + + 0.00 + 1.500 + + + 0.25 + 1.789 + + + 0.50 + 1.938 + + + 0.75 + 1.992 + + + 1.00 + 2.000 + + + 1.25 + 2.008 + + + 1.50 + 2.063 + + + 1.75 + 2.211 + + + 2.00 + 2.500 + + + +
+ +
+ The graph of y = v(t). + +
+ + + + +

+ Using the grid, graph, + and given data appropriately, + estimate the distance traveled by the walker during the two hour interval from t = 0 to t = 2. + You should use time intervals of width \Delta t = 0.5, + choosing a way to use the function consistently to determine the height of each rectangle in order to approximate distance traveled. +

+ + + +

+ How could you get a better approximation of the distance traveled on [0,2]? + Explain, and then find this new estimate. +

+ + + +

+ Now suppose that you know that v is given by v(t) = 0.5t^3-1.5t^2+1.5t+1.5. + Remember that v is the derivative of the walker's position function, + s. + Find a formula for s so that s' = v. +

+ + + +

+ Based on your work in (c), + what is the value of s(2) - s(0)? + What is the meaning of this quantity? +

+ + + + + +

+Explain how to approximate the area under the curve +f(x)=-9 \, x^{3} + 3 \, x - 9 on the +interval [4,10] using a right Riemann +sum with 3 rectangles of uniform width. +

+ + + + + + + + Videos +
+
+ +
\ No newline at end of file diff --git a/calculus/source/04-IN/03.ptx b/calculus/source/04-IN/03.ptx new file mode 100644 index 00000000..9cc031e5 --- /dev/null +++ b/calculus/source/04-IN/03.ptx @@ -0,0 +1,248 @@ + + +
+ Elementary antiderivatives (IN3) + + + + + Activities + + + + +

+ If g and G are functions such that G' = g, + we say that G is an antiderivative + antiderivative + of g. +

+

+ The collection of all antiderivatives of g is called the + general antiderivative or + indefinite integral, + denoted by \int g(x)\,dx. + All antiderivatives differ by a constant C (since + \frac{d}{dx}[C]=0), so + we may write: \int g(x)\,dx=G(x)+C. +

+ + + + + +

Consider the function f(x)=\cos x. Which of the following could be F(x), an antiderivative of f(x)?

+
    +

  1. \sin x

    2. +

  2. \cos x

    3. +

  3. \tan x

    4. +

  4. \sec x

    5. + +
+ + + + + + +

Consider the function f(x)=x^2. Which of the following could be F(x), an antiderivative of f(x)?

+
    +

  1. 2x

    2. +

  2. \frac{1}{3}x^3

    3. +

  3. x^3

    4. +

  4. \frac{2}{3}x^3

    5. + +
+ + + + + +

+ We now note that whenever we know the derivative of a function, + we have a function-derivative pair, + so we also know the antiderivative of a function. + For instance, in we could use our prior knowledge that + + \frac{d}{dx}[\sin(x)] = \cos(x) + , + to determine that F(x) = \sin(x) is an antiderivative of f(x) = \cos(x). + F and f together form a function-derivative pair. + Every elementary derivative rule leads us to such a pair, + and thus to a known antiderivative. +

+ + +

+ In the following activity, + we work to build a list of basic functions whose antiderivatives we already know. +

+ + + + +

+ Use your knowledge of derivatives of basic functions to complete Table of antiderivatives. + For each entry, + your task is to find a function F whose derivative is the given function f. + +

+ + + Familiar basic functions and their antiderivatives. + + + given function, f(x) + antiderivative, F(x) + + + k, (k is constant) + + + + x^n, n \ne -1 + + + + \frac{1}{x}, x \gt 0 + + + + \sin(x) + + + + \cos(x) + + + + \sec(x) \tan(x) + + + + \csc(x) \cot(x) + + + + \sec^2 (x) + + + + \csc^2 (x) + + + + e^x + + + + a^x (a \gt 1) + + + + \frac{1}{1+x^2} + + + + \frac{1}{\sqrt{1-x^2}} + + + + +
+ + + + + + + + + + + + +

+ In Activity, + we constructed a list of the basic antiderivatives we know at this time. + Those rules will help us antidifferentiate sums and constant multiples of basic functions. For example, + we found that -\cos(x) is an antiderivative of \sin(x) and + \frac{1}{3}x^3 is an antiderivative of x^2.

 + + + +

Using this information, which of the following is an antiderivative for f(x) = 5\sin(x) - 4x^2?

+
    + +

  1. F(x) = -5\cos(x) +\frac{4}{3}x^3.

    2. +

  2. F(x) = 5\cos(x) + \frac{4}{3}x^3.

    3. +

  3. F(x) = -5\cos(x) - \frac{4}{3}x^3.

    4. +

  4. F(x) = 5\cos(x) - \frac{4}{3}x^3.

    5. + +
+ + + + + + + + + + + + + + +

+Find the general antiderivative for each function. +

+ + + + f(x) = -4 \, \sec^2\left(x\right) + f(x) = \frac{8}{\sqrt{x}} + + + + + + + + +

+Find each indefinite integral. +

+ + + + \int (-9 \, x^{4} - 7 \, x^{2} + 4) \, dx + \int 3 \, e^{x}\, dx + + + + Videos +
+
+ +
+ + + + + + + + + + + + + + diff --git a/calculus/source/04-IN/04.ptx b/calculus/source/04-IN/04.ptx new file mode 100644 index 00000000..7263b2c2 --- /dev/null +++ b/calculus/source/04-IN/04.ptx @@ -0,0 +1,137 @@ + + +
+ Initial Value Problems (IN4) + + + + + Activities + + +

+ In this section we will discuss the relationship between antiderivatives and solving simple differential equations. A differential equation is an equation that has a derivative. For this section we will focus on differential equations of the form + \frac{dy}{dx} = f(x). + Our goal is to find a relationship of y(x) that satisfies the differential equation. We can solve for y(x) by finding the antiderivative of f(x). +

+ + + + +

Which of the following equations for y(x) satisfies the differential equation \frac{dy}{dx} = x^2+2x.

+
    +

  1. y(x) = \frac{x^3}{3} + x^2 + 4

    2. +

  2. y(x) = 2x + 2

    3. +

  3. y(x) = \frac{x^3}{3} + x^2 + 10

    4. +

  4. y(x) = \frac{x^3}{3} + x^2

    5. +

  5. y(x) = 2x

    6. + +
+ + + + + +

In there are more than one solution that satisfies the differential equation. In fact their is a family of functions that satisfies the differential equation, that is + f(x) = \frac{x^3}{3} + x^2 + c_1, + where c_1 is an arbitrary constant yet to be defined. To find c_1 we have to have some initial value for the differential equation, y(x_0) = y_0, where the point (x_0,y_0) is the starting point for the differential equation. In general this section we will focus on solving initial value problems (differential equation with an initial condition) of the form, + \frac{dy}{dx} = f(x), \;\;\; y(x_0) = y_0. +

+ + + + + +

Which of the following equations for y(x) satisfies the differential equation and initial condition, \frac{dy}{dx} = x^2+2x, \,\,\,y(3) = 16.

+
    +

  1. y(x) = \frac{x^3}{3} + x^2 - 4

    2. +

  2. y(x) = \frac{x^3}{3} + x^2 + 2

    3. +

  3. y(x) = \frac{x^3}{3} + x^2 -2

    4. +

  4. y(x) = \frac{x^3}{3} + x^2 + 16

    5. + +
+ + + + + + +

Which of the following functions satisfies the initial value problem, + \frac{dy}{dx} = \sin(x), \,\,\, y(0) = 1. +

+
    +

  1. y(x) = \cos(x)

    2. +

  2. y(x) = \cos(x) + 2

    3. +

  3. y(x) = \cos(x) + 1

    4. +

  4. y(x) = -\cos(x)

    5. +

  5. y(x) = -\cos(x) + 2

    6. + +
+ + + + + + +

One of the applications of initial value problems is calculating the distance traveled from a point based on the velocity of the object. Given that the velocity of the of an object in km/hr is approximated by v(t) = \cos(t) + 1, what is the approximate distance travelled by the object after 1 hour?

+
    +

  1. s(1) \approx 1 km

    2. +

  2. s(1) \approx 0.1585 km

    3. +

  3. s(1) \approx 1.8415 km

    4. +

  4. s(1) \approx 2.3415 km

    5. +
+ + + + + +

So far we have only been going from velocity to position of an object. Recall that to find the acceleration of an object, you can take the derivative of the velocity of an object. Let use say we have the acceleration of a falling object in m/s^2 given by a(t) = -9.8.

+ +

What is the velocity of the falling object, if the initial velocity is given by v(0) = 0 m/s.

+
    +

  1. v(t) = -9.8t m

    2. +

  2. v(t) = -9.8t m/s

    3. +

  3. v(t) = 9.8t m/s

    4. +

  4. v(t) = 9.8t+1 m

    5. +
+ + +

What is the position of the object, if the initial position is given by s(0) = 10 m.

+
    +

  1. s(t) = 4.9t + 10 m

    2. +

  2. s(t) = -4.9t^2 + 14.9 m

    3. +

  3. s(t) = -4.9t^2 + 10 m

    4. +

  4. s(t) = 4.9t + 5.1 m

    5. +
+ + + + + + + +

+ Let f'(x)=-12 \, x - 6 . Find f(x) such that f(5)=-179. +

+ + + + + + + Videos +
+
+ +
+ diff --git a/calculus/source/04-IN/05.ptx b/calculus/source/04-IN/05.ptx new file mode 100644 index 00000000..66075efc --- /dev/null +++ b/calculus/source/04-IN/05.ptx @@ -0,0 +1,226 @@ + + +
+ FTC for definite integrals (IN5) + + + + + Activities + + + +

Find the area beteween f(x)=\frac{1}{2}x+2 and the x-axis from x=2 to x=6.

+ + +
+ +
+ + + + +

Approximate the area under the curve f(x)=(x-1)^2+2 on the interval [1,5] using a left Riemann sum with four uniform subdivisions. Draw your rectangles on the graph.

+ +
+ +
+ + + + +

Let f(x) be a continuous function on the interval [a,b]. Divide the interval into n subdivisions of equal width, \Delta x, and choose a point x_i in each interval. Then, the definite integral of f(x) from a to b is

+ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x = \int_a^bf(x)dx + + + + + +

How does \displaystyle \int_2^6 \left(\frac{1}{2}x+2\right) \, dx relate to ? Could you use to find \displaystyle \int_0^4 \left(\frac{1}{2}x+2\right) \, dx? What about \displaystyle \int_1^7 \left(\frac{1}{2}x+2\right) \, dx?

+ + + + Properties of Definite Integrals +

+

    +
  1. If f is defined at x=a, then \displaystyle \int_a^a f(x) \, dx =0.
    2. +
  2. If f is integrable on [a,b], then \displaystyle \int_a^b f(x) \, dx = - \displaystyle \int_b^a f(x) \, dx.
    3. +
  3. If f is integrable on [a,b] and c is in [a,b], then \displaystyle \int_a^b f(x) \, dx = \displaystyle \int_a^c f(x) \, dx + \displaystyle \int_c^b f(x) \, dx.
    4. +
  4. If f is integrable on [a,b] and k is a constant, then kf is integrable on [a,b] and \displaystyle \int_a^b kf(x) \, dx = k\displaystyle \int_a^b f(x) \, dx.
    5. +
  5. If f and g are integrable on [a,b], then f\pm g are integrable on [a,b] and \displaystyle \int_a^b [f(x) \pm g(x)] \, dx =\displaystyle \int_a^b f(x) \, dx \pm \displaystyle \int_a^b g(x) \, dx.
    6. +

+ + + + +

Suppose that \displaystyle\int_1^5 f(x)\, dx = 10 and \displaystyle\int_5^7 f(x)\, dx = 4 .Find each of the following.

+ + +

\displaystyle\int_1^7 f(x)\, dx

+ + + +

\displaystyle\int_5^1 f(x)\, dx

+ + + +

\displaystyle\int_7^7 f(x)\, dx

+ + + +

3 \displaystyle\int_5^7 f(x)\, dx

+ + + + + +

We've been looking at two big things in this chapter: antiderivatives and the area under a curve. In the early days of the development of calculus, they were not known to be connected to one another. The integral sign wasn't originally used in both instances. (Gottfried Leibniz introduced it as an elongated S to represent the sum when finding the area.) Connecting these two seemingly separate problems is done by the Fundamental Theorem of Calculus

+ + +The Fundamental Theorem of Calculus + +

If a function f is continuous on the closed interval [a,b] and F is an antiderivative of f on the interval [a,b], then

+ \displaystyle \int_a^b f(x) \, dx = F(b)-F(a) + + + + + +

Evaluate the following definite integrals. Include a sketch of the graph with the area you've found shaded in. Approximate the area to check to see if your definite integral answer makes sense. (Note: Just a guess, you don't have to use Riemann sums. Use the grid to help.)

+ + + +

\displaystyle \int_0^2 \left(x^2+3\right) \, dx

+
+ +
+ + + +

\displaystyle \int_1^4 \left(\sqrt{x}\right) \, dx

+
+ +
+ + + +

\displaystyle \int_{-\pi/4}^{\pi/2} \left(\cos x\right) \, dx

+
+ +
+ + + + + +

Find the area between f(x)=2x-6 on the interval [0,8] using +

    +
  1. geometry
    2. +
  2. the definite integral
    3. +
+

+
+ +
+

What do you notice?

+ + + + + +

Find the area bounded by the curves f(x)=e^x-2, the x-axis, x=0, and x=1.

+
+ +
+ + + + + +

Set up a definite integral that represents the shaded area. Then find the area of the given region using the definite integral.

+ + +

y=\frac{1}{x^2}

+
+ +
+ + +

y=3x^2-x^3

+
+ +
+ + + + + +

+Explain how to compute the exact value of each of the following +definite integrals using the Fundamental Theorem of Calculus. +Leave all answers in exact form, with no decimal approximations. +

+ + +

+ +\int_{ -3 }^{ -2 } \left( -9 \, x^{3} - 9 \, x^{2} + 1 \right) dx + +

+ + +

+ +\int_{ \frac{7}{6} \, \pi }^{ \frac{5}{4} \, \pi } \left( -3 \, \sin\left(x\right) \right) dx + +

+ + +

+ +\int_{ 2 }^{ 6 } \left( 3 \, e^{x} \right) dx + +

+ + + + + + + + Videos +
+
+ +
+ + + + + + + + + + + + + + + + + diff --git a/calculus/source/04-IN/06.ptx b/calculus/source/04-IN/06.ptx new file mode 100644 index 00000000..6e68341c --- /dev/null +++ b/calculus/source/04-IN/06.ptx @@ -0,0 +1,135 @@ + + +
+ FTC for derivatives of integrals (IN6) + + + + + Activities + + +

+ In this section we extend the Fundamental Theorem of Calculus discussed in to include taking the derivatives of integrals. We will call this addition to the Fundamental Theorem of Calculus (FTC) part II. First we will introduce part II and then discuss the implications of this addition. +

+ + + The Fundamental Theorem of Calculus (Part II) + +

If a function f is continuous on the closed interval [a,b], then the area function + A(x) = \int_a^x f(t)\,dt \,\,\,\,\, \mathrm{for}\,\,\, a\leq x\leq b, + is continuous on [a,b] and differentiable on (a,b). The area function satisfies A'(x) = f(x). Equivalently, + A'(x) = \frac{d}{dx}\int_a^x f(t)\,dt = f(x), + which means that the area function of f is an antiderivative of f on [a,b]. +

+ + + + + +

For the following activity we will explore the Fundamental Theorem of Calculus Part II.

+ + +

Given that A(x) = \int_a^xt^3\,dt, then by the Fundamental Theorem of Calculus Part I,

+
    +

  1. A(x) = x^3-a^3

    2. +

  2. A(x) = a^4 - x^4

    3. +

  3. A(x) = \frac{1}{4}(x^4 - a^4)

    4. +

  4. A(x) = 3x^2

    5. +
+ + +

Using what you found for A(x), what is A'(x)

+
    +

  1. A'(x) = 3x^2

    2. +

  2. A'(x) = 4a^3 - 4x^3

    3. +

  3. A'(x) = x^3

    4. +

  4. A'(x) = 6x

    5. +
+ + +

Use the Fundamental Theorem of Calculus Part II to find A'(x). What do you notice between what you got above and using FTC Part II? Which method do you prefer?

+
    +

  1. A'(x) = 3x^2

    2. +

  2. A'(x) = 4a^3 - 4x^3

    3. +

  3. A'(x) = x^3

    4. +

  4. A'(x) = 6x

    5. +
+ + + + + +

Given A(x) = \int_x^be^t\,dt, what is A'(x)?

+
    +

  1. A'(x) = -e^x

    2. +

  2. A'(x) = e^x

    3. +

  3. A'(x) = e^b-e^x

    4. +

  4. A'(x) = e^x-e^b

    5. +
+ + + + + +

For the first two activities we have only explored when the function of the limits of the integrand are x. Now we want to see what happens when the limits are more complicated. To do this we will follow a similar procedure as that done in activity 1.

+ + + + +

Recall that by the Fundamental Theorem of Calculus Part I, \int_a^bf(t)\,dt = F(b)-F(a).

+ + +

Let A(x) = \int_x^{x^2}f(t)\,dt and re-write using FTC Part I.

+ + +

Using what you got find A'(x). Explain what derivative rule(s) you used.

+ + +

Using what you found what is the derivative of A(x) = \int_x^{x^2}(t+2)\,dt?

+
    +

  1. A'(x) = 2x(x+2)-(x+2)

    2. +

  2. A'(x) = (x+2)-2x(x^2+2)

    3. +

  3. A'(x) = (x^2+2)-(x+2)

    4. +

  4. A'(x) = 2x(x^2+2)-(x+2)

    5. +
+ + + + +

+ Now we have some thoughts of how to generalize the FTC Part II when the limts are more complicated. +

+ + + The Fundamental Theorem of Calculus (Part II): Modified + +

If + A(x) = \int_{g(x)}^{h(x)} f(t)\,dt + then + A'(x) = \frac{d}{dx}\int_{g(x)}^{h(x)} f(t)\,dt = f(h(x))h'(x) - f(g(x))g'(x), + where g(x) and h(x) are continous differentiable functions. +

+ + + + + + +

Given A(x) = \int_{x^3}^{x^5}(\sin(t) - 2)\,dt, what is A'(x)?

+ + + + + + Videos +
+
+ +
diff --git a/calculus/source/04-IN/07.ptx b/calculus/source/04-IN/07.ptx new file mode 100644 index 00000000..31212e4a --- /dev/null +++ b/calculus/source/04-IN/07.ptx @@ -0,0 +1,158 @@ + + +
+ Area under curves (IN7) + + + + + Activities + + + +

+A geometrical interpretation of \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x = \int_a^bf(x)dx +() defines \int_a^bf(x)dx as the net area +between the graph of y=f(x) and the x-axis. By net area, we mean the area above the x-axis +(when f(x) is positive) minus the area below the x-axis (when f(x) is negative). +

+
+ + + As the number of subdivisions increases, the Riemann sum more closely appears to measure the + net area between a curve and the x-axis. + + Improving approximations of \int_0^5(x-2)(x-4)dx +
+ + + + + + +

+Write the net area between f(x)=6 \, x^{2} - 18 \, x and the x-axis +from x=2 to x=7 as a definite integral. +

+ + + + +

+Evaluate this definite integral to verify the net area is equal to 265 square units. +

+ + + + + + +

+In order to find the total area between a curve and the x-axis, one +must break up the definite integral at points where f(x)=0, that is, +wherever f(x) may change from positive to negative, or vice versa. +

+
+ + + The total area is illustrated by breaking up the integral from 0 to 5 at 2 and 4 + where (x-2) and (x-4) are equal to 0. + + Partitioning \int_0^5(x-2)(x-4)dx at x=2 and x=4. +
+

+ Since f(x)=(x-2)(x-4) is zero when x=2 and x=4, we may compute the total + area between y=(x-2)(x-4) and the x-axis using absolute values as follows: + \text{Area} = {\color{blue} \left|\int_0^2(x-2)(x-4)dx\right|}+ + {\color{red} \left|\int_2^4(x-2)(x-4)dx\right|}+{\color{blue}\left|\int_4^5(x-2)(x-4)dx\right|} +

+ + + + + +

+Follow these steps to find the total area between f(x)=6 \, x^{2} - 18 \, x and the x-axis +from x=2 to x=7. +

+ + + +

+Find all values for x where f(x)=6 \, x^{2} - 18 \, x is equal to 0. +

+ + + + +

+Only one such value is between x=2 and x=7. Use this value to fill in the \unknown below, +then verify that its value is 279 square units. + +\text{Area} = +\left| \int_{ 2 }^{ \unknown } +\left( 6 \, x^{2} - 18 \, x \right) dx \right| + +\left| \int_{ \unknown }^{ 7 } +\left( 6 \, x^{2} - 18 \, x \right) dx \right| + +

+ + + + + + +

+ Answer the following questions concerning + f(x)=6 \, x^{2} - 96. +

+ + +

+What is the total area between f(x)=6 \, x^{2} - 96 and the x-axis +from x=-1 to x=9? +

+ + +

+What is the net area between f(x)=6 \, x^{2} - 96 and the x-axis +from x=-1 to x=9? +

+ + + + + + + Videos +
+
+ +
\ No newline at end of file diff --git a/calculus/source/04-IN/08.ptx b/calculus/source/04-IN/08.ptx new file mode 100644 index 00000000..d0c65d33 --- /dev/null +++ b/calculus/source/04-IN/08.ptx @@ -0,0 +1,187 @@ + + +
+ Area between curves (IN8) + + + + + Activities + +

In , we learned how to find the area between a curve and the x-axis (f(x)=0) using a definite integral. What if we want the area between any two functions? What if the x-axis is not one of the boundaries?

+ +

In this section, we'll investigate how a definite integral may be used to represent the area between two curves.

+ + + + +

Consider the functions given by f(x) = 5-(x-1)^2 and g(x) = 4-x.

+ + + +

Use algebra to find the points where the graphs of f and g intersect.

+ +

Sketch an accurate graph of f and g on the xy plane, + labeling the curves by name and the intersection points with ordered pairs.

+ + +

Find and evaluate exactly an integral expression that represents the area between + y = f(x) and the x-axis on the interval between the intersection points of f and g. Shade this area in your sketch.

 + +

Find and evaluate exactly an integral expression that represents the area between + y = g(x) and the x-axis on the interval between the intersection points of f and g. Shade this area in your sketch.

 + +

Let's denote the area between + y = f(x) and the x-axis as A_f and the area between + y = g(x) and the x-axis as A_g. How could we use A_f and A_g to find exact area between f and g between their intersection points?

+ +
    +

  1. We could find A_f + A_g to find the area between the curves.

    2. +

  2. We could find A_f - A_g to find the area between the curves.

    3. +

  3. We could find A_g - A_f to find the area between the curves.

    4. + +
+ + + + + + + +

We've seen from that a natural way to think about the area between two curve is as the area beneath the upper curve minus the area beneath the lower curve.

+ + + + +

We now look for a general way of writing definite integrals for the area between two given curves, f(x) and g(x). Consider this area, illustrated in .

+ +
+ Area between f(x) and g(x). + +
+ + + +

How could we represent the shaded area in ?

+
    +

  1. \int_b^a f(x) \,dx - \int_b^a g(x) \,dx

    2. +

  2. \int_a^b f(x) \,dx - \int_a^b g(x) \,dx

    3. +

  3. \int_b^a g(x) \,dx - \int_b^a f(x) \,dx

    4. +

  4. \int_a^b g(x) \,dx - \int_a^b f(x) \,dx

    5. + +
+ +

The two definite integrals above can be rewritten as one definite integral using the sum and difference property of definite integrals:

+

If f and g are continuous functions, then \int_a^b ( f(x) \pm g(x)) \,dx = \int_a^b f(x) \,dx \pm \int_a^b g(x) \,dx

+

Use the property above to represent the shaded area in using one definite integral.

+
    +

  1. \int_b^a (f(x) - g(x)) \,dx

    2. +

  2. \int_a^b (f(x) - g(x)) \,dx

    3. +

  3. \int_b^a (g(x) - f(x)) \,dx

    4. +

  4. \int_a^b (g(x) - f(x)) \,dx

    5. + +
+ + + + +

+ If two curves y = f(x) and + y = g(x) intersect at (a,g(a)) and (b,g(b)), + and for all x such that + a \le x \le b, f(x) \ge g(x), + then the area between the curves is A = \int_a^b (f(x) - g(x)) \, dx. +

+ + + + + +

+ In each of the following problems, + our goal is to determine the area of the region described. + For each region, (i) determine the intersection points of the curves, (ii) sketch the region whose area is being found, (iii) draw and label a representative slice, + and (iv) state the area of the representative slice. + Then, state a definite integral whose value is the exact area of the region, + and evaluate the integral to find the numeric value of the region's area. +

+ +

The finite region bounded by + y = \sqrt{x} and y = \frac{1}{4}x.

 + +

The finite region bounded by + y = 12-4x^2 and y = x^2 - 8.

 + +

The area bounded by the y-axis, + f(x) = \cos(x), + and g(x) = \sin(x), + where we consider the region formed by the first positive value of x for which f and g intersect.

 + +

The finite regions between the curves y = x^3-2x and y = x^2.

 + + + + + + +

+Let \mathbf{R} be the finite region bounded by the graphs of +y={\left(x + 5\right)}^{2} - 1 and y=7 \, x + 34. +

+ +

+Sketch an illustration of \mathbf R, and then +explain how to express the area of \mathbf R in the +following two ways:(Do not evaluate either definite integral.) +

+ +
    +
  1. +

    +As a definite integral with respect to x. +

    +
    2. +
  2. +

    +As a definite integral with respect to y. +

    +
    3. +
+

+ +

+ + + + + + + Videos +
+
+ +
+ diff --git a/calculus/source/04-IN/main.ptx b/calculus/source/04-IN/main.ptx new file mode 100644 index 00000000..4fed7f36 --- /dev/null +++ b/calculus/source/04-IN/main.ptx @@ -0,0 +1,15 @@ + + + + Definite and Indefinite Integrals (IN) + + + + + + + + + + + \ No newline at end of file diff --git a/calculus/source/04-IN/outcomes/01.ptx b/calculus/source/04-IN/outcomes/01.ptx new file mode 100644 index 00000000..d4ac4baa --- /dev/null +++ b/calculus/source/04-IN/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Use geometric formulas to compute definite integrals. +

\ No newline at end of file diff --git a/calculus/source/04-IN/outcomes/02.ptx b/calculus/source/04-IN/outcomes/02.ptx new file mode 100644 index 00000000..8734f5ea --- /dev/null +++ b/calculus/source/04-IN/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Approximate definite integrals using Riemann sums. +

\ No newline at end of file diff --git a/calculus/source/04-IN/outcomes/03.ptx b/calculus/source/04-IN/outcomes/03.ptx new file mode 100644 index 00000000..bc764398 --- /dev/null +++ b/calculus/source/04-IN/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+ Determine basic antiderivatives. +

diff --git a/calculus/source/04-IN/outcomes/04.ptx b/calculus/source/04-IN/outcomes/04.ptx new file mode 100644 index 00000000..685967a1 --- /dev/null +++ b/calculus/source/04-IN/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+ Solve basic initial value problems. +

diff --git a/calculus/source/04-IN/outcomes/05.ptx b/calculus/source/04-IN/outcomes/05.ptx new file mode 100644 index 00000000..997453a0 --- /dev/null +++ b/calculus/source/04-IN/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+Evaluate a definite integral using the Fundamental Theorem of Calculus. +

diff --git a/calculus/source/04-IN/outcomes/06.ptx b/calculus/source/04-IN/outcomes/06.ptx new file mode 100644 index 00000000..aff2c486 --- /dev/null +++ b/calculus/source/04-IN/outcomes/06.ptx @@ -0,0 +1,3 @@ + +

+Find the derivative of an integral using the Fundamental Theorem of Calculus.

diff --git a/calculus/source/04-IN/outcomes/07.ptx b/calculus/source/04-IN/outcomes/07.ptx new file mode 100644 index 00000000..1dfad853 --- /dev/null +++ b/calculus/source/04-IN/outcomes/07.ptx @@ -0,0 +1,3 @@ + +

+Use definite integrals to find area under a curve.

diff --git a/calculus/source/04-IN/outcomes/08.ptx b/calculus/source/04-IN/outcomes/08.ptx new file mode 100644 index 00000000..d07984d9 --- /dev/null +++ b/calculus/source/04-IN/outcomes/08.ptx @@ -0,0 +1,3 @@ + +

+Use definite integral(s) to compute the area bounded by several curves.

diff --git a/calculus/source/04-IN/outcomes/main.ptx b/calculus/source/04-IN/outcomes/main.ptx new file mode 100644 index 00000000..9db1a706 --- /dev/null +++ b/calculus/source/04-IN/outcomes/main.ptx @@ -0,0 +1,37 @@ + +> + + +

+ By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
  7. + +
    8. +
  8. + +
    9. +
+ \ No newline at end of file diff --git a/calculus/source/04-IN/readiness.ptx b/calculus/source/04-IN/readiness.ptx new file mode 100644 index 00000000..a6d1ea77 --- /dev/null +++ b/calculus/source/04-IN/readiness.ptx @@ -0,0 +1,21 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Find the derivative of a function using elementary derivative rules. ()

    +
    2. +

  2. Use sigma notation for sums. (Khan Academy and Active Calculus)

    +
    3. +

  3. Find the area of a rectangle in the coordinate plane. (Example at Khan AcademyExample at virtual nerd)

    +
    4. +

  4. Find the intersection of two graphs. (Khan Academy)

    5. +

  5. Find the area of plane shapes, such as rectangles, triangles, circles, and trapezoids. (Math is fun)

    6. +
+ + \ No newline at end of file diff --git a/calculus/source/04-IN/sageplot/integral-as-net-area.sage b/calculus/source/04-IN/sageplot/integral-as-net-area.sage new file mode 100644 index 00000000..fc6fba77 --- /dev/null +++ b/calculus/source/04-IN/sageplot/integral-as-net-area.sage @@ -0,0 +1,17 @@ +def riemann_plot(f,a,b,n): + a,b = sorted([a,b]) + delta = (b-a)/n + p = plot(f,xmin=a-(b-a)/10,xmax=b+(b-a)/10,color='black') + for k in range(n): + x_0 = a+k*delta + y_0 = f(x=x_0) + if y_0 > 0: + color = "#88f" + elif y_0 < 0: + color = "#f88" + else: + color = "gray" + p += polygon([(x_0,f(x=x_0)),(x_0+delta,f(x=x_0)),(x_0+delta,0),(x_0,0)], color=color, edgecolor="black",aspect_ratio=0.5) + return p + +graphics_array([riemann_plot((x-2)*(x-4),0,5,n) for n in [6,16,41]]) \ No newline at end of file diff --git a/calculus/source/04-IN/sageplot/integral-as-total-area.sage b/calculus/source/04-IN/sageplot/integral-as-total-area.sage new file mode 100644 index 00000000..22743c0e --- /dev/null +++ b/calculus/source/04-IN/sageplot/integral-as-total-area.sage @@ -0,0 +1,32 @@ +def riemann_plot(f,a,b,n): + a,b = sorted([a,b]) + delta = (b-a)/n + p = plot(f,xmin=a-(b-a)/10,xmax=b+(b-a)/10,color='black') + for k in range(n): + x_0 = a+k*delta + y_0 = f(x=x_0) + if y_0 > 0: + color = "#88f" + elif y_0 < 0: + color = "#f88" + else: + color = "gray" + p += polygon([(x_0,f(x=x_0)),(x_0+delta,f(x=x_0)),(x_0+delta,0),(x_0,0)], color=color, edgecolor="black",aspect_ratio=0.5) + return p +def abs_riemann_plot(f,a,b,n): + a,b = sorted([a,b]) + delta = (b-a)/n + p = plot(f,xmin=a-(b-a)/10,xmax=b+(b-a)/10,color='black') + for k in range(n): + x_0 = a+k*delta + y_0 = f(x=x_0) + if y_0 > 0: + color = "#88f" + elif y_0 < 0: + color = "#f88" + else: + color = "gray" + p += polygon([(x_0,abs(y_0)),(x_0+delta,abs(y_0)),(x_0+delta,0),(x_0,0)], color=color, edgecolor="black",aspect_ratio=0.5) + return p + +graphics_array([riemann_plot((x-2)*(x-4),0,5,41),abs_riemann_plot((x-2)*(x-4),0,5,41)]) \ No newline at end of file diff --git a/calculus/source/05-TI/01.ptx b/calculus/source/05-TI/01.ptx new file mode 100644 index 00000000..92a4cde5 --- /dev/null +++ b/calculus/source/05-TI/01.ptx @@ -0,0 +1,621 @@ + + +
+ Substitution method (TI1) + + + + + Activities + +

Answer the following.

 + +

+Using the chain rule, which of these is the derivative of e^{x^3} with respect to x? +

+
    +

  1. e^{3x^2}

    2. +

  2. x^3e^{x^3-1}

    3. +

  3. 3x^2e^{x^3}

    4. +

  4. \frac{1}{4}e^{x^4}

    5. +
+ + +

+Based on this result, which of these would you suspect to equal \int x^2e^{x^3}\,dx? +

+
    +

  1. e^{x^3+1}+C

    2. +

  2. \frac{1}{3x}e^{x^3+1}+C

    3. +

  3. 3e^{x^3}+C

    4. +

  4. \frac{1}{3}e^{x^3}+C

    5. +
+ + + + +

Recall that if u is a function of x, + then \frac{d}{dx}[u^7]=7u^6 u' by the Chain Rule.

+

For each question, choose from the following.

+
    +

  1. \frac{1}{7}u^7+C

    2. +

  2. u^7+C

    3. +

  3. 7u^7+C

    4. +

  4. \frac{6}{7}u^7+C

    5. +
+ + + +

What is \int 7u^6 u'\,dx?

+ + + + +

What is \int u^6 u'\,dx?

+ + + + +

What is \int 6u^6 u'\,dx?

+ + + + + +

Based on these activities, which of these choices seems to be a viable strategy for integration?

+
    +

  1. Memorize an integration formula for every possible function.

    2. +

  2. Attempt to rewrite the integral in the form \int g'(u)u'dx=g(u)+C.

    3. +

  3. Keep differentiating functions until you come across the function you want to integrate.

    4. +
+ + + + +

+By the chain rule, \frac{d}{dx}[g(u)+C]=g'(u)u'. There is a dual integration technique reversing this process, known as the substitution method. +

+

+This technique involves choosing an appropriate function u in terms of x +to rewrite the integral as follows: + \int f(x)\,dx=\dots=\int g'(u)u' dx=g(u)+C. +

+ + + + +

+Recall that \frac{du}{dx}=u', and so du=u'\,dx. This allows for the following common notation: + \int f(x)\,dx=\dots=\int g'(u)du=g(u)+C. +

+

+Therefore, rather than dealing with equations like u'=\frac{du}{dx}=x^2, we will prefer to write +du=x^2\,dx. +

+ + + + +

+Consider \int x^2e^{x^3}\,dx, which we conjectured earlier +to be \frac{1}{3}e^{x^3}+C. +

+

+Suppose we decided to let u=x^3. +

+ + +

+Compute \frac{du}{dx}=\unknown, +and rewrite it as du=\unknown\,dx. +

+ + +

+This \unknown\,dx doesn't appear in \int x^2e^{x^3}\,dx +exactly, so use algebra to solve for x^2\,dx in terms of du. +

+ + +

+Replace x^2dx and x^3 with u,du terms to rewrite +\int x^2e^{x^3}\,dx as +\int \frac{1}{3}e^u\,du. +

+ + +

+Solve \int \frac{1}{3}e^u\,du in terms of u, +then replace u with x^3 to confirm our original +conjecture. +

+ + + + +

+Here is how one might write out the explanation +of how to find +\int x^2e^{x^3}\,dx from start to finish: +

+ + +\int x^2e^{x^3}\,dx &&\text{Let }&u=x^3 + + +&&& du = 3x^2\,dx + + +&&& \frac{1}{3}du = x^2\,dx + + +\int x^2e^{x^3}\,dx &= \int e^{(x^3)} (x^2\,dx) + + +&= \int e^{u} \frac{1}{3}du + + +&= \frac{1}{3}e^{u}+C + + +&= \frac{1}{3}e^{x^3}+C + + + + + + +

+Which step of the previous example do you think was +the most important? +

+
    +
  1. +

    +Choosing u=x^3. +

    +
    2. +
  2. +

    +Finding du=3x^2\,dx and \frac{1}{3}du=x^2\,dx. +

    +
    3. +
  3. +

    +Substituting \int x^2e^{x^3}\,dx +with \int\frac{1}{3}e^u\,du. +

    +
    4. +
  4. +

    +Integrating \int\frac{1}{3}e^u\,du=\frac{1}{3}e^u+C. +

    +
    5. +
  5. +

    +Unsubstituting \frac{1}{3}e^u+C +to get \frac{1}{3}e^{x^3}+C. +

    +
    6. +
+ + + + +

+Below are two correct solutions to the same integral, using +two different choices for u. Which method would you prefer +to use yourself? +

+ +

+ + +\int x\sqrt{4x+4}\,dx &\phantom{=}\text{Let }u=x+1 + + +&\phantom{=} 4u=4x+4 + + +&\phantom{=} x=u-1 + + +&\phantom{=} du = dx + + +\int x\sqrt{4x+4}\,dx &= \int (u-1)\sqrt{4u}\,du + + +&= \int (2u^{3/2}-2u^{1/2})\,du + + +&= \frac{4}{5}u^{5/2}-\frac{4}{3}u^{3/2}+C + + +&= \frac{4}{5}(x+1)^{5/2} + + +&\phantom{=}-\frac{4}{3}(x+1)^{3/2}+C + + +

+

+ + +\int x\sqrt{4x+4}\,dx &\phantom{=}\text{Let }u=\sqrt{4x+4} + + +&\phantom{=} u^2=4x+4 + + +&\phantom{=} x=\frac{1}{4}u^2-1 + + +&\phantom{=} dx=\frac{1}{2}u\,du + + +\int x\sqrt{4x+4}\,dx &= \int \left(\frac{1}{4}u^2-1\right)(u)\left(\frac{1}{2}u\,du\right) + + +&= \int \left(\frac{1}{8}u^4-\frac{1}{2}u^2\right)\,du + + +&= \frac{1}{40}u^5-\frac{1}{6}u^3+C + + +&= \frac{1}{40}(4x+4)^{5/2} + + +&\phantom{=}-\frac{1}{6}(4x+4)^{3/2}+C + + +

+ + + + + +

+Suppose we wanted to try the substitution method to find \int e^x\cos(e^x+3)\,dx. Which of these choices for u appears to be most useful? +

+
    +
  1. +

    +u=x, so du=dx +

    +
    2. +
  2. +

    +u=e^x, so du=e^x\,dx +

    +
    3. +
  3. +

    +u=e^x+3, so du=e^x\,dx +

    +
    4. +
  4. +

    +u=\cos(x), so du=-\sin(x)\,dx +

    +
    5. +
  5. +

    +u=\cos(e^x+3), so du=e^x\sin(e^x+3)\,dx +

    +
    6. +
+ + + + +

+Complete the following solution using your choice from +the previous activity to find \int e^x\cos(e^x+3)\,dx. +

+ + +\int e^x\cos(e^x+3)\,dx &&\text{Let }&u=\unknown + + +&&& du = \unknown\,dx + + +\int e^x\cos(e^x+3)\,dx &= \int \unknown\, du + + +&= \cdots + + +&= \sin(e^x+3)+C + + + + + + +

+Complete the following integration by substitution +to find \int \frac{x^3}{x^4+4}\,dx. +

+ + +\int \frac{x^3}{x^4+4}\,dx &&\text{Let }&u=\unknown + + +&&& du = \unknown\,dx + + +&&& \unknown du = \unknown\,dx + + +\int \frac{x^3}{x^4+4}\,dx &= \int \frac{\unknown}{\unknown}\, du + + +&= \cdots + + +&= \frac{1}{4}\ln|x^4+4|+C + + + + + + +

+Given that +\int \frac{x^3}{x^4+4}\,dx += \frac{1}{4}\ln|x^4+4|+C +, what is the value of +\int_0^2 \frac{x^3}{x^4+4}\,dx +? +

+
    +

  1. \frac{8}{20}

    2. +

  2. -\frac{8}{20}

    3. +

  3. \frac{1}{4}\ln(20)-\frac{1}{4}\ln(4)

    4. +

  4. \frac{1}{4}\ln(4)-\frac{1}{4}\ln(20)

    5. +
+ + + + +

+What's wrong with the following computation? +

+ + +\int_0^2 \frac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4+4 + + +&&& du = 4x^3\,dx + + +&&& \frac{1}{4} du = x^3\,dx + + +\int_0^2 \frac{x^3}{x^4+4}\,dx &= \int_0^2 \frac{1/4}{u}\, du + + +&= \left[\frac{1}{4}\ln|u|\right]_0^2 + + +&= \frac{1}{4}\ln 2-\frac{1}{4}\ln 0 + + +
    +
  1. +

    +The wrong u substitution was made. +

    +
    2. +
  2. +

    +The antiderivative of \frac{1/4}{u} was wrong. +

    +
    3. +
  3. +

    +The x values 0,2 were plugged in for the variable u. +

    +
    4. +
+ + + + +

+Here's one way to show the computation of this definite integral by +tracking x values in the bounds. +

+ + +\int_0^2 \frac{x^3}{x^4+4}\,dx &&\text{Let }&u=x^4+4 + + +&&& du = 4x^3\,dx + + +&&& \frac{1}{4} du = x^3\,dx + + +\int_{x=0}^{x=2} \frac{x^3}{x^4+4}\,dx &= \int_{x=0}^{x=2} \frac{1/4}{u}\, du + + +&= \left[ \frac{1}{4}\ln|u|\right]_{x=0}^{x=2} + + +&= \left[ \frac{1}{4}\ln|x^4+4|\right]_{x=0}^{x=2} + + +&= \frac{1}{4}\ln(20)-\frac{1}{4}\ln(4) + + + + + + +

+Instead of unsubstituting u values for x values, +definite intergrals may be computed by also substituting x values in +the bounds with u values. Use this idea to complete +the following solution: +

+ + +\int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=\unknown + + +&&&du = 3x^2\,dx + + +&&&\frac{1}{3}du = x^2\,dx + + +\int_1^3 x^2e^{x^3}\,dx &= \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx) + + +&= \int_{u=\unknown}^{u=\unknown} e^{u} \frac{1}{3}du + --> + +&= \left[\frac{1}{3}e^{u}\right]_{\unknown}^{\unknown} + + +&= \unknown + + + + + + +

+Here is how one might write out the explanation +of how to find +\int_1^3 x^2e^{x^3}\,dx from start to finish by +leaving bounds in terms of x instead: +

+ + +\int_1^3 x^2e^{x^3}\,dx &&\text{Let }&u=x^3 + + +&&& du = 3x^2\,dx + + +&&& \frac{1}{3}du = x^2\,dx + + +\int_1^3 x^2e^{x^3}\,dx &= \int_{x=1}^{x=3} e^{(x^3)} (x^2\,dx) + + +&= \int_{x=1}^{x=3} e^{u} \frac{1}{3}du + + +&= \left[\frac{1}{3}e^{u}\right]_{x=1}^{x=3} + + +&= \left[\frac{1}{3}e^{x^3}\right]_{x=1}^{x=3} + + +&= \frac{1}{3}e^{3^3} - \frac{1}{3}e^{1^3} + + +&= \frac{1}{3}e^{27} - \frac{1}{3}e + + + + + + +

+Use substitution to show that +\int_1^4 \frac{e^{\sqrt x}}{\sqrt x}\,dx=2e^2-2e. +

+ + + + +

+Use substitution to show that +\int_0^{\pi/4} \sin(2\theta)\,d\theta=\frac{1}{2}. +

+ + + + +

+Use substitution to show that + +\int u^5(u^3+1)^{1/3}\,du= +\frac{1}{7}(u^3+1)^{7/3}- +\frac{1}{4}(u^3+1)^{4/3}+C + . +

+ + + + +

+Consider \int (3x-5)^2\,dx. +

+ + +

+Solve this integral using substitution. +

+ + +

+Replace (3x-5)^2 with (9x^2-30x+25) +in the original integral, the solve using +the reverse power rule. +

+ + +

+Which method did you prefer? +

+ + + + +

+Consider \int \tan(x)\,dx. +

+ + +

+Replace \tan(x) in the integral with +a fraction involving sine and cosine. +

+ + +

+Use substitution to solve the integral. +

+ + + + + + Videos +
+
+

Note: a 1/6 was accidentally forgotten in the last example shown in the video above.

+ + + + + + +
+ diff --git a/calculus/source/05-TI/02.ptx b/calculus/source/05-TI/02.ptx new file mode 100644 index 00000000..15f4b8cd --- /dev/null +++ b/calculus/source/05-TI/02.ptx @@ -0,0 +1,380 @@ + + +
+ Integration by Parts (TI2) + + + + + Activities + +

Answer the following.

 + +

Using the product rule, which of these is derivative of x^3e^x with respect to x?

+
    +

  1. 3x^2e^x

    2. +

  2. 3x^2e^{x}+x^3e^x

    3. +

  3. 3x^2e^{x-1}

    4. +

  4. \frac{1}{4}x^4 e^x

    5. +
+ + +

Based on this result, which of these would you suspect to equal \int 3x^2e^x+x^3e^x\,dx?

+
    +

  1. x^3e^x+C

    2. +

  2. x^3e^x+\frac{1}{4}x^4e^x+C

    3. +

  3. 6xe^x+3x^2e^x+C

    4. +

  4. 6xe^x+3x^2e^x+3x^2e^x+x^3e^x+C

    5. +
+ + + +

Answer the following.

 + +

Which differentiation rule is easier to implement?

+
    +

  1. Product Rule

    2. +

  2. Chain Rule

    3. +
+ + +

Which differentiation strategy do expect to be easier to reverse?

+
    +

  1. Product Rule

    2. +

  2. Chain Rule

    3. +
+ + + +

Answer the following.

 + +

Which of the following equations is equivalent to the formula \frac{d}{dx}[uv]=u'v+uv'?

+
    +

  1. uv'=-\frac{d}{dx}(uv)-vu'

    2. +

  2. uv'=-\frac{d}{dx}(uv)+vu'

    3. +

  3. uv'=\frac{d}{dx}(uv)+vu'

    4. +

  4. uv'=\frac{d}{dx}(uv)-vu'

    5. +
+ + +

Which of these is the most concise result of integrating both sides with respect to x?

+
    +

  1. \int(uv')\,dx=uv-\int(vu')\,dx

    2. +

  2. \int(u)\,dv=uv-\int(v)\,du

    3. +

  3. \int(uv')\,dx=uv-\int(vu')\,dx+C

    4. +

  4. \int(u)\,dv=uv-\int(v)\,du+C

    5. +
+ + + + +

By the product rule, \frac{d}{dx}[uv]=u'v+uv' and, subsequently, uv'=\frac{d}{dx}[uv]-u'v. There is a dual integration technique reversing this process, known as integration by parts.

+

This technique involves using algebra to rewrite an integral of a product of functions in the form \int (u)\,dv and then using the equality + \int (u)\,dv=uv-\int(v)\,du. +

+ + + + + +

Consider \int xe^{x}\,dx. Suppose we decided to let u=x.

+ + +

Compute \frac{du}{dx}=\unknown, and rewrite it as du=\unknown\,dx.

+ + +

What is the best candidate for dv?

+
    +

  1. dv=x\,dx

    2. +

  2. dv=e^x

    3. +

  3. dv=x

    4. +

  4. dv=e^x\,dx

    5. +
+ + +

Given that dv=e^x\,dx, find v=\unknown.

+ + + +

Show why \int xe^{x}\,dx may now be rewritten as xe^x-\int e^x\,dx.

+ + +

Solve \int e^x\,dx, and then give the most general antiderivative of \int xe^{x}\,dx.

+ + + + +

Here is how one might write out the explanation of how to find \int xe^{x}\,dx from start to finish: +

+ + +\int xe^{x}\,dx + + +& u=x & dv = e^x\,dx + + +& du=1\cdot\,dx & v=e^x + + +\int xe^x \,dx &= xe^x-\int e^x \,dx + + +&= xe^{x}-e^x+C + + + + + + +

Which step of the previous example do you think was the most important?

+
    +
  1. +

    Choosing u=x and dv=e^x\,dx.

    +
    2. +
  2. +

    Finding du=1\,dx and v=e^x\,dx.

    +
    3. +
  3. +

    Applying integration by parts to rewrite \int xe^x\,dx as xe^x-\int e^x\,dx.

    +
    4. +
  4. +

    Integrating \int e^x\,dx to get xe^{x}-e^x+C. +

    +
    5. +
+ + + + +

Consider the integral \int x^9\ln(x) \,dx. Suppose we proceed using integration by parts. We choose u=\ln(x) and dv=x^9\,dx.

+ +

What is du?

+ + +

What is v?

+ + +

What do you get when plugging these pieces into integration by parts?

+ + +

Does the new integral \int v\,du seem easier or harder to compute than the original integral \int x^9\ln(x) \,dx?

+
    +

  1. The original integral is easier to compute.

    2. +

  2. The new integral is easier to compute.

    3. +

  3. Neither integral seems harder than the other one.

    4. +
+ + + + + +

Consider the integral \int x^9\ln(x) \,dx once more. Suppose we still proceed using integration by parts. However, this time we choose u=x^9 and dv=\ln(x)\,dx. Do you prefer this choice or the choice we made in ?

+
    +

  1. We prefer the substitution choice of u=\ln(x) and dv=x^9\,dx.

    2. +

  2. We prefer the substitution choice of u=x^9 and dv=\ln(x)\,dx.

    3. +

  3. We do not have a strong preference, since these choices are of the same difficulty.

    4. +
+ + + + +

Consider the integral \int x\cos(x)\,dx. Suppose we proceed using integration by parts. Which of the following candidates for u and dv would best allow you to evaluate this integral?

+
    +
  1. +

    u=\cos(x), dv=xdx

    +
    2. +
  2. +

    u=\cos(x)\,dx, dv=x

    +
    3. +
  3. +

    u=x\,dx, dv=\cos(x)

    +
    4. +
  4. +

    u=x, dv=\cos(x)\,dx

    +
    5. +
+ + + + +

Evaluate the integral \int x\cos(x)\,dx using integration by parts.

+ + + + +

Now use integration by parts to evaluate the integral \displaystyle \int_{\frac{\pi}{6}}^{\pi} x\cos(x)\,dx.

+ + + + +

Consider the integral \int x\arctan(x)\,dx. Suppose we proceed using integration by parts. Which of the following candidates for u and dv would best allow you to evaluate this integral?

+
    +
  1. +

    u=x\,dx, dv=\arctan(x)

    +
    2. +
  2. +

    u=\arctan(x), dv=x\,dx

    +
    3. +
  3. +

    u=x\arctan(x), dv=\,dx

    +
    4. +
  4. +

    u=x, dv=\arctan(x)\,dx

    +
    5. +
+ + + + +

Consider the integral \int e^x\cos(x)\,dx. Suppose we proceed using integration by parts. Which of the following candidates for u and dv would best allow you to evaluate this integral?

+
    +
  1. +

    u=e^x, dv=\cos(x)\,dx

    +
    2. +
  2. +

    u=\cos(x), dv=e^x\,dx

    +
    3. +
  3. +

    u=e^x\,dx, dv=\cos(x)

    +
    4. +
  4. +

    u=\cos(x)\,dx, dv=e^x

    +
    5. +
+ + + + +

Suppose we started using integration by parts to solve the integral \int e^x\cos(x)\,dx as follows:

+ + +\int e^x\cos(x)\,dx + + +& u=\cos(x) & dv = e^x\,dx + + +& du=-\sin(x) \,dx & v=e^x + + +\int e^x\cos(x)\,dx &= \cos(x)e^x-\int e^x(-\sin(x) \,dx) + + +&= \cos(x)e^x+\int e^x\sin(x) \,dx + + +

We will have to use integration by parts a second time to evaluate the integral \int e^x\sin(x) \,dx. Which of the following candidates for u and dv would best allow you to continue evaluating the original integral \int e^x\cos(x)\,dx?

+
    +
  1. +

    u=e^x, dv=\sin(x)\,dx

    +
    2. +
  2. +

    u=\sin(x), dv=e^x\,dx

    +
    3. +
  3. +

    u=e^x\,dx, dv=\sin(x)

    +
    4. +
  4. +

    u=\sin(x)\,dx, dv=e^x

    +
    5. +
+ + + + +

Use integration by parts to show that \displaystyle \int_0^{\frac{\pi}{4}} x\sin(2x)\,dx=\frac{1}{4}.

+ + + + +

Consider the integral \int t^5 \sin(t^3)\,dt.

+ + +

Use the substitution x=t^3 to rewrite the integral in terms of x.

+ + +

Use integration by parts to evaluate the integral in terms of x.

+ + +

Replace x with t^3 to finish evaluating the original integral.

+ + + + +

Use integration by parts to show that \displaystyle \int \ln(z)\,dz=z \ln(z) - z + C.

+ + + + +

Given that that \displaystyle \int \ln(z)\,dz=z \ln(z) - z + C, evaluate \displaystyle \int (\ln(z))^2\,dz.

+ + + + + + +

+ Consider the antiderivative \displaystyle\int (\sin(x))^2dx. +

+ + +

+ Noting that \displaystyle\int (\sin(x))^2dx=\displaystyle\int (\sin(x))(\sin(x))dx and letting u=\sin(x), dv=\sin(x)dx, what equality does integration by parts yield? +

    +
  1. \displaystyle\int (\sin(x))^2dx=\sin(x)\cos(x)+\int (\cos(x))^2 dx.
    2. +
  2. \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)+\int (\cos(x))^2 dx.
    3. +
  3. \displaystyle\int (\sin(x))^2dx=\sin(x)\cos(x)-\int (\cos(x))^2 dx.
    4. +
  4. \displaystyle\int (\sin(x))^2dx=-\sin(x)\cos(x)-\int (\cos(x))^2 dx.
    5. +
+

+ + + +

+ Using the fact that (\cos(x))^2=1-(\sin(x))^2 to rewrite the above equality. +

+ + + +

+ Solve algebraically for \displaystyle\int (\sin(x))^2dx. +

+ + + + + + + + +

+ Modifying the approach from , use parts to find \displaystyle\int (\cos(x))^2dx. +

+ + + + + + + Videos +
+ + Video: Compute integrals using integration by parts +
+ + + + +
diff --git a/calculus/source/05-TI/03.ptx b/calculus/source/05-TI/03.ptx new file mode 100644 index 00000000..45f64a70 --- /dev/null +++ b/calculus/source/05-TI/03.ptx @@ -0,0 +1,276 @@ + + +
+ Integration of trigonometry (TI3) + + + + + Activities + + + +

+ Consider \displaystyle\int \sin(x)\cos(x) \, dx. Which substitution would you choose to evaluate this integral? +

    +
  1. u=\sin(x)
    2. +
  2. u=\cos(x)
    3. +
  3. u=\sin(x)\cos(x)
    4. +
  4. Substitution is not effective
    5. +
+

+ + + + + +

+ Consider \displaystyle\int \sin^4(x)\cos(x) \, dx. Which substitution would you choose to evaluate this integral? +

    +
  1. u=\sin(x)
    2. +
  2. u=\sin^4(x)
    3. +
  3. u=\cos(x)
    4. +
  4. Substitution is not effective
    5. +
+

+ + + + + +

+ Consider \displaystyle\int \sin^4(x)\cos^3(x) \, dx. Which substitution would you choose to evaluate this integral? +

    +
  1. u=\sin(x)
    2. +
  2. u=\cos^3(x)
    3. +
  3. u=\cos(x)
    4. +
  4. Substitution is not effective
    5. +
+

+ + + + + +

+ It's possible to use subtitution to evaluate \displaystyle\int \sin^4(x)\cos^3(x) \, dx, + by taking advantage of the trigonometric identity \sin^2(x)+\cos^2(x)=1. +

+

+ Complete the following substitution of u=\sin(x),du=\cos(x)\,dx + by filling in the missing \unknowns. +

+ + +\int \sin^4(x)\cos^3(x)\,dx &=\int\sin^4(x)(\,\unknown\,)\cos(x)\,dx + + +&=\int\sin^4(x)(1-\unknown)\cos(x)\,dx + + +&= \int\unknown(1-\unknown)\,du + + +&= \int (u^4-u^6)\,du + + +&= \frac{1}{5}u^5-\frac{1}{7}u^7+C + + +&= \unknown + + + + + + + +

+Trying to substitute u=\cos(x),du=-\sin(x)\,dx in the previous example is less successful. +

+ + +\int \sin^4(x)\cos^3(x)\,dx &=-\int\sin^3(x)\cos^3(x)(-\sin(x)\,dx) + + +&=-\int\sin^3(x)u^3\,du + + +&= \cdots? + + +

+Which feature of \sin^4(x)\cos^3(x) made u=\sin(x) the better choice? +

    +
  1. The even power of \sin^4(x)
    2. +
  2. The odd power of \cos^3(x)
    3. +
+

+ + + + + +

+Try to show \int \sin^5(x)\cos^2(x)\,dx= +-\frac{1}{7} \, \cos^{7}\left(x\right) + \frac{2}{5} \, \cos^{5}\left(x\right) - \frac{1}{3} \, \cos^{3}\left(x\right)+C +by first trying u=\sin(x), and then trying u=\cos(x) instead. +

+

+Which substitution worked better and why? +

    +
  1. u=\sin(x) due to \sin^5(x)'s odd power.
    2. +
  2. u=\sin(x) due to \cos^2(x)'s even power.
    3. +
  3. u=\cos(x) due to \sin^5(x)'s odd power.
    4. +
  4. u=\cos(x) due to \cos^2(x)'s even power.
    5. +
+

+ + + + + +

+When integrating the form \int \sin^m(x)\cos^n(x)\,dx: +

+
    +
  • +

    +If \sin's power is odd, rewrite the integral as +\int g(\cos(x))\sin(x)\,dx and use u=\cos(x). +

    +
    • +
  • +

    +If \cos's power is odd, rewrite the integral as +\int h(\sin(x))\cos(x)\,dx and use u=\sin(x). +

    +
    • +
+ + + + + + +

Let's consider \displaystyle\int \sin^2(x) \, dx.

+ + + +

Use the fact that \sin^2(\theta)=\displaystyle\frac{1-\cos(2\theta)}{2} to rewrite the integrand using the above identities as an integral involving \cos(2x).

+ + + + +

Show that the integral evaluates to \frac{1}{2} \, x - \frac{1}{4} \, \sin\left(2 \, x\right)+C.

+ + + + + + +

Let's consider \displaystyle\int \sin^2(x)\cos^2(x) \, dx.

+ + + +

Use the fact that \cos^2(\theta)=\displaystyle\frac{1+\cos(2\theta)}{2} and \sin^2(\theta)=\displaystyle\frac{1-\cos(2\theta)}{2} to rewrite the integrand using the above identities as an integral involving \cos^2(2x).

+ + + + +

Use the above identities to rewrite this new integrand as one involving \cos(4x).

+ + + + +

Show that integral evaluates to \frac{1}{8} \, x - \frac{1}{32} \, \sin\left(4 \, x\right)+C.

+ + + + + + + +

+ Consider \displaystyle\int \sin^4(x)\cos^4(x) \, dx. Which would be the most useful way to rewrite the integral? +

    +
  1. \displaystyle\int (1-\cos^2(x))^2\cos^4(x) \, dx
    2. +
  2. \displaystyle\int \sin^4(x)(1-\sin^2(x))^2 \, dx
    3. +
  3. \displaystyle\int \left(\frac{1-\cos(2x)}{2}\right)^2\left(\frac{1+\cos(2x)}{2}\right)^2 \, dx
    4. +
+

+ + + + + + +

+ Consider \displaystyle\int \sin^3(x)\cos^5(x) \, dx. Which would be the most useful way to rewrite the integral? +

    +
  1. \displaystyle\int (1-\cos^2(x))\cos^5(x) \sin(x)\, dx
    2. +
  2. \displaystyle\int \sin^3(x)\left(\frac{1+\cos(2x)}{2}\right)^2\cos(x) \, dx
    3. +
  3. \displaystyle\int \sin^3(x)(1-\sin^2(x))^2\cos(x) \, dx
    4. +
+

+ + + + + + + +

+We might also use some other trigonometric identities to manipulate our integrands, listed in +. +

+ + + + + + + + +

+ Consider \displaystyle\int \sin(\theta)\sin(3\theta) \, d\theta. +

+ + +

Find an identity from which could be used to transform our integrand.

+ + +

Rewrite the integrand using the selected identity.

+ + +

Evaluate the integral.

+ + + + + + + + + Videos +
+ + Video: Compute integrals involving products of trigonometric functions +
+ + + + + + + +
diff --git a/calculus/source/05-TI/04.ptx b/calculus/source/05-TI/04.ptx new file mode 100644 index 00000000..7c34e1fe --- /dev/null +++ b/calculus/source/05-TI/04.ptx @@ -0,0 +1,378 @@ + + +
+ Trigonometric Substitution (TI4) + + + + + Activities + + +

+ Consider \displaystyle \int \sqrt{9-4x^2} \,dx. Which substitution would you choose to evaluate this integral? +

    +
  1. u=9-4x^2
    2. +
  2. u=\sqrt{9-4x^2}
    3. +
  3. u=3-2x
    4. +
  4. Substitution is not effective
    5. +
+

+ + + + + +

+ To find \displaystyle \int \sqrt{9-4x^2} \,dx, we will need a more advanced substitution. + Which of these candidates is most reasonable? +

    +
  1. Let v satisfy 9-4x^2=9-9e^{2v}=9e^{-2v}.
    2. +
  2. Let \theta satisfy 9-4x^2=9-9\sin^2\theta=9\cos^2\theta.
    3. +
  3. Let w satisfy 9-4x^2=4-8\ln|w|=4\ln|2w|.
    4. +
  4. Let \phi satisfy 9-4x^2=4-4\cos^2\phi=4\sin^2\phi.
    5. +
+

+ + + + + +

+ Fill in the missing \unknowns for the following calculation. + + +\text{Let }9-4x^2&=9-9\sin^2\theta=9\cos^2\theta + + +4x^2&=\unknown + + +x&=\unknown + + +dx&=\unknown\,d\theta + + + + +\int\sqrt{9-4x^2} \,dx&=\int\sqrt{\unknown}\,(\unknown\,d\theta) + + +&= \int\frac{9}{2}\cos^2 \theta\,d\theta + + +

+ + + + + +

+ From we may find \int\cos^2 \theta\,d\theta=\frac{1}{2}\theta+\frac{1}{2}\sin\theta\cos\theta+C. +

+

+ Use this to continue your work in the previous activity and complete the integration by trigonometric substitution. +

+ + +\sin(\theta)&=\unknown + + +\theta&=\arcsin(\unknown) + + +\cos(\theta)&=\unknown\sqrt{\unknown} + + + + +\int\sqrt{9-4x^2} \,dx&= \cdots = \int\frac{9}{2}\cos^2 \theta\,d\theta + + +&= \frac{9}{2}\left(\frac{1}{2}\theta+\frac{1}{2}\sin\theta\cos\theta\right)+C + + +&= \frac{9}{4}(\unknown)+\frac{9}{4}(\unknown)(\unknown)+C + + + + + + + +

+Use similar reasoning to complete the following proof that \frac{d}{dx}\left[\arcsin(x)\right]=\frac{1}{\sqrt{1-x^2}}. + + +\text{Let }1-x^2&=1-\unknown\theta=\unknown\theta + + +x^2&=\unknown + + +x&=\unknown + + +dx&=\unknown\,d\theta + + +\theta&=\unknown + + + + +\int \frac{1}{\sqrt{1-x^2}} \,dx&=\int\frac{1}{\sqrt{\unknown}}\,(\unknown\,d\theta) + + +&= \int d\theta + + +&= \unknown + C + + +&= \arcsin(x) + C + + +

+ + + + + +

+Substitutions of the form 16-25x^2=16-16\sin^2x=16\cos^2x are made possible due to the +Pythagorean identity \sin^2(x)+\cos^2(x)=1. +

+

+Which two of these four identities can be obtained from dividing both sides of +\sin^2(x)+\cos^2(x)=1 by \cos^2(x) and rearranging? +

    +
  1. \tan^2(x)-1=\sec^2(x)
    2. +
  2. \tan^2(x)+1=\sec^2(x)
    3. +
  3. \sec^2(x)-1=\tan^2(x)
    4. +
  4. \sec^2(x)+1=\tan^2(x)
    5. +
+

+ + + + + +

+In summary, certain quadratic expressions inside an integral may be substituted with +trigonometric functions to take advantage of trigonometric identities and simplify the integrand: + + +\text{Let } b^2-a^2x^2&=b^2-b^2\sin^2(\theta)=b^2\cos^2(\theta) + + +\text{So } x&=\frac{b}{a}\sin(\theta) + + + + +\text{Let } b^2+a^2x^2&=b^2+b^2\tan^2(\theta)=b^2\sec^2(\theta) + + +\text{So } x&=\frac{b}{a}\tan(\theta) + + + + +\text{Let } a^2x^2-b^2&=b^2\sec^2(\theta)-b^2=b^2\tan^2(\theta) + + +\text{So } x&=\frac{b}{a}\sec(\theta) + + +

+ + + + + + +

+Complete the following trignometric substitution to find \displaystyle\int\frac{3}{4+25x^2}\,dx. + + +\text{Let }4+25x^2&=2+\unknown\theta=\unknown\theta + + +25x^2&=\unknown + + +x&=\unknown + + +dx&=\unknown\,d\theta + + +\theta&=\unknown + + + + +\int\frac{3}{4+25x^2}\,dx &=\int\frac{3}{\unknown}\,(\unknown\,d\theta) + + +&= \int \unknown d\theta + + +&= \unknown + C + + +&= \frac{3}{10}\arctan(\frac{5}{2}x) + C + + +

+ + + + + +

+Complete the following trignometric substitution to find \displaystyle\int\frac{7}{x\sqrt{9x^2-16}}\,dx. + + +\text{Let }9x^2-16&=\unknown\theta-16=\unknown\theta + + +9x^2&=\unknown + + +x&=\unknown + + +dx&=\unknown\,d\theta + + +\theta&=\unknown + + + + +\int\frac{7}{x\sqrt{9x^2-16}}\,dx &=\int\frac{7}{\unknown\sqrt{\unknown}}\,(\unknown\,d\theta) + + +&= \int \unknown d\theta + + +&= \unknown + C + + +&= \frac{7}{4}\arcsec(\frac{3}{4}x) + C + + +

+ + + + + +

Use appropriate trignometric substitutions and the given trigonometric integrals to find each of the following.

+ + + +

+ +\int \frac{\sqrt{-9 \, x^{2} + 16}}{x^{2}}\,dx &= \cdots + + +&=\int \frac{3\cos^2\theta}{\sin^2\theta}\,d\theta + + +&=-3\theta-3\frac{\cos\theta}{\sin\theta}+C + + +&= - 3 \, \arcsin\left(\unknown\right)-\frac{\sqrt{\unknown}}{\unknown} +C + +

+ + + + +

+ +\int \frac{2 \, \sqrt{9 \, x^{2} - 16}}{x}\,dx &= \cdots + + +&=\int 8\tan^2\theta\,d\theta + + +&=8\tan\theta-8\theta+C + + +&= \unknown \, \sqrt{\unknown} - 8 \, \operatorname{arcsec}\left(\unknown\right) +C + +

+ + + + +

+ +\int \frac{1}{\sqrt{81 \, x^{2} + 4}}\,dx&= \cdots + + +&=\int \frac{1}{9}\sec\theta\,d\theta + + +&=\frac{1}{9}\log|\sec\theta+\tan\theta|+C + + +&= \frac{1}{9} \, \log\left| \unknown + \frac{1}{2} \, \sqrt{\unknown} \right| +C + +

+ + + + + + + +

+ Consider the unit circle x^2+y^2=1. Find a function f(x) so that y=f(x) is the graph of the upper-half semicircle of the unit circle. +

+ + + + + +

+ Find the area under the curve y=f(x) from . +

+ + + +

+ How does this value compare to what we know about areas of circles? +

+ + + + + + + Videos +
+ + Video: Use trigonometric substitution to compute indefinite integrals +
+ + + + + + +
diff --git a/calculus/source/05-TI/05.ptx b/calculus/source/05-TI/05.ptx new file mode 100644 index 00000000..7cc1b905 --- /dev/null +++ b/calculus/source/05-TI/05.ptx @@ -0,0 +1,180 @@ + + +
+ Tables of Integrals (TI5) + + + + + Activities + +

Consider the integral \displaystyle\int \sqrt{16-9x^2} \,dx. Which of the following substitutions appears most promising to find an antiderivaitve for this integral?

 +
    +

  1. u=16-9x^2

    2. +

  2. u=9x^2

    3. +

  3. u=3x

    4. +

  4. u=x

    5. +
+ + +

The form of which entry from best matches the form of the integral \displaystyle\int \sqrt{16-9x^2} \,dx?

 +
    +

  1. b.

    2. +

  2. c.

    3. +

  3. g.

    4. +

  4. h.

    5. +
+ + +

For each of the following integrals, identify which entry from best matches the form of that integral.

 + +

\displaystyle\int \frac{25x^2}{\sqrt{25x^2-9}} \,dx

+ + +

\displaystyle\int \frac{81x^2}{\sqrt{16-x^2}} \,dx

+ + +

\displaystyle\int \frac{1}{10x \sqrt{100-x^2}} \,dx

+ + +

\displaystyle\int \frac{7}{\sqrt{25x^2-9}} \,dx

+ + +

\displaystyle\int \frac{1}{\sqrt{25x^2+16}} \,dx

+ + + + + +

Here is how one might write out the explanation of how to find \displaystyle\int \frac{3}{x\sqrt{49x^2-4}} \,dx from start to finish: +

+ + +\int \frac{3}{x\sqrt{49x^2-4}} \,dx &&\text{Let }&u^2=49x^2 + + +&&\text{Let }&a^2=4 + + +&&& u = 7x + + +&&& \,du = 7\,dx + + +&&& \frac{1}{7}\,du = \,dx + + +&&& a = 2 + + +\int \frac{3}{x\sqrt{49x^2-4}} \,dx &= 3\int \frac{1}{x\sqrt{49x^2-4}} (\,dx) + + +&= 3\int \frac{1}{\frac{u}{7}\sqrt{u^2-a^2}} \bigg(\frac{1}{7}\,du\bigg) + + +&= 3\int \frac{1}{u\sqrt{u^2-a^2}} \,du & \text{which best matches f.} + + +&= 3\bigg(\frac{1}{a}\arcsec \bigg(\frac{u}{a}\bigg)\bigg)+C + + +&= \frac{3}{2}\arcsec \bigg(\frac{7x}{2}\bigg)+C + + + + + + +

Which step of the previous example do you think was the most important?

+
    +
  1. +

    Choosing u^2=49x^2 and a^2=4.

    +
    2. +
  2. +

    Finding u=7x, du=7\,dx, \displaystyle\frac{1}{7}\,du=\,dx, and a=2.

    +
    3. +
  3. +

    Substituting \displaystyle \frac{3}{x\sqrt{49x^2-4}} \,dx with \displaystyle3\int \frac{1}{u\sqrt{u^2-a^2}} \,du and finding the best match of f from .

    +
    4. +
  4. +

    Integrating \displaystyle3\int \frac{1}{u\sqrt{u^2-a^2}} \,du=3(\frac{1}{a}\arcsec(\frac{u}{a}))+C. +

    +
    5. +
  5. +

    Unsubstituting \displaystyle3(\frac{1}{a}\arcsec(\frac{u}{a}))+C to get \displaystyle\frac{3}{2}\arcsec(\frac{7x}{2})+C.

    +
    6. +
+ + + + + +

Consider the integral \displaystyle\int \frac{1}{\sqrt{64-9x^2}} \,dx. Suppose we proceed using . We choose u^2=9x^2 and a^2=64.

+ + +

What is u?

+ + +

What is du?

+ + +

What is a?

+ + +

What do you get when plugging these pieces into the integral \displaystyle\int \frac{1}{\sqrt{64-9x^2}} \,dx?

+ + +

Is this a good substitution choice or a bad substitution choice?

+ + + + +

Consider the integral \displaystyle\int \frac{1}{\sqrt{64-9x^2}} \,dx once more. Suppose we still proceed using . However, this time we choose u^2=x^2 and a^2=64. Do you prefer this choice of substitution or the choice we made in ?

+
    +

  1. We prefer the substitution choice of u^2=x^2 and a^2=64.

    2. +

  2. We prefer the substitution choice of u^2=9x^2 and a^2=64.

    3. +

  3. We do not have a strong preference, since these substitution choices are of the same difficulty.

    4. +
+ + + + +

Use the appropriate substitution and entry from to show that \displaystyle\int \frac{7}{x\sqrt{4+49x^2}} \,dx=-\frac{7}{2}\ln\big|\frac{2+\sqrt{49x^2+4}}{7x}\big|+C.

+ + + + +

Use the appropriate substitution and entry from to show that \displaystyle\int \frac{3}{5x^2\sqrt{36-49x^2}} \,dx=-\frac{\sqrt{36-49x^2}}{60x}+C.

+ + + + +

Evaluate the integral \displaystyle\int 8\sqrt{4x^2-81} \,dx. Be sure to specify which entry is used from at the corresponding step.

+ + + + + + + Videos +
+ + Video: I can integrate functions using a table of integrals +
+ + + + + +
diff --git a/calculus/source/05-TI/06.ptx b/calculus/source/05-TI/06.ptx new file mode 100644 index 00000000..1f8bcf7d --- /dev/null +++ b/calculus/source/05-TI/06.ptx @@ -0,0 +1,354 @@ + + +
+ Partial fractions (TI6) + + + + + Activities + + +

+ Consider \displaystyle \int \frac{x^2+x+1}{x^3+x} \,dx. Which substitution would you choose to evaluate this integral? +

    +
  1. u=x^3
    2. +
  2. u=x^3+x
    3. +
  3. u=x^2+x+1
    4. +
  4. Substitution is not effective
    5. +
+

+ + + + +

+ Using the method of substitution, which of these is equal to \displaystyle\int \frac{5}{x+7} dx? +

    +
  1. 5\ln|x+7| +C
    2. +
  2. \frac{5}{7}\ln|x+7| +C
    3. +
  3. 5\ln|x|+5\ln|7|+C
    4. +
  4. \frac{5}{7}\ln|x|+C
    5. +
+

+ + + + +

+ To avoid repetitive substitution, the following integral formulas will be useful. + \int\frac{1}{x+b}dx=\ln|x+b|+C + \int\frac{1}{(x+b)^2}dx=-\frac{1}{x+b}+C + \int\frac{1}{x^2+b^2}dx=\frac{1}{b}\arctan\left(\frac{x}{b}\right)+C +

+ + + + +

+ Which of the following is equal to \displaystyle\frac{1}{x}+\frac{1}{x^2+1}? +

    +
  1. \frac{2x}{x^2+x+1}
    2. +
  2. \frac{x^3+x}{x^2+x+1}
    3. +
  3. \frac{2x}{x^3+x}
    4. +
  4. \frac{x^2+x+1}{x^3+x}
    5. +
+

+ + + + +

+ Based on the previous activities, + which of these is equal to \displaystyle\int \frac{x^2+x+1}{x^3+x} dx? +

    +
  1. \ln|x|+\arctan(x)+C
    2. +
  2. \ln|x^2+x+1|+C
    3. +
  3. \ln|x^3+x|+C
    4. +
  4. \arctan(x^3+x)+C
    5. +
+

+ + + + + +

+ Suppose we know \frac{10x-11}{x^2+x-2}=\frac{7}{x-1}+\frac{3}{x+2}. + Which of these is equal to \int\frac{10x-11}{x^2+x-2} dx? +

    +
  1. 7\ln|x-1|+3\arctan(x+2)+C
    2. +
  2. 7\ln|x-1|+3\ln|x+2|+C
    3. +
  3. 7\arctan(x-1)+3\arctan(x+2)+C
    4. +
  4. 7\arctan(x-1)+3\ln|x+2|+C
    5. +
+

+ + + + + +

+ To find integrals like + \int \frac{x^2+x+1}{x^3+x} dx and + \int\frac{10x-11}{x^2+x-2} dx, we'd like to decompose the fractions + into simpler partial fractions that may be integrated with these formulas + \int\frac{1}{x+b}dx=\ln|x+b|+C + \int\frac{1}{(x+b)^2}dx=-\frac{1}{x+b}+C + \int\frac{1}{x^2+b^2}dx=\frac{1}{b}\arctan\left(\frac{x}{b}\right)+C +

+ + + + + Partial Fraction Decomposition + +

+ Let \displaystyle \frac{p(x)}{q(x)} be a rational function, + where the degree of p is less than the degree of q. + integrationpartial fraction decomp. +

+ +

+

    +
  1. +

    + Linear Terms: Let (x-a)^n divide q(x), + Then the decomposition of \frac{p(x)}{q(x)} will contain the terms + + \frac{A_1}{(x-a)} + \frac{A_2}{(x-a)^2} + \cdots +\frac{A_n}{(x-a)^n} + . +

    +
    2. + +
  2. +

    + Quadratic Terms: Let + (x^2+bx+c)^n divide q(x), + where x^2+bx+c is irreducable. + Then the decomposition of \frac{p(x)}{q(x)} will contain the terms + + \frac{B_1x+C_1}{x^2+bx+c}+\frac{B_2x+C_2}{(x^2+bx+c)^2}+\cdots+\frac{B_nx+C_n}{(x^2+bx+c)^n} + . +

    +
    3. +
+

+ + + + + + +

+Following is an example of a rather involved partial fraction decomposition. + + +&\frac{7 \, x^{6} - 4 \, x^{5} + 41 \, x^{4} - 20 \, x^{3} + 24 \, x^{2} + 11 \, x + 16}{x(x-1)^2(x^2+4)^2} + + +=& \frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}+\frac{Dx+E}{x^2+4}+\frac{Fx+G}{(x^2+4)^2} + + +Using some algebra, it's possible to find values for A through G to determine + + +&\frac{7 \, x^{6} - 4 \, x^{5} + 41 \, x^{4} - 20 \, x^{3} + 24 \, x^{2} + 11 \, x + 16}{x(x-1)^2(x^2+4)^2} + + +=& \frac{1}{x}+\frac{2}{x-1}+\frac{3}{(x-1)^2}+\frac{4x+5}{x^2+4}+\frac{6x+7}{(x^2+4)^2} + +. +

+ + + + + +

+ Which of the following is the form of the partial fraction decomposition of \displaystyle\frac{x^3-7x^2-7x+15}{x^3(x+5)}? + +

    +
  1. \frac{A}{x}+\frac{B}{x+5}
    2. +
  2. \frac{A}{x^3}+\frac{B}{x+5}
    3. +
  3. \frac{A}{x}+\frac{B}{x^2}+ \frac{C}{x^3}+\frac{D}{x+5}
    4. +
  4. \frac{A}{x}+\frac{B}{x^2}+ \frac{C}{x^3}+\frac{Dx+E}{x+5}
    5. +
+

+ + + + +

+ Which of the following is the form of the partial fraction decomposition of \displaystyle\frac{x^2+1}{(x-3)^2(x^2+4)^2}? + +

    +
  1. \frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{C}{x^2+4}+\frac{D}{(x^2+4)^2}
    2. +
  2. \frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{Cx+D}{(x^2+4)^2}
    3. +
  3. \frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{C}{x^2+4}+\frac{Dx+E}{(x^2+4)^2}
    4. +
  4. \frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{Cx+D}{x^2+4}+\frac{Ex+F}{(x^2+4)^2}
    5. +
+

+ + + + + +

+ Consider that the partial decomposition of \displaystyle \frac{x^2+5x+3}{(x+1)^2x} is + \displaystyle \frac{x^2+5x+3}{(x+1)^2x}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x}. + What equality do we obtain if we multiply both sides of the above equation by (x+1)^2x? +

    +
  1. x^2+5x+3=Ax(x+1)+Bx+C(x+1)^2
    2. +
  2. x^2+5x+3=A(x+1)+B(x+1)^2+Cx
    3. +
  3. x^2+5x+3=Ax(x+1)+Bx+C(x+1)
    4. +
  4. x^2+5x+3=Ax(x+1)+Bx^2+C(x+1)^2
    5. +
+

+ + + +

Use your choice in (which must hold for any x value) to answer the following.

 + +

By substituting x=0 into the equation, we may find:

+
    +
  1. A=1
    2. +
  2. B=-2
    3. +
  3. C=3
    4. +
+ + +

By substituting x=-1 into the equation, we may find:

+
    +
  1. A=-4
    2. +
  2. B=1
    3. +
  3. C=5
    4. +
+ + + +

+ Using the results of , show how to rewrite our choice from + + \unknown x^2+\unknown x=Ax^2+Ax. +

+

+ What value of A satisfies this equation? +

+
    +
  1. -2
    2. +
  2. 3
    3. +
  3. 4
    4. +
  4. -5
    5. +
+ + + + +

+ By using the form of the decomposition \displaystyle \frac{x^2+5x+3}{(x+1)^2x}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x} + and the coefficients found in and , + evaluate \displaystyle \int \frac{x^2+5x+3}{(x+1)^2x} dx. +

+ + + + + + +

+ Given that \displaystyle\frac{x^3-7x^2-7x+15}{x^3(x+5)}=\frac{A}{x}+\frac{B}{x^2}+ \frac{C}{x^3}+\frac{D}{x+5} do the following + to find A, B, C, and D. +

+ + +

+ Eliminate the fractions to obtain x^3-7x^2-7x+15=A(\unknown)(\unknown)+B(\unknown)(\unknown)+C(\unknown)+D(\unknown). +

+ + +

+ Plug in an x value that lets you find the value of C. +

+ + +

+ Plug in an x value that lets you find the value of D. +

+ + +

+ Use other algebra techniques to find the values of A and B. +

+ + + + +

+ Given your choice in Find \displaystyle\int \frac{x^3-7x^2-7x+15}{x^3(x+5)} dx. +

+ + + + +

+ Consider the rational expression \displaystyle\frac{2x^3+2x+4}{x^4+2x^3+4x^2}. + Which of the following is the partial fraction decomposition of this rational expression? + +

    +
  1. \frac{1}{x}+\frac{1}{x^2}+\frac{2x-1}{x^2+2x+4}
    2. +
  2. \frac{2}{x}+\frac{0}{x^2}+\frac{-1}{x^2+2x+4}
    3. +
  3. \frac{0}{x}+\frac{1}{x^2}+\frac{-1}{x^2+2x+4}
    4. +
  4. \frac{0}{x}+\frac{1}{x^2}+\frac{2x-1}{x^2+2x+4}
    5. +

+ + + + +

+ Given your choice in Find \displaystyle\int \frac{2x^3+2x+4}{x^4+2x^3+4x^2} dx. +

+ + + + + +

+ Given that \displaystyle \frac{2x+5}{x^2+3x+2}=\frac{-1}{x+2}+\frac{3}{x+1}, find \displaystyle\int_0^3 \frac{2x+5}{x^2+3x+2} dx. +

+ + + + + + +

+ Evaluate \displaystyle \int \frac{4x^2-3x+1}{(2x+1)(x+2)(x-3)}dx. +

+ + + + + + + + Videos +
+ + Video: I can integrate functions using the method of partial fractions +
+ + + + + +
diff --git a/calculus/source/05-TI/07.ptx b/calculus/source/05-TI/07.ptx new file mode 100644 index 00000000..83f340b2 --- /dev/null +++ b/calculus/source/05-TI/07.ptx @@ -0,0 +1,104 @@ + + +
+ Integration strategy (TI7) + + + + + Activities + + +

Consider the integral \displaystyle\int e^t \tan(e^t) \sec^2(e^t)\,dt. Which strategy is a reasonable first step to make progress towards evaluating this integral?

+
    +

  1. The method of substitution

    2. +

  2. The method of integration by parts

    3. +

  3. Trigonometric substitution

    4. +

  4. Using a table of integrals

    5. +

  5. The method of partial fractions

    6. +
+ + + + +

Consider the integral \displaystyle\int \frac{2x+3}{1+x^2}\,dx. Which strategy is a reasonable first step to make progress towards evaluating this integral?

+
    +

  1. The method of substitution

    2. +

  2. The method of integration by parts

    3. +

  3. Trigonometric substitution

    4. +

  4. Using a table of integrals

    5. +

  5. The method of partial fractions

    6. +
+ + + + +

Consider the integral \displaystyle\int \frac{x}{\sqrt[3]{1-x^2}}\,dx. Which strategy is a reasonable first step to make progress towards evaluating this integral?

+
    +

  1. The method of substitution

    2. +

  2. The method of integration by parts

    3. +

  3. Trigonometric substitution

    4. +

  4. Using a table of integrals

    5. +

  5. The method of partial fractions

    6. +
+ + + + +

Consider the integral \displaystyle\int \frac{1}{2x\sqrt{1-36x^2}}\,dx. Which strategy is a reasonable first step to make progress towards evaluating this integral?

+
    +

  1. The method of substitution

    2. +

  2. The method of integration by parts

    3. +

  3. Trigonometric substitution

    4. +

  4. Using a table of integrals

    5. +

  5. The method of partial fractions

    6. +
+ + + + +

Consider the integral \displaystyle\int t^5\cos(t^3)\,dt. Which strategy is a reasonable first step to make progress towards evaluating this integral?

+
    +

  1. The method of substitution

    2. +

  2. The method of integration by parts

    3. +

  3. Trigonometric substitution

    4. +

  4. Using a table of integrals

    5. +

  5. The method of partial fractions

    6. +
+ + + + +

Consider the integral \displaystyle\int \frac{1}{1+e^x}\,dx. Which strategy is a reasonable first step to make progress towards evaluating this integral?

+
    +

  1. The method of substitution

    2. +

  2. The method of integration by parts

    3. +

  3. Trigonometric substitution

    4. +

  4. Using a table of integrals

    5. +

  5. The method of partial fractions

    6. +
+ + + + + + Videos +
+ + Video: I can select appropriate strategies for integration +
+ + + + +
+ diff --git a/calculus/source/05-TI/08.ptx b/calculus/source/05-TI/08.ptx new file mode 100644 index 00000000..17865638 --- /dev/null +++ b/calculus/source/05-TI/08.ptx @@ -0,0 +1,572 @@ + + +
+ Improper integrals (TI8) + + + + + Activities + +

Recall \displaystyle \int\frac{1}{x^2}dx=-\frac{1}{x}+C. Compute the following definite integrals.

 + +

\displaystyle \int_{1/100}^1 \frac{1}{x^2} dx=\left[-\frac{1}{x}\right]_{1/100}^1

+ + +

\displaystyle \int_{1/10000}^1 \frac{1}{x^2} dx

+ + +

\displaystyle \int_{1/1000000}^1 \frac{1}{x^2} dx

+ + + +

What do you notice about \displaystyle \int_{a}^1 \frac{1}{x^2} dx as a approached 0 in ?

+
    +

  1. \displaystyle \int_{a}^1 \frac{1}{x^2} dx approaches 0.

    2. +

  2. \displaystyle \int_{a}^1 \frac{1}{x^2} dx approaches a finite constant greater than 0.

    3. +

  3. \displaystyle \int_{a}^1 \frac{1}{x^2} dx approaches \infty.

    4. +

  4. There is not enough information.

    5. +
+ + +

Compute the following definite integrals, again using \displaystyle \int\frac{1}{x^2}dx=-\frac{1}{x}+C.

 + +

\displaystyle \int_{1}^{100} \frac{1}{x^2} dx=\left[-\frac{1}{x}\right]_{1}^{100}

+ + +

\displaystyle \int_{1}^{10000} \frac{1}{x^2} dx

+ + +

\displaystyle \int_{1}^{1000000} \frac{1}{x^2} dx

+ + + +

What do you notice about \displaystyle \int_{1}^b \frac{1}{x^2} dx as b approached \infty in ?

+
    +

  1. \displaystyle \int_{1}^b \frac{1}{x^2} dx approaches 0.

    2. +

  2. \displaystyle \int_{1}^b \frac{1}{x^2} dx approaches a finite constant greater than 0.

    3. +

  3. \displaystyle \int_{1}^b \frac{1}{x^2} dx approaches \infty.

    4. +

  4. There is not enough information.

    5. +
+ + +

Recall \displaystyle \int\frac{1}{\sqrt x}dx=2\sqrt{x}+C. Compute the following definite integrals.

 + +

\displaystyle \int_{1/100}^1 \frac{1}{\sqrt{x}} dx=\left[2\sqrt{x}\right]_{1/100}^1

+ + +

\displaystyle \int_{1/10000}^1 \frac{1}{\sqrt{x}} dx

+ + +

\displaystyle \int_{1/1000000}^1 \frac{1}{\sqrt{x}} dx

+ + + + +

What do you notice about the integral \displaystyle \int_{a}^1 \frac{1}{\sqrt{x}} dx as a approached 0 in ?

+
    +

  1. \displaystyle \int_{a}^1 \frac{1}{\sqrt{x}} dx approaches 0.

    2. +

  2. \displaystyle \int_{a}^1 \frac{1}{\sqrt{x}} dx approaches a finite constant greater than 0.

    3. +

  3. \displaystyle \int_{a}^1 \frac{1}{\sqrt{x}} dx approaches \infty.

    4. +

  4. There is not enough information.

    5. +
+ + +

How does this compare to what you found in ?

+ + + +

Compute the following definite integrals using \displaystyle \int\frac{1}{\sqrt x}dx=2\sqrt{x}+C.

 + +

\displaystyle \int_{1}^{100} \frac{1}{\sqrt{x}} dx = \left[2\sqrt{x}\right]_{1}^{100}

+ + +

\displaystyle \int_{1}^{10000} \frac{1}{\sqrt{x}} dx

+ + +

\displaystyle \int_{1}^{1000000} \frac{1}{\sqrt{x}} dx

+ + + + +

What do you notice the integral \displaystyle \int_{1}^b \frac{1}{\sqrt{x}} dx as b approached \infty in ?

+
    +

  1. \displaystyle \int_{1}^b \frac{1}{\sqrt{x}} dx approaches 0.

    2. +

  2. \displaystyle \int_{1}^b \frac{1}{\sqrt{x}} dx approaches a finite constant greater than 0.

    3. +

  3. \displaystyle \int_{1}^b \frac{1}{\sqrt{x}} dx approaches \infty.

    4. +

  4. There is not enough information.

    5. +
+ + +

How does this compare to what you found in ?

+ + + +

+For a function f(x) and a constant a, we let \displaystyle \int_a^\infty f(x) dx denote +\int_a^\infty f(x) dx=\lim_{b\to\infty}\left( \int_a^b f(x) dx\right). +If this limit is a defined real number, then we say \displaystyle \int_a^\infty f(x) dx is convergent. +Otherwise, it is divergent. +

+

Similarly, \int_{-\infty}^b f(x) dx=\lim_{a\to-\infty}\left( \int_a^b f(x) dx\right).

+ + +

Which of these limits is equal to \displaystyle\int_1^\infty \frac{1}{x^2} dx?

+
    +
  1. \displaystyle\lim_{b\to\infty}\int_1^b\frac{1}{x^2}dx
    2. +
  2. \displaystyle\lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b
    3. +
  3. \displaystyle\lim_{b\to\infty}\left[-\frac{1}{b}+1\right]
    4. +
  4. All of these.
    5. +
+ + +

Given the result of , what is \displaystyle\int_1^\infty \frac{1}{x^2} dx?

+
    +
  1. 0
    2. +
  2. 1
    3. +
  3. \infty
    4. +
  4. -\infty
    5. +
+ + +

Does \displaystyle\int_1^\infty \frac{1}{\sqrt x} dx converge or diverge?

+
    +
  1. Converges because \displaystyle\lim_{b\to 0^+}\left[2\sqrt b-2\right] converges.
    2. +
  2. Diverges because \displaystyle\lim_{b\to 0^+}\left[2\sqrt b-2\right] diverges.
    3. +
  3. Converges because \displaystyle\lim_{b\to \infty}\left[2\sqrt b-2\right] converges.
    4. +
  4. Diverges because \displaystyle\lim_{b\to \infty}\left[2\sqrt b-2\right] diverges.
    5. +
+ + +

+For a function f(x) with a vertical asymptote at x=c>a, +we let \displaystyle \int_a^c f(x) dx denote \int_a^c f(x) dx=\lim_{b\to c^{-}}\left( \int_a^b f(x) dx\right). +

+

+For a function f(x) with a vertical asymptote at x=c<b, +we let \displaystyle \int_c^b f(x) dx denote \int_c^b f(x) dx=\lim_{a\to c^{+}}\left( \int_a^b f(x) dx\right). +

+ + +

Which of these limits is equal to \displaystyle\int_0^1 \frac{1}{\sqrt x} dx?

+
    +
  1. \displaystyle\lim_{a\to0^+}\int_a^1\frac{1}{\sqrt x}dx
    2. +
  2. \displaystyle\lim_{a\to0^+}\left[2\sqrt x\right]_a^1
    3. +
  3. \displaystyle\lim_{a\to0^+}\left[2-2\sqrt a\right]
    4. +
  4. All of these.
    5. +
+ + +

Given the this result, what is \displaystyle\int_0^1 \frac{1}{\sqrt x} dx?

+
    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. \infty
    5. +
+ + +

Does \displaystyle\int_0^1 \frac{1}{x^2} dx converge or diverge?

+
    +
  1. Converges because \displaystyle\lim_{a\to 0^+}\left[-1+\frac{1}{a}\right] converges.
    2. +
  2. Diverges because \displaystyle\lim_{a\to 0^+}\left[-1+\frac{1}{a}\right] diverges.
    3. +
  3. Converges because \displaystyle\lim_{a\to 1^-}\left[-1+\frac{1}{a}\right] converges.
    4. +
  4. Diverges because \displaystyle\lim_{a\to 1^-}\left[-1+\frac{1}{a}\right] diverges.
    5. +
+ + + +

Explain and demonstrate how to write each of the following improper integrals as a limit, and why this limit converges or diverges.

+ + + +

+ \displaystyle\int_{ -2 }^{ +\infty } \frac{1}{\sqrt{x + 6}} dx. +

+ + + + +

+ \displaystyle\int_{ -4 }^{ -2 } \frac{1}{{\left(x + 4\right)}^{\frac{4}{3}}} dx. +

+ + + + +

+ \displaystyle\int_{ -5 }^{ 0 } \frac{1}{{\left(x + 5\right)}^{\frac{5}{9}}} dx. +

+ + + + +

+ \displaystyle\int_{ 10 }^{ +\infty } \frac{1}{{\left(x - 8\right)}^{\frac{4}{3}}} dx. +

+ + + + + +

Suppose that 0< p and p\neq 1. Applying the integration power rule gives us the indefinite integral \displaystyle \int \frac{1}{x^p} dx=\frac{1}{(1-p)}x^{1-p}+C.

+ + + + +

If 0< p< 1, which of the following statements must be true? Select all that apply.

+
    +

  1. 1-p< 0

    2. +

  2. 1-p> 0

    3. +

  3. 1-p< 1

    4. +

  4. \displaystyle \int_1^\infty \frac{1}{x^p} dx converges.

    5. +

  5. \displaystyle \int_1^\infty \frac{1}{x^p} dx diverges.

    6. +
+ + +

If p>1, which of the following statements must be true? Select all that apply.

+
    +

  1. 1-p< 0

    2. +

  2. 1-p> 0

    3. +

  3. 1-p< 1

    4. +

  4. \displaystyle \int_1^\infty \frac{1}{x^p} dx converges.

    5. +

  5. \displaystyle \int_1^\infty \frac{1}{x^p} dx diverges.

    6. +
+ + + + +

If 0< p< 1, which of the following statements must be true?

+
    +

  1. \displaystyle \int_0^1 \frac{1}{x^p} dx converges.

    2. +

  2. \displaystyle \int_0^1 \frac{1}{x^p} dx diverges.

    3. +
+ + +

If p>1, which of the following statements must be true?

+
    +

  1. \displaystyle \int_0^1 \frac{1}{x^p} dx converges.

    2. +

  2. \displaystyle \int_0^1 \frac{1}{x^p} dx diverges.

    3. +
+ + + +

Consider when p=1. Then \frac{1}{x^p}=\frac{1}{x} and \displaystyle \int \frac{1}{x^p}\ dx=\displaystyle \int \frac{1}{x}\ dx=\ln|x|+C.

 + +

What can we conclude about \displaystyle \int_1^\infty \frac{1}{x} dx?

+
    +

  1. \displaystyle \int_1^\infty \frac{1}{x} dx converges.

    2. +

  2. \displaystyle \int_1^\infty \frac{1}{x} dx diverges.

    3. +

  3. There is not enough information to determine whether this integral converges or diverges.

    4. +
+ + +

What can we conclude about \displaystyle \int_0^1 \frac{1}{x} dx?

+
    +

  1. \displaystyle \int_0^1 \frac{1}{x} dx converges.

    2. +

  2. \displaystyle \int_0^1 \frac{1}{x} dx diverges.

    3. +

  3. There is not enough information to determine whether this integral converges or diverges.

    4. +
+ + + + + +

+ Let c, p>0. +

    +
  • + \displaystyle \int_0^c \frac{1}{x^p} dx converges if and only if p< 1. +
    • +
+
    +
  • + \displaystyle \int_c^\infty \frac{1}{x^p} dx converges if and only if p> 1. +
    • +
+

+ + + + +

+ Consider the plots of f(x), g(x), h(x) where 0 < g(x) < f(x) < h(x). +

+
+ Plots of f(x), g(x), h(x) + + Plots of positive functions f(x), g(x) where f(x) is an upper bound of g(x). + + \begin{tikzpicture} + \draw[->] (-1, 0) -- (4.2, 0) node[below right] {$x$}; + \draw[->] (0, -1) -- (0, 2.2) node[above] {$y$}; + \draw[orange] plot [ + samples=100, + domain=1:4, + ] (\x,{1/\x }); + \draw[dashed] (1,-1) -- (1,2); + \fill [orange, domain=1:4, variable=\x, opacity=0.2] + (1, 0) + -- plot ({\x}, {1/\x}) + -- (4, 0) + -- cycle; + \draw[red] plot [ + samples=100, + domain=1:4, + ] (\x,{1/(\x)^2 }); + \draw[blue] plot [ + samples=100, + domain=1:4, + ] (\x,{1/(\x^(1/2)) }); + \draw[orange] (4, 1/4) --node[ right]{\tiny $f(x)$} (4,1/4); + \draw[red] (4, 1/16) --node[ right]{\tiny $g(x)$} (4,1/16); + \draw[blue] (4, 1/2) --node[ right]{\tiny $h(x)$} (4,1/2); + \end{tikzpicture} + + +
+

+ If \displaystyle \int_1^\infty f(x) dx is convergent, what can we say about g(x), h(x)? +

    +
  1. \displaystyle \int_1^\infty g(x) dx and \displaystyle \int_1^\infty h(x) dx are both convergent.
    2. +
  2. \displaystyle \int_1^\infty g(x) dx and \displaystyle \int_1^\infty h(x) dx are both divergent.
    3. +
  3. Whether or not \displaystyle \int_1^\infty g(x) dx and \displaystyle \int_1^\infty h(x) dx are convergent or divergent cannot be determined.
    4. +
  4. \displaystyle \int_1^\infty g(x) dx is convergent and \displaystyle \int_1^\infty h(x) dx is divergent.
    5. +
  5. \displaystyle \int_1^\infty g(x) dx is convergent and \displaystyle \int_1^\infty h(x) dx could be either convergent or divergent.
    6. +
+

+ + + +

+ Consider the plots of f(x), g(x), h(x) where 0 < g(x) < f(x) < h(x). +

+ Plots of f(x), g(x), h(x) + + Plots of positive functions f(x), g(x) where f(x) is an upper bound of g(x). + + \begin{tikzpicture} + \draw[->] (-1, 0) -- (4.2, 0) node[below right] {$x$}; + \draw[->] (0, -1) -- (0, 2.2) node[above] {$y$}; + \draw[orange] plot [ + samples=100, + domain=1:4, + ] (\x,{1/\x }); + \draw[dashed] (1,-1) -- (1,2); + \fill [orange, domain=1:4, variable=\x, opacity=0.2] + (1, 0) + -- plot ({\x}, {1/\x}) + -- (4, 0) + -- cycle; + \draw[red] plot [ + samples=100, + domain=1:4, + ] (\x,{1/(\x)^2 }); + \draw[blue] plot [ + samples=100, + domain=1:4, + ] (\x,{1/(\x^(1/2)) }); + \draw[orange] (4, 1/4) --node[ right]{\tiny $f(x)$} (4,1/4); + \draw[red] (4, 1/16) --node[ right]{\tiny $g(x)$} (4,1/16); + \draw[blue] (4, 1/2) --node[ right]{\tiny $h(x)$} (4,1/2); + \end{tikzpicture} + + +
+ If \displaystyle \int_1^\infty f(x) dx is divergent, what can we say about g(x), h(x)? +
    +
  1. \displaystyle \int_1^\infty g(x) dx and \displaystyle \int_1^\infty h(x) dx are both convergent.
    2. +
  2. \displaystyle \int_1^\infty g(x) dx and \displaystyle \int_1^\infty h(x) dx are both divergent.
    3. +
  3. Whether or not \displaystyle \int_1^\infty g(x) dx and \displaystyle \int_1^\infty h(x) dx are convergent or divergent cannot be determined.
    4. +
  4. \displaystyle \int_1^\infty g(x) dx could be either convergent or dicergent and \displaystyle \int_1^\infty h(x) dx is divergent.
    5. +
  5. \displaystyle \int_1^\infty g(x) dx is convergent and \displaystyle \int_1^\infty h(x) dx is divergent.
    6. +
+

+ + + + +

+ Consider the plots of f(x), g(x), h(x) where 0 < g(x) < f(x) < h(x). +

+ Plots of f(x), g(x), h(x) + + Plots of positive functions f(x), g(x) and h(x). + + \begin{tikzpicture} + \draw[->] (-1, 0) -- (2.2, 0) node[below right] {$x$}; + \draw[->] (0, -1) -- (0, 4.2) node[above] {$y$}; + \draw[orange] plot [ + samples=100, + domain=0.25:1, + ] (\x,{1/\x }); + \draw[dashed] (1,-1) -- (1,4); + \fill [orange, domain=0.25:1, variable=\x, opacity=0.2] + (0, 0) + -- (0, 4) + -- plot ({\x}, {1/\x}) + --(1,0) + -- cycle; + \draw[orange] (1, 1) --node[ right]{\tiny $f(x)$} (1,1); + \draw[red] plot [ + samples=100, + domain=0.175:1, + ] (\x,{0.7/\x }); + \draw[blue] plot [ + samples=100, + domain=0.325:1, + ] (\x,{1.3/\x }); + \draw[blue] (1, 1.3) --node[ right]{\tiny $h(x)$} (1,1.3); + \draw[red] (1, 0.7) --node[ right]{\tiny $g(x)$} (1,.7); + + \end{tikzpicture} + + +
+ If \displaystyle \int_0^1 f(x) dx is convergent, what can we say about g(x) and h(x)? +
    +
  1. \displaystyle \int_0^1 g(x) dx and \displaystyle \int_0^1 h(x) dx are both convergent.
    2. +
  2. \displaystyle \int_0^1 g(x) dx and \displaystyle \int_0^1 h(x) dx are both divergent.
    3. +
  3. Whether or not \displaystyle \int_0^1 g(x) dx and \displaystyle \int_0^1 h(x) dx are convergent or divergent cannot be determined.
    4. +
  4. \displaystyle \int_0^1 g(x) dx is convergent and \displaystyle \int_0^1 h(x) dx is divergent.
    5. +
  5. \displaystyle \int_0^1 g(x) dx is convergent and \displaystyle \int_0^1 h(x) dx can either be convergent or divergent.
    6. +
+

+ + + +

+ Consider the plots of f(x), g(x), h(x) where 0 < g(x) < f(x) < h(x). +

+ Plots of f(x), g(x), h(x) + + Plots of positive functions f(x), g(x) and h(x). + + \begin{tikzpicture} + \draw[->] (-1, 0) -- (2.2, 0) node[below right] {$x$}; + \draw[->] (0, -1) -- (0, 4.2) node[above] {$y$}; + \draw[orange] plot [ + samples=100, + domain=0.25:1, + ] (\x,{1/\x }); + \draw[dashed] (1,-1) -- (1,4); + \fill [orange, domain=0.25:1, variable=\x, opacity=0.2] + (0, 0) + -- (0, 4) + -- plot ({\x}, {1/\x}) + --(1,0) + -- cycle; + \draw[orange] (1, 1) --node[ right]{\tiny$f(x)$} (1,1); + \draw[red] plot [ + samples=100, + domain=0.175:1, + ] (\x,{0.7/\x }); + \draw[blue] plot [ + samples=100, + domain=0.325:1, + ] (\x,{1.3/\x }); + \draw[blue] (1, 1.3) --node[ right]{\tiny $h(x)$} (1,1.3); + \draw[red] (1, 0.7) --node[ right]{\tiny $g(x)$} (1,.7); + + \end{tikzpicture} + + +
+ If \displaystyle \int_0^1 f(x) dx is dinvergent, what can we say about g(x) and h(x)? +
    +
  1. \displaystyle \int_0^1 g(x) dx and \displaystyle \int_0^1 h(x) dx are both convergent.
    2. +
  2. \displaystyle \int_0^1 g(x) dx and \displaystyle \int_0^1 h(x) dx are both divergent.
    3. +
  3. Whether or not \displaystyle \int_0^1 g(x) dx and \displaystyle \int_0^1 h(x) dx are convergent or divergent cannot be determined.
    4. +
  4. \displaystyle \int_0^1 g(x) dx can be either convergent or divergent and \displaystyle \int_0^1 h(x) dx is divergent.
    5. +
  5. \displaystyle \int_0^1 g(x) dx is convergent and \displaystyle \int_0^1 h(x) dx is divergent.
    6. +
+

+ + + + +

+ Let f(x), g(x) be functions such that for a<x<b, 0\leq f(x) \leq g(x). + Then \displaystyle 0\leq \int_a^b f(x) dx\leq \int_a^b g(x) dx. In particular: + +

    +

  • If \int_a^b g(x)dx converges, so does the smaller \int_a^b f(x)dx.

    • +

  • If \int_a^b f(x)dx diverges, so does the bigger \int_a^b g(x)dx.

    • +
+

+ + + + + + +

+ Compare \frac{1}{x^3+1} to one of the following functions where x>2 and use this to determine if \displaystyle \int_2^\infty \frac{1}{x^3+1}dx is convergent or divergent. +

    +
  1. \displaystyle\frac{1}{x}
    2. +
  2. \displaystyle\frac{1}{\sqrt{x}}
    3. +
  3. \displaystyle\frac{1}{x^2}
    4. +
  4. \displaystyle\frac{1}{x^3}
    5. +
+

+ + + + + +

+ Comparing \frac{1}{x^3-4} to which of the following functions where x>3 allows you to determine that \displaystyle\int_3^{\infty} \dfrac{1}{x^3-4}\ dx converges? +

    +
  1. \displaystyle\frac{1}{x^3+x}
    2. +
  2. \displaystyle\frac{1}{4x^3}
    3. +
  3. \displaystyle\frac{1}{x^3}
    4. +
  4. \displaystyle\frac{1}{x^3-x^3/2}
    5. +
+

+ + + + +

Find \displaystyle \int_{\pi/2}^a \cos(x)dx.

+ + +

Which of the following is true about \int_{\pi/2}^\infty \cos(x)dx?

+
    +

  1. \displaystyle \int_{\pi/2}^\infty \cos(x)dx is convergent.

    2. +

  2. \displaystyle \int_{\pi/2}^\infty \cos(x)dx is divergent.

    3. +

  3. More information is needed.

    4. +
+ + + + + + Videos +
+ + Video: I can compute improper integrals, p>1 +
+ +
+ + Video: I can compute improper integrals, p < 1 +
+ + + + + + + +
diff --git a/calculus/source/05-TI/main.ptx b/calculus/source/05-TI/main.ptx new file mode 100644 index 00000000..75458552 --- /dev/null +++ b/calculus/source/05-TI/main.ptx @@ -0,0 +1,15 @@ + + + + Techniques of Integration (TI) + + + + + + + + + + + diff --git a/calculus/source/05-TI/outcomes/01.ptx b/calculus/source/05-TI/outcomes/01.ptx new file mode 100644 index 00000000..ce770c0a --- /dev/null +++ b/calculus/source/05-TI/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Evaluate various integrals via the substitution method. +

\ No newline at end of file diff --git a/calculus/source/05-TI/outcomes/02.ptx b/calculus/source/05-TI/outcomes/02.ptx new file mode 100644 index 00000000..07053268 --- /dev/null +++ b/calculus/source/05-TI/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Compute integrals using integration by parts. +

\ No newline at end of file diff --git a/calculus/source/05-TI/outcomes/03.ptx b/calculus/source/05-TI/outcomes/03.ptx new file mode 100644 index 00000000..8bc98d27 --- /dev/null +++ b/calculus/source/05-TI/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Compute integrals involving products of trigonometric functions. +

\ No newline at end of file diff --git a/calculus/source/05-TI/outcomes/04.ptx b/calculus/source/05-TI/outcomes/04.ptx new file mode 100644 index 00000000..b56f936f --- /dev/null +++ b/calculus/source/05-TI/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Use trigonometric substitution to compute indefinite integrals. +

\ No newline at end of file diff --git a/calculus/source/05-TI/outcomes/05.ptx b/calculus/source/05-TI/outcomes/05.ptx new file mode 100644 index 00000000..43c21778 --- /dev/null +++ b/calculus/source/05-TI/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+I can integrate functions using a table of integrals. +

\ No newline at end of file diff --git a/calculus/source/05-TI/outcomes/06.ptx b/calculus/source/05-TI/outcomes/06.ptx new file mode 100644 index 00000000..21a9945c --- /dev/null +++ b/calculus/source/05-TI/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+I can integrate functions using the method of partial fractions. +

\ No newline at end of file diff --git a/calculus/source/05-TI/outcomes/07.ptx b/calculus/source/05-TI/outcomes/07.ptx new file mode 100644 index 00000000..0ab8eac1 --- /dev/null +++ b/calculus/source/05-TI/outcomes/07.ptx @@ -0,0 +1,4 @@ + +

+I can select appropriate strategies for integration. +

\ No newline at end of file diff --git a/calculus/source/05-TI/outcomes/08.ptx b/calculus/source/05-TI/outcomes/08.ptx new file mode 100644 index 00000000..92e3f361 --- /dev/null +++ b/calculus/source/05-TI/outcomes/08.ptx @@ -0,0 +1,4 @@ + +

+I can compute improper integrals. +

\ No newline at end of file diff --git a/calculus/source/05-TI/outcomes/main.ptx b/calculus/source/05-TI/outcomes/main.ptx new file mode 100644 index 00000000..cedc86e4 --- /dev/null +++ b/calculus/source/05-TI/outcomes/main.ptx @@ -0,0 +1,37 @@ + +> + +

+How do we use various techniques to integrate less simple functions? +

+

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
  7. + +
    8. +
  8. + +
    9. +
+ \ No newline at end of file diff --git a/calculus/source/05-TI/readiness.ptx b/calculus/source/05-TI/readiness.ptx new file mode 100644 index 00000000..b1942374 --- /dev/null +++ b/calculus/source/05-TI/readiness.ptx @@ -0,0 +1,71 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Recognize when to apply the chain rule, and use it successfully.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    2. +
  2. +

    Recognize when to apply the product rule, and use it successfully.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    3. +
  3. +

    Perform basic definite integrals.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    4. +
  4. +

    Compute basic antiderivatives.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    5. +
  5. +

    Differentiate trigonometric functions.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    6. +
  6. +

    Compute limits.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    • +

      Review: Khan Academy

      +
      • +
    +
    7. +
  7. +

    Apply the pythagorean identity.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    8. +
+ + \ No newline at end of file diff --git a/calculus/source/06-AI/01.ptx b/calculus/source/06-AI/01.ptx new file mode 100644 index 00000000..a1fce027 --- /dev/null +++ b/calculus/source/06-AI/01.ptx @@ -0,0 +1,276 @@ + + +
+ Average Value (AI1) + + + + + Activities + + +

+ Suppose a car drives due east at 70 miles per hour for 2 hours, and then slows down to 40 miles per hour for an additional hour. +

+ + +

How far did the car travel in these 3 hours?

+
    +
  1. + 110 miles +
    2. +
  2. + 150 miles +
    3. +
  3. + 180 miles +
    4. +
  4. + 220 miles +
    5. +
+ + +

What was its average velocity over these 3 hours?

+
    +
  1. + 55 miles per hour +
    2. +
  2. + 60 miles per hour +
    3. +
  3. + 70 miles per hour +
    4. +
  4. + 75 miles per hour +
    5. +
+ + + + +

+ Suppose instead the car starts with a velocity of 30 miles per hour, and increases velocity + linearly according to the function v(t)=30+20t so its velocity after three hours is 90 miles per hour. +

+ + +

How can we model the car's distance traveled using calculus?

+
    +
  1. + Integrate velocity, because position is the rate of change of velocity. +
    2. +
  2. + Integrate velocity, because velocity is the rate of change of position. +
    3. +
  3. + Differentiate velocity, because position is the rate of change of velocity. +
    4. +
  4. + Differentiate velocity, because velocity is the rate of change of position. +
    5. +
+ + +

Then, which of these expressions is a mathematical model for the car's distance traveled after 3 hours?

+
    +
  1. + \int (30+20t)\,dt +
    2. +
  2. + \int (30t+10t^2)\,dt +
    3. +
  3. + \int_0^3 (30+20t)\,dt +
    4. +
  4. + \int_0^3 (30t+10t^2)\,dt +
    5. +
+ + +

How far did the car travel in these 3 hours?

+
    +
  1. + 110 miles +
    2. +
  2. + 150 miles +
    3. +
  3. + 180 miles +
    4. +
  4. + 220 miles +
    5. +
+ + +

Thus, what was its average velocity over three hours?

+
    +
  1. + 55 miles per hour +
    2. +
  2. + 60 miles per hour +
    3. +
  3. + 70 miles per hour +
    4. +
  4. + 75 miles per hour +
    5. +
+ + + + + +

+To obtain the average velocity of an object traveling with velocity v(t) for a\leq t\leq b, +we may find its distance traveled by calculating \int_a^b v(t). Thus, the average velocity is obtained +by dividing by the time b-a elapsed: + \frac{1}{b-a}\int_a^b v(t)\,dt. +

+

+For example, the following calculuation confirms the previous activity: + \frac{1}{3-0}\int_0^3 (30+20t)\,dt. +

+ + + + + +

+ Given a function f(x) defined on [a,b], it's average value is + defined to be \frac{1}{b-a}\int_a^b f(x) \,dx. +

+ + + + + + +

+ Which of the following expressions represent the average value of f(x)=-12 \, x^{2} + 8 \, x + 4 over the interval [-1, 2]? + +

    +
  1. \displaystyle \frac{1}{3}\int_{-1}^{2}\left(-12 \,x^{2} + 8 \, x + 4 \right) dx
    2. +
  2. \displaystyle \frac{-1}{1}\int_{1}^{2}\left(-12 \,x^{2} + 8 \, x + 4 \right) dx
    3. +
  3. \displaystyle \frac{1}{2}\int_{1}^{2}\left(-12 \,x^{2} + 8 \, x + 4 \right) dx
    4. +
  4. \displaystyle \frac{-1}{4}\int_{-1}^{2}\left(-12 \,x^{2} + 8 \, x + 4 \right) dx
    5. +
+

+ + +

Show that the average value of f(x)=-12 \, x^{2} + 8 \, x + 4 over the interval [-1, 2] is -4.

+ + + + + +

+ Which of the following expressions represent the average value of f(x)=x\cos(x^2)+x on the interval [\pi, 4\pi]? + +

    +
  1. \displaystyle \frac{1}{3\pi}\int_{0}^{4\pi}\Big(x\cos(x^2)+x\Big) dx
    2. +
  2. \displaystyle \frac{1}{4\pi}\int_{0}^{4\pi}\Big(x\cos(x^2)+x\Big) dx
    3. +
  3. \displaystyle \frac{1}{3\pi}\int_{\pi}^{4\pi}\Big(x\cos(x^2)+x\Big) dx
    4. +
  4. \displaystyle \frac{1}{4\pi}\int_{\pi}^{4\pi}\Big(x\cos(x^2)+x\Big) dx
    5. + +
+

+ + +

+ Find the average value of f(x)=x\cos(x^2)+x on the interval [\pi, 4\pi] using the chosen expression. +

+ + + + + +

+ Find the average value of \displaystyle g(t)=\frac{t}{t^2+1} on the interval [0, 4]. +

+ + + + + +

+ A shot of a drug is administered to a patient and the quantity of the drug in the bloodstream over time is q(t)=3te^{-0.25t}, where t is measured in hours and q is measured in milligrams. What is the average quantity of this drug in the patient's bloodstream over the first 6 hours after injection? + +

+ + + + + +

+ Which of the following is the average value of f(x) over the interval [0,8]? + +

+ Plot of f(x). + + A plot of f(x). + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-1.5,-1.5) grid (8.5,4.5); + %axes + \draw[very thick,->] (-1,0) -- (8,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,-1) -- (0,4) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1,2,3,4,5,6, 7, 8} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {1,2,3,4} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] (0,1) -- (3,1); + \draw[blue] (3,4) -- (6,4); + \draw[blue] (6,2) -- (8,2); + \draw[blue] (8,2)--node[right]{\tiny$y=f(x)$} (8,2); + + \end{tikzpicture} + + +
+

+

+ Note f(x)=\begin{cases} 1, & 0\leq x\leq 3 \\ 4, & 3 < x \leq 6 \\ 2, & 6 < x \leq 8 \end{cases}. +

+ +
    +
  1. 4
    2. +
  2. 2
    3. +
  3. \displaystyle \frac{7}{3}
    4. +
  4. 19
    5. +
  5. 2.375
    6. +
+ + + + + + + + Videos +
+ + Video: Compute the average value of a function on an interval +
+ + + +
+ diff --git a/calculus/source/06-AI/01point5.ptx b/calculus/source/06-AI/01point5.ptx new file mode 100644 index 00000000..661d4573 --- /dev/null +++ b/calculus/source/06-AI/01point5.ptx @@ -0,0 +1,303 @@ + + +
+ Arclength (AI1.5) + + + + + +

+ Suppose we wanted to find the arclength of the parabola y=-x^2+6x over the interval [0,4]. +

+ Plot of y=-x^2+6x over [0,4]. + + Plot of y=-x^2+6x over [0,4]. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-.5,-0.5) grid (5.5,10.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,0) -- (0,10) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1,2,3,4,5} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {0,...,10} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{-1*\x^2+6*\x }); + + + + \end{tikzpicture} + + +
+

+ + + +

+ Suppose we wished to estimate this length with two line segments, where \Delta x=2. +

+ Plot of y=-x^2+6x over [0,4]. + + Plot of y=-x^2+6x over [0,4]. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-.5,-0.5) grid (5.5,10.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,0) -- (0,10) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1,2,3,4,5} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {0,...,10} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{-1*\x^2+6*\x }); + + \draw[dashed] (0,0)--(2,8)--(4,8); + + \draw[fill] (0,0) circle (2pt); + \draw[fill] (2,8) circle (2pt); + \draw[fill] (4,8) circle (2pt); + + + + \end{tikzpicture} + + +
+ Which of the following expressions represents the sum of the lengths of the line segments with endpoints (0,0), (2,8) and (4,8). +
    +
  1. \sqrt{4+8}.
    2. +
  2. \sqrt{4^2+8^2}.
    3. +
  3. \sqrt{2^2+8^2}+\sqrt{4^2+8^2}.
    4. +
  4. \sqrt{2^2+8^2}+\sqrt{(4-2)^2+(8-2)^2}.
    5. +
+

+ + + +

+ Suppose we wished to estimate this length with four line segments, where \Delta x=1. +

+ Plot of y=-x^2+6x over [0,4]. + + Plot of y=-x^2+6x over [0,4]. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-.5,-0.5) grid (5.5,10.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,0) -- (0,10) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1,2,3,4,5} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {0,...,10} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{-1*\x^2+6*\x }); + + \draw[dashed] (0,0)--(1,5)--(2,8)--(3,9)--(4,8); + + \draw[fill] (0,0) circle (2pt); + \draw[fill] (1,5) circle (2pt); + \draw[fill] (2,8) circle (2pt); + \draw[fill] (3,9) circle (2pt); + \draw[fill] (4,8) circle (2pt); + + + + \end{tikzpicture} + + +
+ Which of the following expressions represents the sum of the lengths of the line segments with endpoints (0,0), (1,5), (2,8), (3,9) and (4,8). +
    +
  1. \sqrt{4^2+8^2}.
    2. +
  2. \sqrt{1^2+(5-0)^2}+\sqrt{1^2+(8-5)^2}+\sqrt{1^2+(9-8)^2}+\sqrt{1^2+(8-9)^2}.
    3. +
  3. \sqrt{1^2+5^2}+\sqrt{2^2+8^2}+\sqrt{3^2+9^2}+\sqrt{4^2+8^2}.
    4. + +
+

+ + + +

+ Suppose we wished to estimate this length with n line segments \Delta x=4/n. Let f(x)=-x^2+6x. +

+ Plot of y=-x^2+6x over [0,4]. + + Plot of y=-x^2+6x over [0,4]. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-.5,-0.5) grid (5.5,10.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,0) -- (0,10) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1,2,3,4,5} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {0,...,10} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{-1*\x^2+6*\x }); + + \draw[dashed, red] (2.1,8.19)--(2.65,8.8775)--(2.65,8.19)--(2.1,8.19); + + \draw[fill, red] (2.1,8.19) circle (2pt); + \draw[red] (2.1,8.19)--node[above left]{$(x_0,f(x_0))$} (2.1,8.19); + \draw[red] (2.65,8.8775)--node[above right]{$(x_0+\Delta x,f(x_0+\Delta x))$} (2.65,8.8775); + \draw[fill, red] (2.65,8.8775) circle (2pt); + + \draw [decorate, decoration={brace,amplitude=10pt,mirror},yshift=-0.4pt, red](2.1,8.19)--(2.65,8.19) node[red,midway,yshift=-0.5cm] {\footnotesize $\Delta x$}; + + + + \end{tikzpicture} + + +
+ Which of the following expressions represents the length of the line segment from (x_0, f(x_0)) to (\Delta x, f(x_0+\Delta x))? +
    +
  1. \sqrt{x_0^2+f(x_0)^2}.
    2. +
  2. \sqrt{(x_0+\Delta x)^2+f(x_0+\Delta x)^2}.
    3. +
  3. \sqrt{(\Delta x)^2+(f(x_0+\Delta)-f(x_0))^2}.
    4. +
  4. \sqrt{(\Delta x)^2+f(\Delta x)^2}.
    5. + +
+

+ + + +

+ Which of the following Riemann sums best estimates the arclength of the parabola y=-x^2+6x over the interval [0,4]? Let f(x)=-x^2+6x. +

    +
  1. \displaystyle \sum \sqrt{x_i^2+f(x_i)^2}.
    2. +
  2. \displaystyle \sum \sqrt{(x_i+\Delta x)^2+f(x_i+\Delta x)^2}.
    3. +
  3. \displaystyle \sum \sqrt{(\Delta x)^2+(f(x_i+\Delta)-f(x_i))^2}.
    4. +
  4. \displaystyle \sum \sqrt{(\Delta x)^2+f(\Delta x)^2}.
    5. + +
+

+ + + +

+ Note that + + + \sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2} & = \sqrt{(\Delta x)^2\left(1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2\right)} + &=\sqrt{1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2}\Delta x + + + Which of the following best describes \displaystyle\lim_{\Delta x\to 0} \frac{f(x_i+\Delta x)-f(x_i)}{\Delta x}? +

    +
  1. 0.
    2. +
  2. 1.
    3. +
  3. f'(x_i).
    4. +
  4. This limit is undefined.
    5. +
+

+ + + + + + + +

+ Given a differentiable function f(x), the arclength of y=f(x) defined on [a,b] is computed by the integral: + + + \lim_{n\to \infty}\sum \sqrt{(\Delta x)^2+(f(x_i+\Delta)-f(x_i))^2} & =\lim_{n\to \infty}\sum \sqrt{1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2}\Delta x + & = \int_a^b \sqrt{1+(f'(x))^2}dx. + + +

+ + + + +

+ Use to find an integral which measures the arclength of the parabola y=-x^2+6x over the interval [0,4]. +

+ + + + + +

+ Consider the curve y=2^x-1 defined on [1,5]. +

+ + + +

+ Estimate the arclength of this curve with two line segments with \Delta x=2. + + \begin{array}{|c|c|c|c|} + \hline + x_i & (x_i, f(x_i)) & (x_i+\Delta x, f(x_i+\Delta x)) & \text{Length of segment}\\ + \hline + 1 & & & \\ + \hline + 3 & & & \\ + \hline + \end{array} + +

+ + + +

+ Estimate the arclength of this curve with four line segments with \Delta x=1. + + \begin{array}{|c|c|c|c|} + \hline + x_i & (x_i, f(x_i)) & (x_i+\Delta x, f(x_i+\Delta x)) & \text{Length of segment}\\ + \hline + 1 & & & \\ + \hline + 2 & & & \\ + \hline + 3 & & & \\ + \hline + 4 & & & \\ + \hline + \end{array} + +

+ + +

+ Find an integral which computes the arclength of the curve y=2^x-1 defined on [1,5]. +

+ + + + + + + + +
+ diff --git a/calculus/source/06-AI/02.ptx b/calculus/source/06-AI/02.ptx new file mode 100644 index 00000000..97d5cb54 --- /dev/null +++ b/calculus/source/06-AI/02.ptx @@ -0,0 +1,381 @@ + + +
+ Arclength (AI2) + + + + + Activities + + + +

+ Suppose we wanted to find the arclength of the parabola y=-x^2+6x over the interval [0,4]. +

+ Plot of y=-x^2+6x over [0,4]. + + Plot of y=-x^2+6x over [0,4]. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-.5,-0.5) grid (5.5,10.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,0) -- (0,10) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1,2,3,4,5} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {0,...,10} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{-1*\x^2+6*\x }); + \end{tikzpicture} + + +
+

+ + + +

+ Suppose we wished to estimate this length with two line segments where \Delta x=2. +

+ Plot of y=-x^2+6x over [0,4] with two line segments where \Delta x=2. + + Plot of y=-x^2+6x over [0,4]. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-.5,-0.5) grid (5.5,10.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,0) -- (0,10) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1,2,3,4,5} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {0,...,10} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{-1*\x^2+6*\x }); + + \draw[dashed] (0,0)--(2,8)--(4,8); + + \draw[fill] (0,0) circle (2pt); + \draw[fill] (2,8) circle (2pt); + \draw[fill] (4,8) circle (2pt); + + + + \end{tikzpicture} + + +
+

+

Which of the following expressions represents the sum of the lengths of the line segments with endpoints (0,0), (2,8) and (4,8)?

+ +
    +
  1. \sqrt{4+8}
    2. +
  2. \sqrt{2^2+8^2}+\sqrt{(4-2)^2+(8-8)^2}
    3. +
  3. \sqrt{4^2+8^2}
    4. +
  4. \sqrt{2^2+8^2}+\sqrt{4^2+8^2}
    5. +
+ + + +

+ Suppose we wished to estimate this length with four line segments where \Delta x=1. +

+ Plot of y=-x^2+6x over [0,4] with four line segments where \Delta x=1. + + Plot of y=-x^2+6x over [0,4]. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-.5,-0.5) grid (5.5,10.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,0) -- (0,10) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1,2,3,4,5} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {0,...,10} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{-1*\x^2+6*\x }); + + \draw[dashed] (0,0)--(1,5)--(2,8)--(3,9)--(4,8); + + \draw[fill] (0,0) circle (2pt); + \draw[fill] (1,5) circle (2pt); + \draw[fill] (2,8) circle (2pt); + \draw[fill] (3,9) circle (2pt); + \draw[fill] (4,8) circle (2pt); + + + + \end{tikzpicture} + + +
+

+

Which of the following expressions represents the sum of the lengths of the line segments with endpoints (0,0), (1,5), (2,8), (3,9) and (4,8)?

+
    +
  1. \sqrt{4^2+8^2}
    2. +
  2. \sqrt{1^2+(5-0)^2}+\sqrt{1^2+(8-5)^2}+\sqrt{1^2+(9-8)^2}+\sqrt{1^2+(8-9)^2}
    3. +
  3. \sqrt{1^2+5^2}+\sqrt{2^2+8^2}+\sqrt{3^2+9^2}+\sqrt{4^2+8^2}
    4. +
+ + + +

+ Suppose we wished to estimate this length with n line segments where \displaystyle \Delta x=\frac{4}{n}. Let f(x)=-x^2+6x. +

+ Plot of y=-x^2+6x over [0,4] with n line segments where \displaystyle \Delta x=\frac{4}{n}. + + Plot of y=-x^2+6x over [0,4]. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-.5,-0.5) grid (5.5,10.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,0) -- (0,10) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1,2,3,4,5} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {0,...,10} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{-1*\x^2+6*\x }); + + \draw[dashed, red] (2.1,8.19)--(2.65,8.8775)--(2.65,8.19)--(2.1,8.19); + + \draw[fill, red] (2.1,8.19) circle (2pt); + \draw[red] (2.1,8.19)--node[above left]{$(x_0,f(x_0))$} (2.1,8.19); + \draw[red] (2.65,8.8775)--node[above right]{$(x_0+\Delta x,f(x_0+\Delta x))$} (2.65,8.8775); + \draw[fill, red] (2.65,8.8775) circle (2pt); + + \draw [decorate, decoration={brace,amplitude=10pt,mirror},yshift=-0.4pt, red](2.1,8.19)--(2.65,8.19) node[red,midway,yshift=-0.5cm] {\footnotesize $\Delta x$}; + + \end{tikzpicture} + + +
+

+

Which of the following expressions represents the length of the line segment from (x_0, f(x_0)) to (x_0+\Delta x, f(x_0+\Delta x))?

+
    +
  1. \sqrt{x_0^2+f(x_0)^2}
    2. +
  2. \sqrt{(x_0+\Delta x)^2+f(x_0+\Delta x)^2}
    3. +
  3. \sqrt{(\Delta x)^2+f(\Delta x)^2}
    4. +
  4. \sqrt{(\Delta x)^2+(f(x_0+\Delta x)-f(x_0))^2}
    5. +
+ + + +

Which of the following Riemann sums best estimates the arclength of the parabola y=-x^2+6x over the interval [0,4]? Let f(x)=-x^2+6x.

+
    +
  1. \displaystyle \sum \sqrt{(\Delta x)^2+f(\Delta x)^2}
    2. +
  2. \displaystyle \sum \sqrt{(x_i+\Delta x)^2+f(x_i+\Delta x)^2}
    3. +
  3. \displaystyle \sum \sqrt{x_i^2+f(x_i)^2}
    4. +
  4. \displaystyle \sum \sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2}
    5. +
+ + + +

+ Note that + + \sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2} & = \sqrt{(\Delta x)^2\left(1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2\right)} + &=\sqrt{1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2}\Delta x + . +

+

+ Which of the following best describes \displaystyle\lim_{\Delta x\to 0} \frac{f(x_i+\Delta x)-f(x_i)}{\Delta x}? +

    +
  1. 0
    2. +
  2. 1
    3. +
  3. f'(x_i)
    4. +
  4. This limit is undefined.
    5. +
+

+ + + + +

+ Given a differentiable function f(x), the arclengtharclength of y=f(x) defined on [a,b] is computed by the integral + + + \lim_{n\to \infty}\sum \sqrt{(\Delta x)^2+(f(x_i+\Delta)-f(x_i))^2} & =\lim_{n\to \infty}\sum \sqrt{1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2}\Delta x + & = \int_a^b \sqrt{1+(f'(x))^2}dx + . +

+ + + + + + +

+ Use to find an integral which measures the arclength of the parabola y=-x^2+6x over the interval [0,4]. +

+ + + + + +

+ Consider the curve y=2^x-1 defined on [1,5]. +

+ + + +

+ Estimate the arclength of this curve with two line segments where \Delta x=2. +

+ + + + x_i + + + (x_i,f(x_i)) + + + (x_i+\Delta x,f(x_i+\Delta x)) + + + Length of segment + + + + + 1 + + + + + + + + 3 + + + + + + + + + +

+ Estimate the arclength of this curve with four line segments where \Delta x=1. +

+ + + + x_i + + + (x_i,f(x_i)) + + + (x_i+\Delta x,f(x_i+\Delta x)) + + + Length of segment + + + + + 1 + + + + + + + + 2 + + + + + + + + 3 + + + + + + + + 4 + + + + + + + + +

+ Find an integral which computes the arclength of the curve y=2^x-1 defined on [1,5]. +

+ + + + + + +

+ Consider the curve y=5e^{-x^2} over the interval [-1,4]. +

+ + +

+ Estimate this arclength with 5 line segments where \Delta x=1. +

+ + +

+ Find an integral which computes this arclength. +

+ + + + + + Videos +
+ + Video: Estimate the arclength of a curve with Riemann sums and find an integral which computes the arclength +
+ + + + + +
+ diff --git a/calculus/source/06-AI/03.ptx b/calculus/source/06-AI/03.ptx new file mode 100644 index 00000000..a491e33d --- /dev/null +++ b/calculus/source/06-AI/03.ptx @@ -0,0 +1,374 @@ + + +
+ Volumes of Revolution (AI3) + + + + + Activities + + + +

+Consider the following visualization to decide which of these statements is most appropriate for +describing the relationship of lengths and areas. +

+ + + + + +
    +
  1. Length is the integral of areas.
    2. +
  2. Area is the integral of lengths.
    3. +
  3. Length is the derivative of areas.
    4. +
  4. None of these.
    5. +
+ + + + + +

+We define the volume of a solid with cross sectional area given by A(x) +laying between a\leq x\leq b to be the definite integral +\mathrm{Volume}=\int_a^b A(x)\,dx. +

+ + + + + + + + + + +

+We will be focused on the volumes of solids obtained by revolving a region around an axis. +Let's use the running example of the region bounded by the curves x=0,y=4,y=x^2. +

+ + + + + + + +

+Consider the below illustrated revolution of this region, and the cross-section drawn +from a horizontal line segment. Choose the most appropriate description of this illustration. +

+ + + + + +
    +
  1. Region is rotated around the x-axis; the cross-sectional area is determined by the line segment's x-value.
    2. +
  2. Region is rotated around the x-axis; the cross-sectional area is determined by the line segment's y-value.
    3. +
  3. Region is rotated around the y-axis; the cross-sectional area is determined by the line segment's x-value.
    4. +
  4. Region is rotated around the y-axis; the cross-sectional area is determined by the line segment's y-value.
    5. +
+ + +

+Which of these formulas is most appropriate to find this illustration's cross-sectional area? +

+
    +
  1. \pi r^2
    2. +
  2. 2\pi rh
    3. +
  3. \pi R^2-\pi r^2
    4. +
  4. \frac{1}{2}bh
    5. +
+ + +

+Consider the below illustrated revolution of this region, and the cross-section drawn +from a vertical line segment. Choose the most appropriate description of this illustration. +

+ + + + + +
    +
  1. Region is rotated around the x-axis; the cross-sectional area is determined by the line segment's x-value.
    2. +
  2. Region is rotated around the x-axis; the cross-sectional area is determined by the line segment's y-value.
    3. +
  3. Region is rotated around the y-axis; the cross-sectional area is determined by the line segment's x-value.
    4. +
  4. Region is rotated around the y-axis; the cross-sectional area is determined by the line segment's y-value.
    5. +
+ + +

+Which of these formulas is most appropriate to find this illustration's cross-sectional area? +

+
    +
  1. \pi r^2
    2. +
  2. 2\pi rh
    3. +
  3. \pi R^2-\pi r^2
    4. +
  4. \frac{1}{2}bh
    5. +
+ + +

+Consider the below illustrated revolution of this region, and the cross-section drawn +from a horizontal line segment. Choose the most appropriate description of this illustration. +

+ + + + + +
    +
  1. Region is rotated around the x-axis; the cross-sectional area is determined by the line segment's x-value.
    2. +
  2. Region is rotated around the x-axis; the cross-sectional area is determined by the line segment's y-value.
    3. +
  3. Region is rotated around the y-axis; the cross-sectional area is determined by the line segment's x-value.
    4. +
  4. Region is rotated around the y-axis; the cross-sectional area is determined by the line segment's y-value.
    5. +
+ + +

+Which of these formulas is most appropriate to find this illustration's cross-sectional area? +

+
    +
  1. \pi r^2
    2. +
  2. 2\pi rh
    3. +
  3. \pi R^2-\pi r^2
    4. +
  4. \frac{1}{2}bh
    5. +
+ + +

+Consider the below illustrated revolution of this region, and the cross-section drawn +from a vertical line segment. Choose the most appropriate description of this illustration. +

+ + + + + +
    +
  1. Region is rotated around the x-axis; the cross-sectional area is determined by the line segment's x-value.
    2. +
  2. Region is rotated around the x-axis; the cross-sectional area is determined by the line segment's y-value.
    3. +
  3. Region is rotated around the y-axis; the cross-sectional area is determined by the line segment's x-value.
    4. +
  4. Region is rotated around the y-axis; the cross-sectional area is determined by the line segment's y-value.
    5. +
+ + +

+Which of these formulas is most appropriate to find this illustration's cross-sectional area? +

+
    +
  1. \pi r^2
    2. +
  2. 2\pi rh
    3. +
  3. \pi R^2-\pi r^2
    4. +
  4. \frac{1}{2}bh
    5. +
+ + + + + +

+Generally when solving problems without the aid of technology, it's useful to draw your region in +two dimensions, choose whether to use a horizontal or vertical line segment, and draw its rotation +to determine the cross-sectional shape. +

+

+When the shape is a disk, this is called the disk method and we use one of these formulas +depending on whether the cross-sectional area depends on x or y. +V=\int_a^b \pi r(x)^2\,dx,\hspace{2em}V=\int_a^b \pi r(y)^2\,dy. +

+

+When the shape is a washer, this is called the washer method and we use one of these formulas +depending on whether the cross-sectional area depends on x or y. +V=\int_a^b\left(\pi R(x)^2- \pi r(x)^2\right)\,dx,\hspace{2em}V=\int_a^b\left(\pi R(y)^2- \pi r(y)^2\right)\,dy. +

+

+When the shape is a cylindrical shell, this is called the shell method and we use one of these formulas +depending on whether the cross-sectional area depends on x or y. +V=\int_a^b 2\pi r(x)h(x)\,dx,\hspace{2em}V=\int_a^b 2\pi r(y)h(y)\,dy. +

+ + + + + +

+Let's now consider the region bounded by the curves x=0,x=1,y=0,y=5e^x, +rotated about the x-axis. +

+ + +

+Sketch two copies of this region in the xy plane. +

+ + +

+Draw a vertical line segment in one region and its rotation around the x-axis. +Draw a horizontal line segment in the other region and its rotation around the x-axis. +

+ + +

+Consider the method required for each cross-section drawn. Which would be the easiest +strategy to proceed with? +

+
    +
  1. The horizontal line segment, using the disk/washer method.
    2. +
  2. The horizontal line segment, using the shell method.
    3. +
  3. The vertical line segment, using the disk/washer method.
    4. +
  4. The vertical line segment, using the shell method.
    5. +
+ + +

+Let's proceed with the vertical segment. Which formula is most appropriate for the radius? +

+
    +
  1. r(x)=x
    2. +
  2. r(x)=5e^x
    3. +
  3. r(x)=5\ln(x)
    4. +
  4. r(x)=\frac{1}{5}\ln(x)
    5. +
+ + +

+Which of these integrals is equal to the volume of the solid of revolution? +

+
    +
  1. \int_0^1 25\pi e^{2x}\,dx
    2. +
  2. \int_0^1 5\pi^2 e^{x}\,dx
    3. +
  3. \int_0^2 25\pi e^{x}\,dx
    4. +
  4. \int_0^2 5\pi^2 e^{2x}\,dx
    5. +
+ + + + + +

+Let's now consider the same region, bounded by the curves x=0,x=1,y=0,y=5e^x, +but this time rotated about the y-axis. +

+ + +

+Sketch two copies of this region in the xy plane. +

+ + +

+Draw a vertical line segment in one region and its rotation around the y-axis. +Draw a horizontal line segment in the other region and its rotation around the y-axis. +

+ + +

+Consider the method required for each cross-section drawn. Which would be the easiest +strategy to proceed with? +

+
    +
  1. The horizontal line segment, using the disk/washer method.
    2. +
  2. The horizontal line segment, using the shell method.
    3. +
  3. The vertical line segment, using the disk/washer method.
    4. +
  4. The vertical line segment, using the shell method.
    5. +
+ + +

+Let's proceed with the vertical segment. Which formula is most appropriate for the radius? +

+
    +
  1. r(x)=x
    2. +
  2. r(x)=5e^x
    3. +
  3. r(x)=5\ln(x)
    4. +
  4. r(x)=\frac{1}{5}\ln(x)
    5. +
+ + +

+Which formula is most appropriate for the height? +

+
    +
  1. h(x)=x
    2. +
  2. h(x)=5e^x
    3. +
  3. h(x)=5\ln(x)
    4. +
  4. h(x)=\frac{1}{5}\ln(x)
    5. +
+ + +

+Which of these integrals is equal to the volume of the solid of revolution? +

+
    +
  1. \int_0^1 5\pi^2 xe^{x}\,dx
    2. +
  2. \int_0^1 10\pi xe^{x}\,dx
    3. +
  3. \int_0^2 5\pi xe^{x}\,dx
    4. +
  4. \int_0^2 10\pi x^2e^{x}\,dx
    5. +
+ + + + + + +

Consider the region bounded by y=2 \, x + 3, y=0, x=4, x=7.

+ + + +

Find an integral which computes the volume of the solid formed by rotating this region about the x-axis.

+ + + + +

Find an integral which computes the volume of the solid formed by rotating this region about the y-axis.

+ + + + + + + Videos +
+ + Video: Compute volumes of solids of revolution, washer x-axis +
+
+ + Video: Compute volumes of solids of revolution, shell x-axis +
+ +
+ + Video: Compute volumes of solids of revolution, washer y-axis +
+
+ + Video: Compute volumes of solids of revolution, shell y-axis +
+ + + + + + + + +
+ diff --git a/calculus/source/06-AI/04.ptx b/calculus/source/06-AI/04.ptx new file mode 100644 index 00000000..f0a47b19 --- /dev/null +++ b/calculus/source/06-AI/04.ptx @@ -0,0 +1,465 @@ + + +
+ Surface Areas of Revolution (AI4) + + + + + Activities + + + + + +

+ A frustumfrustum is the portion of a cone that lies between one or two parallel planes. + +

+ Plot of a frustum. + + frustum + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-0.5,-2.5) grid (4.5,2.5); + %axes + \draw[very thick,->] (-0,0) -- (4,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,-2) -- (0,2) node[anchor=south east] {\tiny$y$}; + \foreach \x in {0, 1, ...,4} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {-2, -1, ...,2} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw (0,0) --(4,2)--(4,-2)--(0,0); + + + + \fill [blue, domain=0:4, variable=\x, opacity=0.2] + (2, 1) + --(3,1.5) + --(3,-1.5) + --(2,-1) + -- cycle; + + \draw [red, decorate, decoration={brace,amplitude=10pt},yshift=0.2pt](2,1) -- (3,1.5) node[midway, left,yshift=0.5cm, xshift=-0.07cm, red] {\footnotesize $l$}; + + \draw [decorate, decoration={brace,amplitude=10pt},xshift=0pt](2,0) -- (2,1) node[black, midway,xshift=-0.65cm] {\footnotesize $r$}; + + \draw [decorate, decoration={brace,amplitude=10pt, mirror},xshift=0pt](3,0) -- (3,1.5) node[black, midway,xshift=0.65cm] {\footnotesize $R$}; + + + \end{tikzpicture} + + +
+

+ +

The surface area of the side of the frustum is: 2\pi \frac{r+R}{2}\cdot l where r and R are the radii of the bases, and l is the length of the side.

+ +

Note that if r=R, this reduces to the surface area of a side of a cylinder.

+ + + + + +

+ Suppose we wanted to find the surface area of the the solid of revolution generated by rotating + y=\sqrt{x}, 0\leq x\leq 4 + about the y-axis. + +

+ Plot of bounded region rotated about x-axis. + + Bounded region rotated about x-axis. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-0.5,-3.5) grid (5.5,3.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,-3) -- (0,3) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1, 2, ..., 4} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {-3, -2,...,3} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{(\x^(1/2) }); + + \draw[blue, dashed] plot [ + samples=100, + domain=0:4, + ] (\x,{(-1*\x^(1/2) }); + + + \fill [blue, domain=0:4, variable=\x, opacity=0.2] + (0, 0) + -- plot ({\x}, {(\x^(1/2)}) + --(4,0) + -- cycle; + + \fill [blue, domain=0:4, variable=\x, opacity=0.2] + (0, 0) + -- plot ({\x}, {-1*(\x)^(1/2)}) + --(4,0) + -- cycle; + + \end{tikzpicture} + + +
+

+ + + +

+ Suppose we wanted to estimate the surface area with two frustums with \Delta x=2. + +

+ Plot of bounded region rotated about x-axis. + + Bounded region rotated about x-axis. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-0.5,-3.5) grid (5.5,3.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,-3) -- (0,3) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1, 2, ..., 4} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {-3, -2,...,3} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{(\x^(1/2) }); + + \draw[blue, dashed] plot [ + samples=100, + domain=0:4, + ] (\x,{(-1*\x^(1/2) }); + + + \fill [blue, domain=0:4, variable=\x, opacity=0.2] + (0, 0) + -- plot ({\x}, {(\x^(1/2)}) + --(4,0) + -- cycle; + + \fill [blue, domain=0:4, variable=\x, opacity=0.2] + (0, 0) + -- plot ({\x}, {-1*(\x)^(1/2)}) + --(4,0) + -- cycle; + + \draw[dashed] (0,0)--(2,1.4142)--(4,2); + + \draw[fill] (0,0) circle (2pt); + \draw[fill] (2,1.4142) circle (2pt); + \draw[fill] (4,2) circle (2pt); + + \end{tikzpicture} + + +
+ What is the surface area of the frustum formed by rotating the line segment from (0,0) to (2, \sqrt{2}) about the x-axis? + +
    +
  1. 2\pi \frac{0+\sqrt{2}}{2}\cdot2
    2. +
  2. 2\pi \frac{0+\sqrt{2}}{2}\cdot\sqrt{2^2+\sqrt{2}^2}
    3. +
  3. \pi \sqrt{2}^2\cdot2
    4. +
  4. \pi \sqrt{2}^2\cdot\sqrt{2^2+\sqrt{2}^2}
    5. +
+

+ + + +

+

+ Plot of bounded region rotated about the x-axis. + + Bounded region rotated about x-axis. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-0.5,-3.5) grid (5.5,3.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,-3) -- (0,3) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1, 2, ..., 4} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {-3, -2,...,3} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{(\x^(1/2) }); + + \draw[blue, dashed] plot [ + samples=100, + domain=0:4, + ] (\x,{(-1*\x^(1/2) }); + + + \fill [blue, domain=0:4, variable=\x, opacity=0.2] + (4,-2)--(2,-1.4142)--(0,0)--(2,1.4142)--(4,2)--cycle; + + \draw[dashed] (4,-2)--(2,-1.4142)--(0,0)--(2,1.4142)--(4,2); + + \draw[fill] (0,0) circle (2pt); + \draw[fill] (2,1.4142) circle (2pt); + \draw[fill] (4,2) circle (2pt); + + \end{tikzpicture} + + +
+ What is the surface area of the frustum formed by rotating the line segment from (2,\sqrt{2}) to (4, 2) about the x-axis? + +
    +
  1. 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{2}
    2. +
  2. 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{6}
    3. +
  3. 2\pi \frac{4+\sqrt{2}}{2}\cdot\sqrt{6-2\sqrt{2}}
    4. +
+

+ + + + +

+ Suppose we wanted to estimate the surface area with four frustums with \Delta x=1. + +

+ Plot of bounded region rotated about x-axis. + + Bounded region rotated about x-axis. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-0.5,-3.5) grid (5.5,3.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,-3) -- (0,3) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1, 2, ..., 4} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {-3, -2,...,3} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{(\x^(1/2) }); + + \draw[blue, dashed] plot [ + samples=100, + domain=0:4, + ] (\x,{(-1*\x^(1/2) }); + + + \fill [blue, domain=0:4, variable=\x, opacity=0.2] + (0,0)--(1,1)--(2,1.4142)--(3,1.732)--(4,2) + --(4,-2)--(3,-1.732)--(2,-1.4142)--(1,-1)--(0,0); + + \draw[dashed] (0,0)--(1,1)--(2,1.4142)--(3,1.732)--(4,2); + \draw[dashed] (0,0)--(1,-1)--(2,-1.4142)--(3,-1.732)--(4,-2); + + \draw[fill] (0,0) circle (2pt); + \draw[fill] (1,1) circle (2pt); + \draw[fill] (2,1.4142) circle (2pt); + \draw[fill] (3,1.732) circle (2pt); + \draw[fill] (4,2) circle (2pt); + + \end{tikzpicture} + + +
+ + + + \begin{array}{|c|c|c|c|c|c|} + \hline + x_i & \Delta x & r_i & R_i & l & \text{Estimated Surface Area}\\ + \hline + x_1=0 & 1 & 0 & 1 & \sqrt{1^2+1^2} &\\ + \hline + x_2=1 & 1& 1 & \sqrt{2} & \sqrt{1^2+(\sqrt{2}-1)^2} & \\ + \hline + x_3=2 & 1& \sqrt{2} & \sqrt{3} & \\ + \hline + x_4=3 & 1 & 3 & 2 & \\ + \hline + + \end{array} + + + +

+ + + + +

+ Suppose we wanted to estimate the surface area with n frustums. + +

+ Plot of bounded region rotated about x-axis. + + Bounded region rotated about x-axis. + + \begin{tikzpicture} + %grid + \draw[step=1cm,gray!50,very thin] (-0.5,-3.5) grid (5.5,3.5); + %axes + \draw[very thick,->] (0,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw[very thick,->] (0,-3) -- (0,3) node[anchor=south east] {\tiny$y$}; + \foreach \x in {1, 2, ..., 4} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \foreach \y in {-3, -2,...,3} + \draw (1pt,\y cm) -- (-1pt,\y cm) node[anchor=east] {\tiny $\mathbf{\y}$}; + %function + \draw[blue] plot [ + samples=100, + domain=0:4, + ] (\x,{(\x^(1/2) }); + + \draw[blue, dashed] plot [ + samples=100, + domain=0:4, + ] (\x,{(-1*\x^(1/2) }); + + + \fill [blue, domain=0:4, variable=\x, opacity=0.2] + (1.8, 1.34164)--(2.1, 1.449)--(2.1,-1.449)--(1.8,-1.34164)--(1.8, 1.34164); + + \draw[dashed, red] (1.8, 1.34164)--(2.1, 1.449)--(2.1,1.134164)--(1.8, 1.34164); + \draw[fill, red] (1.8, 1.34164) circle (2pt); + \draw[fill, red] (2.1, 1.449) circle (2pt); + + \draw [decorate, decoration={brace,amplitude=10pt,mirror},yshift=-0.4pt, red](1.8, 1.34164)--(2.1, 1.34164) node[red,midway,yshift=-0.5cm] {\footnotesize $\Delta x$}; + + + \end{tikzpicture} + + +
+ + + Let f(x)=\sqrt{x}. Which of the following expressions represents the surface area generated bo rotating the line segment from (x_0, f(x_0)) to (\Delta x, f(x_0+\Delta x)) about the x-axis? +
    +
  1. \displaystyle \pi \left(\frac{f(x_0)+f(x_0+\Delta x)}{2}\right)^2\sqrt{(\Delta x)^2+(f(x_0+\Delta x)-f(x_0))^2}.
    2. +
  2. \displaystyle 2\pi\frac{f(x_0)+f(x_0+\Delta x)}{2}\sqrt{(\Delta x)^2+(f(x_0+\Delta x)-f(x_0))^2}.
    3. +
  3. \displaystyle 2\pi\frac{f(x_0)+f(x_0+\Delta x)}{2}\Delta x.
    4. +
+ + + +

+ + + + +

+ Which of the following Riemann sums best estimates the surface area of the solid generated by rotating y=\sqrt{x} over [0,4] about the x-axis ? Let f(x)=\sqrt{x}. +

    +
  1. \displaystyle \sum \pi \left(\frac{f(x_i)+f(x_i+\Delta x)}{2}\right)^2\sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2}.
    2. +
  2. \displaystyle \sum 2\pi\frac{f(x_i)+f(x_i+\Delta x)}{2}\sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2}.
    3. +
  3. \displaystyle \sum 2\pi\frac{f(x_i)+f(x_0+\Delta x)}{2}\Delta x.
    4. +
+

+ + + + + + + + + +

+ Recall from that + + \lim_{\Delta x\to 0}\sqrt{(\Delta x)^2+(f(x_i+\Delta x)-f(x_i))^2} & = \lim_{\Delta x\to 0} \sqrt{(\Delta x)^2\left(1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2\right)} + &= \lim_{\Delta x\to 0} \sqrt{1+\left(\frac{f(x_i+\Delta x)-f(x_i)}{\Delta x} \right)^2}\Delta x + &=\sqrt{1+(f'(x))^2}dx, + + and that + + \lim_{\Delta x\to 0} \frac{f(x_i)+f(x_i+\Delta x)}{2}=f(x_i). + +

+ +

+ Thus given a function f(x)\geq 0 over [a,b], the surface area of the solid generated by rotating this function about the x-axis is SA=\int_a^b 2\pi f(x)\sqrt{1+(f'(x))^2}dx. +

+ + + + + +

+ Consider again the solid generated by rotating y=\sqrt{x} over [0,4] about the x-axis. +

+ + +

+ Find an integral which computes the surface area of this solid. +

+ + +

+ If we instead rotate y=\sqrt{x} over [0,4] about the y-axis, what is an integral which computes the surface area for this solid? +

+ + + + + + +

+ Consider again the function f(x)=\ln(x)+1 over [1,5]. +

+ + +

+ Find an integral which computes the surface area of the solid generated by rotating the above curve about the x-axis. +

+ + +

+ Find an integral which computes the surface area of the solid generated by rotating the above curve about the y-axis. +

+ + + + + + Videos +
+ + Video: Compute surface areas of surfaces of revolution +
+ + + + + + + +
+ diff --git a/calculus/source/06-AI/04comments.ptx b/calculus/source/06-AI/04comments.ptx new file mode 100644 index 00000000..58b71ef6 --- /dev/null +++ b/calculus/source/06-AI/04comments.ptx @@ -0,0 +1,759 @@ + + \ No newline at end of file diff --git a/calculus/source/06-AI/05.ptx b/calculus/source/06-AI/05.ptx new file mode 100644 index 00000000..cfac546f --- /dev/null +++ b/calculus/source/06-AI/05.ptx @@ -0,0 +1,919 @@ + + +
+ Density, Mass, and Center of Mass (AI5) + + + + + Activities + + +

+ Consider a rectangular prism with a 10 meters \times 10 meters square base and height 20 meters. Suppose the density of the material in the prism increases with height, following the function \delta(h)=10+h kg/m^3, where h is the height in meters. +

+ + +

If one were to cut this prism, parallel to the base, into 4 pieces with height 5 meters, what would the volume of each piece be?

+ + +

Consider the piece sitting on top of the slice made at height h=5. Using a density of \delta(5)=15 kg/m^3, and the volume you found in (a), estimate the mass of this piece.

+
    +
  1. 500\cdot 5=2500 kg
    2. +
  2. 500\cdot 15=7500 kg
    3. +
  3. 500\cdot 15\cdot 5=37500 kg
    4. +
+ + +

+ Is this estimate the actual mass of this piece? +

+ + + + + +

+ Consider all 4 slices from . +

+ 10\times 10\times 20 prism sliced into 4 pieces. + + 10\times 10\times 20 prism sliced into 4 pieces. + + \begin{tikzpicture} + \draw [fill=blue, opacity=0.2] (0,0) rectangle (2,1); + \draw [fill=blue, opacity=0.2] (0,1) rectangle (2,2); + \draw [fill=blue, opacity=0.2] (0,2) rectangle (2,3); + \draw [fill=blue, opacity=0.2] (0,3) rectangle (2,4); + \draw (0,0) -- (2,0) -- (2,4) -- (0,4)--(0,0); + + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (2,1) -- (2,2) node [black,midway,xshift=1.2cm] {\footnotesize + $\Delta h= 5$ m}; + \draw (0,0)--node[left]{\tiny $h_1=0$ m} (0,0); + \draw (0,1)--node[left]{\tiny $h_2=5$ m} (0,1); + \draw (0,2)--node[left]{\tiny $h_3=10$ m} (0,2); + \draw (0,3)--node[left]{\tiny $h_4=15$ m} (0,3); + + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-1.25,0) -- (-1.25,4) node [black,midway,xshift=-1cm] {\footnotesize + $20$ m}; + \end{tikzpicture} + + +
+

+ + +

Fill out the following table. + + \begin{array}{|c|c|c|c|} + \hline + h_i & \delta(h_i) & \text{Volume} & \text{Estimated Mass}\\ + \hline + h_4=15\ \text{m} & \delta(15)=25\ \text{kg/m}^3 & 500\ \text{m}^3 & \\ + \hline + h_3=10\ \text{m} & \delta(10)=20\ \text{kg/m}^3 & 500\ \text{m}^3 & \\ + \hline + h_2=5\ \text{m} & \delta(5)=15\ \text{kg/m}^3 & 500\ \text{m}^3 & 7500\ \text{kg}\\ + \hline + h_1=0\ \text{m} & \delta(0)=10\ \text{kg/m}^3 & 500\ \text{m}^3 & \\ + \hline + \end{array} + +

+ + +

+ What is the estimated mass of the rectangular prism? +

+ + + + + + +

+ Suppose instead that we sliced the prism from into 5 pices of height 4 meters. +

+ 10\times 10\times 20 prism sliced into 5 pieces. + + 10\times 10\times 20 prism sliced into 5 pieces. + + \begin{tikzpicture} + \draw [fill=blue, opacity=0.2] (0,0) rectangle (2,0.8); + \draw [fill=blue, opacity=0.2] (0,0.8) rectangle (2,1.6); + \draw [fill=blue, opacity=0.2] (0,1.6) rectangle (2,2.4); + \draw [fill=blue, opacity=0.2] (0,2.4) rectangle (2,3.2); + \draw [fill=blue, opacity=0.2] (0,3.2) rectangle (2,4); + \draw (0,0) -- (2,0) -- (2,4) -- (0,4)--(0,0); + + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (2,0.8) -- (2,1.6) node [black,midway,xshift=1.2cm] {\footnotesize + $\Delta h= 4$ m}; + \draw (0,0)--node[left]{\tiny $h_1=0$ m} (0,0); + \draw (0,0.8)--node[left]{\tiny $h_2=4$ m} (0,0.8); + \draw (0,1.6)--node[left]{\tiny $h_3=8$ m} (0,1.6); + \draw (0,2.4)--node[left]{\tiny $h_4=12$ m} (0,2.4); + \draw (0,3.2)--node[left]{\tiny $h_5=16$ m} (0,3.2); + + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-1.25,0) -- (-1.25,4) node [black,midway,xshift=-1cm] {\footnotesize + $20$ m}; + \end{tikzpicture} + + +
+

+ + +

Fill out the following table. + + \begin{array}{|c|c|c|c|} + \hline + h_i & \delta(h_i) & \text{Volume} & \text{Estimated Mass}\\ + \hline + h_5=16\ \text{m} & \delta(16)=26\ \text{kg/m}^3 & 400\ \text{m}^3 & \\ + \hline + h_4=12\ \text{m} & \delta(12)=22\ \text{kg/m}^3 & 400\ \text{m}^3 & \\ + \hline + h_3=8\ \text{m} & \delta(8)=18\ \text{kg/m}^3 & 400\ \text{m}^3 & \\ + \hline + h_2=4\ \text{m} & \delta(4)=14\ \text{kg/m}^3 & 400\ \text{m}^3 & \phantom{7500\ \text{kg}}\\ + \hline + h_1=0\ \text{m} & \delta(0)=10\ \text{kg/m}^3 & 400\ \text{m}^3 & \\ + \hline + \end{array} + + + +

+ + +

+ What is the estimated mass of the rectangular prism? +

+ + + + + + +

+ Which of the estimates computed in and is a better estimate of the mass of the prism? +

    +
  1. , 4 pieces is a better estimate.
    2. +
  2. , 5 pieces is a better estimate.
    3. +
+

+ + + + + +

+ Suppose now that we slice the prism from into slices of height \Delta h meters. + +

+ 10\times 10\times 20 prism sliced into many pieces. + + 10\times 10\times 20 prism sliced into many pieces. + + \begin{tikzpicture} + \draw [fill=blue, opacity=0.2] (0,1.9) rectangle (2,2.1); + \draw (0,0)--(2,0) -- (2,4)--(0,4)--(0,0); + \node at (1, 0.95){$\vdots$}; + \node at (1, 3.05){$\vdots$}; + + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (2,1.9) -- (2,2.1) node [black,midway,xshift=0.7cm] {\footnotesize + $\Delta h$}; + \draw (0,1.9)--node[left]{\tiny $h_i$} (0,1.9); + + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-.5,0) -- (-.5,4) node [black,midway,xshift=-1cm] {\footnotesize + $20$ m}; + \end{tikzpicture} + + +
+

+ + +

+ Consider the piece sitting atop the slice made at height h_i. Using \delta(h_i)=10+h_i as the estimate for the density of this piece, what is the mass of this piece? +

    +
  1. (10+h)100\cdot h_i
    2. +
  2. (10+\Delta h)100\cdot h_i
    3. +
  3. (10+h_i)100\cdot \Delta h
    4. +
  4. (10+h_i)100\cdot h
    5. +
+

+ + + + + + +

+ Consider a cylindrical cone with a base radius of 15 inches and a height of 60 inches. Suppose the density of the cone is \delta(h)= 15+\sqrt{h} oz/in^3. +

+ 15\times 60 cylindrical cone sliced into two pieces. + + 15\times 60 cylindrical cone sliced into two pieces. + + \begin{tikzpicture} + \draw [fill=blue, opacity=0.2] (-2,0) rectangle (2,4); + \draw[blue] (-2,0) -- (2,0) -- (2,4) -- (-2, 4) -- (-2,0); + \draw [fill=blue, opacity=0.2] (-1,4) rectangle (1,8); + \draw[blue] (-1,4) -- (1,4) -- (1,8) -- (-1, 8) -- (-1,4); + \draw (-2,0) -- (2,0) -- (0,8) -- (-2,0); + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (2,0) -- (2,4) node [black,midway,xshift=1.4cm] {\footnotesize + $\Delta h=30$\ in}; + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (1,4) -- (1,8) node [black,midway,xshift=1.4cm] {\footnotesize + $\Delta h=30$\ in}; + \draw[dashed] (0,8) -- (0,0); + \draw[dashed] (0,0) --node[below]{\tiny $r_1=15$\ in} (2,0); + \draw[dashed] (0,4) --node[below]{\tiny $r_2$} (1,4); + \draw[fill] (0,0) circle (2pt); + \draw (0,0) --node[below]{\tiny $B$} (0,0); + \draw[fill] (2,0) circle (2pt); + \draw (2,0) --node[below right]{\tiny $C$} (2,0); + \draw[fill] (0,4) circle (2pt); + \draw (0,4) --node[below left]{\tiny $B'$} (0,4); + \draw[fill] (1,4) circle (2pt); + \draw (1,4) --node[below right]{\tiny $C'$} (1,4); + \draw[fill] (0,8) circle (2pt); + \draw (0,8) --node[below right]{\tiny $C'$} (0,8); + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-2.25,0) -- (-2.25,8) node [black,midway,xshift=-1cm] {\footnotesize + $60$ in}; + \end{tikzpicture} + + +
+ +

+ + +

+ Let r_2 be the radius of the circular cross section of the cone, made at height 30 inches. Recall that \Delta ABC, \Delta AB'C' are similar triangles, what is r_2? +

    +
  1. 15 inches.
    2. +
  2. 7.5 inches.
    3. +
  3. 30 inches.
    4. +
  4. 60 inches.
    5. +
+

+ + + +

+ What is the volume of a cylinder with radius r_1=15 inches and height 30 inches? +

+ + + +

+ What is the volume of a cylinder with radius r_2 inches and height 30 inches? +

+ + + + + + +

+ Suppose that we estimate the mass of the cone from with 2 cylinders of height 30 inches. +

+ 15\times 60 cylindrical cone sliced into two pieces. + + 15\times 60 cylindrical cone sliced into two pieces. + + \begin{tikzpicture} + \draw [fill=blue, opacity=0.2] (-2,0) rectangle (2,4); + \draw[blue] (-2,0) -- (2,0) -- (2,4) -- (-2, 4) -- (-2,0); + \draw [fill=blue, opacity=0.2] (-1,4) rectangle (1,8); + \draw[blue] (-1,4) -- (1,4) -- (1,8) -- (-1, 8) -- (-1,4); + \draw (-2,0) -- (2,0) -- (0,8) -- (-2,0); + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (2,0) -- (2,4) node [black,midway,xshift=1.4cm] {\footnotesize + $\Delta h=30$\ in}; + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (1,4) -- (1,8) node [black,midway,xshift=1.4cm] {\footnotesize + $\Delta h=30$\ in}; + \draw[dashed] (0,8) -- (0,0); + \draw[dashed] (0,0) --node[below]{\tiny $r_1=15$\ in} (2,0); + \draw[dashed] (0,4) --node[below]{\tiny $r_2$} (1,4); + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-3.25,0) -- (-3.25,8) node [black,midway,xshift=-1cm] {\footnotesize + $60$ in}; + \draw (-2,0) -- node[left]{\tiny $h_1=0$\ in} (-2,0); + \draw (-2,4) -- node[left]{\tiny $h_2=30$\ in} (-2,4); + \end{tikzpicture} + + +
+

+ + +

Fill out the following table. + + \begin{array}{|c|c|c|c|} + \hline + h_i & \delta(h_i) & \text{Volume} & \text{Estimated Mass}\\ + \hline + h_2=30\ \text{m} & \delta(30)=15+\sqrt{30}\ \text{oz/in}^3 & \pi(7.5)^2\cdot30\ \text{in}^3 & \phantom{6500\ \text{kg}}\\ + \hline + h_1=0\ \text{m} & \delta(0)=15\ \text{oz/in}^3 & \pi(15)^2\cdot30\ \text{in}^3 & \\ + \hline + \end{array} + + + +

+ + +

+ What is the estimated mass of the cone? +

+ + + + + + +

+ Suppose that we estimate the mass of the cone from with 3 cylinders of height 20 inches. +

+ 15\times 60 cylindrical cone sliced into three pieces. + + 15\times 60 cylindrical cone sliced into three pieces. + + \begin{tikzpicture} + \draw [fill=blue, opacity=0.2] (-2,0) rectangle (2,8/3); + \draw[blue] (-2,0) -- (2,0) -- (2,8/3) -- (-2, 8/3) -- (-2,0); + \draw [fill=blue, opacity=0.2] (-4/3,8/3) rectangle (4/3,16/3); + \draw[blue] (-4/3,8/3) -- (4/3,8/3) -- (4/3,16/3) -- (-4/3, 16/3) -- (-4/3,8/3); + \draw [fill=blue, opacity=0.2] (-2/3,16/3) rectangle (2/3,8); + \draw[blue] (-2/3,16/3) -- (2/3,16/3) -- (2/3,8) -- (-2/3, 8) -- (-2/3,16/3); + \draw (-2,0) -- (2,0) -- (0,8) -- (-2,0); + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (2,0) -- (2,8/3) node [black,midway,xshift=1.4cm] {\footnotesize + $\Delta h=20$\ in}; + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (4/3,8/3) -- (4/3,16/3) node [black,midway,xshift=1.4cm] {\footnotesize + $\Delta h=20$\ in}; + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (2/3,16/3) -- (2/3,8) node [black,midway,xshift=1.4cm] {\footnotesize + $\Delta h=20$\ in}; + \draw[dashed] (0,8) -- (0,0); + \draw[dashed] (0,0) --node[below]{\tiny $r_1=15$\ in} (2,0); + \draw[dashed] (0,8/3) --node[below]{\tiny $r_2=10$\ in} (4/3,8/3); + \draw[dashed] (0,16/3) --node[below]{\tiny $r_3=5$\ in} (2/3,16/3); + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-3.25,0) -- (-3.25,8) node [black,midway,xshift=-1cm] {\footnotesize + $60$ in}; + \draw (-2,0) -- node[left]{\tiny $h_1=0$\ in} (-2,0); + \draw (-2,8/3) -- node[left]{\tiny $h_2=20$\ in} (-2,8/3); + \draw (-4/3,16/3) -- node[left]{\tiny $h_3=40$\ in} (-4/3,16/3); + \end{tikzpicture} + + +
+

+ + +

Fill out the following table. + + \begin{array}{|c|c|c|c|} + \hline + h_i & \delta(h_i) & \text{Volume} & \text{Estimated Mass}\\ + \hline + h_2=40\ \text{m} & \delta(40)=15+\sqrt{40}\ \text{oz/in}^3 & \pi(5)^2\cdot20\ \text{in}^3 & \phantom{6500\ \text{kg}}\\ + \hline + h_2=20\ \text{m} & \delta(20)=15+\sqrt{20}\ \text{oz/in}^3 & \pi(10)^2\cdot20\ \text{in}^3 & \phantom{6500\ \text{kg}}\\ + \hline + h_1=0\ \text{m} & \delta(0)=15\ \text{oz/in}^3 & \pi(15)^2\cdot20\ \text{in}^3 & \\ + \hline + \end{array} + +

+ + +

+ What is the estimated mass of the cone? +

+ + + + + + +

+ Suppose that we estimate the mass of the cone from with cylinders of height \Delta h. +

+ 15\times 60 cylindrical cone sliced into many pieces. + + 15\times 60 cylindrical cone sliced into many pieces. + + \begin{tikzpicture} + \draw [fill=blue, opacity=0.2] (-1.025,3.9) rectangle (1.025, 4.1); + \draw[blue] (-1.025,3.9) -- (1.025,3.9) -- (1.025,4.1) -- (-1.025, 4.1) -- (-1.025,3.9); + + \draw (-2,0) -- (2,0) -- (0,8) -- (-2,0); + \draw [decorate,decoration={brace,amplitude=10pt,mirror,raise=4pt},yshift=0pt] + (1.025,3.9) -- (1.025,4.1) node [black,midway,xshift=1.2cm] {\footnotesize + $\Delta h$\ in}; + \node at (0, 2){$\vdots$}; + \node at (0, 6){$\vdots$}; + \draw[dashed] (0,3.9) --node[below]{\tiny $r_i$} (1.025,3.9); + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-3.25,0) -- (-3.25,8) node [black,midway,xshift=-1cm] {\footnotesize + $60$ in}; + \draw (-1.025,3.9) -- node[left]{\tiny $h_i$} (-1.025,3.9); + + \end{tikzpicture} + + +
+

+ + + +

+ Consider the piece sitting atop the slice made at height h_i. Using \delta(h_i)=15+\sqrt{h_i} as the estimate for the density of this cylinder, what is the mass of this cylinder? +

    +
  1. (15+\sqrt{h})\pi r_i^2\cdot \Delta h
    2. +
  2. (15+\sqrt{h_i})\pi r_i^2\cdot \Delta h
    3. +
  3. (15+\Delta h)\pi r_i^2\cdot \Delta h_i
    4. +
  4. (15+\sqrt{h_i})\pi r^2\cdot \Delta h
    5. +
+

+ + + + + +

+ Consider a solid where the cross section of the solid at x=x_i has area A(x_i), and the density when x=x_i is \delta(x_i). +

+ Solid approximated with prisms of width \Delta x. + + Solid approximated with prisms of width \Delta x. + + \begin{tikzpicture} + \draw[thick,->] (-1,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw plot [ + samples=100, + domain=0:4, + ] (\x,{(\x-1)^3-3*(\x-1)^2+5 }); + \draw plot [ + samples=100, + domain=0:4, + ] (\x,{\x-5 }); + \draw (0,-5)--(0,1); + \draw (4,-1) -- (4,5); + + \draw[blue] (2,-3)--(2,3)--(2.2,3)--(2.2,-3)--(2,-3); + \draw [fill=blue, opacity=0.2] (2,-3) rectangle (2.2,3); + + \draw(2,-.1)--node[above left]{\tiny$x_i$}(2,.1); + \draw [decorate, decoration={brace,amplitude=10pt,mirror},yshift=-0.4pt](2,-3) -- (2.2,-3) node[black,midway,yshift=-0.5cm] {\footnotesize $\Delta x$}; + + \end{tikzpicture} + + +
+

+ + + +

If we used prisms of width \Delta x to approximate this solid, what is the mass of the slice associated with x_i? +

    +
  1. A(x)\delta(x)\Delta x
    2. +
  2. \pi A(x)^2\delta(x_i)\Delta x
    3. +
  3. A(x_i)\delta(x_i)\Delta x
    4. +
  4. A(x_i)\delta(x_i)\Delta x_i
    5. +
+

+ + + + + +

+ Consider a solid where the cross section of the solid at x=x_i has area A(x_i), and the density when x=x_i is \delta(x_i). Suppose the interval [a,b] represents the x values of this solid. If one slices the solid into n pieces of width \Delta x=\frac{b-a}{n}, then one can approximate the mass of the solid by \sum_{i=1}^n \delta(x_i)A(x_i)\Delta x. +

+ Solid approximated with prisms of width \Delta x. + + Solid approximated with prisms of width \Delta x. + + \begin{tikzpicture} + \draw[thick,->] (-1,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw plot [ + samples=100, + domain=0:4, + ] (\x,{(\x)^(0.5) }); + \draw plot [ + samples=100, + domain=0:4, + ] (\x,{-1*(\x)^(0.5) }); + \draw (4,-2) -- (4,2); + + + \foreach \i in {0,...,20}{ + \draw (0.2*\i, {-1*sqrt(0.2*\i)})--(0.2*\i+0.2, {-1*sqrt(0.2*\i)})--(0.2*\i+0.2, {sqrt(0.2*\i)})--(0.2*\i, {sqrt(0.2*\i)})--(0.2*\i, {-1*sqrt(0.2*\i)}); + \draw [fill=blue, opacity=0.2] (0.2*\i,{-1*sqrt(0.2*\i)}) rectangle (0.2*\i+0.2,{sqrt(0.2*\i)}); + }; + + \draw [decorate, decoration={brace,amplitude=10pt,mirror},yshift=-0.4pt](2,-1.414) -- (2.2,-1.414) node[black,midway,yshift=-0.5cm] {\footnotesize $\Delta x$}; + + \end{tikzpicture} + + +
+

+

+ We can then find actual mass by taking the limit as n\to\infty: \lim_{n\to\infty} \left(\sum_{i=1}^n \delta(x_i)A(x_i)\Delta x\right)=\int_a^b \delta(x)A(x) dx. +

+ Solid mass. + + Solid mass. + + \begin{tikzpicture} + \draw[thick,->] (-1,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw plot [ + samples=100, + domain=0:4, + ] (\x,{(\x)^(0.5) }); + \draw plot [ + samples=100, + domain=0:4, + ] (\x,{-1*(\x)^(0.5) }); + \draw (4,-2) -- (4,2); + + + \fill [blue, domain=0:4, variable=\x, opacity=0.2] + (0, 0) + -- plot ({\x}, {sqrt(\x)}) + --(4,-2) + --plot ({\x}, {-1*sqrt(\x)}) + -- cycle; + + \end{tikzpicture} + + +
+

+ + + + + +

+ Consider that for the prism from , a cross section of height h is A(h)=10^2=100 m^2. Also recall that the density of the prism is \delta(h)=10+h kg/m^3, where h is the height in meters. +

+

+ Use to find the mass of the prism. +

+ + + + + +

+ Consider that for the cone from , a cross section of height h is A(h)=\pi r^2 in^2, where r is the radius of the circular cross-section at height h inches. Also recall that the density of the cone is \delta(h)=15+\sqrt{h} oz/in^3, where h is the height in inches. +

+ + +

+ When the height is h inches, what is r? +

+ +

+ Use similar triangles: +

+ The right triangles in this figure are similar. + + a 15 by 60 triangle, with a similar r by 60-h triangle. + + \begin{tikzpicture} + \draw (-2,0)--(2,0)--(0,8)--(-2,0); + \draw[dashed] (-1.5,2)--(1.5,2); + \draw[dashed] (0,0)--(0,8); + + \draw [decorate, decoration={brace,amplitude=10pt,mirror},yshift=-0.4pt](0,0) -- (2,0) node[black,midway,yshift=-0.5cm] {\footnotesize $15$\ in}; + + \draw [decorate, decoration={brace,amplitude=10pt,mirror},yshift=-0.4pt](0,2) -- (1.5,2) node[black,midway,yshift=-0.5cm] {\footnotesize $r$}; + + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-2,0) -- (-2,8) node [black,midway,xshift=-1cm] {\footnotesize + $60$ in}; + + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (0,0) -- (0,2) node [black,midway,xshift=-1cm] {\footnotesize + $h$}; + \draw [decorate,decoration={brace, mirror, amplitude=10pt,raise=4pt},yshift=0pt] + (1.5,2) -- (1.5,8) node [black,midway,xshift=1cm] {\footnotesize + $60-h$}; + + \end{tikzpicture} + + +
+

+ + + +

+ Find A(h) as a function of h using this information. +

+ + +

+ Use to find the mass of the cone. +

+ + + + + + Consider a pyramid with a 8\times 8 ft square base and a height of 16 feet. Suppose the density of the pyramid is \delta(h)=10+\cos(\pi h) lb/ft^3 where h is the height in feet. + + + + +

+ When the height is h feet, what is the area of the square cross section at that height, A(h)? +

+ +

+ Use similar triangles: +

+ The triangles in this figure are similar. + + a 8 by 60 triangle, with a similar r by 16-h triangle. + + \begin{tikzpicture} + \draw (-2,0)--(2,0)--(0,8)--(-2,0); + \draw[dashed] (-1.5,2)--(1.5,2); + + \draw [decorate, decoration={brace,amplitude=10pt,mirror},yshift=-0.4pt](-2,0) -- (2,0) node[black,midway,yshift=-0.5cm] {\footnotesize $8$\ ft}; + + \draw [decorate, decoration={brace,amplitude=10pt,mirror},yshift=-0.4pt](-1.5,2) -- (1.5,2) node[black,midway,yshift=-0.5cm] {\footnotesize $b$}; + + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-2,0) -- (-2,8) node [black,midway,xshift=-1cm] {\footnotesize + $16$ ft}; + + \draw [decorate,decoration={brace,mirror, amplitude=10pt,raise=4pt},yshift=0pt] + (2,0) -- (2,2) node [black,midway,xshift=-1cm] {\footnotesize + $h$}; + \draw [decorate,decoration={brace, mirror, amplitude=10pt,raise=4pt},yshift=0pt] + (1.5,2) -- (1.5,8) node [black,midway,xshift=1cm] {\footnotesize + $16-h$}; + + \end{tikzpicture} + + +
+

+ + + + +

+ Use to find the mass of the pyramid. +

+ + + + + + +

+ Consider a board sitting atop the x-axis with six 1\times 1 blocks each weighing 1 kg placed upon it in the following way: two blocks are atop the 1, three blocks are atop the 2, and one block is atop the 6. + +

+ Six 1 kg blocks atop the x-axis. + + Six 1 by 1 blocks on the x-axis, each weighing a kilogram. Two blocks atop the 1, three blocks atop the 2, and one block atop the 6. + + \begin{tikzpicture} + \draw[very thick,->] (0,0) -- (8.5,0) node[anchor=north west] {\tiny$x$}; + \foreach \x in {1,2,3,4,5,6,7,8} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \draw (0.5,0)--(1.5,0)--(1.5,1)--(0.5,1)--(0.5,0); + \draw [fill=blue, opacity=0.2] (0.5,0) rectangle (1.5,1); + \draw (0.5,1)--(1.5,1)--(1.5,2)--(0.5,2)--(0.5,1); + \draw [fill=blue, opacity=0.2] (0.5,1) rectangle (1.5,2); + + \draw (1.5,0)--(2.5,0)--(2.5,1)--(1.5,1)--(1.5,0); + \draw [fill=blue, opacity=0.2] (1.5,0) rectangle (2.5,1); + \draw (1.5,1)--(2.5,1)--(2.5,2)--(1.5,2)--(1.5,1); + \draw [fill=blue, opacity=0.2] (1.5,1) rectangle (2.5,2); + \draw (1.5,2)--(2.5,2)--(2.5,3)--(1.5,3)--(1.5,2); + \draw [fill=blue, opacity=0.2] (1.5,2) rectangle (2.5,3); + + \draw (5.5,0)--(6.5,0)--(6.5,1)--(5.5,1)--(5.5,0); + \draw [fill=blue, opacity=0.2] (5.5,0) rectangle (6.5,1); + + \end{tikzpicture} + + +
+ +

+

+ Which of the following describes the x-value of the center of gravity of the board with the blocks? +

    +
  1. \displaystyle\frac{1+6}{2}=3.5.
    2. +
  2. \displaystyle\frac{1+2+6}{3}=3.
    3. +
  3. \displaystyle\frac{2\cdot1+3\cdot2+1\cdot6}{6}\approx 2.3333.
    4. +
+

+ + + + + +

+ Consider a board sitting atop the x-axis with six 1\times 1 blocks each weighing 1 kg placed upon it in the following way: two blocks are atop the 1, three blocks are atop the 2, and one block is atop the 8. + +

+ Six 1 kg blocks atop the x-axis. + + Six 1 by 1 blocks on the x-axis, each weighing a kilogram. Two blocks atop the 1, three blocks atop the 2, and one block atop the 8. + + \begin{tikzpicture} + \draw[very thick,->] (0,0) -- (8.5,0) node[anchor=north west] {\tiny$x$}; + \foreach \x in {1,2,3,4,5,6,7,8} + \draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {\tiny $\mathbf{\x}$}; + \draw (0.5,0)--(1.5,0)--(1.5,1)--(0.5,1)--(0.5,0); + \draw [fill=blue, opacity=0.2] (0.5,0) rectangle (1.5,1); + \draw (0.5,1)--(1.5,1)--(1.5,2)--(0.5,2)--(0.5,1); + \draw [fill=blue, opacity=0.2] (0.5,1) rectangle (1.5,2); + + \draw (1.5,0)--(2.5,0)--(2.5,1)--(1.5,1)--(1.5,0); + \draw [fill=blue, opacity=0.2] (1.5,0) rectangle (2.5,1); + \draw (1.5,1)--(2.5,1)--(2.5,2)--(1.5,2)--(1.5,1); + \draw [fill=blue, opacity=0.2] (1.5,1) rectangle (2.5,2); + \draw (1.5,2)--(2.5,2)--(2.5,3)--(1.5,3)--(1.5,2); + \draw [fill=blue, opacity=0.2] (1.5,2) rectangle (2.5,3); + + \draw (7.5,0)--(8.5,0)--(8.5,1)--(7.5,1)--(7.5,0); + \draw [fill=blue, opacity=0.2] (7.5,0) rectangle (8.5,1); + + \end{tikzpicture} + + +
+ +

+

+ Find the x-value of the center of gravity of the board with the blocks. +

+ + + + + + + +

+ Consider a solid where the cross section of the solid at x=x_i has area A(x_i), and the density when x=x_i is \delta(x_i). Suppose the interval [a,b] represents the x values of this solid. Since each slice has approximate mass \delta(x_i)A(x_i)\delta(x_i), we can approximate the center of mass by taking the weighted average of the x_i-values weighted by the associated mass: \frac{\sum_{i=1}^n x_i\delta(x_i)A(x_i)\Delta x}{\sum_{i=1}^n \delta(x_i)A(x_i)\Delta x}. +

+ Solid approximated with prisms of width \Delta x. + + Solid approximated with prisms of width \Delta x. + + \begin{tikzpicture} + \draw[thick,->] (-1,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw plot [ + samples=100, + domain=0:4, + ] (\x,{(\x)^(0.5) }); + \draw plot [ + samples=100, + domain=0:4, + ] (\x,{-1*(\x)^(0.5) }); + \draw (4,-2) -- (4,2); + + + \foreach \i in {0,...,20}{ + \draw (0.2*\i, {-1*sqrt(0.2*\i)})--(0.2*\i+0.2, {-1*sqrt(0.2*\i)})--(0.2*\i+0.2, {sqrt(0.2*\i)})--(0.2*\i, {sqrt(0.2*\i)})--(0.2*\i, {-1*sqrt(0.2*\i)}); + \draw [fill=blue, opacity=0.2] (0.2*\i,{-1*sqrt(0.2*\i)}) rectangle (0.2*\i+0.2,{sqrt(0.2*\i)}); + }; + + \draw [decorate, decoration={brace,amplitude=10pt,mirror},yshift=-0.4pt](2,-1.414) -- (2.2,-1.414) node[black,midway,yshift=-0.5cm] {\footnotesize $\Delta x$}; + + \end{tikzpicture} + + +
+

+

+ We can then find actual center of mass by taking the limit as n\to\infty: \lim_{n\to\infty} \left(\frac{\sum_{i=1}^n x_i\delta(x_i)A(x_i)\Delta x}{\sum_{i=1}^n \delta(x_i)A(x_i)\Delta x}\right)=\frac{\int_a^b x\delta(x)A(x)dx}{\int_a^b \delta(x)A(x)dx}=\frac{\int_a^b x\delta(x)A(x)dx}{\text{The Total Mass}}. +

+ Solid mass. + + Solid mass. + + \begin{tikzpicture} + \draw[thick,->] (-1,0) -- (5,0) node[anchor=north west] {\tiny$x$}; + \draw plot [ + samples=100, + domain=0:4, + ] (\x,{(\x)^(0.5) }); + \draw plot [ + samples=100, + domain=0:4, + ] (\x,{-1*(\x)^(0.5) }); + \draw (4,-2) -- (4,2); + + + \fill [blue, domain=0:4, variable=\x, opacity=0.2] + (0, 0) + -- plot ({\x}, {sqrt(\x)}) + --(4,-2) + --plot ({\x}, {-1*sqrt(\x)}) + -- cycle; + + \end{tikzpicture} + + +
+

+ + + + + +

+ Consider that for the prism from , a cross section of height h is A(h)=10^2=100 m^2. Also recall that the density of the prism is \delta(h)=10+h kg/m^3, where h is the height in meters, and that we found the total mass to be 40000 kg. +

+

+ Use to find the height where the center of mass occurs. +

+ + + + + + +

+ Consider that for the prism from , a cross section of height h is A(h)=\pi\cdot \left( \frac{60-h}{4}\right)^2 in^2. Also recall that the density of the cone is \delta(h)=15+\sqrt{h} oz/in^3, where h is the height in inches, and that we found the total mass to be about 142492.6 oz. +

+

+ Use to find the height where the center of mass occurs. +

+ + + + + + +

+ Consider that for the pyramid from , a cross section of height h is A(h)=\pi\cdot \left( \frac{16-h}{2}\right)^2 ft^2. Also recall that the density of the pyramid is \delta(h)=10+\cos{\pi h} lb/feet^3, where h is the height in feet, and that we found the total mass to be about 3414.14.6 lbs. +

+

+ Use to find the height where the center of mass occurs. +

+ + + + + + + + + + + + Videos +
+ + Video: Set up integrals to solve problems involving density, mass, and center of mass +
+ + + + + + + + +
+ diff --git a/calculus/source/06-AI/06.ptx b/calculus/source/06-AI/06.ptx new file mode 100644 index 00000000..1df49ec8 --- /dev/null +++ b/calculus/source/06-AI/06.ptx @@ -0,0 +1,452 @@ + + +
+ Work (AI6) + + + + + Activities + + +

+ Given a physical object m, the workwork done on that object is W=Fd=mad, where F is the force applied to the object over a distance of d. Recall that force F=ma, where m is the mass of the object, and a is the acceleration applied to it. +

+ + + + +

+ Consider a bucket with 10 kg of water being pulled against the acceleration of gravity, g=9.8 m/s^2, at a constant speed for 20 meters. Using , what is the work needed to pull this bucket up 20 meters in kgm^2/s^2 (or Nm)? +

    +
  1. 10 kgm^2/s^2
    2. +
  2. 20 kgm^2/s^2
    3. +
  3. 98 kgm^2/s^2
    4. +
  4. 200 kgm^2/s^2
    5. +
  5. 1960 kgm^2/s^2
    6. +
+

+ + + + + +

+ Consider the bucket from with 10 kg of water, being pulled against the acceleration of gravity, g=9.8 m/s^2, at a constant speed for 20 meters. Suppose that halfway up at a height of 10m, 5kg of water spilled out, leaving 5kg left. How much total work does it take to get this bucket to a height of 20m? +

    +
  1. 980 kgm^2/s^2 or Nm
    2. +
  2. 1470 kgm^2/s^2 or Nm
    3. +
  3. 1960 kgm^2/s^2 or Nm
    4. +
+

+ + + + + +

+ Suppose a 10 kg bucket of water is constantly losing water as it's pulled up, so at a height of h meters, the mass of the bucket is m(h)=2+8e^{-0.2h} kg. +

+ Bucket 5 m in the air, to be hoisted by another 5 meters. + + Bucket 5 m in the air, to be hoisted by another 5 meters. + + \begin{tikzpicture} + \draw (1,3)--(2,3)--(1.8,2)--(1.2,2)--(1,3); + \draw[dashed] (1,5)--(2,5)--(1.8,4)--(1.2,4)--(1,5); + \draw (1.5,3)--(1.5,9)--(3,9)--(3,0); + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (1,2) -- (1,4) node [black,midway,xshift=-1.2cm] {\footnotesize + $\Delta h=5$ m}; + + \draw[dotted] (1.8,2) -- (-1,2); + \draw (-1,2)--node[left]{\tiny$h_i=5$ m} (-1,2); + \end{tikzpicture} + + +
+

+ + +

+ What is the mass of the bucket at height h_i=5 m? +

+ + +

+ Assuming that the bucket does not lose water, estimate the amount of work needed to lift this bucket up \Delta h=5 meters. + +

+ + + + + + +

+ using the same the bucket from , consider the bucket's mass at heights h_i=0, 5, 10, 15 meters. +

+ Bucket lifted 5 m at a time. + + Bucket raised 5 m at a time, h_i are 0, 5, 10, 15 meters. + + \begin{tikzpicture} + \draw (1,1)--(2,1)--(1.8,0)--(1.2,0)--(1,1); + \draw[dashed] (1,3)--(2,3)--(1.8,2)--(1.2,2)--(1,3); + \draw[dashed] (1,5)--(2,5)--(1.8,4)--(1.2,4)--(1,5); + \draw[dashed] (1,7)--(2,7)--(1.8,6)--(1.2,6)--(1,7); + \draw[dashed] (1,9)--(2,9)--(1.8,8)--(1.2,8)--(1,9); + \draw (1.5,1)--(1.5,9)--(3,9)--(3,0); + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (1,0) -- (1,2) node [black,midway,xshift=-1.2cm] {\footnotesize + $\Delta h=5$ m}; + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (1,2) -- (1,4) node [black,midway,xshift=-1.2cm] {\footnotesize + $\Delta h=5$ m}; + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (1,4) -- (1,6) node [black,midway,xshift=-1.2cm] {\footnotesize + $\Delta h=5$ m}; + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (1,6) -- (1,8) node [black,midway,xshift=-1.2cm] {\footnotesize + $\Delta h=5$ m}; + \draw[dotted] (1.2,0) -- (-1,0); + \draw (-1,0)--node[left]{\tiny$h_1=0$ m} (-1,0); + \draw[dotted] (1.2,2) -- (-1,2); + \draw (-1,2)--node[left]{\tiny$h_2=5$ m} (-1,2); + \draw[dotted] (1.2,4) -- (-1,4); + \draw (-1,4)--node[left]{\tiny$h_3=10$ m} (-1,4); + \draw[dotted] (1.2,6) -- (-1,6); + \draw (-1,6)--node[left]{\tiny$h_4=15$ m} (-1,6); + \end{tikzpicture} + + +
+

+ + +

Fill out the following table, estimating the work it would take to lift the bucket 20 meters. + + \begin{array}{|c|c|c|c|} + \hline + h_i & \text{Mass } m(h_i) & \text{Distance} & \text{Estimated Work}\\ + \hline + h_4=15\ \text{m} & m(15)=2+8e^{-0.2\cdot 15}\approx 2.398\ \text{kg} & 5\ \text{m} & \\ + \hline + h_3=10\ \text{m} & m(10)=2+8e^{-0.2\cdot 10}\approx 3.083\ \text{kg} & 5\ \text{m} & \\ + \hline + h_2=5\ \text{m} & m(5)=2+8e^{-0.2\cdot 5}\approx 4.943\ \text{kg} & 5\ \text{m} & 242.207\ \text{Nm}\\ + \hline + h_1=0\ \text{m} & m(5)=2+8e^{-0.2\cdot 0}= 10\ \text{kg} & 5\ \text{m} & \\ + \hline + \end{array} + + + +

+ + +

+ What is the total estimated work to lift this bucket 20 meters? +

+ + + + + +

+ If we estimate the mass and work of the bucket from at height h_i with intervals of length \Delta h meters, which of the following best represents the Riemann sum of the work it would take to lift this bucket 20 meters? +

    +
  1. \displaystyle \sum h_i\cdot 9.8\Delta h. Nm
    2. +
  2. \displaystyle \sum \left( 2+8e^{-0.02h}\right)\cdot 9.8\Delta m Nm
    3. +
  3. \displaystyle \sum \left( 2+8e^{-0.02h_i}\right)\cdot 9.8\Delta h Nm
    4. +
  4. \displaystyle \sum \left( 2+8e^{-0.02h_i}\right)\cdot 9.8\Delta m Nm
    5. +
+

+ + + + + + +

+ Based on the Riemann sum chosen in , which of the following integrals computes the work it would take to lift this bucket 20 meters? +

    +
  1. \displaystyle \int_0^{20} h_i\cdot 9.8 dh. Nm
    2. +
  2. \displaystyle \int_0^{20} \left( 2+8e^{-0.02h}\right)\cdot 9.8 dm Nm
    3. +
  3. \displaystyle \int_0^{20} \left( 2+8e^{-0.02h}\right)\cdot 9.8 dh Nm
    4. +
  4. \displaystyle \int_0^{20} \left( 2+8e^{-0.02h_i}\right)\cdot 9.8 dh Nm
    5. +
+

+ + + + + + +

+ Based on the integral chosen in , compute the work it would take to lift this bucket 20 meters. + +

+ + + + + A <q>how to</q> for applying integrals to physics + +

+

    +
  1. Estimate the value over a piece of the problem with x value x_i over interval of length \Delta x.
    2. +
  2. Find a Riemann sum using (1) which estimates the value in question.
    3. +
  3. Convert the Riemann sum to an integral and solve.
    4. +
+

+ + + + + +

+ Consider a cylindrical tank filled with water, where the base of the cylinder has a radius of 3 meters and a height of 10 meters. Consider a 2 meter thick slice of water sitting 6 meters high in the tank. Using the fact that the mass of this water is 1000\cdot \pi (3)^2\cdot 2=18000\pi kg, estimate how much work is needed to lift this slice 4 more meters to the top of the tank. + +

+ 2m thick slice of water lifted 4m. + + 2m thick slice of water lifted 4m. + + \begin{tikzpicture} + \draw (-1.5,0)--(1.5,0)--(1.5,5)--(-1.5,5)--(-1.5,0); + \draw[dashed] (-1.5,3)--(1.5,3)--(1.5,4)--(-1.5,4)--(-1.5,3); + \draw[fill, blue, opacity=0.2] (-1.5,3) rectangle (1.5,4); + + + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-1.5,3) -- (-1.5,4) node [black,midway,xshift=-1.2cm] {\footnotesize + $\Delta h=2$ m}; + \draw (1.5, 3) --node[right]{\tiny$h_i=6$ m} (1.5,3); + \end{tikzpicture} + + +
+
    +
  1. 18000\pi\cdot 4 Nm
    2. +
  2. 18000\pi\cdot 9.8 Nm
    3. +
  3. 18000\pi\cdot 4\cdot 9.8 Nm
    4. +
  4. 18000\pi\cdot 6 Nm
    5. +
  5. 18000\pi\cdot 6\cdot 9.8 Nm
    6. +
+

+ + + + + + +

+ Consider the cylindrical tank filled with water from . We wish to estimate the amount of work required to pump all the water out of the tank. Suppose we slice the water into 5 pieces and estimate the work it would take to lift each piece out of the tank. + +

+ 2m thick slices of water. + + 2m thick slice of water at 0, 2, 4, 6, 8m. + + \begin{tikzpicture} + \draw (-1.5,0)--(1.5,0)--(1.5,5)--(-1.5,5)--(-1.5,0); + + \foreach \x in {1,2,3,4,5}{ + \draw[dashed] (-1.5,\x-1)--(1.5,\x-1)--(1.5,\x)--(-1.5,\x)--(-1.5,\x-1); + \draw[fill, blue, opacity=0.2] (-1.5,\x-1) rectangle (1.5,\x); + + + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-1.5,\x-1) -- (-1.5,\x) node [black,midway,xshift=-1.2cm] {\footnotesize + $\Delta h=2$ m}; + }; + \end{tikzpicture} + + +
+

+ + +

Fill out the following table, estimating the work it would take to pump all the water out. + + \begin{array}{|c|c|c|c|} + \hline + h_i & \text{Mass} & \text{Distance} & \text{Estimated Work}\\ + \hline + h_5=8\ \text{m} & 18000\pi\ \text{kg} & & \\ + \hline + h_4=6\ \text{m} & 18000\pi\ \text{kg} & 4\ \text{m} & 705600\pi\ \text{Nm}\\ + \hline + h_3=4\ \text{m} & 18000\pi\ \text{kg} & & \\ + \hline + h_2=2\ \text{m} & 18000\pi\ \text{kg} & & \\ + \hline + h_1=0\ \text{m} & 18000\pi\ \text{kg} & 10\ \text{m} & \\ + \hline + \end{array} + + + +

+ + +

+ What is the total estimated work to pump out all the water? +

+ + + + + + +

+ Recall . If we estimate the work needed to lift slices of thickness \Delta h m at heights h_i m, which of the following Riemann sums best estimates the total work needed to pump all the water from the tank? +

    +
  1. \displaystyle \sum 1000\cdot \pi3^2 \cdot 9.8(10-h)\Delta h Nm
    2. +
  2. \displaystyle \sum 1000\cdot\pi3^2 \cdot 9.8(10-h_i)\Delta h Nm
    3. +
  3. \displaystyle \sum 1000\cdot\pi(h_i)^2 \cdot 9.8(10-h)\Delta h Nm
    4. +
  4. \displaystyle \sum 1000\cdot\pi(h_i)^2 \cdot 9.8(10-h_i)\Delta h Nm
    5. + +
+

+ + + + + + +

+ Based on the Riemann sum chosen in , which of the following integrals computes the work it would take to pump all the water from the tank? +

    +
  1. \displaystyle \int_0^{10} 9000\pi\cdot 9.8 (10-h) dh Nm
    2. +
  2. \displaystyle \int_0^{10} 1000\pi\cdot 9.8 h^2(10-h) dh Nm
    3. +
+

+ + + + + + +

+ Based on the integral chosen in , compute the work it would take to pump all the water out of the tank. +

+ + + + + +

+ Consider a cylindrical truncated-cone tank where the radius on the bottom of the tank is 10 m, the radius at the top of the tank is 100 m, and the height of the tank is 100m. + +

+ A slice at height h_i of width \Delta h. + + A slice at height h_i of width \Delta h, with radius r_i. + + \begin{tikzpicture} + \draw (-0.5,0)--(0.5,0)--(1,5)--(-1,5)--(-0.5,0); + + \draw[dashed] (-0.7,2)--(0.7,2)--(0.7, 2.2)--(-0.7,2.2)--(-0.7,2); + + \draw[fill, blue, opacity=0.2] (-0.7,2) rectangle (0.7,2.2); + + \draw [decorate,decoration={brace,amplitude=10pt,raise=2pt, mirror},yshift=0pt] + (0,2) -- (0.7,2) node [black,midway,yshift=-0.6cm] {\footnotesize + $r_i$ }; + + + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-0.7,2) -- (-0.7,2.2) node [black,midway,xshift=-0.8cm] {\footnotesize + $\Delta h$ }; + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-1.5,0) -- (-1.5,5) node [black,midway,xshift=-1cm] {\footnotesize + $100$ m}; + \draw (0.7, 2) --node[right]{\tiny$h_{i}$} (0.7,2); + \draw [decorate,decoration={brace,amplitude=10pt,raise=2pt, mirror},yshift=0pt] + (0,0) -- (0.5,0) node [black,midway,yshift=-0.6cm] {\footnotesize + $10$ m}; + \draw [decorate,decoration={brace,amplitude=10pt,raise=2pt},yshift=0pt] + (0,5) -- (1,5) node [black,midway,yshift=0.6cm] {\footnotesize + $20$ m}; + \end{tikzpicture} + + +
+

+ + +

+ What is the radius r_i in meters of the cross section made at height h_i meters? +

+ + +

+ What is the volume of a cylinder with radius r_i meters with width \Delta h meters? +

+ + +

+ Using the fact that water has density 1000 kg/m^3, what is the mass of the volume of water you found in (b)? +

+ + +

+ How far must this cylinder of water be lifted to be out of the tank? +

+ + + + + +

+ Recall the computations done in . +

+ + +

+ Find a Riemann sum which estimates the total work needed to pump all the water out of this tank, using slices at heights h_i m, of width \Delta h m. +

+ + +

+ Use (a) to find an integral expression which computes the amount of work needed to pump all the water out of this tank. +

+ + +

+ Evaluate the integral found in (b). +

+ + + + + + + + + + Videos +
+ + Video: Set up integrals to solve problems involving work, force, and/or pressure +
+ + + + + + + +
+ diff --git a/calculus/source/06-AI/07.ptx b/calculus/source/06-AI/07.ptx new file mode 100644 index 00000000..c509add4 --- /dev/null +++ b/calculus/source/06-AI/07.ptx @@ -0,0 +1,134 @@ + + +
+ Force and Pressure (AI7) + + + + + Activities + + + +

+ Recall that pressure is measured as force over area: P=F/A. Rewriting this, we have that F=PA. +

+ + + + + +

+ Consider a trapezoid-shaped dam that is 60 feet wide at its base and 90 feet wide at its top. Assume the dam is 20 feet tall with water that rises to its top. Water weighs 62.4 pounds per cubic foot and exerts P=62.4d lbs/ft^2 of pressure at depth d ft. Consider a rectangular slice of this dam at height h_i feet and width b_i. +

+ A slice at height h_i of width \Delta h. + + A slice at height h_i of width \Delta h, with base b_i of a damn with base 60 ft, top 90 ft, 20 ft tall. + + \begin{tikzpicture} + \draw (-3,0)--(3,0)--(4.5,2)--(-4.5,2)--(-3,0); + + \draw[dashed] (-3.75,1)--(3.75,1)--(3.75, 1.2)--(-3.75,1.2)--(-3.75,1); + + \draw[fill, blue, opacity=0.2] (-3.75,1) rectangle (3.75,1.2); + + \draw [decorate,decoration={brace,amplitude=10pt,raise=2pt, mirror},yshift=0pt] + (-3.75,1) -- (3.75,1) node [black,midway,yshift=-0.6cm] {\footnotesize + $b_i$ }; + + + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-3.75,1) -- (-3.75,1.2) node [black,midway,xshift=-0.8cm] {\footnotesize + $\Delta h$ }; + \draw [decorate,decoration={brace,amplitude=10pt,raise=4pt},yshift=0pt] + (-5,0) -- (-5,2) node [black,midway,xshift=-1cm] {\footnotesize + $20$ ft}; + \draw (3.75, 1) --node[right]{\tiny$h_{i}$} (3.75,1); + \draw [decorate,decoration={brace,amplitude=10pt,raise=2pt, mirror},yshift=0pt] + (-3,0) -- (3,0) node [black,midway,yshift=-0.6cm] {\footnotesize + $60$ ft}; + \draw [decorate,decoration={brace,amplitude=10pt,raise=2pt},yshift=0pt] + (-4.5,2) -- (4.5,2) node [black,midway,yshift=0.6cm] {\footnotesize + $90$ ft}; + \end{tikzpicture} + + +
+ + +

+ + +

+ At a height of h_i feet, what is the base of the rectangle b_i? +

+ + +

+ What is the area of a rectangle with base b_i feet and height \Delta h feet? +

+ + +

+ Using a depth of 20-h_i feet, how much pressure is exerted on this rectangle? +

+ + +

+ Using the pressure found in (c), the area in (b), and , how much force is exerted on this rectangle? +

+ + + + + +

+ Recall the computations done in . +

+ + +

+ Find a Riemann sum which estimates the total force exerted on the dam, using slices at heights h_i m, of width \Delta h m. +

+ + +

+ Use (a) to find an integral expression which computes the amount of force exerted on this dam. +

+ + +

+ Evaluate the integral found in (b). +

+ + + + + + + + + + Videos +
+ + Video: Set up integrals to solve problems involving work, force, and/or pressure +
+ + + + + + + +
+ diff --git a/calculus/source/06-AI/main.ptx b/calculus/source/06-AI/main.ptx new file mode 100644 index 00000000..7c50f3e2 --- /dev/null +++ b/calculus/source/06-AI/main.ptx @@ -0,0 +1,16 @@ + + + + Applications of Integration (AI) + + + + + + + + + + + + diff --git a/calculus/source/06-AI/outcomes/01.ptx b/calculus/source/06-AI/outcomes/01.ptx new file mode 100644 index 00000000..74958139 --- /dev/null +++ b/calculus/source/06-AI/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Compute the average value of a function on an interval. +

\ No newline at end of file diff --git a/calculus/source/06-AI/outcomes/02.ptx b/calculus/source/06-AI/outcomes/02.ptx new file mode 100644 index 00000000..c8846955 --- /dev/null +++ b/calculus/source/06-AI/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Estimate the arclength of a curve with Riemann sums and find an integral which computes the arclength. +

\ No newline at end of file diff --git a/calculus/source/06-AI/outcomes/03.ptx b/calculus/source/06-AI/outcomes/03.ptx new file mode 100644 index 00000000..9cfc2ef7 --- /dev/null +++ b/calculus/source/06-AI/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Compute volumes of solids of revolution. +

\ No newline at end of file diff --git a/calculus/source/06-AI/outcomes/04.ptx b/calculus/source/06-AI/outcomes/04.ptx new file mode 100644 index 00000000..8e0aa1a8 --- /dev/null +++ b/calculus/source/06-AI/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Compute surface areas of surfaces of revolution. +

\ No newline at end of file diff --git a/calculus/source/06-AI/outcomes/05.ptx b/calculus/source/06-AI/outcomes/05.ptx new file mode 100644 index 00000000..9f01a49a --- /dev/null +++ b/calculus/source/06-AI/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+Set up integrals to solve problems involving density, mass, and center of mass. +

\ No newline at end of file diff --git a/calculus/source/06-AI/outcomes/06.ptx b/calculus/source/06-AI/outcomes/06.ptx new file mode 100644 index 00000000..d4bc4c12 --- /dev/null +++ b/calculus/source/06-AI/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+Set up integrals to solve problems involving work. +

\ No newline at end of file diff --git a/calculus/source/06-AI/outcomes/07.ptx b/calculus/source/06-AI/outcomes/07.ptx new file mode 100644 index 00000000..d8a8f72a --- /dev/null +++ b/calculus/source/06-AI/outcomes/07.ptx @@ -0,0 +1,4 @@ + +

+Set up integrals to solve problems involving force and/or pressure. +

\ No newline at end of file diff --git a/calculus/source/06-AI/outcomes/main.ptx b/calculus/source/06-AI/outcomes/main.ptx new file mode 100644 index 00000000..677c483f --- /dev/null +++ b/calculus/source/06-AI/outcomes/main.ptx @@ -0,0 +1,35 @@ + +> + +

+How can we use integrals to solve application questions? +

+

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
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    6. +
  6. + +
    7. +
  7. + +
    8. + +
+ \ No newline at end of file diff --git a/calculus/source/06-AI/readiness.ptx b/calculus/source/06-AI/readiness.ptx new file mode 100644 index 00000000..9d0b1857 --- /dev/null +++ b/calculus/source/06-AI/readiness.ptx @@ -0,0 +1,57 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Estimate a definite integral with a Riemann sum.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    2. + +
  2. +

    Evaluate the limit of a Riemann sum to find an integral.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    3. + +
  3. +

    Solve accumulation problems with definite integrals.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    4. + +
  4. +

    Find the distance between points on a curve.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    5. + +
  5. +

    Express the sum of indexed values using summation notation.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    6. + +
+ + \ No newline at end of file diff --git a/calculus/source/07-CO/01.ptx b/calculus/source/07-CO/01.ptx new file mode 100644 index 00000000..a6db358c --- /dev/null +++ b/calculus/source/07-CO/01.ptx @@ -0,0 +1,263 @@ + + +
+ Parametric/vector equations (CO1) + + + + + Activities + + + +

+Consider how we might graph the equation y=2-x^2 in the xy-plane. +

+ + +

+Complete the following chart of xy values by plugging each x value +into the equation to produce its y value. +

+ + + Chart of <m>x</m> and <m>y</m> values to graph + + xy + -2 + -11 + 0 + 1 + 2 + +
+ + +

Plot each point (x,y) in your chart in the xy plane.

+ + +

Connect the dots to obtain a reasonable sketch of the equation's graph.

+ + + + + +

+Suppose that we are told that at after t seconds, an object is located at the +x-coordinate given by x=t-2 and the y-coordinate given by +y=-t^2+4t-2. +

+ + +

+Complete the following chart of txy values by plugging each t value +into the equations to produce its x and y values. +

+ + Chart of <m>x</m> and <m>y</m> values for each <m>t</m> + + txy + 0 + 1-11 + 2 + 3 + 4 + +
+ + +

Plot each point (x,y) in your chart in the xy plane, labeling it with its t value.

+ + +

Connect the dots to obtain a reasonable sketch of the equation's graph.

+ + + + + +

+Graphs in the xy plane can be described by parametric equationsparametric equations +x=f(t) and y=g(t), where plugging in different values of t into the functions +f and g produces different points of the graph. +

+

+The t-values may be thought of representing the moment of time when an object is located +at a particular position, and the graph may be thought of as the path the object travels throughout +time. +

+ + + + + +

+Earlier we obtained the same graphs for the xy equation y=2-x^2 +and the parametric equations x=t-2 and y=-t^2+4t-2. Do the following +steps to find out why. +

+ + +

+Which of the following equations describes t in terms of x? +

+
    +

  1. t=x-2

    2. +

  2. t=x+2

    3. +

  3. t=2x

    4. +

  4. t=-2x

    5. +
+ + +

+Which of these is the result of plugging this choice in for t in the +parametric equation for y? +

+
    +

  1. y=-x+2^2+4x+2-2

    2. +

  2. y=-(x+2)^2+4(x+2)-2

    3. +

  3. y=-x^2+2^2+4x+4\cdot2-2

    4. +
+ + +

+Show how to simplify this choice to obtain the equation y=2-x^2. +

+ + + + + +

+One method of graphing parametric equations x=f(t) and y=g(t) +is to combine them into a single equation only involving x and y, +and using your usual graphing techniques. +

+ + + + + +

+Parametric equations have the advantage of describing paths that cannot be +described by a function y=h(x). One such example is the graph of +x=3\sin(\pi t) and y=-3\cos(\pi t). (Use technology or the +approximation \sqrt 2\approx 0.707 to approximate coordinates as needed.) +

+ + +

Complete the folowing table.

+ + Chart of approximate <m>x</m> and <m>y</m> values + + txy + 0 + 1/4 + 1/2 + 3/42.122.12 + 1 + 5/4 + 3/2 + 7/4 + 2 + +
+ + +

+Plot these (x,y) points in the xy plane and connect the dots to draw a +sketch of the graph. +

+ + +

+What do you obtain by plugging the parametric equations into the expression x^2+y^2? +

+
    +

  1. x^2+y^2=-6\sin(\pi x)\cos(\pi x)

    2. +

  2. x^2+y^2=9

    3. +

  3. x^2+y^2=6\sin(\pi x)\cos(\pi x)

    4. +

  4. x^2+y^2=0

    5. +
+ + +

+Which of these describes the xy equation and graph given by these parametric equations? +

+
    +

  1. a parabola

    2. +

  2. a line

    3. +

  3. a circle

    4. +

  4. a square

    5. +
+ + +

+The graph of these parametric equations cannot be described by a function. Why? +

+
    +

  1. The graph fails the vertical line test.

    2. +

  2. The graph fails the horizontal line test.

    3. +

  3. The graph doesn't extend vertically to +\infty.

    4. +

  4. The graph doesn't extend horizontally to -\infty.

    5. +
+ + + + + +

+The parametric equations x=f(t) and y=g(t) are sometimes written in the form of +the vector equationvector equation \vec r=\tuple{f(t),g(t)}. +

+

+For example, the parametric equations x=3\sin(\pi t) and y=-3\cos(\pi t) +may be combined into the single vector equation \vec r=\tuple{3\sin(\pi t),-3\cos(\pi t)}. +

+ + + + + +

+Consider the vector equation \vec r=\tuple{2t-3,-6t+13}. +

+ + +

What are the corresponding parametric equations?

+
    +

  1. x=2t-3 and y=-6t+13

    2. +

  2. y=2t-3 and x=-6t+13

    3. +

  3. xy=2t-3-6t+13

    4. +

  4. Vector equations cannot be converted into parametric equations.

    5. +
+ + +

Draw a table of t, x, and y values with t=0,1,2,3,4.

+ + +

+Plot these (x,y) points in the plane and connect the dots to sketch the graph +of this vector equation. +

+ + +

+Solve for t in terms of x and plug into the y parametric equation to +show that this is the vector equation for the line y=-3x+4. +

+ + + + + Videos +
+
+ +
+ diff --git a/calculus/source/07-CO/02.ptx b/calculus/source/07-CO/02.ptx new file mode 100644 index 00000000..5ed3ed32 --- /dev/null +++ b/calculus/source/07-CO/02.ptx @@ -0,0 +1,204 @@ + + +
+ Parametric/vector derivatives (CO2) + + + + + Activities + + + +

+Consider the parametric equations x=2t-1 and y=(2t-1)(2t-5). +The coordinate on this graph at t=2 is (3,-3). +

+ + +

+Which of the following equations of x,y describes the graph of +these paramteric equations? +

+
    +

  1. y=2x(x+2)=2x^2+2x

    2. +

  2. y=2x(x-2)=2x^2-2x

    3. +

  3. y=x(x+4)=x^2+4x

    4. +

  4. y=x(x-4)=x^2-4x

    5. +
+ + +

+Which of the following describes the slope of the line tangent to the +graph at the point (3,-3)? +

+
    +

  1. \frac{dy}{dx}=2x+4, which is 10 when x=3.

    2. +

  2. \frac{dy}{dx}=2x+4, which is 8 when t=2.

    3. +

  3. \frac{dy}{dx}=2x-4, which is 2 when x=3.

    4. +

  4. \frac{dy}{dx}=2x-4, which is 0 when t=2.

    5. +
+ + +

+Note that the parametric equation for y simplifies to y=4t^2-12t+5. +What do we get for the derivatives \frac{dx}{dt} of x=2t-1 and \frac{dy}{dt} +for y=4t^2-12t+5? +

+
    +

  1. \frac{dx}{dt}=2 and \frac{dy}{dt}=8t-12.

    2. +

  2. \frac{dx}{dt}=-1 and \frac{dy}{dt}=8t-12.

    3. +

  3. \frac{dx}{dt}=2 and \frac{dy}{dt}=6t+5.

    4. +

  4. \frac{dx}{dt}=-1 and \frac{dy}{dt}=6t+5.

    5. +
+ + +

+It follows that when t=2, \frac{dx}{dt}=2 and \frac{dy}{dt}=4. +Which of the following conjectures seems most likely? +

+
    +
  1. +

    +The slope \frac{dy}{dx} could also be found by computing \frac{dx}{dt}+\frac{dy}{dt}. +

    +
    2. +
  2. +

    +The slope \frac{dy}{dx} could also be found by computing \frac{dy/dt}{dx/dt}. +

    +
    3. +
  3. +

    +The slope \frac{dy}{dx} is always equal to \frac{dx}{dt}. +

    +
    4. +
  4. +

    +The slope \frac{dy}{dx} is always equal to \frac{dy}{dt}. +

    +
    5. +
+ + + + + +

+Suppose x is a function of t, and y may be thought of +as a function of either x or t. Then the Chain Rule requires that +\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}. This provides the slope formula +for parametric equations: +\frac{dy}{dx}=\frac{dy/dt}{dx/dt}. +

+ + + + + +

+Let's draw the picture of the line tangent to the parametric equations +x=2t-1 and y=(2t-1)(2t-5) when t=2. +

+ + +

+Use a t,x,y chart to sketch the parabola given by these parametric +equations for 0\leq t\leq 3, including the point (3,-3) +when t=2. +

+ + +

+Earlier we determined that the slope of the tangent line was 2. Draw a line +with slope 2 passing through (3,-3) and confirm that it +appears to be tangent. +

+ + +

+Use the point-slope +formula y-y_0=m(x-x_0) along with the slope 2 and point +(3,-3) to find the exact equation for this tangent line. +

+
    +

  1. y=2x-10

    2. +

  2. y=2x-9

    3. +

  3. y=2x-8

    4. +

  4. y=2x-7

    5. +
+ + + + + +

+Consider the vector equation \vec{r}(t)=\tuple{3t^2-9,t^3-3t}. +

+ + +

+What are the corresponding parametric equations and their derivatives? +

+
    +
  1. +

    +y=3t^2-9 and x=t^3-3t; \frac{dy}{dt}=9t and \frac{dx}{dt}=3t-6 +

    +
    2. +
  2. +

    +x=3t^2-9 and y=t^3-3t; \frac{dx}{dt}=9t and \frac{dy}{dt}=3t-6 +

    +
    3. +
  3. +

    +y=3t^2-9 and x=t^3-3t; \frac{dy}{dt}=6t and \frac{dx}{dt}=3t^2-3 +

    +
    4. +
  4. +

    +x=3t^2-9 and y=t^3-3t; \frac{dx}{dt}=6t and \frac{dy}{dt}=3t^2-3 +

    +
    5. +
+ + +

+The formula \frac{dy}{dx}=\frac{dy/dt}{dx/dt} allows us to compute slopes +as which of the following functions of t? +

+
    +

  1. \frac{6t}{t^2+3}

    2. +

  2. \frac{6t}{t^2+1}

    3. +

  3. \frac{t^2-1}{2t}

    4. +

  4. \frac{2t}{3t^2-1}

    5. +
+ + +

+Find the point, tangent slope, and tangent line equation (recall y-y_0=m(x-x_0)) +corresponding to the parameter t=-3. +

+
    +

  1. Point (-12,9), slope -\frac{4}{3}, EQ y=-\frac{4}{3}x-7

    2. +

  2. Point (18,-18), slope -\frac{4}{3}, EQ y=-\frac{4}{3}x+6

    3. +

  3. Point (-12,9), slope \frac{3}{4}, EQ y=\frac{3}{4}x-8

    4. +

  4. Point (18,-18), slope \frac{3}{4}, EQ y=\frac{3}{4}x+5

    5. +
+ + + + + Videos +
+
+ +
+ diff --git a/calculus/source/07-CO/03.ptx b/calculus/source/07-CO/03.ptx new file mode 100644 index 00000000..76081a19 --- /dev/null +++ b/calculus/source/07-CO/03.ptx @@ -0,0 +1,311 @@ + + +
+ Parametric/vector arclength (CO3) + + + + + Activities + + + +

+In , the blue curve is the graph of the parametric +equations x=t^2 and y=t^3 for 1\leq t\leq 2. +This curve connects the point (1,1) to the point (4,8). +The red dashed line is the straight line segment connecting these points. +

+
+ A parametric curve and segment from (1,1) to (4,8) + + A parametric curve and segment from (1,1) to (4,8). + +g = line([(1,1),(4,8)],color="red",linestyle="dashed")+parametric_plot((x^2,x^3),(x,1,2)) +g.set_aspect_ratio(0.5) +g.set_axes_range(xmin=0,xmax=5,ymin=0,ymax=10) +g + + +
+ + + + + +

+Let's first investigate the length of the dashed red line segment +in . +

+ + +

+Draw a right triangle with the red dashed line segment as its hypotenuse, +one leg parallel to the x-axis, and the other parallel to the y-axis. +

+

+How long are these legs? +

+
    +

  1. 3 and 7.

    2. +

  2. 4 and 8.

    3. +

  3. 3 and 8.

    4. +

  4. 4 and 7.

    5. +
+ + +

+The Pythagorean theorem states that for a right triangle with leg lengths a,b +and hypotenuse length c, we have... +

+
    +

  1. a=b=c.

    2. +

  2. a+b=c.

    3. +

  3. a^2=b^2=c^2.

    4. +

  4. a^2+b^2=c^2.

    5. +
+ + +

+Using the leg lengths and Pythagorean theorem, how long must the red dashed hypotenuse be? +

+
    +

  1. \sqrt{20}\approx 4.47.

    2. +

  2. \sqrt{58}\approx 7.62.

    3. +

  3. \sqrt{67}\approx 8.19.

    4. +

  4. \sqrt{100}=10.

    5. +
+ + +

+Compared with the blue parametric curve connecting the same two points, is the red dashed line +segement length an overestimate or underestimate? +

+
    +

  1. Overestimate: the blue curve is shorter than the red line.

    2. +

  2. Underestimate: the blue curve is longer than the red line.

    3. +

  3. Exact: the blue curve is exactly as long as the red line.

    4. +
+ + + + + +

+Recall that the linear distance between two points (x_1,y_1) and +(x_2,y_2) may be computed by the distance formula +\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. Note that \Delta x=|x_2-x_1| and +\Delta y=|y_2-y_1| measure leg lengths of a right triangle whose hypotenuse +is the distance we want to measure, so we may rewrite this formula as +\sqrt{(\Delta x)^2+(\Delta y)^2}. +

+

+This formula will need to be modified to measure a curved path between two points. +

+ + + + + +

+By approximating the curve by several (say N) segements connecting +points along the curve, we obtain a better approximation than a single line segment. +For example, the illustration shown in gives +three segments whose distances sum to about 7.6315, while the actual length of the curve turns +out to be about 7.6337. +

+
+ Subdividing a parametric curve where N=3 + + Subdividing a parametric curve with three segments + +points = [[n(1+t/3)^2,n(1+t/3)^3] for t in range(4)] +segs = line(points,color="red",linestyle="dashed") +ppoints = list_plot(points,color="black") +para = parametric_plot((x^2,x^3),(x,1,2)) +g = para+segs+ppoints +g.set_aspect_ratio(0.5) +g.set_axes_range(xmin=0,xmax=5,ymin=0,ymax=10) +g + + +
+ + + + + +

+How should we modify the distance formula \sqrt{(\Delta x)^2+(\Delta y)^2} +to measure arclength as illustrated +in ? +

+ + +

+Let \Delta L_1,\Delta L_2,\Delta L_3 describe the lengths of each +of the three segements. Which expression describes the total length of these +segments? +

+
    +

  1. \Delta L_1\times \Delta L_2\times \Delta L_3

    2. +

  2. \Delta L_1+ 2\Delta L_2+ 3\Delta L_3

    3. +

  3. \sum_{i=1}^{3} \Delta L_i

    4. +
+ + +

+We can let each \Delta L_i=\sqrt{(\Delta x_i)^2+(\Delta y_i)^2}. +But we will find it useful to involve the parameter t as well, or more +accurately, the change \Delta t_i of t between each point of the subdivision. +

+

+Which of these is algebraically the same as the above formula for \Delta L_i? +

+
    +

  1. \sqrt{\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2}

    2. +

  2. \sqrt{\left[\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2\right]\Delta t_i}

    3. +

  3. \sqrt{\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2}\Delta t_i

    4. +
+ + +

+Finally, we'll want to increase N from 3 so that it limits to \infty. +What can we conclude when that happens? +

+
    +

  1. Each segment is infintely small.

    2. +

  2. \Delta x_i\to 0

    3. +

  3. \frac{\Delta x_i}{\Delta t_i}\to\frac{dx}{dt}

    4. +

  4. All of the above.

    5. +
+ + + + + +

+Put together, and limiting the subdivisions of the curve N\to \infty, we obtain +the Riemann sum +\lim_{N\to\infty}\sum_{i=1}^N \sqrt{\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2}\Delta t_i. +

+

+Thus arclengtharclength formula along a parametric curve from a\leq t\leq b may be calculated by using the corresponding +definite integral +\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt. +

+ + + + + +

+Let's gain confidence in the arclength formula +\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt +by checking to make sure it matches the distance formula for line segments. +

+

+The parametric equations x=3t-1 and y=2-4t for 1\leq t\leq 3 represent +the segment of the line y=-\frac{4}{3}x-\frac{2}{3} connecting (2,-2) to +(8,-10). +

+ + +

+Find dx/dt and dy/dt, and substitute them into the formula above along with +a=1 and b=3. +

+ + +

+Show that the value of this formula is 10. +

+ + +

+Show that the length of the line segment connecting (2,-2) to +(8,-10) is 10 by applying the distance formula directly instead. +

+ + + + + +

+For each of these parametric equations, use +\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt +to write a definite integral that computes the given length. (Do not evaluate the +integral.) +

+ + +

+The portion of x=\sin 3t, y=\cos 3t where 0\leq t\leq \pi/6. +

+ + +

+The portion of x=e^t, y=\ln t where 1\leq t\leq e. +

+ + +

+The portion of x=t+1, y=t^2 between the points (3,4) and (5,16). +

+ + + + + +

+Let's see how to modify +\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt +to produce the arclength of the graph of a function y=f(x). +

+ + +

+Let x=t. How can \frac{dx}{dt} be simplified? +

+
    +

  1. dx

    2. +

  2. dt

    3. +

  3. 1

    4. +

  4. 0

    5. +
+ + +

+Given x=t, how should \frac{dy}{dt} and dt be rewritten? +

+
    +

  1. \frac{dy}{dt}=\frac{dy}{dx} and dt=dx.

    2. +

  2. \frac{dy}{dt}=\frac{dx}{dt} and dt=dx.

    3. +

  3. \frac{dy}{dt}=\frac{dy}{dx} and dt=1.

    4. +

  4. \frac{dy}{dt}=\frac{dy}{dt} and dt=1.

    5. +
+ + +

+Write a modified, simplified formula for +\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt +with t replaced with x. +

+ + + + + + Videos +
+
+ + +
+ diff --git a/calculus/source/07-CO/04.ptx b/calculus/source/07-CO/04.ptx new file mode 100644 index 00000000..9ba6d99a --- /dev/null +++ b/calculus/source/07-CO/04.ptx @@ -0,0 +1,222 @@ + + +
+ Polar coordinates (CO4) + + + + + Activities + + + +

+ + As the crow flies is an idiom used to describe the most direct path between two points. + The polar coordinate systempolar coordinate system is a useful parametrization of the plane that, rather than describing horizontal and vertical position relative to the origin in the usual way, describes a point in terms of distance from the origin and direction. + The origin is also known as the polepole (hence polar coordinates).

+

Let \overline{OP} be a line segment from the origin to a given point P in the plane. + The length of \overline{OP} is the distance (or radiuspolar coordinate systemradius) r from the origin to P. + The polar axispolar coordinate systempolar axis is a ray starting at the origin.

+

To define the "direction" of P, we form an angle \theta by letting the polar axis serve as the initial ray and \overrightarrow{OP} as the terminal ray. + We will set the positive x-axis as the polar axis and assume the movement in the positive direction is counter-clockwise (as in trigonometry). + Notice that, unlike in the rectangular (or Cartesian) coordinate system, the polar coordinates (r,\theta) for a point are not unique, as we could turn either way to face a given point (or even spin around a number of times before facing that direction).

+

Furthermore, by allowing r to be negative, we can also "walk backwards" to get to a point by facing in the opposite direction. + Rather than the grid lines defined by specific values for x and y in the rectangular coordinate system, specific values of r correspond to circles of radius r centered about the origin, and specific values of \theta correspond to lines going through the pole (called radial linespolar coordinate systemradial lines). + + +

+ + + +
+ A point in the polar coordinate system. + + A point in the polar coordinate system. + + +
+ + + +
+ The polar grid. + + The polar grid + + + +
+ + + + + + +

+ Plot the Cartesian point P=(x,y)=(\sqrt{3},-1) and draw line segments connecting the origin to P, the origin to (x,y)=(\sqrt{3},0), and P to (x,y)=(\sqrt{3},0). +

+ + +

+ Solve the triangle formed by the line segments you just drew (i.e. find the lengths of all sides and the measures of each angle). +

+ + +

+ Find all polar coordinates for the Cartesian point (x,y)=(\sqrt{3},-1). +

+ + +

+ Find Cartesian coordinates for the polar point (r,\theta)=\left(-\sqrt{2},\displaystyle\frac{3\pi}{4}\right). +

+ + + + + + +

+Graph each of the following. +

+ + +

+ r=1 +

+ + +

+ r=-1 +

+ + +

+ \theta=\displaystyle\frac{\pi}{6} +

+ + +

+ \theta=\displaystyle\frac{7\pi}{6} +

+ + +

+ \theta=\displaystyle\frac{-5\pi}{6} +

+ + +

+ 1\leq r < -1, 0\leq\theta\leq\displaystyle\frac{\pi}{2} +

+ + +

+ -3\leq r \leq 2, \theta=\displaystyle\frac{\pi}{4} +

+ + +

+ r \leq 0, \theta=\displaystyle\frac{-\pi}{2} +

+ + +

+ \displaystyle\frac{2\pi}{3}\leq\theta\leq\displaystyle\frac{5\pi}{6} +

+ + +

+ r=3\sec(\theta) +

+ + + + + +

+ If a polar graph is symmetric about the x-axis, then if the point (r,\theta) lies on the graph, then the point (r,-\theta) or (-r, \pi-\theta) also lies on the graph. + +

+ + + + + +

+ If a polar graph is symmetric about the y-axis, then if the point (r,\theta) lies on the graph, then the point (r,\pi-\theta) or (-r,-\theta) also lies on the graph. + +

+ + + + + +

+ If a polar graph is rotationally symmetric about the origin, then if the point (r,\theta) lies on the graph, then the point (-r,\theta) or (r,\pi+\theta) also lies on the graph. + +

+ + + + + + +

+ Find a polar form of the the Cartesian equation x^2+(y-3)^2=9. +

+ + + + + + +

+Find a Cartesian form of each of the given polar equations. +

+ + +

+ r^2=4r\cos(\theta) +

+ + +

+ r=\displaystyle\frac{4}{2\cos(\theta)-\sin(\theta)} +

+ + + + +

+ The slope of a polar curve r=f(\theta)is \displaystyle\frac{dy}{dx}=\displaystyle\frac{dy/d\theta}{dx/d\theta}=\displaystyle\frac{f^\prime(\theta)\sin(\theta)+f(\theta)\cos(\theta)}{f^\prime(\theta)\cos(\theta)-f(\theta)\sin(\theta)}, provided that dx/d\theta\neq 0 at (r,\theta). + Vertical tangents occur when dy/d\theta=0 and dx/d\theta\neq 0; + horizontal tangents occur when dx/d\theta=0 and dy/d\theta\neq 0. + +

+ + + + + + + + Videos +
+
+ + +
+ diff --git a/calculus/source/07-CO/05.ptx b/calculus/source/07-CO/05.ptx new file mode 100644 index 00000000..9f522a86 --- /dev/null +++ b/calculus/source/07-CO/05.ptx @@ -0,0 +1,349 @@ + + +
+ Polar Arclength (CO5) + + + + + Activities + + + + +

+Recall that the length of a parametric curve is given by \int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt. +

+ + + + + + +

+Let x(t)=r\cos(\theta) and y(t)=r\sin(\theta) and show that the length of a polar curve r=f(\theta) with \alpha\leq\theta\leq\beta is given by + \int_{\theta=\alpha}^{\theta=\beta} \sqrt{\left(r\right)^2+\left(\frac{dr}{d\theta}\right)^2}d\theta. + +

+ + + +

Find an integral computing the arclength of the polar curve defined by r=3\cos(\theta)-2 on \pi/3\leq\theta\leq\pi.

+ + +

Find the length of the cardioid r=1-\cos(\theta).

+ + + + + + + + + + Videos +
+
+ + +
+ diff --git a/calculus/source/07-CO/06.ptx b/calculus/source/07-CO/06.ptx new file mode 100644 index 00000000..bc927300 --- /dev/null +++ b/calculus/source/07-CO/06.ptx @@ -0,0 +1,420 @@ + + +
+ Polar area (CO6) + + + + + Activities + + + + + + + +

+ The area of the fan-shaped region between the pole and r=f(\theta) as the angle \theta ranges from \alpha to \beta is given by + \int_{\theta=\alpha}^{\theta=\beta} \frac{r^2}{2}d\theta. + +

+ + + +
+ Finding the polar area differential + + Illustration of polar area differential. + + \begin{tikzpicture}[>=latex] + +% Draw the lines at multiples of pi/12 +\foreach \ang in {0,...,31} { + \draw [lightgray] (0,0) -- (\ang * 180 / 16:4); +} + +% Concentric circles and radius labels +\foreach \s in {0, 1, 2, 3} { + \draw [lightgray] (0,0) circle (\s + 0.5); + \draw (0,0) circle (\s); + \node [fill=white] at (\s, 0) [below] {\scriptsize $\s$}; +} + +% Add the labels at multiples of pi/4 +\foreach \ang/\lab/\dir in { + 0/0/right, + 1/{\pi/4}/{above right}, + 2/{\pi/2}/above, + 3/{3\pi/4}/{above left}, + 4/{\pi}/left, + 5/{5\pi/4}/{below left}, + 7/{7\pi/4}/{below right}, + 6/{3\pi/2}/below} { + \draw (0,0) -- (\ang * 180 / 4:4.1); + \node [fill=white] at (\ang * 180 / 4:4.2) [\dir] {\scriptsize $\lab$}; +} + +% The double-lined circle around the whole diagram +\draw [style=double] (0,0) circle (4); + +\draw [fill=red!50!black, opacity=0.4] ({cos(30)},{sin(30)})--plot [domain=pi/6:pi/3] + (xy polar cs:angle=\x r, radius= {2+2*cos(\x r)})--plot [domain=pi/3:pi/6] + (xy polar cs:angle=\x r, radius= {1})--({cos(30)},{sin(30)}); +\draw [thick, color=red, domain=0:2*pi, samples=200, smooth] + plot (xy polar cs:angle=\x r, radius={2+2*cos(\x r)}); +\draw [thick, color=blue, domain=0:2*pi, samples=200, smooth] + plot (xy polar cs:angle=\x r, radius={1}); + +\draw [fill=blue!50!black, opacity=0.4] ({cos(30)},{sin(30)})--plot [domain=pi/6:4*pi/18] + (xy polar cs:angle=\x r, radius= {2+2*cos(pi/6 r)})--plot [domain=4*pi/18:pi/6] + (xy polar cs:angle=\x r, radius= {1})--({cos(30)},{sin(30)}); + +\draw [fill=blue!50!black, opacity=0.4] ({cos(40)},{sin(40)})--plot [domain=4*pi/18:5*pi/18] + (xy polar cs:angle=\x r, radius= {2+2*cos(4*pi/18 r)})--plot [domain=5*pi/18:4*pi/18] + (xy polar cs:angle=\x r, radius= {1})--({cos(40)},{sin(40)}); + +\draw [fill=blue!50!black, opacity=0.4] ({cos(50)},{sin(50)})--plot [domain=5*pi/18:6*pi/18] + (xy polar cs:angle=\x r, radius= {2+2*cos(5*pi/18 r)})--plot [domain=6*pi/18:5*pi/18] + (xy polar cs:angle=\x r, radius= {1})--({cos(50)},{sin(50)}); +%\node [fill=white] at (2,1) {$r=1+\cos\theta$}; + +\draw[purple, dashed] (0,0)--({cos(30)*(2+2*cos(30))}, {sin(30)*(2+2*cos(30))}); +\draw[purple, dashed] (0,0)--({cos(40)*(2+2*cos(30))}, {sin(40)*(2+2*cos(30))}); +\draw[purple, dashed] (0,0)--({cos(50)*(2+2*cos(40))}, {sin(50)*(2+2*cos(40))}); +\draw[purple, dashed] (0,0)--({cos(60)*(2+2*cos(50))}, {sin(60)*(2+2*cos(50))}); + +\end{tikzpicture} + + +
+ + + + + + + + + +

Find an integral computing the area of the region defined by 0\leq r\leq-\cos(\theta)+5 and \pi/2\leq \theta\leq 3\pi/4.

+ + +

Find the area enclosed by the cardioid r=2(1+\cos(\theta).

+ + +

Find the area enclosed by one loop of the 4-petaled rose r=\cos(2\theta).

+ + + + + + + + + Videos +
+
+ + +
+ diff --git a/calculus/source/07-CO/main.ptx b/calculus/source/07-CO/main.ptx new file mode 100644 index 00000000..785db1be --- /dev/null +++ b/calculus/source/07-CO/main.ptx @@ -0,0 +1,13 @@ + + + + Coordinates and Vectors (CO) + + + + + + + + + diff --git a/calculus/source/07-CO/outcomes/01.ptx b/calculus/source/07-CO/outcomes/01.ptx new file mode 100644 index 00000000..6080ddd4 --- /dev/null +++ b/calculus/source/07-CO/outcomes/01.ptx @@ -0,0 +1,5 @@ + +

+Sketch the graph of a two-dimensional parametric/vector equation, +and convert such equations into equations of only x and y. +

\ No newline at end of file diff --git a/calculus/source/07-CO/outcomes/02.ptx b/calculus/source/07-CO/outcomes/02.ptx new file mode 100644 index 00000000..9ac7ecbf --- /dev/null +++ b/calculus/source/07-CO/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Compute derivatives and tangents related to two-dimensional parametric/vector equations. +

\ No newline at end of file diff --git a/calculus/source/07-CO/outcomes/03.ptx b/calculus/source/07-CO/outcomes/03.ptx new file mode 100644 index 00000000..0cd34cc7 --- /dev/null +++ b/calculus/source/07-CO/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Compute arclengths related to two-dimensional parametric/vector equations. +

\ No newline at end of file diff --git a/calculus/source/07-CO/outcomes/04.ptx b/calculus/source/07-CO/outcomes/04.ptx new file mode 100644 index 00000000..f8cb0fe1 --- /dev/null +++ b/calculus/source/07-CO/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Convert points and equations between polar and Cartesian coordinates and equations. +

\ No newline at end of file diff --git a/calculus/source/07-CO/outcomes/05.ptx b/calculus/source/07-CO/outcomes/05.ptx new file mode 100644 index 00000000..db6040ea --- /dev/null +++ b/calculus/source/07-CO/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+Compute arclengths of curves given in polar coordinates. +

\ No newline at end of file diff --git a/calculus/source/07-CO/outcomes/06.ptx b/calculus/source/07-CO/outcomes/06.ptx new file mode 100644 index 00000000..6be8b9b6 --- /dev/null +++ b/calculus/source/07-CO/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+Compute areas bounded by curves given in polar coordinates. +

\ No newline at end of file diff --git a/calculus/source/07-CO/outcomes/main.ptx b/calculus/source/07-CO/outcomes/main.ptx new file mode 100644 index 00000000..65e79453 --- /dev/null +++ b/calculus/source/07-CO/outcomes/main.ptx @@ -0,0 +1,31 @@ + +> + +

+How do we use alternative coordinates and vectors to describe points in the plane? +

+

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
+ \ No newline at end of file diff --git a/calculus/source/07-CO/readiness.ptx b/calculus/source/07-CO/readiness.ptx new file mode 100644 index 00000000..72c3b901 --- /dev/null +++ b/calculus/source/07-CO/readiness.ptx @@ -0,0 +1,27 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Write equations of lines using slope-intercept and/or point-slope form (Math is Fun)

    +
    2. +
  2. +

    Recall special trig values on the unit circle (Khan Academy (1) and Khan Academy (2) )

    +
    3. +
  3. +

    Evaluate functions at variable expressions (Purple Math)

    +
    4. +
  4. +

    Find the derivative of a function using methods from , , , and .

    +
    5. +
  5. +

    Find an integral which computes the arclength of a curve using methods from .

    +
    6. +
+ + diff --git a/calculus/source/07-CO/tikz/polarCoord.tex b/calculus/source/07-CO/tikz/polarCoord.tex new file mode 100644 index 00000000..86a0fffa --- /dev/null +++ b/calculus/source/07-CO/tikz/polarCoord.tex @@ -0,0 +1,31 @@ + +\begin{tikzpicture} +%Circles +\foreach \r in {1, 2,...,7} +\draw[blue, thick] (0,0) circle (\r); +\foreach \r in {0.5, 1.5,...,7} +\draw[blue, thin] (0,0) circle (\r); +%1° Rays +\foreach \a in {0, 1,...,359} +\draw[blue] (\a:7.7) -- (\a:8); +%5° Rays +\foreach \a in {0, 5,...,359} +\draw[blue] (\a:7.5) -- (\a:8); +%15° Rays +\foreach \a in {0, 15,...,359} +\draw[thick,blue] (\a:1) -- (\a:8); +%30° Rays +\foreach \a in {0, 30,...,359} +\draw[thick,blue] (0, 0) -- (\a:8); +%Radius labels (background filled white) +\foreach \r in {1, 2,...,7} +\draw (\r,0) node[inner sep=1pt,below=3pt,rectangle,fill=white] {$\r$}; +%Main rays +\foreach \a in {0, 90,...,359} +\draw[very thick] (0, 0) -- (\a:8); +%Angle labels +\foreach \a in {0, 1,...,23} +\draw (\a*15: 8.5) node {$\frac{\a \pi}{12}$}; +%Central point +\draw[fill=red] (0,0) circle(0.7mm); +\end{tikzpicture} \ No newline at end of file diff --git a/calculus/source/07-CO/tikz/polarPoint.tex b/calculus/source/07-CO/tikz/polarPoint.tex new file mode 100644 index 00000000..1447dbba --- /dev/null +++ b/calculus/source/07-CO/tikz/polarPoint.tex @@ -0,0 +1,28 @@ +\begin{tikzpicture} + %Coordinates +\coordinate (A) at (6,0); +\coordinate (B) at (0,0); +\coordinate (C) at (2.5,2.5); +\coordinate (B') at (2.5,3.5); +\coordinate (A') at (1.8,3.2); + +%Axis + +\draw[thick,-latex] (-1ex,0) -- (6,0) node [below] {$x$}; +\draw[thick,-latex] (0,-1ex) -- (0,5) node [left] {$y$}; + +%Vectors +\draw[thick,-latex] (0,0) -- (2.5,2.5) node[pos=0.6, above left] {$r$}; + +%Help Lines +\draw[dashed, red] (0,2.5) -- (2.5,2.5) -- (2.5,0); +\draw[dashed] (3.53,0) arc (0:90:3.53); +\draw[red] (2.5,0)--node[below]{$x$}(2.5,0); +\draw[red] (0,2.5)--node[left]{$y$}(0,2.5); +\draw (2.5, 2.5) --node[above right]{$(r,\theta)$} (2.5, 2.5); + +%Angle +\draw (0.3,0) arc[start angle=0, end angle=45, radius=0.3]; +\node[anchor=south west] at (0.3,0) {$\theta$}; + +\end{tikzpicture} \ No newline at end of file diff --git a/calculus/source/08-SQ/01.ptx b/calculus/source/08-SQ/01.ptx new file mode 100644 index 00000000..2f4acf0c --- /dev/null +++ b/calculus/source/08-SQ/01.ptx @@ -0,0 +1,295 @@ + + +
+ Sequence Formulas (SQ1) + + + + + Activities + + +

Which of the following are sequences?

+
    +
  1. monthly gas bill
    2. +
  2. days in the year
    3. +
  3. how long you wash dishes
    4. +
  4. 1, 1, 2, 3, 5, 8, \ldots
    5. +
  5. how much you spend on groceries
    6. +
+ + + +

Consider the sequence 1, 2, 4, \ldots.

 + +

Which of the choices below reasonably continues this sequence of numbers?

+
    +
  1. 7, 12, 24, \ldots
    2. +
  2. 7, 11, 16, \ldots
    3. +
  3. 8, 16, 32, \ldots
    4. +
  4. 1, 2, 4, \ldots
    5. +
  5. 7, 12, 20, \ldots
    6. +
+ + +

Where possible, find a formula that allows us to move from one term to the next one.

+ + + + + + +

+ As seen in the previous activity, having too few terms may prevent us from finding a unique way to continue creating a sequence of numbers. In fact, we need sufficiently many terms to uniquely continue a sequence of numbers (and how many terms is sufficient depends on which sequence of numbers you are trying to generate). Sometimes, we do not want to write out all of the terms needed to allow for this. Therefore, we will want to find short-hand notation that allows us to do so. +

+ + + + + +

+ A sequence is a list of real numbers. Let a_n denote the nth term in a sequence. We will use the notation \displaystyle \{a_n\}_{n=1}^\infty=a_1, a_2, \ldots, a_n, \ldots. A general formula that indicates how to explicitly find the n-th term of a sequence is the closed form of the sequence. +

+ + + + + + +

Consider the sequence \displaystyle 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots. Which of the following choices gives a closed formula for this sequence? Select all that apply. +

    +
  1. \displaystyle \left\{\left(\frac{1}{3}\right)^{n-1}\right\}_{n=1}^\infty
    2. +
  2. \displaystyle \left\{\left(\frac{1}{3}\right)^{n}\right\}_{n=1}^\infty
    3. +
  3. \displaystyle \left\{\left(\frac{1}{3}\right)^{n-1}\right\}_{n=2}^\infty
    4. +
  4. \displaystyle \left\{\left(\frac{1}{3}\right)^{n+1}\right\}_{n=0}^\infty
    5. +
  5. \displaystyle \left\{\left(\frac{1}{3}\right)^{n}\right\}_{n=0}^\infty
    6. +
+

+ + + + + +

Let a_n be the nth term in the sequence \displaystyle \left\{\frac{n+1}{n}\right\}_{n=1}^\infty. Which of the following terms corresponds to the 27^{th} term of this sequence? +

    +
  1. \frac{27}{26}
    2. +
  2. \frac{26}{27}
    3. +
  3. \frac{27}{28}
    4. +
  4. \frac{28}{27}
    5. +
  5. \frac{29}{28}
    6. +
+

+ + + + + +

Let a_n be the nth term in the sequence \displaystyle \left\{\frac{n+1}{n}\right\}_{n=2}^\infty. Which of the following terms corresponds to the 27^{th} term of this sequence? +

    +
  1. \frac{27}{26}
    2. +
  2. \frac{26}{27}
    3. +
  3. \frac{27}{28}
    4. +
  4. \frac{28}{27}
    5. +
  5. \frac{29}{28}
    6. +
+ +

+ + + + + +

Let a_n be the nth term in the sequence \displaystyle 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots. Identify the 81st term of this sequence. +

    +
  1. \frac{1}{79}
    2. +
  2. \frac{1}{80}
    3. +
  3. \frac{1}{81}
    4. +
  4. \frac{1}{82}
    5. +
  5. \frac{1}{83}
    6. +
+

+ + + + + + +

Find a closed form for the sequence 0, 3, 8, 15, 24, \ldots.

+ + + + + +

Find a closed form for the sequence \displaystyle \frac{12}{1}, \frac{16}{2}, \frac{20}{3}, \frac{24}{4}, \frac{28}{5}, \ldots.

+ + + + + +

Let a_n be the nth term in the sequence 1, 1, 2, 3, 5, 8, \ldots. Find a formula for a_n.

+ + + + + +

+ A sequence is recursive if the terms are defined as a function of previous terms (with the necessary initial terms provided). +

+ + + +

Consider the sequence defined by a_1=6 and a_{k+1}=4a_k-7 for k\geq 1. What are the first four terms?

 + + + + +

Consider the sequence 2, 7, 22, 67, 202, \ldots. Which of the following offers the best recursive formula for this sequence? +

    +
  1. a_{n+1} = 3a_n+1
    2. +
  2. a_1=2, a_k=3a_{k-1}+1 for k>1
    3. +
  3. a_1=2, a_2=7, a_k=3a_{k-1}+1 for k>2
    4. +
+

+ + + + +

Once more, consider the sequence 1, 1, 2, 3, 5, 8, \ldots from . Suppose a_1=1 and a_2=1. Give a recursive formula for a_n for all n\geq 3.

+ + + +

Give a recursive formula that generates the sequence 1, 2, 4, 8, 16, 32, \ldots.

 + + + + + + +

Find the first 5 terms of the following sequence:

+
    +
  • a_n=3 \cdot 2^{n}.
    • + , +
+ + +

Find a closed form for the following sequence:

+
    +
  • 4, 5, 8, 13, 20,\ldots
    • + , +
+ + +

Find a recursive form for the following sequence:

+
    +
  • -3, 2, 7, 12, 17,\ldots
    • + , +
+ + + + + + + + + + +

Find the first 5 terms of the following sequence:

+
    +
  • a_n=5 \, n + 4.
    • + , +
+ + +

Find a closed form for the following sequence:

+
    +
  • 0, 1, 4, 9, 16,\ldots
    • + , +
+ + +

Find a recursive form for the following sequence:

+
    +
  • 2, -1, \frac{1}{2}, -\frac{1}{4}, \frac{1}{8},\ldots
    • + , +
+ + + + + + + + + + + + + Videos +
+ + Video: Define and use explicit and recursive formulas for sequences +
+ + + +
\ No newline at end of file diff --git a/calculus/source/08-SQ/02.ptx b/calculus/source/08-SQ/02.ptx new file mode 100644 index 00000000..9dc8ee99 --- /dev/null +++ b/calculus/source/08-SQ/02.ptx @@ -0,0 +1,419 @@ + + +
+ Sequence Properties and Limits (SQ2) + + + + + Activities + + +

We will consider the function f(x) = \displaystyle \frac{4x+8}{x}.

 + +

Compute the limit \displaystyle \lim_{x\to \infty} \frac{4x+8}{x}.

+
    +
  1. 0
    2. +
  2. 8
    3. +
  3. 1
    4. +
  4. 4
    5. +
+ + + +

Determine on which intervals f(x) is increasing and/or decreasing. (Hint: compute f'(x) first.)

+ + +

Which statement best describes f(x) for x>0?

+
    +
  1. f(x) is bounded above by 4
    2. +
  2. f(x) is bounded below by 4
    3. +
  3. f(x) is bounded above and below by 4
    4. +
  4. f(x) is not bounded above
    5. +
  5. f(x) is not bounded below
    6. +
+ + + + + + + +

+ Given a sequence \{x_n\}: +

    +
  • \{x_n\} is monotonically increasing if x_{n+1}>x_n for every choice of n.
    • +
  • \{x_n\} is monotonically non-decreasing if x_{n+1}\geq x_n for every choice of n.
    • +
  • \{x_n\} is monotonically decreasing if x_{n+1} < x_n for every choice of n.
    • +
  • \{x_n\} is monotonically non-increasing if x_{n+1}\leq x_n for every choice of n.
    • +
+ + All of these sequences would be monotonic. +

+ + + + + + +

+ Consider the sequence \left\{\displaystyle \frac{(-1)^n}{n}\right\}_{n=1}^\infty. +

+ + +

+ Compute x_{n+1}-x_n. +

+ + +

+ Which of the following is true about x_{n+1}-x_n? There can be more or less than one answer. +

    +
  1. x_{n+1}-x_n> 0 for every choice of n.
    2. +
  2. x_{n+1}-x_n\geq 0 for every choice of n.
    3. +
  3. x_{n+1}-x_n < 0 for every choice of n.
    4. +
  4. x_{n+1}-x_n\leq 0 for every choice of n.
    5. +
+

+ + +

+ Which of the following (if any) describe \left\{\displaystyle \frac{(-1)^n}{n}\right\}_{n=1}^\infty? +

    +
  1. Monotonically increasing.
    2. +
  2. Monotonically non-decreasing.
    3. +
  3. Monotonically decreasing.
    4. +
  4. Monotonically non-increasing.
    5. +
+

+ + + + + + +

+ Consider the sequence \left\{\displaystyle \frac{n^2+1}{n}\right\}_{n=1}^\infty. +

+ + +

+ Compute x_{n+1}-x_n. +

+ + +

+ Which of the following is true about x_{n+1}-x_n? There can be more or less than one answer. +

    +
  1. x_{n+1}-x_n> 0 for every choice of n.
    2. +
  2. x_{n+1}-x_n\geq 0 for every choice of n.
    3. +
  3. x_{n+1}-x_n < 0 for every choice of n.
    4. +
  4. x_{n+1}-x_n\leq 0 for every choice of n.
    5. +
+

+ + +

+ Which of the following (if any) describe \left\{\displaystyle \frac{n^2+1}{n}\right\}_{n=1}^\infty? +

    +
  1. Monotonically increasing.
    2. +
  2. Monotonically non-decreasing.
    3. +
  3. Monotonically decreasing.
    4. +
  4. Monotonically non-increasing.
    5. +
+

+ + + + + + +

+ Consider the sequence \left\{\displaystyle \frac{n+1}{n}\right\}_{n=1}^\infty. +

+ + +

+ Compute x_{n+1}-x_n. +

+ + +

+ Which of the following is true about x_{n+1}-x_n? There can be more or less than one answer. +

    +
  1. x_{n+1}-x_n> 0 for every choice of n.
    2. +
  2. x_{n+1}-x_n\geq 0 for every choice of n.
    3. +
  3. x_{n+1}-x_n < 0 for every choice of n.
    4. +
  4. x_{n+1}-x_n\leq 0 for every choice of n.
    5. +
+

+ + +

+ Which of the following (if any) describe \left\{\displaystyle \frac{n+1}{n}\right\}_{n=1}^\infty? +

    +
  1. Monotonically increasing.
    2. +
  2. Monotonically non-decreasing.
    3. +
  3. Monotonically decreasing.
    4. +
  4. Monotonically non-increasing.
    5. +
+

+ + + + + + +

+ Consider the sequence \left\{\displaystyle \frac{2}{3^n}\right\}_{n=0}^\infty. +

+ + +

+ Compute x_{n+1}-x_n. +

+ + +

+ Which of the following is true about x_{n+1}-x_n? There can be more or less than one answer. +

    +
  1. x_{n+1}-x_n> 0 for every choice of n.
    2. +
  2. x_{n+1}-x_n\geq 0 for every choice of n.
    3. +
  3. x_{n+1}-x_n < 0 for every choice of n.
    4. +
  4. x_{n+1}-x_n\leq 0 for every choice of n.
    5. +
+

+ + +

+ Which of the following (if any) describe \left\{\displaystyle \frac{2}{3^n}\right\}_{n=0}^\infty? +

    +
  1. Monotonically increasing.
    2. +
  2. Monotonically non-decreasing.
    3. +
  3. Monotonically decreasing.
    4. +
  4. Monotonically non-increasing.
    5. +
+

+ + + + + + +

+ A sequence \{x_n\} is bounded if there are real numbers b_u, b_{\ell} such that b_{\ell}\leq x_n \leq b_u for every n. +

+ + + + + +

+ Consider the sequence \left\{\displaystyle \frac{(-1)^n}{n}\right\}_{n=1}^\infty from . +

+ + +

+ Is there a b_u such that x_n\leq b_u for every n? If so, what would be one such b_u? +

+ + + +

+ Is there a b_\ell such that b_\ell \leq x_n for every n? If so, what would be one such b_\ell? +

+ + + +

+ Is \left\{\displaystyle \frac{(-1)^n}{n}\right\}_{n=1}^\infty bounded? +

+ + + + + + +

+ Consider the sequence \left\{\displaystyle \frac{n^2+1}{n}\right\}_{n=1}^\infty from . +

+ + +

+ Is there a b_u such that x_n\leq b_u for every n? If so, what would be one such b_u? +

+ + + +

+ Is there a b_\ell such that b_\ell \leq x_n for every n? If so, what would be one such b_\ell? +

+ + + +

+ Is \left\{\displaystyle \frac{n^2+1}{n}\right\}_{n=1}^\infty bounded? +

+ + + + + + +

+ Consider the sequence \left\{\displaystyle \frac{n+1}{n}\right\}_{n=1}^\infty from . +

+ + +

+ Is there a b_u such that x_n\leq b_u for every n? If so, what would be one such b_u? +

+ + + +

+ Is there a b_\ell such that b_\ell \leq x_n for every n? If so, what would be one such b_\ell? +

+ + + +

+ Is \left\{\displaystyle \frac{n+1}{n}\right\}_{n=1}^\infty bounded? +

+ + + + + + +

+ Consider the sequence \left\{\displaystyle \frac{2}{3^n}\right\}_{n=1}^\infty from . +

+ + +

+ Is there a b_u such that x_n\leq b_u for every n? If so, what would be one such b_u? +

+ + + +

+ Is there a b_\ell such that b_\ell \leq x_n for every n? If so, what would be one such b_\ell? +

+ + + +

+ Is \left\{\displaystyle \frac{2}{3^n}\right\}_{n=1}^\infty bounded? +

+ + + + + + +

+ Given a sequence \{x_n\}, we say x_n has limit L, denoted \lim_{n\to\infty} x_n=L if we can make x_n as close to L as we like by making n sufficiently large. If such an L exists, we say \{x_n\} converges to L. If no such L exists, we say \{x_n\} does not converge. +

+ + + + + + +

+ For each of the following, determine if the sequence converges. +

    +
  1. \left\{\displaystyle \frac{(-1)^n}{n}\right\}_{n=1}^\infty.
    2. +
  2. \left\{\displaystyle \frac{n^2+1}{n}\right\}_{n=1}^\infty.
    3. +
  3. \left\{\displaystyle \frac{n+1}{n}\right\}_{n=1}^\infty.
    4. +
  4. \left\{\displaystyle \frac{2}{3^n}\right\}_{n=0}^\infty.
    5. +
+

+ + +

+ Where possible, find the limit of the sequence. +

+ + + + + +

+ Determine to what value \left\{\displaystyle \frac{4n}{n+1}\right\}_{n=0}^\infty converges. +

+ + +

+ Which of the following ia most likely true about \left\{\displaystyle \frac{4n(-1)^n}{n+1}\right\}_{n=0}^\infty? +

    +
  1. \left\{\displaystyle \frac{4n(-1)^n}{n+1}\right\}_{n=0}^\infty converges to 4.
    2. +
  2. \left\{\displaystyle \frac{4n(-1)^n}{n+1}\right\}_{n=0}^\infty converges to 0.
    3. +
  3. \left\{\displaystyle \frac{4n(-1)^n}{n+1}\right\}_{n=0}^\infty converges to -4.
    4. +
  4. \left\{\displaystyle \frac{4n(-1)^n}{n+1}\right\}_{n=0}^\infty does not converge.
    5. +
+

+ + + + + + +

+ For each of the following sequences, determine which of the properties: monotonic, bounded and convergent, the sequence satisfies. If a sequence is convergent, determine to what it converges. + + \left\{\displaystyle 3n\right\}_{n=0}^\infty. + + + \left\{\displaystyle \frac{n^3}{3^n}\right\}_{n=0}^\infty. + + + \left\{\displaystyle \frac{n}{n+3}\right\}_{n=1}^\infty. + + + \left\{\displaystyle \frac{(-1)^n}{n+3}\right\}_{n=1}^\infty. + +

+ + + + + + +

+ If a sequence is monotonic and bounded, then it is convergent. +

+ + + + + + + + + + Videos +
+ + Video: Determine if a sequence is convergent, divergent, monotonic, or bounded, and compute limits of convergent sequences +
+ + + + + + + +
+ + diff --git a/calculus/source/08-SQ/03.ptx b/calculus/source/08-SQ/03.ptx new file mode 100644 index 00000000..de223d65 --- /dev/null +++ b/calculus/source/08-SQ/03.ptx @@ -0,0 +1,440 @@ + + +
+ Partial Sum Sequence (SQ3) + + + + + Activities + + +

+ Consider the sequence \displaystyle \{a_n\}_{n=0}^\infty=\left\{\frac{1}{2^n}\right\}_{n=0}^\infty. +

+ + +

+ Find the first 5 terms of this sequence. +

+ + +

+ Compute the following: +

    +
  1. a_0.
    2. +
  2. a_0+a_1.
    3. +
  3. a_0+a_1+a_2.
    4. +
  4. a_0+a_1+a_2+a_3.
    5. +
  5. a_0+a_1+a_2+a_3+a_4.
    6. +
+

+ + + + + +

+ Consider the sequence \displaystyle \{a_n\}_{n=1}^\infty=\left\{\frac{1}{n}\right\}_{n=1}^\infty. +

+ + +

+ Find the first 5 terms of this sequence. +

+ + +

+ Compute the following: +

    +
  1. a_1.
    2. +
  2. a_1+a_2.
    3. +
  3. a_1+a_2+a_3.
    4. +
  4. a_1+a_2+a_3+a_4.
    5. +
  5. a_1+a_2+a_3+a_4+a_5.
    6. +
+

+ + + + + +

+Given a sequence \{a_n\}_{n=0}^\infty define the k^{\text{th}} partial sum for this sequence to be +A_k=\sum_{i=0}^k a_i=a_0+a_1+a_2+\cdots+a_k. +Note that \{A_n\}_{n=0}^\infty=A_0, A_1, A_2, \ldots is itself a sequence called the partial sum sequence. +

+

+More generally, partial sums may be defined for any starting index. Given \{a_n\}_{n=N}^\infty, let +A_k=\sum_{i=N}^k a_i=a_N+a_{N+1}+a_{N+2}+\cdots+a_k. +

+ + + + + +

+ Let a_n=\frac{2}{3^n}. Find the following partial sums of the sequence \{a_n\}_{n=0}^\infty. +

+ + A_0. + A_1. + A_2. + A_3. + A_{100}. + + + + +

Consider the sequence a_n=\frac{2}{3^n}. What is the best way to find the 100th partial sum A_{100}? +

    +
  1. Sum the first 101 terms of the sequence \{a_n\}.
    2. +
  2. Find a closed form for the partial sum sequence \{A_n\}.
    3. +
+

+ + + + + +

Expand the following polynomial products, and then reduce to as few summands as possible. +

    +
  1. (1-x)(1+x+x^2).
    2. +
  2. (1-x)(1+x+x^2+x^3).
    3. +
  3. (1-x)(1+x+x^2+x^3+x^4).
    4. +
  4. (1-x)(1+x+x^2+\cdots+x^n), where n is any nonnegative integer.
    5. +
+

+ + + + + +

Suppose \displaystyle S_5=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}. Without actually computing this sum, which of the following is equal to \left(1-\frac{1}{2}\right)S_5? +

    +
  1. \displaystyle\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}-\frac{1}{64}.
    2. +
  2. \displaystyle1-\frac{1}{64}.
    3. +
  3. \displaystyle1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}.
    4. +
+

+ + + + + +

+ Recall from that \displaystyle A_{100}=2+\frac{2}{3}+\frac{2}{3^2}+\frac{2}{3^3}+\frac{2}{3^4}+\cdots+\frac{2}{3^{100}}=2\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\cdots+\frac{1}{3^{100}}\right). +

+ + +

+ Which of the following is equal to \displaystyle\left(1-\frac{1}{3}\right)A_{100}? +

    +
  1. \displaystyle1-\frac{1}{3^{101}}.
    2. +
  2. \displaystyle1-\frac{1}{3^{100}}.
    3. +
  3. \displaystyle2\left(1-\frac{1}{3^{101}}\right).
    4. +
  4. \displaystyle2\left(1-\frac{1}{3^{100}}\right).
    5. +
+

+ + +

+ Based on your previous choice, write out an expression for A_{100}. +

+ + + + + + +

+ Suppose that \displaystyle \{b_n\}_{n=0}^\infty=\{(-2)^n\}_{n=0}^\infty=\{1,-2,4,-8,\ldots\}. Let B_n=\displaystyle\sum_{i=0}^n b_i be the nth partial sum of \{b_n\}. +

+ + +

+ Find simple expressions for the following: +

    +
  1. (1-(-2))B_{10}.
    2. +
  2. (1-(-2))B_{30}.
    3. +
  3. +

    (1-(-2))B_{n}. Choose from the following:

    +
      +
    1. 1+(-2)^n.
      2. +
    2. 1-(-2)^n.
      3. +
    3. 1+(-2)^{n+1}.
      4. +
    4. 1-(-2)^{n+1}.
      5. +
    5. 1-2^n.
      6. +
    +
    4. +
+

+ + +

+ Based on your previous answers, solve for the following: +

    +
  1. B_{10}.
    2. +
  2. B_{30}.
    3. +
  3. +

    B_{n}. Choose from the following:

    +
      +
    1. \displaystyle \frac{1-(-2)^{n+1}}{1-(-2)}
      2. +
    2. \displaystyle \frac{1-(-2)^{n+1}}{1-2}
      3. +
    3. \displaystyle \frac{1-(-2)^{n+1}}{1+(-2)}
      4. +
    4. \displaystyle \frac{1-(-2)^{n}}{1-2}
      5. +
    5. \displaystyle \frac{1-(-2)^{n}}{1-(-2)}
      6. +
    +
    4. +
+

+ + + + + + +

+ Consider the following sequences: +

    +
  1. \displaystyle\{a_n\}_{n=0}^\infty=\left\{\left(-\frac{2}{3}\right)^n\right\}_{n=0}^\infty.
    2. +
  2. \displaystyle\{b_n\}_{n=0}^\infty=\left\{2\cdot\left(-1\right)^n\right\}_{n=0}^\infty.
    3. +
  3. \displaystyle\{c_n\}_{n=0}^\infty=\left\{-3\cdot \left(1.2\right)^n\right\}_{n=0}^\infty.
    4. +
+

+ + +

+ Find the closed form for the nth partial sum for the geometric sequence A_n=\displaystyle\sum_{i=0}^n a_i=\displaystyle\sum_{i=0}^n \left(-\frac{2}{3}\right)^n. +

    +
  1. \displaystyle \frac{3}{5}\left(1-\left(-\frac{2}{3}\right)^{n+1}\right).
    2. +
  2. \displaystyle \frac{5}{3}\left(1-\left(-\frac{2}{3}\right)^{n+1}\right).
    3. +
  3. \displaystyle \frac{5}{3}\left(1+\frac{2}{3}\left(\frac{2}{3}\right)^{n}\right).
    4. +
  4. \displaystyle \frac{3}{5}\left(1+\frac{2}{3}\left(\frac{2}{3}\right)^{n}\right).
    5. +
  5. \displaystyle 1-\left(-\frac{2}{3}\right)^{n+1}.
    6. +
+

+ + +

+ Find the closed form for the nth partial sum for the geometric sequence B_n=\displaystyle\sum_{i=0}^n b_i=\displaystyle\sum_{i=0}^n 2\cdot\left(-1\right)^n. +

    +
  1. \displaystyle 2^{n+1}.
    2. +
  2. \displaystyle 1-(-1)^{n+1}.
    3. +
  3. \displaystyle 1+(-1)^{n}.
    4. +
  4. \displaystyle 2(1+(-1)^{n}).
    5. +
  5. \displaystyle 2(1-(-1)^{n+1}).
    6. +
+

+ + +

+ Find the closed form for the nth partial sum for the geometric sequence C_n=\displaystyle\sum_{i=0}^n c_i=\displaystyle\sum_{i=0}^n -3\cdot \left(1.2\right)^n. +

+ + + +

+ Given the closed forms you found in , which of the following limits are defined? If defined, what is the limit? +

    +
  1. \displaystyle\lim_{n\to\infty}A_n.
    2. +
  2. \displaystyle\lim_{n\to\infty}B_n.
    3. +
  3. \displaystyle\lim_{n\to\infty}C_n.
    4. +
+

+ + + + +

+ Given a sequence a_n, we define the limit of the series \displaystyle\sum_{n=k}^\infty a_n:=\lim_{n\to \infty} A_n where A_n=\displaystyle \sum_{i=k}^n a_i. We call \displaystyle\sum_{n=k}^\infty a_n an infinite series. +

+ + + + + +

+ Which of the following series are infinite? +

    +
  1. \displaystyle\sum_{n=0}^\infty 3(0.8)^n.
    2. +
  2. \displaystyle\sum_{n=0}^\infty 2\left(\frac{5}{4}\right)^n.
    3. +
  3. \displaystyle\sum_{n=0}^\infty \left(\frac{5}{6}\right)^n.
    4. +
  4. \displaystyle\sum_{n=0}^\infty \frac{1}{2}\left(81\right)^n.
    5. +
  5. \displaystyle\sum_{n=0}^\infty 10\left(-\frac{1}{5}\right)^n.
    6. +
+

+ + + + + + +

+ Let \displaystyle\{a_n\}_{n=1}^\infty=\left\{\frac{1}{n}-\frac{1}{n+1}\right\}=1-\frac{1}{2}, \frac{1}{2}-\frac{1}{3}, \frac{1}{3}-\frac{1}{4},\ldots. Let \displaystyle A_n=\sum_{i=1}^na_i=\sum_{i=1}^n \left(\frac{1}{i}-\frac{1}{i+1} \right). +

+

+ Which of the following is the best strategy for evaluating \displaystyle A_{4}=\left(1-\frac{1}{2} \right)+\left(\frac{1}{2}-\frac{1}{3} \right)+\left(\frac{1}{3}-\frac{1}{4} \right)+\left(\frac{1}{4}-\frac{1}{5} \right)? +

    +
  1. Compute \displaystyle A_{4}=\left(1-\frac{1}{2} \right)+\left(\frac{1}{2}-\frac{1}{3} \right)+\left(\frac{1}{3}-\frac{1}{4} \right)+\left(\frac{1}{4}-\frac{1}{5} \right)=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}, then evaluate the sum.
    2. +
  2. Rewrite \displaystyle A_{4}=\left(1-\frac{1}{2} \right)+\left(\frac{1}{2}-\frac{1}{3} \right)+\left(\frac{1}{3}-\frac{1}{4} \right)+\left(\frac{1}{4}-\frac{1}{5} \right)=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)-\frac{1}{5}, then simplify.
    3. +
+

+ + + + + + +

+ Recall from that \displaystyle\{a_n\}_{n=1}^\infty=\left\{\frac{1}{n}-\frac{1}{n+1}\right\} and \displaystyle A_n=\sum_{i=1}^na_i=\sum_{i=1}^n \left(\frac{1}{i}-\frac{1}{i+1} \right). +

+

+ Compute the following partial sums: +

    +
  1. A_3.
    2. +
  2. A_{10}.
    3. +
  3. A_{100}.
    4. +
+

+ + + + + + +

+ Recall from that \displaystyle\{a_n\}_{n=1}^\infty=\left\{\frac{1}{n}-\frac{1}{n+1}\right\} and \displaystyle A_n=\sum_{i=1}^na_i=\sum_{i=1}^n \left(\frac{1}{i}-\frac{1}{i+1} \right). +

+

+ Which of the following is equal to A_n? +

    +
  1. n-\frac{1}{n+1}.
    2. +
  2. 1-\frac{1}{n}.
    3. +
  3. 1-\frac{1}{n+1}.
    4. +
  4. 1-\frac{1}{i}.
    5. +
  5. 1-\frac{1}{i+1}.
    6. +
+

+ + + + + + +

+ Given a sequence \{x_n\}_1^\infty and a sequence of the form \{s_n\}_1^\infty:=\{x_n-x_{n+1}\}_1^\infty we call the series S_n=\displaystyle\sum_{i=1}^n s_i=\sum_{i=1}^n(x_i-x_{i+1}) to be a telescoping series. +

+ + + + + +

+ Given a telescoping series S_n=\displaystyle\sum_{i=1}^n s_i=\sum_{i=1}^n(x_i-x_{i+1}), find: +

    +
  1. S_2.
    2. +
  2. S_{10}.
    3. +
  3. +

    Choose S_{n} from the following options:

    +
      +
    1. x_1-x_n
      2. +
    2. x_1-x_{n+1}
      3. +
    3. x_1-x_{n-1}
      4. +
    4. x_1-x_n+1
      5. +
    5. x_1-x_n-1
      6. +
    +
    4. +
+

+ + + + + + +

+ For each of the following telescoping series, find the closed form for the nth partial sum. +

    +
  1. S_n=\displaystyle\sum_{i=1}^n (2^{-i}-(2^{-i-1})).
    2. +
  2. S_n=\displaystyle\sum_{i=1}^n (i^2-(i+1)^2).
    3. +
  3. S_n=\displaystyle\sum_{i=1}^n \left( \frac{1}{2i+1}-\frac{1}{2i+3}\right).
    4. +
+

+ + + +

Given the closed forms you found in , determine which of the following telescoping series converge. If so, to what value does it converge?

+
    +
  1. \displaystyle\sum_{i=1}^\infty (2^{-i}-(2^{-i-1})).
    2. +
  2. \displaystyle\sum_{i=1}^\infty (i^2-(i+1)^2).
    3. +
  3. \displaystyle\sum_{i=1}^\infty \left( \frac{1}{2i+1}-\frac{1}{2i+3}\right).
    4. +
+ + + + + +

Consider the partial sum sequence \displaystyle A_n=\left(-2\right)+\left(\frac{2}{3}\right)+\left(-\frac{2}{9}\right)+\cdots+\left(-2\cdot \left( -\frac{1}{3}\right)^n \right).

 + +

Find a closed form for A_n.

 +

Does \{A_n\} converge? If so, to what value?

 + + + + +

Consider the partial sum sequence \displaystyle B_n=\sum_{i=1}^n \left( \frac{1}{5 \, i + 2}-\frac{1}{5 \, i + 7} \right).

 + +

Find a closed form for B_n.

 +

Does \{B_n\} converge? If so, to what value?

 + + + + + + + + Videos +
+ + Video: Compute the first few terms of a telescoping or geometric partial sum sequence, and find a closed form for this sequence, and compute its limit. + + +
+ + + + + + +
diff --git a/calculus/source/08-SQ/04.ptx b/calculus/source/08-SQ/04.ptx new file mode 100644 index 00000000..94de1862 --- /dev/null +++ b/calculus/source/08-SQ/04.ptx @@ -0,0 +1,207 @@ + + +
+ Geometric Series (SQ4) + + + + + Activities + + + + + +

+ Recall from that for any real numbers a, r and \displaystyle S_n=\sum_{i=0}^n ar^i that: + + S_n=\sum_{i=0}^n ar^i &= a+ar+ar^2+\cdots ar^n + (1-r)S_n=(1-r)\sum_{i=0}^n ar^i&= (1-r)(a+ar+ar^2+\cdots ar^n) + (1-r)S_n=(1-r)\sum_{i=0}^n ar^i&= a-ar^{n+1} + S_n&=a\frac{1-r^{n+1}}{1-r}. + +

+ + +

+ Using , for which values of r does \displaystyle \sum_{n=0}^\infty ar^n converges? +

    +
  1. |r|>1.
    2. +
  2. |r|=1.
    3. +
  3. |r|<1.
    4. +
  4. The series converges for every value of r.
    5. +
+

+ + + +

+ Where possible, determine what value \displaystyle \sum_{n=0}^\infty ar^n converges to. +

+ + + + + +

+Geometric series are sums of the form \sum_{n=0}^\infty ar^n=a+ar+ar^2+ar^3+\dots, +where a and r are real numbers. When |r|<1 this series converges to the value +\displaystyle\frac{a}{1-r}. Otherwise, the geometric series diverges. +

+ + + + +

+ Consider the infinite series 5+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\cdots. + +

+ + +

+ Complete the following rearrangement of terms. + +5+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\cdots & = \unknown + \left(3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\cdots\right) +& = \unknown + \sum_{n=0}^\infty \unknown \cdot \left(\frac{1}{\unknown}\right)^n + +

+ + +

+Since |\frac{1}{\unknown}|<1, this series converges. Use the formula \sum_{n=0}^\infty ar^n=\frac{a}{1-r} +to find the value of this series. +

+
    +
  1. \frac{7}{2}
    2. +
  2. \frac{13}{2}
    3. +
  3. 8
    4. +
  4. 10
    5. +
+ + + + + + +

+ Complete the following calculation, noting |0.6|<1: + +\sum_{n=2}^\infty 2(0.6)^n &=\left(\sum_{n=0}^\infty 2(0.6)^n\right) - \unknown - \unknown + & = \left(\frac{\unknown}{1-\unknown}\right)- \unknown - \unknown + + What does this simplify to? +

+
    +
  1. 1.1
    2. +
  2. 1.4
    3. +
  3. 1.8
    4. +
  4. 2.1
    5. +
+ + + + + +

+ Given a series that appears to be mostly geometric such as + 3+(1.1)^3+(1.1)^4+\cdots(1.1)^n+\cdots + we can always rewrite it as the sum of a standard geometric series + with some finite modification, in this case: + -0.31 + \sum_{n=0}^\infty (1.1)^n +

+

+Thus the original series converges if and only if \displaystyle \sum_{n=0}^\infty (1.1)^n converges. +

+

+When the series diverges as in this example, then the reason why (|1.1|\geq 1) can be +seen without any modification of the original series. +

+ + + + + + +

+ For each of the following modified geometric series, determine without rewriting if they converge or diverge. +

+ + +

\displaystyle -7+\left( -\frac{3}{7}\right)^2+\left( -\frac{3}{7}\right)^3+\cdots.

+ + +

-6+\left(\frac{5}{4}\right)^3+\left(\frac{5}{4}\right)^4+\cdots.

+ + +

\displaystyle 4+\sum_{n=4}^\infty \left(\frac{2}{3}\right)^n.

+ + +

8-1+1-1+1-1+\cdots.

+ + + + + +

+ Find the value of each of the following convergent series. +

+ + +

-1 + \sum_{n = 1 }^\infty 2\cdot\left(\frac{1}{2}\right)^n.

+ + +

\displaystyle -7+\left( -\frac{3}{7}\right)^2+\left( -\frac{3}{7}\right)^3+\cdots.

+ + +

\displaystyle 4+\sum_{n=4}^\infty \left(\frac{2}{3}\right)^n.

+ + + + + + + + + + Videos +
+ + Video: Determine if a geometric series converges, and if so, the value it converges to. + + + + +
+ + + + + +
diff --git a/calculus/source/08-SQ/05.ptx b/calculus/source/08-SQ/05.ptx new file mode 100644 index 00000000..111f5abf --- /dev/null +++ b/calculus/source/08-SQ/05.ptx @@ -0,0 +1,389 @@ + + +
+ Basic Convergence Tests (SQ5) + + + + + Activities + + + + +

+ Which of the following series seem(s) to diverge? It might be helpful to write out the first several terms. +

+ +
    +
  1. \displaystyle \sum_{n=0}^\infty n^2.
    2. +
  2. \displaystyle \sum_{n=1}^\infty \displaystyle\frac{n+1}{n}.
    3. +
  3. \displaystyle \sum_{n=0}^\infty (-1)^n.
    4. +
  4. \displaystyle \sum_{n=1}^\infty \frac{1}{n}.
    5. +
  5. \displaystyle \sum_{n=1}^\infty \frac{1}{n^2}.
    6. +
+ + + + + + +

+ If the series \displaystyle\sum a_n is convergent, + then \displaystyle\lim_{n\rightarrow\infty} a_n=0. +

+ + + + + + + The Divergence (<m>n^{th}</m> term) Test + +

+ If the \displaystyle\lim_{n\rightarrow\infty} a_n\neq 0, + then \displaystyle\sum a_n diverges. +

+ + + + +

Which of the series from diverge by ?

+ + + + +

+ If a_n>0 for all n, then \displaystyle\sum a_n is convergent if and only if the sequence of partial sums is bounded from above. +

+ + + + + +

+ Consider the so-called harmonic series, \displaystyle \sum_{n=1}^\infty \displaystyle\frac{1}{n}, and let S_n be its n^{th} partial sum. +

+ + +

+ Determine which of the following inequalities hold(s). +

+
    +
  1. \displaystyle\frac{1}{3}+\frac{1}{4}\lt \frac{1}{2}.
    2. +
  2. \displaystyle\frac{1}{3}+\frac{1}{4}\gt \frac{1}{2}.
    3. +
  3. S_4\geq S_2+\displaystyle\frac{1}{2}.
    4. +
  4. S_4\leq S_2+\displaystyle\frac{1}{2}.
    5. +
  5. S_4= S_2+\displaystyle\frac{1}{2}.
    6. +
+ + +

+ Determine which of the following inequalities hold(s). +

+
    +
  1. \displaystyle\frac{1}{2}\lt \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}.
    2. +
  2. \displaystyle\frac{1}{2}\gt \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}.
    3. +
  3. S_8=S_4+\displaystyle\frac{1}{2}.
    4. +
  4. S_8\geq S_4+\displaystyle\frac{1}{2}.
    5. +
  5. S_8\leq S_4+\displaystyle\frac{1}{2}.
    6. +
+ + + + +

+ In , we found that S_4\geq S_2+\displaystyle\frac{1}{2} and S_8\geq S_4+\displaystyle\frac{1}{2}. Based on these inequalities, which statement seems most likely to hold? +

+
    +

  1. The harmonic series converges.

    2. +

  2. The harmonic series diverges.

    3. +
+ + + + +

+ Consider the series \displaystyle \sum_{n=1}^\infty \displaystyle\frac{1}{n^2}. +

+ + +

+ We want to compare this series to an improper integral. Which of the following is the best candidate? +

+
    +
  1. \displaystyle\int_1^\infty x^2 \, dx.
    2. +
  2. \displaystyle\int_1^\infty \displaystyle\frac{1}{x^3} \, dx.
    3. +
  3. \displaystyle\int_1^\infty \displaystyle\frac{1}{x^2} \, dx.
    4. +
  4. \displaystyle\int_1^\infty \displaystyle\frac{1}{x} \, dx.
    5. +
  5. \displaystyle\int_1^\infty x \, dx.
    6. +
+ + + +

+ Select the true statements below. +

    +
  1. The sum \displaystyle \sum_{n=1}^\infty \frac{1}{n^2} corresponds to approximating the integral chosen above using left Riemann sums where \Delta x=1.
    2. +
  2. The sum \displaystyle \sum_{n=1}^\infty \frac{1}{n^2} corresponds to approximating the integral chosen above using right Riemann sums where \Delta x=1.
    3. +
  3. The sum \displaystyle \sum_{n=2}^\infty \frac{1}{n^2} corresponds to approximating the integral chosen above using left Riemann sums where \Delta x=1.
    4. +
  4. The sum \displaystyle \sum_{n=2}^\infty \frac{1}{n^2} corresponds to approximating the integral chosen above using right Riemann sums where \Delta x=1.
    5. +
+

+ + + +

+ Using the Riemann sum interpretation of the series, identify which of the following inequalities holds. +

+
    +
  1. \displaystyle \sum_{n=1}^\infty \frac{1}{n^2} \leq \displaystyle\int_1^\infty \displaystyle\frac{1}{x^2} \, dx.
    2. +
  2. \displaystyle \sum_{n=1}^\infty \frac{1}{n^2} \geq \displaystyle\int_1^\infty \displaystyle\frac{1}{x^2} \, dx.
    3. +
  3. \displaystyle \sum_{n=2}^\infty \frac{1}{n^2} \geq \displaystyle\int_1^\infty \displaystyle\frac{1}{x^2} \, dx.
    4. +
  4. \displaystyle \sum_{n=2}^\infty \frac{1}{n^2} \leq \displaystyle\int_1^\infty \displaystyle\frac{1}{x^2} \, dx.
    5. +
+ + + +

+ What can we say about the improper integral \displaystyle\int_1^\infty \displaystyle\frac{1}{x^2} \, dx? +

    +
  1. This improper integral converges.
    2. +
  2. This improper integral diverges.
    3. +
+

+ + + +

+ What do you think is true about the series \displaystyle \sum_{n=1}^\infty \frac{1}{n^2}? +

    +
  1. The series converges.
    2. +
  2. The series diverges.
    3. +
+

+ + + + + The Integral Test + +

+ Let \{a_n\} be a sequence of positive numbers. + If f(x) is continuous, positive, and decreasing, + and there is some positive integer N such that f(n)=a_n for all n\geq N, + then \displaystyle \sum_{n=N}^\infty a_n and \displaystyle\int_N^\infty \displaystyle f(x) \, dx both converge or both diverge. +

+ + + + + +

Consider the p-series \displaystyle \sum_{n=1}^\infty \frac{1}{n^p}.

+ + +

+ Recall that the harmonic series diverges. What value of p corresponds to the harmonic series? +

+
    +
  1. p=-1.
    2. +
  2. p=1.
    3. +
  3. p=-2.
    4. +
  4. p=2.
    5. +
  5. p=0.
    6. +
+ + +

+ From , what can we conclude about the p-series with p=2? +

    +
  1. There is not enough information to draw a conclusion.
    2. +
  2. This series converges.
    3. +
  3. This series diverges.
    4. +
+

+ + + + + The <m>p</m>-Test + +

+The series \displaystyle \sum_{n=1}^\infty \displaystyle\frac{1}{n^p} converges for p\gt 1, and diverges otherwise. +

+ + + + + +

+ Consider the series \displaystyle \sum_{n=1}^\infty \displaystyle\frac{1}{n^2+1}. +

+ + +

+ If we aim to use the integral test, what is an appropriate choice for f(x)? +

+
    +
  1. \displaystyle \frac{1}{x^2}.
    2. +
  2. x^2+1.
    3. +
  3. \displaystyle \frac{1}{x^2+1}.
    4. +
  4. x^2.
    5. +
  5. \displaystyle \frac{1}{x}.
    6. +
+ + + +

+ Does the series converge or diverge by ? +

+ + + + +

Prove .

+ + + +

Which of the following statements seem(s) most likely to be true? +

    +
  1. \displaystyle \sum_{n=1}^\infty (-1)^n \frac{1}{n} diverges.
    2. +
  2. \displaystyle \sum_{n=1}^\infty (-1)^n \frac{1}{n} converges.
    3. +
  3. \displaystyle \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} converges.
    4. +
  4. \displaystyle \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} diverges.
    5. +
+

+ + + + The Alternating Series Test (Leibniz's Theorem) + +

+ The series \displaystyle\sum (-1)^{n+1}u_n converges if all of the following conditions are satisfied: +

    +
  1. u_n is always positive,
    2. +
  2. there is an integer N such that u_n\geq u_{n+1} for all n\geq N, and
    3. +
  3. \displaystyle\lim_{n\rightarrow\infty}u_n=0.
    4. +
+

+ + + + +

What conclusions can you now make? +

    +
  1. \displaystyle \sum_{n=1}^\infty (-1)^n \frac{1}{n} diverges.
    2. +
  2. \displaystyle \sum_{n=1}^\infty (-1)^n \frac{1}{n} converges.
    3. +
  3. \displaystyle \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} converges.
    4. +
  4. \displaystyle \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} diverges.
    5. +
+

+ + + + +

+ For each of the following series, use the Divergence, Alternating Summation or Integral test to determine if the series converges. +

+ + + \displaystyle \sum_{n=1}^\infty \frac{2 \, {\left(n^{2} + 2\right)}}{n^{2}}. + + + \displaystyle \sum_{n=1}^\infty \frac{1}{n^{4}}. + + + \displaystyle \sum_{n=1}^\infty \frac{3 \, \left(-1\right)^{n}}{4 \, n}. + + + + + + + The Alternating Series Estimation Theorem + + +

+ If the alternating series \displaystyle\sum a_n=\displaystyle\sum (-1)^{n+1}u_n converges to L and has n^{th} partial sum S_n, + then for n\geq N (as in the alternating series test): +

    +
  1. |L-S_n| is less than |a_{n+1}|, and
    2. +
  2. (L-S_n) has the same sign as a_{n+1}.
    3. +
+

+ + + + + + + + +

+ Consider the so-called alternating harmonic series, \displaystyle \sum_{n=1}^\infty \displaystyle\frac{(-1)^{n+1}}{n}. +

+ + +

+ Use the alternating series test to determine if the series converges. +

+ + + +

+ If so, estimate the series using the first 3 terms. +

+ + + + + + + + + Videos +
+ + Video: Use the divergence, alternating series, and integral tests to determine if a series converges or diverges + + + + +
+ + + + + + +
diff --git a/calculus/source/08-SQ/06.ptx b/calculus/source/08-SQ/06.ptx new file mode 100644 index 00000000..03ef0984 --- /dev/null +++ b/calculus/source/08-SQ/06.ptx @@ -0,0 +1,660 @@ + + +
+ Comparison Tests (SQ6) + + + + + Activities + + + +

+ Let \{a_n\}_{n=1}^\infty be a sequence, with infinite series \displaystyle \sum_{n=1}^\infty a_n=a_1+a_2+\cdots . Suppose \{b_n\}_{n=1}^\infty is a sequence where each b_n=3a_n, whith infinite series \displaystyle \sum_{n=1}^\infty b_n=\sum_{n=1}^\infty 3a_n=3a_1+3a_2+\cdots . +

+ + +

+ If \displaystyle \sum_{n=1}^\infty a_n=5 what can be said about \displaystyle\sum_{n=1}^\infty b_n? +

    +
  1. \displaystyle\sum_{n=1}^\infty b_n converges but the value cannot be determined.
    2. +
  2. \displaystyle\sum_{n=1}^\infty b_n converges to 3\cdot 5=15.
    3. +
  3. \displaystyle\sum_{n=1}^\infty b_n converges to some value other than 15.
    4. +
  4. \displaystyle\sum_{n=1}^\infty b_n diverges.
    5. +
  5. It cannot be determined whether \displaystyle\sum_{n=1}^\infty b_n converges or diverges.
    6. +
+

+ + + + +

+ If \displaystyle\sum a_n=A, + \displaystyle\sum b_n=B, and + c is a real number, then +

    +
  1. \displaystyle\sum ca_n=c\displaystyle\sum a_n=cA, and
    2. +
  2. \displaystyle\sum (a_n+b_n)=\sum a_n+\displaystyle\sum b_n=A+B.
    3. +
+

+ + + + +

+ If \displaystyle \sum_{n=1}^\infty a_n diverges, what can be said about \displaystyle\sum_{n=1}^\infty b_n? +

    +
  1. \displaystyle\sum_{n=1}^\infty b_n converges but the value cannot be determined.
    2. +
  2. \displaystyle\sum_{n=1}^\infty b_n converges and the value can be determined.
    3. +
  3. \displaystyle\sum_{n=1}^\infty b_n diverges.
    4. +
  4. It cannot be determined whether \displaystyle\sum_{n=1}^\infty b_n converges or diverges.
    5. +
+

+ + + + + + +

+ Using , we know the geometric series \displaystyle \sum_{n=0}^\infty \frac{1}{2^n}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^n}+\cdots=\frac{1}{1-\frac{1}{2}}=2. +

+ + + +

+ What can we say about the series \displaystyle 3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\cdots+\frac{3}{2^n}+\cdots? + +

    +
  1. \displaystyle 3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\cdots+\frac{3}{2^n}+\cdots converges to 3\cdot 2=6.
    2. +
  2. \displaystyle 3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\cdots+\frac{3}{2^n}+\cdots converges to some value other than 6.
    3. +
  3. \displaystyle 3+\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\cdots+\frac{3}{2^n}+\cdots diverges.
    4. + +
+

+ + + +

+ What do you think we can say about the series \displaystyle \frac{3.1}{2}+\frac{3.01}{4}+\frac{3.001}{8}+\cdots+\frac{3+(0.1)^n}{2^n}+\cdots? + +

    +
  1. \displaystyle 3+\frac{3.1}{2}+\frac{3.01}{4}+\frac{3.001}{8}+\cdots+\frac{3+(0.1)^n}{2^n}+\cdots converges to 3\cdot 2=6.
    2. +
  2. \displaystyle 3+\frac{3.1}{2}+\frac{3.01}{4}+\frac{3.001}{8}+\cdots+\frac{3+(0.1)^n}{2^n}+\cdots converges to some value other than 6.
    3. +
  3. \displaystyle 3+\frac{3.1}{2}+\frac{3.01}{4}+\frac{3.001}{8}+\cdots+\frac{3+(0.1)^n}{2^n}+\cdots diverges.
    4. + +
+

+ + + + + + + +

+ From , we know \displaystyle 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}+\cdots diverges. +

+ + + +

+ What can we say about the series \displaystyle 5+\frac{5}{2}+\frac{5}{3}+\frac{5}{4}+\cdots+\frac{5}{n}+\cdots? + +

    +
  1. \displaystyle 5+\frac{5}{2}+\frac{5}{3}+\frac{5}{4}+\cdots+\frac{5}{n}+\cdots converges to a known value we can compute.
    2. +
  2. \displaystyle 5+\frac{5}{2}+\frac{5}{3}+\frac{5}{4}+\cdots+\frac{5}{n}+\cdots converges to some unknown value.
    3. +
  3. \displaystyle 5+\frac{5}{2}+\frac{5}{3}+\frac{5}{4}+\cdots+\frac{5}{n}+\cdots diverges.
    4. + +
+

+ + + +

+ What do you think we can say about the series \displaystyle 4.9+\frac{4.99}{2}+\frac{4.999}{3}+\frac{4.9999}{4}+\cdots+\frac{5-(0.1)^n}{n}+\cdots? + +

    +
  1. \displaystyle 4.9+\frac{4.99}{2}+\frac{4.999}{3}+\frac{4.9999}{4}+\cdots+\frac{5-(0.1)^n}{n}+\cdots converges to a known value we can compute.
    2. +
  2. \displaystyle 4.9+\frac{4.99}{2}+\frac{4.999}{3}+\frac{4.9999}{4}+\cdots+\frac{5-(0.1)^n}{n}+\cdots converges to some unknown value.
    3. +
  3. \displaystyle 4.9+\frac{4.99}{2}+\frac{4.999}{3}+\frac{4.9999}{4}+\cdots+\frac{5-(0.1)^n}{n}+\cdots diverges.
    4. + +
+

+ + + + + The Limit Comparison Test + +

+ Let \displaystyle\sum a_n and \displaystyle\sum b_n be series with positive terms. + If + + \lim_{n \to \infty} \frac{b_n}{a_n} = c + + for some positive (finite) constant c, + then \displaystyle\sum a_n and \displaystyle\sum b_n either both converge or both diverge. +

+ + + + + + +

+ Recall that \displaystyle \sum_{n=1}^\infty \frac{1}{2^n} converges. +

+ + +

+ Let b_n=\frac{1}{n}. Compute \displaystyle \lim_{n\to\infty}\frac{\frac{1}{n}}{\frac{1}{2^n}}. +

+
    +
  1. -\infty.
    2. +
  2. 0.
    3. +
  3. \displaystyle \frac{1}{2}.
    4. +
  4. 1.
    5. +
  5. \infty.
    6. +
+ + +

+ Does \displaystyle\sum_{n=1}^\infty \frac{1}{n} converge or diverge? +

+ + +

+ Let \displaystyle b_n=\frac{1}{n^2}. Compute \displaystyle \lim_{n\to\infty}\frac{\frac{1}{n^2}}{\frac{1}{2^n}}. +

+
    +
  1. \infty.
    2. +
  2. \ln(2).
    3. +
  3. 1.
    4. +
  4. \displaystyle \frac{1}{2}.
    5. +
  5. -\infty.
    6. +
+ + +

+ Does \displaystyle\sum_{n=1}^\infty \frac{1}{n^2} converge or diverge? +

+ + +

+ Let \displaystyle\sum a_n and \displaystyle\sum b_n be series with positive terms. + If + + \lim_{n \to \infty} \frac{b_n}{a_n} + + diverges, can we conclude that \displaystyle \sum b_n converges or diverges? +

+ + + + + + +

+ We wish to determine if \displaystyle \sum_{n=1}^\infty \frac{1}{4^n-1} converges or diverges using . +

+ + +

+ Compute \displaystyle\lim_{n\to\infty}\frac{\frac{1}{4^n-1}}{\frac{1}{4^n}}. +

+ + + +

Does the geometric series \displaystyle \sum_{n=1}^\infty \frac{1}{4^n} converge or diverge by ?

+ + +

+ Does \displaystyle \sum_{n=1}^\infty \frac{1}{4^n-1} converge or diverge? +

+ + + + + + +

+ We wish to determine if \displaystyle \sum_{n=2}^\infty \frac{2}{\sqrt{n+3}} converges or diverges using . +

+ + +

+ To which of the following should we compare \displaystyle \{a_n\}=\left\{\frac{2}{\sqrt{n+3}}\right\}? +

    +
  1. \displaystyle \left\{\frac{1}{n}\right\}.
    2. +
  2. \displaystyle \left\{\frac{1}{\sqrt{n}}\right\}.
    3. +
  3. \displaystyle \left\{\frac{1}{n^2}\right\}.
    4. +
  4. \displaystyle \left\{\frac{1}{2^n}\right\}.
    5. +
+

+ + +

+ Compute \displaystyle \lim_{n\to\infty} \frac{b_n}{a_n}. +

+ + +

+ Compute \displaystyle \lim_{n\to\infty} \frac{a_n}{b_n}. +

+ + +

+ What is true about \displaystyle \lim_{n\to\infty} \frac{b_n}{a_n} and \displaystyle \lim_{n\to\infty} \frac{a_n}{b_n}? +

    +
  1. Their values are reciprocals.
    2. +
  2. Their values negative reciprocals.
    3. +
  3. They are both positive finite constants.
    4. +
  4. Only one value is a finite positive constant.
    5. +
  5. One value is 0 and the other value is infinite.
    6. +
+

+ + +

+ Does the series \displaystyle \sum_{n=2}^\infty \frac{1}{\sqrt{n}} converge or diverge? +

+ + +

+ Using your chosen sequence and , does \displaystyle \sum_{n=2}^\infty \frac{2}{\sqrt{n+3}} converge or diverge? +

+ + + + + + +

+ We wish to determine if \displaystyle \sum_{n=1}^\infty \frac{3}{n^2+8n+5} converges or diverges using . +

+ + +

+ To which of the following should we compare \displaystyle \{x_n\}=\left\{\frac{3}{n^2+8n+5} \right\}? +

    +
  1. \displaystyle \left\{\frac{1}{n}\right\}.
    2. +
  2. \displaystyle \left\{\frac{1}{\sqrt{n}}\right\}.
    3. +
  3. \displaystyle \left\{\frac{1}{n^2}\right\}.
    4. +
  4. \displaystyle \left\{\frac{1}{2^n}\right\}.
    5. +
+

+ + +

+ Using your chosen sequence and , does \displaystyle \frac{3}{n^2+8n+5} converge or diverge? +

+ + + + +

+ Use to determine if the series \displaystyle \sum_{n=5}^\infty \frac{2}{4^n} converges or diverges. +

+ + + + +

+ Consider sequences \{a_n\}, \{b_n\} where a_n\geq b_n\geq 0. +

+ Plots of \{a_n\}, \{b_n\} + + Plots of sequences \{a_n\}, \{b_n\} where a_n\geq b_n\geq 0. + + \begin{tikzpicture} + \draw[->] (-1, 0) -- (10.2, 0) node[below right] {$x$}; + \draw[->] (0, -1) -- (0, 4.2) node[above] {$y$}; + + + \draw[blue] (0,4)\foreach \x in {0,...,9}{ + --({\x+1}, {4/(\x+1)})--({\x+1}, {4/(\x+2)})}; + \draw (0,0)--(0,4); + \foreach \x in {0,...,9}{ + \draw[blue] ({\x+1}, {4/(\x+2)})--({\x+1}, {0});} + + \draw[blue, fill, opacity=0.2] (0,0)--(0,4)\foreach \x in {0,...,9}{ + --({\x+1}, {4/(\x+1)})--({\x+1}, {4/(\x+2)})} + --(10,0) + --cycle; + + \draw[red] (0,4)\foreach \x in {0,...,9}{ + --({\x+1}, {4/(\x+1)^2})--({\x+1}, {4/(\x+2)^2})}; + \draw (0,0)--(0,4); + \foreach \x in {0,...,9}{ + \draw[red] ({\x+1}, {4/(\x+2)^2})--({\x+1}, {0});} + + \draw[red, fill, opacity=0.2] (0,0)--(0,4)\foreach \x in {0,...,9}{ + --({\x+1}, {4/(\x+1)^2})--({\x+1}, {4/(\x+2)^2})} + --(10,0) + --cycle; + + \draw[blue] (0,4)--node[left]{$a_0$} (0,4); + \draw[red] (1,4)--node[right]{$b_0$} (1,4); + + \draw[blue] (2,4/2)--node[ right]{$a_1$} (2,4/2); + \draw[red] (2,4/2^2)--node[ right]{$b_1$} (2,4/2^2); + + \draw[blue] (3,4/3)--node[ right]{$a_2$} (3,4/3); + \draw[red] (3,4/3^2)--node[right]{$b_2$} (3,4/3^2); + + \draw (10,0)-- node[above right]{$\cdots$} (10,0); + + + \end{tikzpicture} + + +
+

+ + + +

+ Suppose that \displaystyle \sum_{n=0}^\infty a_n converges. What could be said about \{b_n\}? +

    +
  1. \displaystyle \sum_{n=0}^\infty b_n converges.
    2. +
  2. \displaystyle \sum_{n=0}^\infty b_n diverges.
    3. +
  3. Whether or not \displaystyle \sum_{n=0}^\infty b_n converges or diverges cannot be determined with this information.
    4. +
+

+ + + +

+ Suppose that \displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{1}{n+1} which diverges. Which of the following statements are true? +

    +
  1. \displaystyle 0\leq \frac{1}{2n^2} \leq \frac{1}{n+1} for each n \geq 1 and \displaystyle \sum_{n=1}^\infty \frac{1}{2n^2} is a convergent p-series where p=2.
    2. +
  2. \displaystyle 0\leq \frac{1}{2n}\leq \frac{1}{n+1} for each n \geq 1 and \displaystyle \sum_{n=1}^\infty \frac{1}{2n} is a divergent p-series where p=1.
    3. +
+

+ + + +

+ Suppose that \displaystyle \sum_{n=0}^\infty a_n was some series that diverges. What could be said about \{b_n\}? +

    +
  1. \displaystyle \sum_{n=0}^\infty b_n converges.
    2. +
  2. \displaystyle \sum_{n=0}^\infty b_n diverges.
    3. +
  3. Whether or not \displaystyle \sum_{n=0}^\infty b_n converges or diverges cannot be determined with this information.
    4. +
+

+ + + + + +

+ Suppose that \displaystyle \sum_{n=0}^\infty b_n diverges. What could be said about \{a_n\}? +

    +
  1. \displaystyle \sum_{n=0}^\infty a_n converges.
    2. +
  2. \displaystyle \sum_{n=0}^\infty a_n diverges.
    3. +
  3. Whether or not \displaystyle \sum_{n=0}^\infty a_n converges or diverges cannot be determined with this information.
    4. +
+

+ + + +

+ Suppose that \displaystyle \sum_{n=0}^\infty b_n=\sum_{n=0}^\infty \frac{1}{3^n} which converges. Which of the following statements are true? +

    +
  1. \displaystyle 0\leq \frac{1}{3^n} \leq \frac{1}{2^n} for each n and \displaystyle \sum_{n=0}^\infty \frac{1}{2^n} is a convergent geometric series where \displaystyle |r|=\frac{1}{2} \lt 1.
    2. +
  2. \displaystyle 0\leq \frac{1}{3^n} \leq 1 for each n and \displaystyle \sum_{n=0}^\infty 1 diverges by the Divergence Test.
    3. +
+

+ + + +

+ Suppose that \displaystyle \sum_{n=0}^\infty b_n was some series that converges. What could be said about \{a_n\}? +

    +
  1. \displaystyle \sum_{n=0}^\infty a_n converges.
    2. +
  2. \displaystyle \sum_{n=0}^\infty a_n diverges.
    3. +
  3. Whether or not \displaystyle \sum_{n=0}^\infty a_n converges or diverges cannot be determined with this information.
    4. +
+

+ + + + + + + +

+ Supppose we have sequences \{a_n\}, \{b_n\} so that for some k we have that 0\leq b_n\leq a_n for each k\geq n. Then we have the following results: +

    +
  • If \displaystyle\sum_{k=n}^\infty a_n converges, then so does \displaystyle\sum_{k=n}^\infty b_n.
    • +
  • If \displaystyle\sum_{k=n}^\infty b_n diverges, then so does \displaystyle\sum_{k=n}^\infty a_n.
    • +
+

+ + + + + + +

+ Suppose that you were handed positive sequences \{a_n\}, \{b_n\}. For the first few values a_n\geq b_n, but after that what happens is unclear until n=100. Then for any n\geq 100 we have that a_n \leq b_n. + +

+ Plots of \{a_n\}, \{b_n\} + + Plots of sequences \{a_n\}, \{b_n\} where a_n\geq b_n\geq 0 initially but eventually a_n\leq b_n\geq 0. + + \begin{tikzpicture} + \draw[->] (-1, 0) -- (10.2, 0) node[below right] {$x$}; + \draw[->] (0, -1) -- (0, 4.2) node[above] {$y$}; + + + \draw[blue] (0,4)\foreach \x in {0,...,4}{ + --({\x+1}, {4*(0.8^(\x))})--({\x+1}, {4*0.8^(\x+1)})}; + \draw (0,0)--(0,4); + \foreach \x in {0,...,4}{ + \draw[blue] ({\x+1}, {4*0.8^(\x+1)})--({\x+1}, {0});} + + \draw[blue, fill, opacity=0.2] (0,0)--(0,4)\foreach \x in {0,...,4}{ + --({\x+1}, {4*0.8^(\x)})--({\x+1}, {4*0.8^(\x+1)})} + --(5,0) + --cycle; + + \draw[red] (0,4)\foreach \x in {0,...,4}{ + --({\x+1}, {4*(0.7^(\x))})--({\x+1}, {4*0.7^(\x+1)})}; + \draw (0,0)--(0,4); + \foreach \x in {0,...,4}{ + \draw[red] ({\x+1}, {4*0.7^(\x+1)})--({\x+1}, {0});} + + \draw[red, fill, opacity=0.2] (0,0)--(0,4)\foreach \x in {0,...,4}{ + --({\x+1}, {4*0.7^(\x)})--({\x+1}, {4*0.7^(\x+1)})} + --(5,0) + --cycle; + + \draw (5.5,0)--node[above]{$\cdots$} (5.5,0); + + \draw[blue] (6,4*0.75^9)\foreach \x in {6,...,9}{ + --({\x+1}, {4*(0.75^(\x+3))})--({\x+1}, {4*0.75^(\x+4)})}; + \draw[blue] (6,0)--(6,4*0.75^9); + \foreach \x in {6,...,9}{ + \draw[blue] ({\x+1}, {4*0.75^(\x+4)})--({\x+1}, {0});} + + \draw[blue, fill, opacity=0.2] (6,0)--(6,4*0.75^9)\foreach \x in {6,...,9}{ + --({\x+1}, {4*0.75^(\x+3)})--({\x+1}, {4*0.75^(\x+4)})} + --(10,0) + --cycle; + + + \draw[red] (6,4*0.8^9)\foreach \x in {6,...,9}{ + --({\x+1}, {4*(0.8^(\x+3))})--({\x+1}, {4*0.8^(\x+4)})}; + \draw[blue] (6,0)--(6,4*0.8^9); + \foreach \x in {6,...,9}{ + \draw[red] ({\x+1}, {4*0.8^(\x+4)})--({\x+1}, {0});} + + \draw[red, fill, opacity=0.2] (6,0)--(6,4*0.8^9)\foreach \x in {6,...,9}{ + --({\x+1}, {4*0.8^(\x+3)})--({\x+1}, {4*0.8^(\x+4)})} + --(10,0) + --cycle; + + \draw (6,-0.1)--(6,0.1); + \draw (6,-0.1)--node[below]{$100$} (6,-0.1); + + \draw (10.5,0)--node[above]{$\cdots$} (10.5,0); + + + \end{tikzpicture} + + +
+

+ + +

+ How might we best utilize to determine the convergence of \displaystyle \sum_{n=0}^\infty a_n or \displaystyle \sum_{n=0}^\infty b_n? +

    +
  1. Since a_n is sometimes greater than, and sometimes less than b_n, there is no way to utilize .
    2. +
  2. Since initially, we have b_n\leq a_n, we can utilize by assuming a_n\geq b_n.
    3. +
  3. Since we can rewrite \displaystyle \sum_{n=0}^\infty a_n=\sum_{n=0}^{99} a_n+\sum_{n=100}^\infty a_n and \displaystyle \sum_{n=0}^\infty b_n=\sum_{n=0}^{99} b_n+\sum_{n=100}^\infty b_n and \displaystyle \sum_{n=0}^{99} a_n, \sum_{n=0}^{99} b_n are necessarily finite, we can compare \displaystyle \sum_{n=100}^\infty a_n, \sum_{n=100}^\infty b_n with .
    4. +
+

+ + + + + + + The Direct Comparison Test + +

+ Let \displaystyle\sum a_n and \displaystyle\sum b_n be series with positive terms. + If there is a k such that b_n\leq a_n for each n\geq k, then: +

    +
  • If \displaystyle \sum a_n converges, then so does \displaystyle \sum b_n.
    • +
  • If \displaystyle \sum b_n diverges, then so does \displaystyle \sum a_n.
    • +
+

+ + + + + + +

+ Suppose we wish to determine if \displaystyle \sum_{n=1}^\infty \frac{1}{2n+3} converged using . +

+ + + +

+ Does \displaystyle \sum_{n=1}^\infty \frac{1}{3n} converge or diverge? +

+ + + + +

+ For which value k is \displaystyle\frac{1}{3n}\leq \frac{1}{2n+3} for each n\geq k? +

    +
  1. \displaystyle\frac{1}{3n}\leq \frac{1}{2n+3} for each n\geq k=0.
    2. +
  2. \displaystyle\frac{1}{3n}\leq \frac{1}{2n+3} for each n\geq k=1.
    3. +
  3. \displaystyle\frac{1}{3n}\leq \frac{1}{2n+3} for each n\geq k=2.
    4. +
  4. \displaystyle\frac{1}{3n}\leq \frac{1}{2n+3} for each n\geq k=3.
    5. +
  5. There is no k for which \displaystyle \frac{1}{3n}\leq \frac{1}{2n+3} for each n\geq k.
    6. +
+

+ + + +

+ Use and compare \displaystyle \sum_{n=1}^\infty \frac{1}{2n+3} to \displaystyle \sum_{n=1}^\infty \frac{1}{3n} to determine if \displaystyle \sum_{n=1}^\infty \frac{1}{2n+3} converges or diverges. +

+ + + + + + + +

+ Suppose we wish to determine if \displaystyle \sum_{n=1}^\infty \frac{1}{n^2+5} converged using . +

+ + + +

+ Which series should we compare \displaystyle \sum_{n=1}^\infty \frac{1}{n^2+5} to best utilize ? +

    +
  1. \displaystyle\sum_{n=1}^\infty \frac{1}{n}.
    2. +
  2. \displaystyle\sum_{n=1}^\infty \frac{1}{n^2}.
    3. +
  3. \displaystyle\sum_{n=1}^\infty \frac{1}{2^n}.
    4. +
  4. \displaystyle\sum_{n=1}^\infty \frac{1}{n+5}.
    5. +
  5. \displaystyle\sum_{n=1}^\infty \frac{1}{n^2+5}.
    6. +
  6. \displaystyle\sum_{n=1}^\infty \frac{1}{2^n+5}.
    7. +
+

+ + + +

+ Using your chosen series and , does \displaystyle \sum_{n=1}^\infty \frac{1}{n^2+5} converge or diverge? +

+ + + + + + +

+ For each of the following series, determine if it converges or diverges, and explain your choice.

 + + + \displaystyle \sum_{n= 4 }^\infty \frac{3}{\log\left(n\right) + 2}. + + + \displaystyle \sum_{n= 3 }^\infty \frac{1}{n^{2} + 2 \, n + 1}. + + + + + + + + + + + Videos +
+ + Video: Use the direct comparison and limit comparison tests to determine if a series converges or diverges + + + + +
+ + + + + + +
diff --git a/calculus/source/08-SQ/07.ptx b/calculus/source/08-SQ/07.ptx new file mode 100644 index 00000000..0f4ea53b --- /dev/null +++ b/calculus/source/08-SQ/07.ptx @@ -0,0 +1,452 @@ + + +
+ Ratio and Root Tests (SQ7) + + + + + Activities + + + + +

+ Consider the series \displaystyle \sum_{n=0}^\infty \frac{2^n}{3^n-2}. +

+ + + +

+ Which of these series most closely resembles \displaystyle \sum_{n=0}^\infty \frac{2^n}{3^n-2}? +

    +
  1. \displaystyle \sum_{n=0}^\infty \frac{2}{3}.
    2. +
  2. \displaystyle \sum_{n=0}^\infty \frac{2}{3}n.
    3. +
  3. \displaystyle \sum_{n=0}^\infty \left(\frac{2}{3}\right)^n.
    4. +
+

+ + +

+ Based on your previous choice, do we think this series is more likely to converge or diverge? +

+ + +

+ Find \displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=\lim_{n\to\infty}\frac{2^{n+1}(3^n-2)}{(3^{n+1}-2)2^n}=\lim_{n\to\infty}\frac{2\cdot 2^{n}(3^n-2)}{3(3^{n}-\frac{2}{3})2^n}. +

    +
  1. \displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=0.
    2. +
  2. \displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=\frac{2}{3}.
    3. +
  3. \displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=1.
    4. +
  4. \displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=2.
    5. +
  5. \displaystyle \lim_{n\to\infty} \frac{\frac{2^{n+1}}{3^{n+1}-2}}{\frac{2^n}{3^n-2}}=3.
    6. +
+

+ + + + + +

+ Consider the series \displaystyle \sum_{n=0}^\infty a_n=\sum_{n=0}^\infty \frac{3}{2^n}. +

+ + +

+ Does \displaystyle \sum_{n=0}^\infty a_n=\sum_{n=0}^\infty \frac{3}{2^n} converge? +

+ + +

+ Find \displaystyle\frac{a_{n+1}}{a_n}. +

    +
  1. 2.
    2. +
  2. \displaystyle \frac{1}{2}.
    3. +
  3. \displaystyle \frac{2^n}{2^n+1}.
    4. +
  4. \displaystyle \frac{9}{2^{2n+1}}.
    5. +
  5. \displaystyle \frac{9}{2^{n+2}}.
    6. +
+

+ + +

+ Find \displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|. +

    +
  1. -\infty.
    2. +
  2. 0.
    3. +
  3. \displaystyle \frac{1}{2}.
    4. +
  4. 2.
    5. +
  5. \infty.
    6. +
+

+ + + + + + + +

+ Consider the series \displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{n^2}{n+1}. +

+ + +

+ Does \displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{n^2}{n+1} converge? +

+ + +

+ Find \displaystyle\frac{a_{n+1}}{a_n}. +

    +
  1. \displaystyle \frac{n+1}{2}.
    2. +
  2. \displaystyle \frac{(n^2+1)(n+1)}{(n+2)n^2}.
    3. +
  3. \displaystyle \frac{(n+1)^2}{n+2}.
    4. +
  4. \displaystyle \frac{1}{2}.
    5. +
  5. \displaystyle \frac{(n+1)n^2}{n+2}.
    6. +
+

+ + +

+ Find \displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|. +

    +
  1. -\infty.
    2. +
  2. 0.
    3. +
  3. \displaystyle \frac{1}{2}.
    4. +
  4. 2.
    5. +
  5. \infty.
    6. +
+

+ + + + + + + +

+ Consider the series \displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{1}{n}. +

+ + +

+ Does \displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{1}{n} converge? +

+ + +

+ Find \displaystyle\frac{a_{n+1}}{a_n}. +

+ + +

+ Find \displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|. +

+ + + + + + + +

+ Consider the series \displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{(-1)^n}{n}. +

+ + +

+ Does \displaystyle \sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \frac{(-1)^n}{n} converge? +

+ + +

+ Find \frac{a_{n+1}}{a_n}. +

+ + +

+ Find \displaystyle \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|. +

+ + + + + + + + The Ratio Test + +

+ Let \displaystyle\sum a_n be a series + and suppose that \displaystyle\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\rho. + Then +

    +
  1. \displaystyle\sum a_n converges if \rho is less than 1, and
    2. +
  2. \displaystyle\sum a_n diverges if \rho is greater than 1.
    3. +
  3. If \rho=1, we cannot determine if \displaystyle\sum a_n converges or diverges with this method.
    4. +
+ +

+ + + + + + + The Root Test + +

+ Let N be an integer and let \displaystyle\sum a_n be a series with a_n\geq 0 for n\geq N, + and suppose that \displaystyle\lim_{n\rightarrow\infty}\sqrt[n]{|a_n|}=\rho. + Then +

    +
  1. \displaystyle\sum a_n converges if \rho is less than 1, and
    2. +
  2. \displaystyle\sum a_n diverges if \rho is greater than 1.
    3. +
  3. If \rho=1, we cannot determine if \displaystyle\sum a_n converges or diverges with this method.
    4. +
+ +

+ + + + + + +

+ Consider the series \displaystyle\sum_{n=0}^\infty \frac{n^2}{n!}. +

+ + +

+ Which of the following is a_n? +

    +
  1. n^2.
    2. +
  2. n!.
    3. +
  3. \displaystyle\frac{n^2}{n!}.
    4. +
+

+ + +

+ Which of the following is a_{n+1}? +

    +
  1. \displaystyle\frac{n^2}{n!}.
    2. +
  2. \displaystyle(n+1)^2.
    3. +
  3. \displaystyle(n+1)!.
    4. +
  4. \displaystyle\frac{(n+1)^2}{(n+1)!}.
    5. +
  5. \displaystyle\frac{n^2+1}{n!+1}.
    6. +
+

+ + +

+ Which of the following is \displaystyle\left|\frac{a_{n+1}}{a_n}\right|? +

    +
  1. \displaystyle\frac{(n+1)^2n^2}{(n+1)!n!}.
    2. +
  2. \displaystyle\frac{(n+1)^2n!}{(n+1)!n^2}.
    3. +
  3. \displaystyle\frac{(n+1)!n!}{(n+1)^2n^2}.
    4. +
  4. \displaystyle\frac{(n+1)!n^2}{(n+1)^2n!}.
    5. +
+

+ + +

+ Using the fact (n+1)!=(n+1)\cdot n!, simplify \displaystyle\left|\frac{a_{n+1}}{a_n}\right| as much as possible. +

+ + +

+ Find \displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|. +

+ + +

Does \displaystyle\sum_{n=0}^\infty \frac{n^2}{n!} converge?

+ + + + +

+ Consider the series \displaystyle \sum_{n=1}^\infty \frac{n^n}{9^n}. +

+ +

What is a_n?

+ + +

+ Which of the following is \displaystyle\sqrt[n]{|a_n|}? +

    +
  1. \displaystyle \frac{n+1}{9}.
    2. +
  2. \displaystyle \frac{n}{9}.
    3. +
  3. n.
    4. +
  4. 9.
    5. +
  5. \displaystyle \frac{1}{9}.
    6. +
+

+ + +

Find \displaystyle\lim_{n\rightarrow\infty}\sqrt[n]{|a_n|}.

+ + +

Does \displaystyle \sum_{n=1}^\infty \frac{n^n}{9^n} converge?

+ + + + + +

+ For each series, use the ratio or root test to determine if the series converges or diverges. +

+ + +

+ \displaystyle \sum_{n=1}^\infty \displaystyle\left(\frac{1}{1+n}\right)^n +

+ + + +

+ \displaystyle \sum_{n=1}^\infty \displaystyle\frac{2^n}{n^n} +

+ + + + +

+ \displaystyle \sum_{n=1}^\infty \displaystyle\frac{(2n)!}{(n!)(n!)} +

+ + + + +

+ \displaystyle \sum_{n=1}^\infty \displaystyle\frac{4^n(n!)(n!)}{(2n)!} +

+ + + + + + +

+ Consider the series \displaystyle \sum_{n=0}^\infty \displaystyle\frac{2^n+5}{3^n}. +

+ + +

+ Use the root test to check for convergence of this series. +

+ + + +

+ Use the ratio test to check for convergence of this series. +

+ + + + +

+ Use the comparison (or limit comparison) test to check for convergence of this series. +

+ + + + +

+ Find the sum of this series. +

+ + + + + + +

+ Consider \displaystyle\sum_{n=1}^\infty \frac{n}{3^n}. Recall that \displaystyle \sqrt[n]{\frac{n}{3^n}}=\left(\frac{n}{3^n}\right)^{1/n}=\frac{n^{1/n}}{(3^n)^{1/n}}. +

+ + +

+ Let \displaystyle \alpha=\lim_{n\to\infty}\ln(n^{1/n})=\lim_{n\to\infty}\frac{1}{n}\ln(n). Find \alpha. +

+ + + +

+ Recall that \displaystyle \lim_{n\to\infty}n^{1/n}=\lim_{n\to\infty} e^{\ln(n^{1/n})}=e^\alpha. Find \displaystyle \lim_{n\to\infty}n^{1/n}. +

+ + + +

+ Find \displaystyle \lim_{n\to\infty} \sqrt[n]{\frac{n}{3^n}}=\lim_{n\to\infty}\left(\frac{n}{3^n}\right)^{1/n}=\lim_{n\to\infty}\frac{n^{1/n}}{(3^n)^{1/n}}. +

+ + + + +

+ Does \displaystyle\sum_{n=1}^\infty \frac{n}{3^n} converge? +

+ + + + + +

+ Consider the series \displaystyle \sum_{n=0}^\infty \displaystyle\frac{n^2}{2^n}. +

+ + +

+ Use the root test to check for convergence of this series. +

+ + + +

+ Use the ratio test to check for convergence of this series. +

+ + + + +

+ Use the comparison (or limit comparison) test to check for convergence of this series. +

+ + + + + + + + + + Videos +
+ + Video: Use the ratio and root tests to determine if a series converges or diverges + + + + +
+ + + + + + +
diff --git a/calculus/source/08-SQ/08.ptx b/calculus/source/08-SQ/08.ptx new file mode 100644 index 00000000..109885eb --- /dev/null +++ b/calculus/source/08-SQ/08.ptx @@ -0,0 +1,184 @@ + + +
+ Absolute Convergence (SQ8) + + + + + Activities + + + +

+ Recall the series \displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n} from . +

+ + +

+ Does the series \displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n} converge or diverge? +

+ + + +

+ Does the series \displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n}{n}\right| converge or diverge? +

+ + + + + +

+ Consider the series \displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n^2}. +

+ + +

+ Does the series \displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{n^2} converge or diverge? +

+ + + +

+ Does the series \displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n}{n^2}\right| converge or diverge? +

+ + + + + +

+ Given a series \sum a_n we say that \displaystyle \sum a_n is absolutely convergent if \displaystyle \sum |a_n| converges. +

+ + + + + +

+ Consider the series: \displaystyle \sum_{n=1}^\infty \frac{(-1)^nn!}{(2n)!}. +

+ + +

+ Does the series \displaystyle \sum_{n=1}^\infty \frac{(-1)^nn!}{(2n)!} converge or diverge? (Recall .) +

+ + + +

Compute |a_n|.

+ + + +

+ Does the series \displaystyle \sum_{n=1}^\infty \frac{(-1)^nn!}{(2n)!} converge absolutely? +

+ + + + +

+ Notice that and both involve taking absolute values to determine convergence. As such, series that are convergent by either the Ratio Test or the Root Test are also absolutely convergent (by applying the same test after taking the absolute value). +

+ + + + +

+ Consider the series: \displaystyle \sum_{n=1}^\infty -n. +

+ + +

+ Does the series \displaystyle \sum_{n=1}^\infty -n converge or diverge? +

+ + + +

Compute |a_n|.

+ + + +

+ Does the series \displaystyle \sum_{n=1}^\infty -n converge absolutely? +

+ + + + + +

+ For each of the following series, determine if the series is convergent, and if the series is absolutely convergent. +

+ + + \displaystyle \sum_{n=1}^\infty \frac{n^2(-1)^n}{n^3+1}. + + + \displaystyle \sum_{n=1}^\infty \frac{1}{n^2}. + + + \displaystyle \sum_{n=1}^\infty (-1)^n \left(\frac{2}{3}\right)^n. + + + + + +

+ If you know a series \displaystyle \sum a_n is absolutely convergent, what can you conclude about whether or not \displaystyle \sum a_n is convergent? +

    +
  1. We cannot determine if \displaystyle \sum a_n is convergent.
    2. +
  2. \displaystyle \sum a_n is convergent since it grows slower than \displaystyle \sum |a_n| (and falls slower than \displaystyle \sum -|a_n|).
    3. +
+

+ + + + + +

+ If \displaystyle \sum a_n is absolutely convergent, then it must be convergent. +

+ + + + + +

+ Find 3 series that are convergent but not absolutely convergent (recall , ). +

+ + + + + + + + + Videos +
+ + Video: Determine if a series converges absolutely or conditionally + + + + +
+ + + + + + + +
diff --git a/calculus/source/08-SQ/09.ptx b/calculus/source/08-SQ/09.ptx new file mode 100644 index 00000000..1e1750fa --- /dev/null +++ b/calculus/source/08-SQ/09.ptx @@ -0,0 +1,409 @@ + + +
+ Series Convergence Strategy (SQ9) + + + + + Activities + + +

Which test for convergence is the best first test to apply to any series \displaystyle \sum_{k=1}^\infty a_k?

+
    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+ + + + +

In which of the following scenarios can we successfully apply the Direct Comparison Test to determine the convergence of the series \displaystyle \sum a_k?

+
    +

  1. When we find a convergent series \displaystyle \sum b_k where 0\leq a_k\leq b_k

    2. +

  2. When we find a divergent series \displaystyle \sum b_k where 0\leq a_k\leq b_k

    3. +

  3. When we find a convergent series \displaystyle \sum b_k where 0\leq b_k\leq a_k

    4. +

  4. When we find a divergent series \displaystyle \sum b_k where 0\leq b_k\leq a_k

    5. +
+ + + + +

Which test(s) for convergence would we use for a series \displaystyle \sum a_k where a_k involves k^{th} powers?

+
    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+ + + + +

Which test(s) for convergence would we use for a series of the form \displaystyle \sum ar^k?

+
    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+ + + + +

Which test(s) for convergence would we use for a series \displaystyle \sum a_k where a_k involves factorials and powers?

+
    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+ + + + +

Which test(s) for convergence would we use for a series \displaystyle \sum a_k where a_k is a rational function?

+
    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+ + + + +

Which test(s) for convergence would we use for a series of the form \displaystyle \sum (-1)^ka_k?

+
    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+ + + + +

Here is a strategy checklist when dealing with series:

+
    +

  1. The divergence test: unless a_n\rightarrow 0, \displaystyle \sum a_n diverges

    2. +

  2. Geometric Series: \displaystyle \sum ar^k converges if -1<r<1 and diverges otherwise

    3. +

  3. p-series: \displaystyle \sum \frac{1}{n^p} converges if p>1 and diverges otherwise

    4. +

  4. Series with no negative terms: try the ratio test, root test, integral test, or try to compare to a known series with the comparison test or limit comparison test

    5. +

  5. Series with some negative terms: check for absolute convergence

    6. +

  6. Alternating series: use the alternating series test (Leibniz's Theorem)

    7. +

  7. Anything else: consider the sequence of partial sums, possibly rewriting the series in a different form, hope for the best

    8. +
+ + + +

Consider the series \displaystyle \sum_{k=3}^\infty \frac{2}{\sqrt{k-2}}.

 +

Which test(s) seem like the most appropriate one(s) to test for convergence or divergence? +

    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+

+

Apply an appropriate test to determine the convergence of this series. +

    +

  1. This series is convergent.

    2. +

  2. This series is divergent.

    3. +
+

+ + + +

Consider the series \displaystyle \sum_{k=1}^\infty \frac{k}{1+2k}.

 +

Which test(s) seem like the most appropriate one(s) to test for convergence or divergence? +

    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+

+

Apply an appropriate test to determine the convergence of this series. +

    +

  1. This series is convergent.

    2. +

  2. This series is divergent.

    3. +
+

+ + + +

Consider the series \displaystyle \sum_{k=0}^\infty \frac{2k^2+1}{k^3+k+1}.

 +

Which test(s) seem like the most appropriate one(s) to test for convergence or divergence? +

    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+

+

Apply an appropriate test to determine the convergence of this series. +

    +

  1. This series is convergent.

    2. +

  2. This series is divergent.

    3. +
+

+ + + +

Consider the series \displaystyle \sum_{k=0}^\infty \frac{100^k}{k!}.

 +

Which test(s) seem like the most appropriate one(s) to test for convergence or divergence? +

    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+

+

Apply an appropriate test to determine the convergence of this series. +

    +

  1. This series is convergent.

    2. +

  2. This series is divergent.

    3. +
+

+ + + +

Consider the series \displaystyle \sum_{k=1}^\infty \frac{2^k}{5^k}.

 +

Which test(s) seem like the most appropriate one(s) to test for convergence or divergence? +

    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+

+

Apply an appropriate test to determine the convergence of this series. +

    +

  1. This series is convergent.

    2. +

  2. This series is divergent.

    3. +
+

+ + + +

Consider the series \displaystyle \sum_{k=1}^\infty \frac{k^3-1}{k^5+1}.

 +

Which test(s) seem like the most appropriate one(s) to test for convergence or divergence? +

    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+

+

Apply an appropriate test to determine the convergence of this series. +

    +

  1. This series is convergent.

    2. +

  2. This series is divergent.

    3. +
+

+ + + +

Consider the series \displaystyle \sum_{k=2}^\infty \frac{3^{k-1}}{7^k}.

 +

Which test(s) seem like the most appropriate one(s) to test for convergence or divergence? +

    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+

+

Apply an appropriate test to determine the convergence of this series. +

    +

  1. This series is convergent.

    2. +

  2. This series is divergent.

    3. +
+

+ + + +

Consider the series \displaystyle \sum_{k=2}^\infty \frac{1}{k^k}.

 +

Which test(s) seem like the most appropriate one(s) to test for convergence or divergence? +

    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+

+

Apply an appropriate test to determine the convergence of this series. +

    +

  1. This series is convergent.

    2. +

  2. This series is divergent.

    3. +
+

+ + + +

Consider the series \displaystyle \sum_{k=1}^\infty \frac{(-1)^{k+1}}{\sqrt{k+1}}.

 +

Which test(s) seem like the most appropriate one(s) to test for convergence or divergence? +

    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+

+

Apply an appropriate test to determine the convergence of this series. +

    +

  1. This series is convergent.

    2. +

  2. This series is divergent.

    3. +
+

+ + + +

Consider the series \displaystyle \sum_{k=2}^\infty \frac{1}{k\ln(k)}.

 +

Which test(s) seem like the most appropriate one(s) to test for convergence or divergence? +

    +

  1. Divergence Test

    2. +

  2. Geometric Series

    3. +

  3. Integral Test

    4. +

  4. Direct Comparison Test

    5. +

  5. Limit Comparison Test

    6. +

  6. Ratio Test

    7. +

  7. Root Test

    8. +

  8. Alternating Series Test

    9. +
+

+

Apply an appropriate test to determine the convergence of this series. +

    +

  1. This series is convergent.

    2. +

  2. This series is divergent.

    3. +
+

+ + + + +

+ Determine which of the following series is convergent and which is divergent. Justify both choices with an appropriate test. +

+ + + \displaystyle \sum_{n=1}^\infty \frac{4 \, \left(-1\right)^{n + 1} n^{2}}{2 \, n^{3} + 4 \, n^{2} + 5}. + + + \displaystyle \sum_{n=1}^\infty \frac{n!}{3 \cdot 3^{n} n^{4}}. + + + + + + + + + + Videos +
+ + Video: Identify appropriate convergence tests for various series + + + + +
+ + + + + + + + + + + +
diff --git a/calculus/source/08-SQ/main.ptx b/calculus/source/08-SQ/main.ptx new file mode 100644 index 00000000..5f3d4162 --- /dev/null +++ b/calculus/source/08-SQ/main.ptx @@ -0,0 +1,17 @@ + + + + Sequences and Series (SQ) + + + + + + + + + + + + + diff --git a/calculus/source/08-SQ/outcomes/01.ptx b/calculus/source/08-SQ/outcomes/01.ptx new file mode 100644 index 00000000..1205f9ea --- /dev/null +++ b/calculus/source/08-SQ/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Define and use explicit and recursive formulas for sequences. +

\ No newline at end of file diff --git a/calculus/source/08-SQ/outcomes/02.ptx b/calculus/source/08-SQ/outcomes/02.ptx new file mode 100644 index 00000000..6d341346 --- /dev/null +++ b/calculus/source/08-SQ/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Determine if a sequence is convergent, divergent, monotonic, or bounded, and compute limits of convergent sequences. +

\ No newline at end of file diff --git a/calculus/source/08-SQ/outcomes/03.ptx b/calculus/source/08-SQ/outcomes/03.ptx new file mode 100644 index 00000000..bea76064 --- /dev/null +++ b/calculus/source/08-SQ/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Compute the first few terms of a telescoping or geometric partial sum sequence, and find a closed form for this sequence, and compute its limit. +

\ No newline at end of file diff --git a/calculus/source/08-SQ/outcomes/04.ptx b/calculus/source/08-SQ/outcomes/04.ptx new file mode 100644 index 00000000..bfb2d4c7 --- /dev/null +++ b/calculus/source/08-SQ/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Determine if a geometric series converges, and if so, the value it converges to. +

\ No newline at end of file diff --git a/calculus/source/08-SQ/outcomes/05.ptx b/calculus/source/08-SQ/outcomes/05.ptx new file mode 100644 index 00000000..53e65848 --- /dev/null +++ b/calculus/source/08-SQ/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+Use the divergence, alternating series, and integral tests to determine if a series converges or diverges. +

\ No newline at end of file diff --git a/calculus/source/08-SQ/outcomes/06.ptx b/calculus/source/08-SQ/outcomes/06.ptx new file mode 100644 index 00000000..47cce552 --- /dev/null +++ b/calculus/source/08-SQ/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+Use the direct comparison and limit comparison tests to determine if a series converges or diverges. +

\ No newline at end of file diff --git a/calculus/source/08-SQ/outcomes/07.ptx b/calculus/source/08-SQ/outcomes/07.ptx new file mode 100644 index 00000000..2b86824f --- /dev/null +++ b/calculus/source/08-SQ/outcomes/07.ptx @@ -0,0 +1,4 @@ + +

+Use the ratio and root tests to determine if a series converges or diverges. +

\ No newline at end of file diff --git a/calculus/source/08-SQ/outcomes/08.ptx b/calculus/source/08-SQ/outcomes/08.ptx new file mode 100644 index 00000000..99b1651c --- /dev/null +++ b/calculus/source/08-SQ/outcomes/08.ptx @@ -0,0 +1,4 @@ + +

+Determine if a series converges absolutely or conditionally. +

\ No newline at end of file diff --git a/calculus/source/08-SQ/outcomes/09.ptx b/calculus/source/08-SQ/outcomes/09.ptx new file mode 100644 index 00000000..06246e2d --- /dev/null +++ b/calculus/source/08-SQ/outcomes/09.ptx @@ -0,0 +1,4 @@ + +

+Identify appropriate convergence tests for various series. +

\ No newline at end of file diff --git a/calculus/source/08-SQ/outcomes/main.ptx b/calculus/source/08-SQ/outcomes/main.ptx new file mode 100644 index 00000000..98e56e61 --- /dev/null +++ b/calculus/source/08-SQ/outcomes/main.ptx @@ -0,0 +1,37 @@ + +> + + +

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
  7. + +
    8. +
  8. + +
    9. +
+ \ No newline at end of file diff --git a/calculus/source/08-SQ/readiness.ptx b/calculus/source/08-SQ/readiness.ptx new file mode 100644 index 00000000..5e13fa5e --- /dev/null +++ b/calculus/source/08-SQ/readiness.ptx @@ -0,0 +1,42 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Compute limits.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    • +

      Review: Khan Academy

      +
      • +
    +
    2. +
  2. +

    Express the sum of indexed values using summation notation.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    3. + + +
  3. +

    Evaluate indefinite integrals.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    4. + +
+ + \ No newline at end of file diff --git a/calculus/source/09-PS/01.ptx b/calculus/source/09-PS/01.ptx new file mode 100644 index 00000000..08eefff1 --- /dev/null +++ b/calculus/source/09-PS/01.ptx @@ -0,0 +1,529 @@ + + +
+ Power Series (PS1) + + + + + Activities + + +

+ Suppose we could define a function as an infinite-length polynomial: + f(x)=1+x+x^2+x^3+x^4+\cdots. +

+ + +

+ Would f(1) be well-defined as a finite real number? +

+
    +
  1. No, the sum would diverge towards \infty.
    2. +
  2. No, the sum would oscillate between 0 and 1.
    3. +
  3. Yes, the sum would be 0.
    4. +
  4. Yes, the sum would be 1.
    5. +
+ + +

+ Would f(-1) be well-defined as a finite real number? +

+
    +
  1. No, the sum would diverge towards \infty.
    2. +
  2. No, the sum would oscillate between 0 and 1.
    3. +
  3. Yes, the sum would be 0.
    4. +
  4. Yes, the sum would be 1.
    5. +
+ + +

+ Would f(1/2) be well-defined as a finite real number? +

+
    +
  1. No, the sum would diverge towards \infty.
    2. +
  2. Yes, the sum would be approximately 1.
    3. +
  3. Yes, the sum would be approximately 2.
    4. +
  4. Yes, the sum would be exactly 2.
    5. +
+ + +

+ When is f(x) well-defined as a finite real number? +

+
    +
  1. Its value is \frac{x}{1-x} when |x|<1.
    2. +
  2. Its value is \frac{x}{1-x} when x<1.
    3. +
  3. Its value is \frac{1}{1-x} when |x|<1.
    4. +
  4. Its value is \frac{1}{1-x} when x<1.
    5. +
+ + + + +

+ Given a sequence of numbers a_n and a number c, + we may define a function f(x) as a power series: + f(x)=\sum_{n=0}^\infty a_n(x-c)^n = a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\cdots. +

+

+ The above power series is said to be centered at c. Often + power series are centered at 0; in this case, they may be written as: + f(x)=\sum_{n=0}^\infty a_n x^n = a_0+a_1x+a_2x^2+a_3x^3+\cdots. +

+

The domain of this function (often referred to as the domain of convergence + or interval of convergence) is exactly the set of x-values for which the series converges.

+ + + + + + +

+ In we will learn how to prove that + \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!} converges for each real value x. + Thus the function f(x)=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\cdots + has the domain of all real numbers. +

+ + + +

+ To estimate f(2), use technology to compute the first few terms as follows: + + f(2)=\sum_{n=0}^\infty \frac{2^n}{n!} \amp = 1+2+\frac{2^2}{2}+\frac{2^3}{6}+\frac{2^4}{24}+\frac{2^5}{120}+\cdots + \amp = \unknown +\cdots + \amp \approx \unknown + + +

+

+ Which of these choices is the closest to this value? +

    +
  1. \sqrt{2}\approx 1.414.
    2. +
  2. e^2\approx 7.389.
    3. +
  3. \sin(2)\approx 0.909.
    4. +
  4. \cos(2)\approx -0.416.
    5. +
+

+ + +

+ Estimate f(-1) in a similar fashion: + + f(-1)=\sum_{n=0}^\infty \frac{\unknown}{n!} \amp = \unknown+\unknown+\unknown+\unknown+\unknown+\unknown+\cdots + \amp = \unknown +\cdots + \amp \approx \unknown + +

+

+ Which of these choices is the closest to this value? +

    +
  1. \frac{1}{\sqrt{1}}\approx 1.000.
    2. +
  2. \frac{1}{e^1}\approx 0.369.
    3. +
  3. \frac{1}{\sin(1)}\approx 1.188.
    4. +
  4. \frac{1}{\cos(1)}\approx 1.851.
    5. +
+

+ + + + + + + + + +

+ The function + f(x)=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}=\sum_{n=0}^\infty \frac{1}{n!}(x-0)^n + is centered at 0. Likewise, graphing the polynomial that uses the first six terms + f_5(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120} + alongside the graph of e^x reveals the illustration given in the following figure. +

+
+ Plots of y=f_5(x), y=e^x. + + Plots of y=f_5(x), y=e^x. + + \begin{tikzpicture} + \begin{axis}[ + axis lines=middle, + grid=major, + xmin=-2, + xmax=2, + ymin=-1, + ymax=8, + xlabel=$x$, + ylabel=$y$, + xtick={-2,-1,...,2}, + ytick={0,1,...,7}, + tick style={very thick}, + legend style={ + at={(rel axis cs:0,1)}, + anchor=north west,draw=none,inner sep=0pt,fill=gray!10} + ] + + \addplot[blue,samples=100] {e^x}; + \addlegendentry{$y=e^x$} + + \addplot[red, dashed,samples=100] {1+x+x^2/2+x^3/6+x^4/24+x^5/120}; + \addlegendentry{$y=f_5(x)$} + \end{axis} + \end{tikzpicture} + + +
+

What might we conclude?

+
    +
  1. e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120} near x=0.
    2. +
  2. e^x= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120} near x=0.
    3. +
  3. e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120} for all x.
    4. +
  4. e^x= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120} for all x.
    5. +
+ + + + + + + +

+ Given a power series + f(x)=\sum_{n=0}^\infty a_n(x-c)^n = a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\cdots, + let f_N(x)=\sum_{n=0}^N a_n(x-c)^n = a_0+a_1(x-c)+a_2(x-c)^2+\cdots+a_N(x-c)^N be its + degree N polynomial approximation for x nearby c. +

+

+ For example, + g_3(x)=\sum_{n=0}^3 n^2 (x-1)^n &= 0+(x-1)+4(x-1)^2+9(x-1)^3 + &= -6+20x-23x^3+9x^3 + is a degree 3 approximation of g(x)=\sum_{n=0}^\infty n^2 (x-1)^n + valid for x values nearby 1. +

+ + + + + + +

+ Consider a function p(x) defined by \displaystyle p(x)=\sum_{n=0}^\infty \frac{2^n}{(2n)!}x^n. +

+ + +

+ Find p_3(x), the degree 3 polynomial approximation for p(x). +

+ + +

+ Use p_3(x) to estimate p(-1). +

+ + + + + + + + Videos +
+ + Video: Approximate functions defined as power series + + + + +
+ + + + + + + + + + + + +
diff --git a/calculus/source/09-PS/02.ptx b/calculus/source/09-PS/02.ptx new file mode 100644 index 00000000..ac23d90e --- /dev/null +++ b/calculus/source/09-PS/02.ptx @@ -0,0 +1,330 @@ + + +
+ Convergence of Power Series (PS2) + + + + + Activities + + +

+ Consider the series \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n where x is a real number. +

+ + +

+ If x=2, then \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{2^n}{n!}. What can be said about this series? +

    +
  1. The techniques we have learned so far allow us to conclude that \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{2^n}{n!} converges.
    2. +
  2. The techniques we have learned so far allow us to conclude that \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{2^n}{n!} diverges.
    3. +
  3. None of the techniques we have learned so far allow us to conclude whether \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{2^n}{n!} converges or diverges.
    4. +
+

+ + +

+ If x=-100, then \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{(-100)^n}{n!}. What can be said about this series? +

    +
  1. The techniques we have learned so far allow us to conclude that \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{(-100)^n}{n!} converges.
    2. +
  2. The techniques we have learned so far allow us to conclude that \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{(-100)^n}{n!} diverges.
    3. +
  3. None of the techniques we have learned so far allow us to conclude whether \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{(-100)^n}{n!} converges or diverges.
    4. +
+

+ + + +

+ Suppose that x were some arbitrary real number. What can be said about this series? +

    +
  1. The techniques we have learned so far allow us to conclude that \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n converges.
    2. +
  2. The techniques we have learned so far allow us to conclude that \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n diverges.
    3. +
  3. None of the techniques we have learned so far allow us to conclude whether \displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n converges or diverges.
    4. +
+

+ + + + + +

Consider a power series \displaystyle\sum c_n(x-a)^n. Recall from that if + + \displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| & < 1 + + + then \displaystyle\sum c_n(x-a)^n converges. +

+ +

Then recall: + + \displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\right| & = \displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}(x-a)}{c_n}\right| + &=\displaystyle \lim_{n\to \infty} |x-a|\left|\frac{c_{n+1}}{c_n}\right| + &=\displaystyle |x-a|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right|. + + + +

+ + + + + + + +

+ Consider \displaystyle\sum_{n=0}^\infty \frac{1}{n^2+1}x^n. +

+ + +

+ Letting c_n=\frac{1}{n^2+1}, find \displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right|. +

+ + + + +

+ For what values of x is \displaystyle |x|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1? +

    +
  1. x < 1.
    2. +
  2. 0\leq x < 1.
    3. +
  3. -1 < x < 1.
    4. +
+

+ + + +

+ If x=1, does \displaystyle\sum_{n=0}^\infty \frac{1}{n^2+1}x^n converge? +

+ + + +

+ If x=-1, does \displaystyle\sum_{n=0}^\infty \frac{1}{n^2+1}x^n converge? +

+ + + + +

+ Which of the following describe the values of x for which \displaystyle\sum_{n=0}^\infty \frac{1}{n^2+1}x^n converges? +

    +
  1. (-1,1).
    2. +
  2. [-1,1).
    3. +
  3. (-1,1].
    4. +
  4. [-1,1].
    5. +
+

+ + + + + + + +

+ Consider \displaystyle\sum_{n=0}^\infty \frac{2^n}{5^n}(x-2)^n. +

+ + +

+ Letting c_n=\frac{2^n}{5^n}, find \displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right|. +

+ + + + +

+ For what values of x is \displaystyle |x-2|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1? +

    +
  1. -\frac{2}{5} < x < \frac{2}{5}.
    2. +
  2. \frac{8}{5} < x < \frac{12}{5}.
    3. +
  3. -\frac{5}{2} < x < \frac{5}{2}.
    4. +
  4. -\frac{1}{2} < x < \frac{9}{2}.
    5. +
+

+ + + +

+ If x=\frac{9}{2}, does \displaystyle\sum_{n=0}^\infty \frac{2^n}{5^n}(x-2)^n converge? +

+ + + +

+ If x=-\frac{1}{2}, does \displaystyle\sum_{n=0}^\infty \frac{2^n}{5^n}(x-2)^n converge? +

+ + + + +

+ Which of the following describe the values of x for which \displaystyle\sum_{n=0}^\infty \frac{2^n}{5^n}(x-2)^n converges? +

    +
  1. (-\frac{1}{2},\frac{9}{2}).
    2. +
  2. [-\frac{1}{2},\frac{9}{2}).
    3. +
  3. (-\frac{1}{2},\frac{9}{2}].
    4. +
  4. [-\frac{1}{2},\frac{9}{2}].
    5. +
+

+ + + + + + + + +

+ Consider \displaystyle\sum_{n=0}^\infty \frac{n^2}{n!}\left(x+\frac{1}{2}\right)^n. +

+ + +

+ Letting c_n=\frac{n^2}{n!}, find \displaystyle \lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right|. +

+ + + + +

+ For what values of x is \displaystyle \left|x+\frac{1}{2}\right|\lim_{n\to \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1? +

    +
  1. 0\leq x < \infty.
    2. +
  2. All real numbers.
    3. +
+

+ + + + +

+ What describes the values of x for which \displaystyle\sum_{n=0}^\infty \frac{n^2}{n!}\left(x+\frac{1}{2}\right)^n converges? + +

+ + + + + + + +

+ Given the power series \displaystyle\sum c_n(x-a)^n, the center of convergencecenter of convergence is x=a. The radius of convergenceradius of convergence is r=\frac{1}{\displaystyle\lim_{n\to\infty} \left| \frac{c_{n+1}}{c_n} \right|}. If \displaystyle\lim_{n\to\infty} \left| \frac{c_{n+1}}{c_n} \right|=0, we say that r=\infty. +

+ +

+ The interval of convergenceinterval of convergence represents all possible values of x for which \displaystyle\sum c_n(x-a)^n converges, which is of the form: +

    +
  • (a-r, a+r)
    • +
  • [a-r, a+r)
    • +
  • (a-r, a+r]
    • +
  • [a-r, a+r]
    • +
+ Depending on if \displaystyle\sum c_n(x-a)^n converges when x=a-r or x=a+r.

+

If r=\infty the interval of convergence is all real numbers. +

+ + + + + +

+ Find the center of convergence, radius of convergence, and interval of convergence for the series: \sum_{n=0}^\infty \frac{3^{n} \left(-1\right)^{n} {\left(x - 1\right)}^{n}}{n!}. +

+ + + + + + + +

+ Find the center of convergence, radius of convergence, and interval of convergence for the series: \sum_{n=0}^\infty \frac{3^{n} {\left(x + 2\right)}^{n}}{n}. +

+ + + + + + + +

Consider the power series \displaystyle \sum_{n=0}^\infty \frac{2^n+1}{n3^n}\left(x+1\right)^n.

+ + +

+ What is the center of convergence for this power series? +

+ + +

+ What is the radius of convergence for this power series? +

+ + + +

+ What is the interval of convergence for this power series? +

+ + + +

+ If x=-0.5, does this series converge? (Use the interval of convergence.) +

+ + + +

+ If x=1, does this series converge? (Use the interval of convergence.) +

+ + + + + + + Videos +
+ + Video: Determine the interval of convergence for a given power series + + + + +
+ + + + + + + + +
diff --git a/calculus/source/09-PS/03.ptx b/calculus/source/09-PS/03.ptx new file mode 100644 index 00000000..e05d0157 --- /dev/null +++ b/calculus/source/09-PS/03.ptx @@ -0,0 +1,611 @@ + + +
+ Manipulation of Power Series (PS3) + + + + + Activities + + + + +

+How might we use the known geometric power series +\frac{1}{1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+x^3+x^4+\dots +to find the value of +\unknown =\sum_{n=0}^\infty nx^{n-1}=0+1+2x+3x^2+4x^3+\dots? +

+ + +

+Which operation describes the relationship between these two series? +

+
    +
  1. Bifurcation
    2. +
  2. Composition
    3. +
  3. Differentiation
    4. +
  4. Multiplication
    5. +
+ + +

+What is the result of applying this operation to \frac{1}{1-x}? +

+
    +
  1. 0
    2. +
  2. \frac{1}{(1-x)^2}
    3. +
  3. 1-\frac{1}{x}
    4. +
  4. \frac{x}{1-x^2}
    5. +
+ + + + + +

+Whenever a function is defined as a power series: +f(x)=\sum_{n=0}^\infty a_n(x-c)^n +then its derivative and general antiderivative are also defined as power series +with the same domain of convergence as f(x), +found by differentiating or integrating term-by-term: + + \frac{d}{dx}[f(x)] \amp=\sum_{n=0}^\infty\frac{d}{dx}\left[a_n(x-c)^n\right] + \amp=\sum_{n=0}^\infty na_n(x-c)^{n-1} + \int f(x)\,dx\amp=C+\sum_{n=0}^\infty\left[\int a_n(x-c)^n\,dx\right] + \amp=C+\sum_{n=0}^\infty\frac{(x-c)^{n+1}}{n+1} + +

+ + + + + +

+Let's investigate the power series +\exp(x)=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\dots. +

+ + +

+ What is the value of \exp(0)? +

+
    +
  1. 0.
    2. +
  2. 1.
    3. +
  3. 2.
    4. +
  4. \infty.
    5. +
+ + +

+ What is the value of \exp'(x)? +

+
    +
  1. 0+1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots.
    2. +
  2. 1+x+\frac{x^2}{6}+\frac{x^3}{24}+\frac{x^4}{120}+\dots.
    3. +
  3. 0+1+x+\frac{x^2}{3}+\frac{x^3}{12}+\frac{x^4}{60}+\dots.
    4. +
  4. 1+x+\frac{x^2}{3}+\frac{x^3}{12}+\frac{x^4}{60}+\dots.
    5. +
+ + + + +

+ What can we conclude from our calculation of f'(x)? +

+
    +
  1. \exp'(x)=[\exp(x)]^2.
    2. +
  2. \exp'(x)=\exp(x^2).
    3. +
  3. \exp'(x)=2\exp(x).
    4. +
  4. \exp'(x)=\exp(x).
    5. +
+ + + + + +

+ What function do we know of that shares each of these properites? +

+
    +
  1. \exp(x)=\frac{1}{1+x}
    2. +
  2. \exp(x)=\cos(x)
    3. +
  3. \exp(x)=e^x
    4. +
  4. \exp(x)=0
    5. +
+ + + + + + + + +

+ We have that \exp(x)=e^x=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{x^n}{n!}. + That is, for any real number x, the series \exp(x)=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n will converge to e^x. +

+ + + + + + + +

+ We may similarly determine that + \cos(x)=\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!} + and + \sin(x)=\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!} + for all real numbers x. However, we will delay until to + prove this fact another way. +

+ + + + + + + + + +

+ Suppose we wish to find the power series for the function f(x)=e^{2x} by modifying + the power series \exp(z)=e^z=\displaystyle\sum_{n=0}^\infty \frac{z^n}{n!}. +

+ + +

+ Substituting z=2x, what is the power series for \exp(2x)? +

    +
  1. \exp(2x)=\displaystyle\sum_{n=0}^\infty \frac{2x^n}{n!}=2+2x+x^2+\frac{1}{3}x^3+\dots.
    2. +
  2. \exp(2x)=\displaystyle\sum_{n=0}^\infty \frac{2x^{n+1}}{n!}=2x+2x^2+x^3+\frac{1}{3}x^4+\dots.
    3. +
  3. \exp(2x)=\displaystyle\sum_{n=0}^\infty \frac{(2x)^n}{n!}=1+2x+2x^2+\frac{4}{3}x^3+\dots.
    4. +
  4. \exp(2x)=\displaystyle\sum_{n=0}^\infty \frac{x^n}{(2n)!}=1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{720}+\dots.
    5. +
+

+ + + +

+ What is the interval of convergence for x for this series? +

+
    +
  1. \left(-\infty,\infty\right).
    2. +
  2. \displaystyle \left(-\frac{1}{2},\frac{1}{2}\right).
    3. +
  3. \displaystyle \left(0,\frac{1}{2}\right).
    4. +
  4. \displaystyle \left(-\frac{1}{2},\frac{1}{2}\right].
    5. +
+ + + + + +

+ If a power series + f(x)=\sum_{n=0}^\infty a_n(x-c)^n + is known, then for any polynomial g(x) the composition f\circ g + has a power series given by + (f\circ g)(x)=f(g(x))=\sum_{n=0}^\infty a_n(g(x)-c)^n + where the domain of convergence is transformed based upon the transformation + given by g(x). +

+

+ For example, if f(x) has the domain of convergence -2\leq x < 2, + then f(2x+4) has the domain of convergence: + -2\leq 2x+4 < 2 + -6\leq 2x < -2 + -3\leq x < -1 +

+ + + + + + +

+ Suppose we wish to find the power series for the function f(x)=\frac{1}{x}. +

+ + +

+ Which of the following represents the power series for g(r)=\frac{1}{1-r}? +

    +
  1. g(r)=\displaystyle\sum_{n=0}^\infty rx^n.
    2. +
  2. g(r)=\displaystyle\sum_{n=0}^\infty (rx)^n.
    3. +
  3. g(r)=\displaystyle\sum_{n=0}^\infty r^n.
    4. +
  4. g(r)=\displaystyle\sum_{r=0}^\infty x^r.
    5. +
+

+ + +

+ For what value of r is \frac{1}{1-r}=\frac{1}{x}? +

    +
  1. r=x-1.
    2. +
  2. r=1-x.
    3. +
  3. r=x+1.
    4. +
  4. r=-x.
    5. +
+

+ + +

+ Substituting r with this value, which of the following is a power series for f(x)=\frac{1}{x}? +

    +
  1. f(x)=\displaystyle\sum_{n=0}^\infty (-x)^n.
    2. +
  2. f(x)=\displaystyle\sum_{n=0}^\infty (1-x)^n.
    3. +
  3. f(x)=\displaystyle\sum_{n=0}^\infty (x-1)^n.
    4. +
  4. f(x)=\displaystyle\sum_{n=0}^\infty (1+x)^n.
    5. +
+

+ + + +

+ Given that the domain of convergence for r in f(r) is -1 < r < 1, + what should be the domain of convergence for x in f(x)? +

    +
  1. -1 < x < 1.
    2. +
  2. -2 < x < 0.
    3. +
  3. -2 < x < 2.
    4. +
  4. 0 < x < 2.
    5. +
+

+ + + + + +

+ Suppose we wish to find the power series for the function f(x)=\frac{1}{3-2x}. + Recall that g(x)=\frac{1}{1-r}=\displaystyle\sum_{n=0}^\infty r^n. +

+ + + +

+ For what value of r is \frac{1}{1-r}=\frac{1}{3-2x}? +

    +
  1. r=2x-2.
    2. +
  2. r=2-2x.
    3. +
  3. r=2x-3.
    4. +
  4. r=3-2x.
    5. +
+

+ + +

+ Evaluating r at the previously found value, + which of the following is the power series of f(x)=\frac{1}{3-2x}? +

    +
  1. f(x)=\displaystyle\sum_{n=0}^\infty (3-2x)^n.
    2. +
  2. f(x)=\displaystyle\sum_{n=0}^\infty (2x-3)^n.
    3. +
  3. f(x)=\displaystyle\sum_{n=0}^\infty (2-2x)^n.
    4. +
  4. f(x)=\displaystyle\sum_{n=0}^\infty (2x-2)^n.
    5. +
+

+ + + +

+ Given that the interval of convergence for r is -1 < r < 1, + what is the interval of convergence for x? +

    +
  1. -1 < x < \frac{3}{2}.
    2. +
  2. -\frac{1}{2} < x < 1.
    3. +
  3. \frac{1}{2} < x < \frac{3}{2}.
    4. +
  4. -\frac{1}{2} < x < \frac{3}{2}.
    5. +
+

+ + + + + + + +

+ Suppose we wish to find the power series for the function f(x)=\frac{1}{1+x^2}. + Recall that g(x)=\frac{1}{1-r}=\displaystyle\sum_{n=0}^\infty r^n. +

+ + + +

+ For what value of r is \frac{1}{1-r}=\frac{1}{1+x^2}? +

    +
  1. r=x^2.
    2. +
  2. r=-x^2.
    3. +
  3. r=1-x^2.
    4. +
  4. r=x^2-1.
    5. +
+

+ + +

+ Evaluating r at the previously found value, which of the following is the power series of + f(x)=\frac{1}{1+x^2}? +

    +
  1. \displaystyle\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}.
    2. +
  2. \displaystyle\frac{1}{1+x^2}=\sum_{n=0}^\infty (1-x^2)^n.
    3. +
  3. \displaystyle\frac{1}{1+x^2}=\sum_{n=0}^\infty x^{2n}.
    4. +
  4. \displaystyle\frac{1}{1+x^2}=\sum_{n=0}^\infty (x^2-1)^n.
    5. +
+

+ + +

+ Given that the interval of convergence for r is -1 < r < 1, + what is the interval of convergence for x? +

    +
  1. -1 < x < 1.
    2. +
  2. -1 < x < 0.
    3. +
  3. 0 < x < 1.
    4. +
  4. 0 < x < 4.
    5. +
+

+ + +

+ How can the power series for \frac{1}{1+x^2} be manipulated to obtain + a power series for \arctan(x)? +

    +
  1. Differentiate each term.
    2. +
  2. Integrate each term.
    3. +
  3. Replace x with x^2 in each term.
    4. +
  4. Replace x with 1/x in each term.
    5. +
+

+ + +

+ Which of these power series is the result of this manipulation? +

    +
  1. \displaystyle\arctan(x)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}.
    2. +
  2. \displaystyle\arctan(x)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n-1}}{2n-1}.
    3. +
  3. \displaystyle\arctan(x)=\sum_{n=0}^\infty (-1)^n(2n)x^{2n-1}.
    4. +
  4. \displaystyle\arctan(x)=\sum_{n=0}^\infty (-1)^n(2n+1)x^{2n}.
    5. +
+

+ + + + + + +

+ What function f(x) has power series f(x)=\displaystyle \sum_{n=0}^\infty \frac{(-1)^nx^n}{n!}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots? +

    +
  1. f(x)=(-1)^ne^x.
    2. +
  2. f(x)=-e^x.
    3. +
  3. f(x)=e^{-x}.
    4. +
  4. f(x)=-e^{-x}.
    5. +
+

+ + + + + + +

+ What function f(x) has power series f(x)=\displaystyle \sum_{n=0}^\infty \frac{x^{n+3}}{n!}=x^3+x^4+\frac{x^5}{2}+\frac{x^6}{6}+\cdots? +

    +
  1. f(x)=e^{x+3}.
    2. +
  2. f(x)=e^{x^3}.
    3. +
  3. f(x)=e^{3x}.
    4. +
  4. f(x)=x^3e^{x}.
    5. +
+

+ + + + + + +

+ If a power series + f(x)=\sum_{n=0}^\infty a_n(x-c)^n + is known, then for any polynomial g(x) the product fg + has a power series given by + (fg)(x)=f(x)g(x)=\sum_{n=0}^\infty a_ng(x)(x-c)^n + where the domain of convergence is the same as f(x). +

+ + + + + +

+ What function f(x) has power series f(x)=\displaystyle \sum_{n=3}^\infty x^n=x^3+x^4+\cdots? +

+
    +
  1. f(x)=\frac{1}{1-3x}.
    2. +
  2. f(x)=\frac{3}{1-x}.
    3. +
  3. f(x)=\frac{1}{1-x}-x^2-x-1.
    4. +
  4. f(x)=\frac{x^3}{1-x}.
    5. +
+ + + + + +

+ The function n(x)=e^{-x^2} is one whose integrals are very important for statistics. However, it does not admit an elementary antiderivative. +

+ + +

+ Which of the following best represents the power series for n(x)=e^{-x^2}? +

    +
  1. n(x)=\displaystyle -x^2\sum_{n=0}^\infty \frac{1}{n!}x^n=\displaystyle \sum_{n=0}^\infty -\frac{1}{n!}x^{n+2}.
    2. +
  2. n(x)=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}(-x^2)^n=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}(-1)^{n}x^{2n}.
    3. +
  3. n(x)=\displaystyle x^{-2}\sum_{n=0}^\infty \frac{1}{n!}(-x)^n=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}(-1)^{n+2}x^{n+2}.
    4. +
+

+ + +

+ Which of the following best represents a degree 10 polynomial that approximates n(x)? +

    +
  1. n_{10}(x)=\displaystyle -x^2-x^3-\frac{1}{2}x^4-\frac{1}{6}x^5-\frac{1}{24}x^6-\frac{1}{120}x^7-\frac{1}{720}x^8-\frac{1}{5040}x^9-\frac{1}{40320}x^{10}.
    2. +
  2. n_{10}(x)=\displaystyle x^2-x^3+\frac{1}{2}x^4-\frac{1}{6}x^5+\frac{1}{24}x^6-\frac{1}{120}x^7+\frac{1}{720}x^8-\frac{1}{5040}x^9+\frac{1}{40320}x^{10}.
    3. +
  3. n_{10}(x)=1-x^2+\frac{1}{2}x^4-\frac{1}{6}x^6+\frac{1}{24}x^8-\frac{1}{120}x^{10}.
    4. +
+

+ + + + +

+ Use your choice of n_{10}(x) to estimate \displaystyle \int_0^1 n(x)dx by computing \displaystyle \int_0^1 n_{10}(x)dx. +

+ + + + + + + + + + + + Recall that g(x)=\displaystyle\sum_{n=0}^\infty x^n=\frac{1}{1-x} for -1< x< 1. + + +

+ Which of the following represents an antiderivative of g(x)=\displaystyle \frac{1}{1-x}? +

    +
  1. G(x)=C+\displaystyle\sum_{n=0}^\infty x^{n+1}.
    2. +
  2. G(x)=C+\displaystyle\sum_{n=1}^\infty \frac{1}{n}x^{n+1}.
    3. +
  3. G(x)=C+\displaystyle\sum_{n=0}^\infty \frac{1}{n+1}x^{n+1}.
    4. +
  4. G(x)=C+\displaystyle\sum_{n=1}^\infty \frac{1}{n+1}x^{n}.
    5. +
+

+ + + +

+ Find the interval of convergence for this series. +

+ + +

+ Recall that \tilde{G}(x)=\ln|1-x| is an antiderivative of g(x)=\displaystyle \frac{1}{1-x}. For which C is your chosen G(x)=\ln|1-x|? +

+ + +

+ Use G_4(x) to estimate \displaystyle \int_2^4 \ln|1-x|dx. +

+ + + + + + +

+ Recall that the power series for f(x)=\sin\left(x\right) is: \sin\left(x\right)=\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n} x^{2 \, n + 1}}{\left(2 \, n + 1\right)!}.

 + + +

Find a power series for \sin\left(-5 \, x^{2}\right).

 +

Find a power series for x^{4} \sin\left(x\right).

 +

Find a power series for F(x), an antiderivative of f(x) such that F(0)=4.

 + + + + + + +

+ Recall that the power series for f(x)=-\frac{1}{x - 1} is: -\frac{1}{x - 1}=\sum_{n=0}^{\infty} x^{n}.

 + + +

Find a power series for \frac{1}{x^{4} + 1}.

 +

Find a power series for -\frac{x^{5}}{x - 1}.

 +

Find a power series for f'(x).

 + + + + + + +

+ Recall that g(x)=\displaystyle\sum_{n=0}^\infty x^n=\frac{1}{1-x} for -1< x< 1 and \frac{d}{dx}[\arctan(x)]=\frac{1}{1+x^2}=g(-x^2). We computed the power series for g(-x^2) in . +

+ + +

+ Integrate this power series and find C to find a power series for H(x)=\arctan(x). Recall that \arctan(0)=0. +

+ + + +

+ Find the interval of convergence for this series. +

+ + + + + + + + + +

Find the power series for \alpha(x)=\ln|x|.

 +

Find the interval of convergence for this series.

 + + + + + + +

Find the power series for \beta(x)=\arctan(-3x^2).

 +

Find the interval of convergence for this series.

 + + + + + + + Videos +
+ + Video: Compute power series by manipulating known exponential/trigonometric/binomial power series + + +
+ + + +
diff --git a/calculus/source/09-PS/04.ptx b/calculus/source/09-PS/04.ptx new file mode 100644 index 00000000..47d1e229 --- /dev/null +++ b/calculus/source/09-PS/04.ptx @@ -0,0 +1,504 @@ + + +
+ Taylor Series (PS4) + + + + + Activities + + + + + +

+ The following tasks will help us find a mechanism to produce + a power series given information about its derivatives. +

+ + +

+ Find the 2nd derivative of x^2. +

+
    +
  1. 2x
    2. +
  2. 2
    3. +
  3. 4x
    4. +
  4. 4
    5. +
+ + + +

+ Find the 3rd derivative of x^3. +

+
    +
  1. 2
    2. +
  2. 3x
    3. +
  3. 6
    4. +
  4. 12x
    5. +
+ + + +

+ Find the 4th derivative of x^4. +

+
    +
  1. 18
    2. +
  2. 24
    3. +
  3. 32
    4. +
  4. 64
    5. +
+ + + +

+ Based on these results, which of the following should always + equal the nth derivative of x^n with respect to x? +

+
    +
  1. n
    2. +
  2. n^2
    3. +
  3. n!
    4. +
  4. n^n
    5. +
+ + + + + + + + +

+ Let's use derivatives to rediscover the sequence a_n which gives + a power series representation for e^x. +

+ + +

+ Let's say that + e^x=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4\dots. +

+

+ What must a_0 be to satisfy e^0=1? +

+ + + +

+ Then, \frac{d}{dx}[e^x]=e^x=a_1+2a_2x+3a_3x^2+4a_4x^3\dots. +

+

+ What must a_1 be to also satisfy e^0=1? +

+ + + + +

+ Then, \frac{d^2}{dx^2}[e^x]=e^x=2a_2+6a_3x+12a_4x^2+\dots. +

+

+ What must a_2 be to also satisfy e^0=1? +

+ + + + +

+ Then, \frac{d^3}{dx^3}[e^x]=e^x=6a_3+24a_4x+\dots. +

+

+ What must a_3 be to also satisfy e^0=1? +

+ + + + +

+ So this 6a_3 term was obtained from the fact that the 3rd + derivative of x^3 is 3!=6. +

+

+ So finally, we may skip ahead to the nth derivative: + \frac{d^n}{dx^n}[e^x]=e^x=n!\cdot a_n+(n+1)!\cdot a_{n+1}\cdot x+\dots. +

+

+ What must a_n be to also satisfy e^0=1? +

+ + + + +

+ This reveals the power series we previously found for e^x: + e^x=\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty \frac{1}{n!}x^n. +

+

+ So in general, if f(x)=a_0+a_1x+a_2x^2+\dots, + then \frac{d^n}{dx^n}[f(x)]=f^{(n)}(x)=n!\cdot a_n+(n+1)!\cdot a_{n+1}\cdot x+\dots. +

+

+ What must a_n be to produce the correct value for f^{(n)}(0)? +

+ + + + + + + + + + + + +

+ If f(x) can be written as a power series, then there is a real number c such that + + f(x)&=\sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!}(x-c)^n + &=f(c)+f^\prime(c)(x-c)+\frac{f^{\prime\prime}(c)}{2!}(x-c)^2+\frac{f^{(3)}(c)}{3!}(x-c)^3+\ldots + + on some interval centered at x=c. +

+

+ In fact, the functions that can be represented as power series are exactly those functions which + are infinitely differentiable on some open interval. +

+ + + + + +

+ The Taylor seriesTaylor Series generated by f(x) and + centered at x=c is given by + + f(x)&=\sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!}(x-c)^n + &=f(c)+f^\prime(c)(x-c)+\frac{f^{\prime\prime}(c)}{2!}(x-c)^2+\frac{f^{(3)}(c)}{3!}(x-c)^3+\ldots + + with an interval of convergence determinable by series convergence rules. +

+

+When c=0, + + f(x)&=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n + &=f(0)+f^\prime(0)x+\frac{f^{\prime\prime}(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\ldots + +is called the Maclaurin seriesMaclaurin Series generated by f. +

+ + + + + + +

+ Observe that f(x)=\sin(x) is a function such that: + + + \begin{array}{c|c|c|c|c|c|c|c} + f(0) & f'(0) & f''(0) & f^{(3)}(0) & f^{(4)}(0) & f^{(5)}(0) & f^{(6)}(0) & f^{(7)}(0) \\ + \hline + \sin(0) & \cos(0) & -\sin(0) & -\cos(0) & \sin(0) & \cos(0) & -\sin(0) & -\cos(0) \\ + \hline + 0 & 1 & 0 & -1 & 0 & 1 & 0 & -1 + \end{array} +

+ + +

Given the zeros appearing for every even derivative above, + which of these is a valid simplification of the Maclarin series + \displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n for \sin(x)?

+
    +
  1. \sum_{n=1}^\infty\frac{f^{(n)}(0)}{n!}x^n
    2. +
  2. \sum_{2n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n
    3. +
  3. \sum_{n=0}^\infty\frac{f^{(2n)}(0)}{(2n)!}x^{2n}
    4. +
  4. \sum_{n=0}^\infty\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1}
    5. +
+ + +

Now consider + the following consolidated chart: + + \begin{array}{c|c|c|c|c|c|c|c} + f^{(1)}(0) & f^{(3)}(0) &f^{(5)}(0) & f^{(7)}(0) \\ + \hline + \cos(0) & -\cos(0) & \cos(0) & -\cos(0) \\ + \hline + 1 & -1 & 1 & -1 + \end{array} +

+

Which formula yields these alternating 1s and -1s + appearing for f^{(2n+1)}(0)? +

+
    +
  1. f^{(2n+1)}(0)=(-1)^n
    2. +
  2. f^{(2n+1)}(0)=(-1)^{n+1}
    3. +
  3. f^{(2n+1)}(0)=(-1)^{2n}
    4. +
  4. f^{(2n+1)}(0)=(-1)^{2n+1}
    5. +
+ + + + + +

+ The power series we've introduced for each of the following functions + are in fact their Maclaurin series (Taylor series centered at 0). +

+ + \frac{1}{1-x} = \sum_{n=0}^\infty \frac{n!}{n!}x^n\amp= 1+x+x^2+x^3+\dots + e^x = \sum_{n=0}^\infty \frac{1}{n!}x^n\amp= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots + \cos(x)= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}\amp =1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\dots + \sin(x)=\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}\amp =x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\dots + + + + + + + +

+ For a function f(x) with a Taylor series centered at x=c, + + f(x) \amp\approx T_k(x) + \amp = \sum_{n=0}^k \frac{f^{(n)}(c)}{n!}(x-c)^n + \amp = f(c)+f^\prime(c)(x-c)+\frac{f^{\prime\prime}(c)}{2!}(x-c)^2+ + + \ldots+\frac{f^{(k)}(c)}{k!}(x-c)^k + + where T_k(x) is called the k^{th} degree + Taylor polynomialTaylor polynomial generated by f and centered at x=c. +

+

+The k^{th} degree Taylor polynomial can be seen as the best polynomial of degree k or less for approximating f(x) for values close to x=c. +Note that the 1^{st} degree Taylor polynomial is also known as the linearizationlinearization of a function of f. +

+ + + + + +

+ Let f(x) be a function such that: + + + \begin{array}{c|c|c|c|c|c|c} + f(4) & f'(4) & f''(4) & f'''(4) & f^{(4)}(4) & f^{(5)}(4) & f^{(6)}(4) \\ + \hline + 0 & 1 & 2 & 3 & 4 & 5 & 6 + \end{array} +

+ + +

Find a Taylor polynomial for f(x) centered at x=4 of degree 3.

 +

Using the table above, find a general closed form for f^{(n)}(4).

 +

Use (b) to find a Taylor series for f(x) centered at x=4.

 + + + + + + +

+ Let f(x) be a function such that: + + + \begin{array}{c|c|c|c|c|c|c} + f(-2) & f'(-2) & f''(-2) & f'''(-2) & f^{(4)}(-2) & f^{(5)}(-2) & f^{(6)}(-2) \\ + \hline + 0 & 2 & -16 & 54 & -128 & 250 & -432 + \end{array} +

+ + +

Find a Taylor polynomial for f(x) centered at x=-2 of degree 3.

 +

Using the table above, find a general closed form for f^{(n)}(-2).

 +

Use (b) to find a Taylor series for f(x) centered at x=-2.

 + + + + + + + + + +

+ You might have seen \sqrt{-1} written as i, and know that z is a complex number if z=a+bi for some real numbers a and b. +Note that i^2=-1, i^3=(i^2)i=-i, i^4=(i^2)^2=1, i^5=(i^4)i=i, and so on. +This gives rise to the following notion. +

+ + + + + + Euler's Identity + +

+ For any real number \theta, + + + + e^{i\theta} & = 1+\displaystyle\frac{i\theta}{1!}+\displaystyle\frac{(i\theta)^2}{2!}+\displaystyle\frac{(i\theta)^3}{3!}+\displaystyle\frac{(i\theta)^4}{4!}+\displaystyle\frac{(i\theta)^5}{5!}+\displaystyle\frac{(i\theta)^6}{6!}+\displaystyle\frac{(i\theta)^7}{7!}+\displaystyle\frac{(i\theta)^8}{8!}+\ldots + & = 1+i\theta-\displaystyle\frac{\theta^2}{2!}-\displaystyle\frac{i\theta^3}{3!}+\displaystyle\frac{\theta^4}{4!}+\displaystyle\frac{i\theta^5}{5!}-\displaystyle\frac{\theta^6}{6!}-\displaystyle\frac{i\theta^7}{7!}+\displaystyle\frac{\theta^8}{8!}+\ldots + & = \left(1-\displaystyle\frac{\theta^2}{2!}+\displaystyle\frac{\theta^4}{4!}-\displaystyle\frac{\theta^6}{6!}+\ldots\right)+i\left(\theta-\displaystyle\frac{\theta^3}{3!}+\displaystyle\frac{\theta^5}{5!}-\displaystyle\frac{\theta^7}{7!}+\ldots\right) + & = \cos(\theta)+i\sin(\theta). + + + +

+ + + + + + + + + +

+ Use Euler's identity to evaluate e^{i\pi}. +

+ + + + + + + + + + Videos +
+ + Video: Determine a Taylor or Maclaurin series for a function + + + + +
+ + + + + + + + + + + + + + + +
+ + \ No newline at end of file diff --git a/calculus/source/09-PS/main.ptx b/calculus/source/09-PS/main.ptx new file mode 100644 index 00000000..d6699c3a --- /dev/null +++ b/calculus/source/09-PS/main.ptx @@ -0,0 +1,13 @@ + + + + Power Series (PS) + + + + + + + + + diff --git a/calculus/source/09-PS/outcomes/01.ptx b/calculus/source/09-PS/outcomes/01.ptx new file mode 100644 index 00000000..cb891a4b --- /dev/null +++ b/calculus/source/09-PS/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Approximate functions defined as power series. +

\ No newline at end of file diff --git a/calculus/source/09-PS/outcomes/02.ptx b/calculus/source/09-PS/outcomes/02.ptx new file mode 100644 index 00000000..b278a04d --- /dev/null +++ b/calculus/source/09-PS/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+ Determine the interval of convergence for a given power series. +

\ No newline at end of file diff --git a/calculus/source/09-PS/outcomes/03.ptx b/calculus/source/09-PS/outcomes/03.ptx new file mode 100644 index 00000000..5a9aa830 --- /dev/null +++ b/calculus/source/09-PS/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Compute power series by manipulating known exponential/trigonometric/binomial power series. +

\ No newline at end of file diff --git a/calculus/source/09-PS/outcomes/04.ptx b/calculus/source/09-PS/outcomes/04.ptx new file mode 100644 index 00000000..423c5797 --- /dev/null +++ b/calculus/source/09-PS/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Determine a Taylor or Maclaurin series for a function. +

\ No newline at end of file diff --git a/calculus/source/09-PS/outcomes/main.ptx b/calculus/source/09-PS/outcomes/main.ptx new file mode 100644 index 00000000..d1f2ecb6 --- /dev/null +++ b/calculus/source/09-PS/outcomes/main.ptx @@ -0,0 +1,26 @@ + +> + +

+How do we use series to understand functions? +

+

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
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    4. +
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    5. + +
+ \ No newline at end of file diff --git a/calculus/source/09-PS/readiness.ptx b/calculus/source/09-PS/readiness.ptx new file mode 100644 index 00000000..a8daa516 --- /dev/null +++ b/calculus/source/09-PS/readiness.ptx @@ -0,0 +1,56 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Compute limits.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    • +

      Review: Khan Academy

      +
      • +
    +
    2. +
  2. +

    Express the sum of indexed values using summation notation.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    3. + +
  3. +

    Employ the ratio test for convergence.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    4. + +
  4. +

    Power rule for differentiating and integrating.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    • +

      Review: Khan Academy

      +
      • +
    +
    5. + + + + +
+ + \ No newline at end of file diff --git a/calculus/source/main.ptx b/calculus/source/main.ptx new file mode 100644 index 00000000..a582dd6f --- /dev/null +++ b/calculus/source/main.ptx @@ -0,0 +1,33 @@ + + + + + + + Calculus for Team-Based Inquiry Learning + 2024 Edition PREVIEW + 2024 Edition PREVIEW (Instructor Version) + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/calculus/source/meta/backmatter.ptx b/calculus/source/meta/backmatter.ptx new file mode 100644 index 00000000..4aaeb617 --- /dev/null +++ b/calculus/source/meta/backmatter.ptx @@ -0,0 +1,16 @@ + + + + + + + + + Backmatter + + + + + + + diff --git a/calculus/source/meta/copyright.ptx b/calculus/source/meta/copyright.ptx new file mode 100644 index 00000000..3f8b5a5d --- /dev/null +++ b/calculus/source/meta/copyright.ptx @@ -0,0 +1,13 @@ + + + + 20212023 + Steven Clontz and Drew Lewis + +This work is freely available for noncommerical, educational purposes. +For specific licensing information, including the terms for licensing of derivative works, please visit + + GitHub.com/TeamBasedInquiryLearning +. + + diff --git a/calculus/source/meta/docinfo.ptx b/calculus/source/meta/docinfo.ptx new file mode 100644 index 00000000..72484c39 --- /dev/null +++ b/calculus/source/meta/docinfo.ptx @@ -0,0 +1,55 @@ + + + + + + + + TBIL-Calc + + + + + + + + + + + + +\newcommand{\N}{\mathbb N} +\newcommand{\Z}{\mathbb Z} +\newcommand{\Q}{\mathbb Q} +\newcommand{\R}{\mathbb R} +\newcommand{\unknown}{{\color{gray} ?}} +\DeclareMathOperator{\arcsec}{arcsec} +\DeclareMathOperator{\arccot}{arccot} +\DeclareMathOperator{\arccsc}{arccsc} +\newcommand{\tuple}[1]{\left\langle#1\right\rangle} + + + + + + \usepackage{tikz, pgfplots} + \usetikzlibrary{positioning,matrix,arrows} + \usetikzlibrary{shapes,decorations,shadows,fadings,patterns} + \usetikzlibrary{decorations.markings} + \tikzset{% + /pgfplots/table/row sep=crcr + } + + + + + Learning Outcomes + + + + + + + + diff --git a/calculus/source/meta/frontmatter.ptx b/calculus/source/meta/frontmatter.ptx new file mode 100644 index 00000000..ec504e8f --- /dev/null +++ b/calculus/source/meta/frontmatter.ptx @@ -0,0 +1,116 @@ + + + + + + + + + TBIL Institute Fellows + sclontz@southalabama.edu + + + + Editors + + Steven Clontz + Department of Mathematics and Statistics + University of South Alabama + sclontz@southalabama.edu + + + Drew Lewis + Department of Mathematics and Statistics + University of South Alabama + drew.lewis@gmail.com + + + + Contributing Authors + + Tien Chih + Department of Mathematics + Montana State University-Billings + tien.chih@msubillings.edu + + + PJ Couch + Department of Mathematics + Lamar University + pjcouch@lamar.edu + + + Francesca Gandini + Department of Mathematics + Kalamazoo College + fra.gandi.phd@gmail.com + + + Joseph Hibdon + Department of Mathematics + Northeastern Illinois University + j-hibdonjr@neiu.edu + + + Sarah A. Nelson + Donald and Helen Schort School of Mathematics and Computing Sciences + Lenoir-Rhyne University + sarah.nelson@lr.edu + + + Abby Noble + Department of Mathematics and Statistics + Middle Georgia State University + abby.noble@mga.edu + + + Dave Rosoff + Department of Mathematics and Physical Sciences + The College of Idaho + drosoff@collegeofidaho.edu + + + Belin Tsinnajinnie + Department of Mathematics + Santa Fe Community College + belin.tsinnajinnie@sfcc.edu + + + + + + + + + Calculus for Team-Based Inquiry Learning + + + +

+This work includes materials used under license from the following works: +

+
    +
  • +

    Active Calculus - single variable

    +
      +
    • +

      +
      • +
    • +

      CC BY-SA 4.0

      +
      • +
    +
    • +
+ + + TBIL Resource Library +

+This work is made available as part of the +TBIL Resource Library, +a product of NSF DUE Award #2011807. +

+ + + + diff --git a/calculus/source/meta/integral-table.ptx b/calculus/source/meta/integral-table.ptx new file mode 100644 index 00000000..9106afd2 --- /dev/null +++ b/calculus/source/meta/integral-table.ptx @@ -0,0 +1,74 @@ + + + + A Short Table of Integrals +

+

    +
  1. +

    + \int \frac{du}{a^2 + u^2} = \frac{1}{a} \arctan \left( \frac{u}{a} \right) + C +

    +
    2. + +
  2. +

    + \int \frac{du}{\sqrt{u^2 \pm a^2}} = \ln |u + \sqrt{u^2 \pm a^2}| + C +

    +
    3. + +
  3. +

    + \int \sqrt{u^2 \pm a^2} \, du = \frac{u}{2}\sqrt{u^2 \pm a^2} \pm \frac{a^2}{2}\ln|u + \sqrt{u^2 \pm a^2}| + C +

    +
    4. + +
  4. +

    + \int \frac{u^2 du}{\sqrt{u^2 \pm a^2}} = \frac{u}{2}\sqrt{u^2 \pm a^2} \mp \frac{a^2}{2}\ln|u + \sqrt{u^2 \pm a^2}| + C +

    +
    5. + +
  5. +

    + \int \frac{du}{u\sqrt{u^2+a^2}} = -\frac{1}{a} \ln \left| \frac{a+\sqrt{u^2 + a^2}}{u} \right| + C +

    +
    6. + +
  6. +

    + \int \frac{du}{u\sqrt{u^2-a^2}} = \frac{1}{a} \arcsec \left( \frac{u}{a} \right) + C +

    +
    7. + +
  7. +

    + \int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin \left( \frac{u}{a} \right) + C +

    +
    8. + +
  8. +

    + \int \sqrt{a^2 - u^2} \, du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2} \arcsin \left( \frac{u}{a} \right) + C +

    +
    9. + +
  9. +

    + \int \frac{u^2}{\sqrt{a^2 - u^2}} \, du = -\frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2} \arcsin \left( \frac{u}{a} \right) + C +

    +
    10. + +
  10. +

    + \int \frac{du}{u\sqrt{a^2 - u^2}} = -\frac{1}{a} \ln \left| \frac{a + \sqrt{a^2 - u^2}}{u} \right| + C +

    +
    11. + +
  11. +

    + \int \frac{du}{u^2\sqrt{a^2 - u^2}} = -\frac{\sqrt{a^2-u^2}}{a^2 u} + C +

    +
    12. +
+

+ \ No newline at end of file diff --git a/calculus/source/meta/sample-solutions.ptx b/calculus/source/meta/sample-solutions.ptx new file mode 100644 index 00000000..3b827910 --- /dev/null +++ b/calculus/source/meta/sample-solutions.ptx @@ -0,0 +1,54 @@ + + + + Sample Solutions + + + LT1 + +

+ Sketch the graph of a function \(f(x)\) that meets all of the following criteria. Be sure to scale your axes and label any important features of your graph. +

    +
  • +

    + \(f(x)\) has a vertical asymptote at \(2\), but \(\displaystyle\lim_{x\to 2^+} f(x)\) is finite. +

    +
    • +
  • +

    + \(\displaystyle\lim_{x\to 0} f(x)=-4\), but \(f(0)=-3\). +

    +
    • +
  • +

    + \(\displaystyle\lim_{x\to 6^-} f(x)=2\) and \(\displaystyle\lim_{x\to 6^+} f(x)=4\). +

    +
    • +
+

+ + +

+ Answers vary. Here are some things to check for: +

    +
  • +

    + The graph should have a vertical asymptote at \(x=2\). As the function approaches the vertical asymptote from the left, the function outputs should grow infinitely large in magnitude (either in the postive or negative \(y\)-direction). As the function approaches the vertical asymptote from the right, the function should bump straight into the vertical asymptote. +

    +
    • +
  • +

    + The graph should approach the point \((0,-4)\), but there should be a removable discontinuity (hole) in the graph right at that point (indicated by an open circle). The actual value of the function at \(x=0\) should be \(-3\), so the point \((0,-3)\) should be included on the graph. +

    +
    • +
  • +

    + There is a jump discontinuity at \(x=6\) with the graph approaching the point \((6,2)\) as \(x\) approaches \(6\) from the left and approaching the point \((6,4)\) as \(x\) approaches \(6\) from the right. +

    +
    • +
+

+ + + + \ No newline at end of file diff --git a/calculus/source/meta/trig-table.ptx b/calculus/source/meta/trig-table.ptx new file mode 100644 index 00000000..e9a1566b --- /dev/null +++ b/calculus/source/meta/trig-table.ptx @@ -0,0 +1,36 @@ + + + + List of Trigonometric Identities +
    +

  1. [\sin(\alpha)]^2+[\cos(\alpha)]^2=1

    2. + +

  2. \cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)

    3. + +

  3. \sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)

    4. + +

  4. \cos(2\alpha) = [\cos(\alpha)]^2-[\sin(\alpha)]^2 = 2[\cos(\alpha)]^2-1 = 1-2[\sin(\alpha)]^2

    5. + +

  5. \sin(2\alpha)=2\sin(\alpha)\cos(\alpha)

    6. + +

  6. [\cos(\alpha)]^2=\displaystyle\frac{1+\cos(2\alpha)}{2}

    7. + +

  7. [\sin(\alpha)]^2=\displaystyle\frac{1-\cos(2\alpha)}{2}

    8. + +

  8. \sin(\alpha)\cos(\beta)=\displaystyle\frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}

    9. + +

  9. \sin(\alpha)\sin(\beta)=\displaystyle\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}

    10. + +

  10. \cos(\alpha)\sin(\beta)=\displaystyle\frac{\sin(\alpha+\beta)-\sin(\alpha-\beta)}{2}

    11. + +

  11. \cos(\alpha)\cos(\beta)=\displaystyle\frac{\cos(\alpha-\beta)+\cos(\alpha+\beta)}{2}

    12. + +

  12. \sin(\alpha)+\sin(\beta)=2\sin\left(\displaystyle\frac{\alpha+\beta}{2}\right)\cos\left(\displaystyle\frac{\alpha-\beta}{2}\right)

    13. + +

  13. \sin(\alpha)-\sin(\beta)=2\cos\left(\displaystyle\frac{\alpha+\beta}{2}\right)\sin\left(\displaystyle\frac{\alpha-\beta}{2}\right)

    14. + +

  14. \cos(\alpha)+\cos(\beta)=2\cos\left(\displaystyle\frac{\alpha+\beta}{2}\right)\cos\left(\displaystyle\frac{\alpha-\beta}{2}\right)

    15. + +

  15. \cos(\alpha)-\cos(\beta)=-2\sin\left(\displaystyle\frac{\alpha+\beta}{2}\right)\sin\left(\displaystyle\frac{\alpha-\beta}{2}\right)

    16. +
+ \ No newline at end of file diff --git a/calculus/source/sage/AI3-length-area-relationship.sage b/calculus/source/sage/AI3-length-area-relationship.sage new file mode 100644 index 00000000..e810e015 --- /dev/null +++ b/calculus/source/sage/AI3-length-area-relationship.sage @@ -0,0 +1,10 @@ + +f(x) = (x-3)*(x-5)*(x-7)+40 +p = line([(2,0),(2,f(2))], color='black') +p += line([(8,0),(8,f(8))], color='black') +p += polygon([(2,0),(2,f(2))] + [(x, f(x)) for x in [2,2.1,..,8]] + [(8,0),(2,0)], rgbcolor=(0.8,0.8,0.8),aspect_ratio='automatic') +p += text("$\\mathrm{Area}=\\int_{a}^b f(x) dx$", (4, 55), fontsize=16, color='black') +p += plot(f, (1, 8.5), thickness=3) +p += line([(6,0),(6,f(6))], color="#8888ff") +p += text("$\\mathrm{Length}=f(x)$", (5.9, 20), fontsize=16, color='black', horizontal_alignment='right') +p \ No newline at end of file diff --git a/calculus/source/sage/AI3-volume-definition.sage b/calculus/source/sage/AI3-volume-definition.sage new file mode 100644 index 00000000..ec45f37b --- /dev/null +++ b/calculus/source/sage/AI3-volume-definition.sage @@ -0,0 +1,9 @@ +x,y,z=var("x y z") +p = revolution_plot3d(x/2,(x,0,4), show_curve=False,parallel_axis="x",opacity=0.5,frame=False) +p += revolution_plot3d((4,z),(z,0,2), show_curve=False,parallel_axis="x",opacity=0.5) +p += revolution_plot3d((2,z),(z,0,1), show_curve=False,parallel_axis="x",opacity=1) +p += line([(0,0,-3), (0,0,3)],color="black",label="x") +p += line([(-1,0,0), (5,0,0)],color="black") +# p += text3d("Volume=∫A(x)dx",(4,0,2.5)) +# p += text3d("Area=A(x)",(2,0,1.25)) +p diff --git a/calculus/source/sage/AI3-x0-y4-yx2-revolution-1.sage b/calculus/source/sage/AI3-x0-y4-yx2-revolution-1.sage new file mode 100644 index 00000000..7adeeb11 --- /dev/null +++ b/calculus/source/sage/AI3-x0-y4-yx2-revolution-1.sage @@ -0,0 +1,7 @@ +x,y,z = var('x y z') +p = revolution_plot3d(x^2,(x,0,2), show_curve=True,parallel_axis="z",opacity=0.5,frame=False) +p += revolution_plot3d(4,(x,0,2), show_curve=True,parallel_axis="z",opacity=0.5) +p += revolution_plot3d(2,(x,0,sqrt(2)), show_curve=True,parallel_axis="z",opacity=1) +p += line([(0,0,-1), (0,0,5)],color="black",label="x") +p += line([(-4,0,0), (4,0,0)],color="black") +p diff --git a/calculus/source/sage/AI3-x0-y4-yx2-revolution-2.sage b/calculus/source/sage/AI3-x0-y4-yx2-revolution-2.sage new file mode 100644 index 00000000..67ca8aa9 --- /dev/null +++ b/calculus/source/sage/AI3-x0-y4-yx2-revolution-2.sage @@ -0,0 +1,7 @@ +x,y,z = var('x y z') +p = revolution_plot3d(x^2,(x,0,2), show_curve=True,parallel_axis="z",opacity=0.5,frame=False) +p += revolution_plot3d(4,(x,0,2), show_curve=True,parallel_axis="z",opacity=0.5) +p += revolution_plot3d((1,x),(x,1,4), show_curve=True,parallel_axis="z",opacity=1) +p += line([(0,0,-1), (0,0,5)],color="black") +p += line([(-4,0,0), (4,0,0)],color="black") +p diff --git a/calculus/source/sage/AI3-x0-y4-yx2-revolution-3.sage b/calculus/source/sage/AI3-x0-y4-yx2-revolution-3.sage new file mode 100644 index 00000000..da94677f --- /dev/null +++ b/calculus/source/sage/AI3-x0-y4-yx2-revolution-3.sage @@ -0,0 +1,8 @@ +x,y,z = var('x y z') +p = revolution_plot3d(x^2,(x,0,2), show_curve=True,parallel_axis="x",opacity=0.5,frame=False) +p += revolution_plot3d(4,(x,0,2), show_curve=True,parallel_axis="x",opacity=0.5) +p += revolution_plot3d((0,z),(z,0,4), show_curve=True,parallel_axis="x",opacity=0.5) +p += revolution_plot3d(2,(x,0,sqrt(2)), show_curve=True,parallel_axis="x",opacity=1) +p += line([(0,0,-5), (0,0,5)],color="black",label="x") +p += line([(-1,0,0), (4,0,0)],color="black") +p diff --git a/calculus/source/sage/AI3-x0-y4-yx2-revolution-4.sage b/calculus/source/sage/AI3-x0-y4-yx2-revolution-4.sage new file mode 100644 index 00000000..7bed40cb --- /dev/null +++ b/calculus/source/sage/AI3-x0-y4-yx2-revolution-4.sage @@ -0,0 +1,8 @@ +x,y,z = var('x y z') +p = revolution_plot3d(x^2,(x,0,2), show_curve=True,parallel_axis="x",opacity=0.5,frame=False) +p += revolution_plot3d(4,(x,0,2), show_curve=True,parallel_axis="x",opacity=0.5) +p += revolution_plot3d((0,z),(z,0,4), show_curve=True,parallel_axis="x",opacity=0.5) +p += revolution_plot3d((1,x),(x,1,4), show_curve=True,parallel_axis="x",opacity=1) +p += line([(0,0,-5), (0,0,5)],color="black",label="x") +p += line([(-1,0,0), (4,0,0)],color="black") +p diff --git a/calculus/source/sage/AI3-x0-y4-yx2.sage b/calculus/source/sage/AI3-x0-y4-yx2.sage new file mode 100644 index 00000000..df8e6e3d --- /dev/null +++ b/calculus/source/sage/AI3-x0-y4-yx2.sage @@ -0,0 +1,4 @@ +x,y,z = var('x y z') +p = plot(x^2,(x,0,2),aspect_ratio=1,color='red') +p += line([(0,0),(0,4),(2,4)],color="red") +p diff --git a/codechat_config_calc.yaml b/codechat_config_calc.yaml new file mode 100644 index 00000000..b1711c23 --- /dev/null +++ b/codechat_config_calc.yaml @@ -0,0 +1,5 @@ +# Replace the contents of `codechat_config.yaml` with this file to edit Calculus +source_path: calculus/source +output_path: output/calc-web-insructor +args: "{sys_executable} -m pretext build calc-web-insructor --xmlid {xml_id} --no-knowls" +project_type: PreTeXt diff --git a/codechat_config_la.yaml b/codechat_config_la.yaml new file mode 100644 index 00000000..a9cfdfb0 --- /dev/null +++ b/codechat_config_la.yaml @@ -0,0 +1,5 @@ +# Replace the contents of `codechat_config.yaml` with this file to edit Linear Algebra +source_path: linear-algebra/source +output_path: output/la-web-insructor +args: "{sys_executable} -m pretext build la-web-insructor --xmlid {xml_id} --no-knowls" +project_type: PreTeXt diff --git a/codechat_config_precal.yaml b/codechat_config_precal.yaml new file mode 100644 index 00000000..a7afd5e0 --- /dev/null +++ b/codechat_config_precal.yaml @@ -0,0 +1,5 @@ +# Replace the contents of `codechat_config.yaml` with this file to edit Precalculus +source_path: precalculus/source +output_path: output/precal-web-insructor +args: "{sys_executable} -m pretext build precal-web-insructor --xmlid {xml_id} --no-knowls" +project_type: PreTeXt diff --git a/linear-algebra/assets/teapot.blend b/linear-algebra/assets/teapot.blend new file mode 100644 index 00000000..c97a81ae Binary files /dev/null and b/linear-algebra/assets/teapot.blend differ diff --git a/linear-algebra/assets/teapot.png b/linear-algebra/assets/teapot.png new file mode 100644 index 00000000..a4adacee Binary files /dev/null and b/linear-algebra/assets/teapot.png differ diff --git a/linear-algebra/assets/truss.jpg b/linear-algebra/assets/truss.jpg new file mode 100644 index 00000000..97f107db Binary files /dev/null and b/linear-algebra/assets/truss.jpg differ diff --git a/linear-algebra/source/01-LE/01.ptx b/linear-algebra/source/01-LE/01.ptx new file mode 100644 index 00000000..09b5199c --- /dev/null +++ b/linear-algebra/source/01-LE/01.ptx @@ -0,0 +1,649 @@ + +
+ Linear Systems, Vector Equations, and Augmented Matrices (LE1) + + + + +Class Activities + + +

+A Euclidean vectorEuclidean vectorvectorEuclidean is an ordered +list of real numbers + + \left[\begin{array}{c} + a_1 \\ + a_2 \\ + \vdots \\ + a_n + \end{array}\right] +. +We will find it useful to almost always typeset Euclidean vectors vertically, but the notation +\left[\begin{array}{cccc}a_1 & a_2 & \cdots & a_n\end{array}\right]^T is also +valid when vertical typesetting is inconvenient. The set of all Euclidean vectors with +n components is denoted as \mathbb R^n, and vectors are often described using +the notation \vec v. +

+

+Each number in the list is called a component, and we use the following +definitions for the sum of two vectors, and the product of a real number and a vector: + + \left[\begin{array}{c} + a_1 \\ + a_2 \\ + \vdots \\ + a_n + \end{array}\right]+ + \left[\begin{array}{c} + b_1 \\ + b_2 \\ + \vdots \\ + b_n + \end{array}\right]= + \left[\begin{array}{c} + a_1+b_1 \\ + a_2+b_2 \\ + \vdots \\ + a_n+b_n + \end{array}\right] + \hspace{3em} + c + \left[\begin{array}{c} + a_1 \\ + a_2 \\ + \vdots \\ + a_n + \end{array}\right]= + \left[\begin{array}{c} + ca_1 \\ + ca_2 \\ + \vdots \\ + ca_n + \end{array}\right] + +

+ + + +

+Following are some examples of addition and scalar multiplication in \mathbb R^4. + + \left[\begin{array}{c} + 3 \\ + -3 \\ + 0 \\ + 4 + \end{array}\right]+ + \left[\begin{array}{c} + 0 \\ + 2 \\ + 7 \\ + 1 + \end{array}\right]= + \left[\begin{array}{c} + 3+0 \\ + -3+2 \\ + 0+7 \\ + 4+1 + \end{array}\right]= + \left[\begin{array}{c} + 3 \\ + -1 \\ + 7 \\ + 5 + \end{array}\right] + + + -4 + \left[\begin{array}{c} + 0 \\ + 2 \\ + -2 \\ + 3 + \end{array}\right]= + \left[\begin{array}{c} + -4(0) \\ + -4(2)\\ + -4(-2) \\ + -4(3) + \end{array}\right]= + \left[\begin{array}{c} + 0 \\ + -8 \\ + 8 \\ + -12 + \end{array}\right] + +

+ + + +

+A linear equationlinear equation is an equation of the variables x_i of the form + +a_1x_1+a_2x_2+\dots+a_nx_n=b +. +

+

+ A solution for a linear equation is a Euclidean vector + + \left[\begin{array}{c} + s_1 \\ + s_2 \\ + \vdots \\ + s_n + \end{array}\right] + +that satisfies + +a_1s_1+a_2s_2+\dots+a_ns_n=b + +(that is, a Euclidean vector whose components can be plugged into the equation). +

+ linear equationsolution + + + + + +

+In previous classes you likely used the variables x,y,z in equations. +However, since this course often deals with equations of four or more +variables, we will often write our variables as x_i, and assume +x=x_1,y=x_2,z=x_3,w=x_4 when convenient. +

+ + + + + +

+ A system of linear equations (or a linear system for short) +is a collection of one or more linear equations. + + + a_{11}x_1 &\,+\,& a_{12}x_2 &\,+\,& \dots &\,+\,& a_{1n}x_n &\,=\,& b_1 + + + a_{21}x_1 &\,+\,& a_{22}x_2 &\,+\,& \dots &\,+\,& a_{2n}x_n &\,=\,& b_2 + + + \vdots& &\vdots& && &\vdots&&\vdots + + + a_{m1}x_1 &\,+\,& a_{m2}x_2 &\,+\,& \dots &\,+\,& a_{mn}x_n &\,=\,& b_m + + +

+ Its solution setsolution set is given by + + \setBuilder + { + \left[\begin{array}{c} + s_1 \\ + s_2 \\ + \vdots \\ + s_n + \end{array}\right] + }{ + \left[\begin{array}{c} + s_1 \\ + s_2 \\ + \vdots \\ + s_n + \end{array}\right] + \text{is a solution to all equations in the system} + } +. +

system of linear equationslinear system + + + +

+ When variables in a large linear system are missing, we prefer to + write the system in one of the following standard forms: +

+ +

+ Original linear system: + + + x_1 + 3x_3 &\,=\,& 3 + + + 3x_1 - 2x_2 + 4x_3 &\,=\,& 0 + + + -x_2 + x_3 &\,=\,& -2 + + +

+

+ Verbose standard form: + + + 1x_1 &\,+\,& 0x_2 &\,+\,& 3x_3 &\,=\,& 3 + + + 3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0 + + + 0x_1 &\,-\,& 1x_2 &\,+\,& 1x_3 &\,=\,& -2 + + +

+

+ Concise standard form: + + + x_1 & & &\,+\,& 3x_3 &\,=\,& 3 + + + 3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0 + + + &\,-\,& x_2 &\,+\,& x_3 &\,=\,& -2 + + +

+ + + + + + +

+It will often be convenient to think of a system of equations as a vector equation.vector equation +

+

+By applying vector operations and equating components, it is straightforward to see that the vector equation + x_1 \left[\begin{array}{c} 1 \\ 3 \\ 0 \end{array}\right]+ x_2 \left[\begin{array}{c} 0 \\ -2 \\ -1 \end{array}\right] + + x_3 \left[\begin{array}{c} 3 \\ 4 \\1 \end{array}\right] = \left[\begin{array}{c} 3 \\ 0 \\ -2 \end{array}\right] +is equivalent to the system of equations + + + x_1 & & &\,+\,& 3x_3 &\,=\,& 3 + + + 3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0 + + + &\,-\,& x_2 &\,+\,& x_3 &\,=\,& -2 + + +

+ + + + + + +

+A linear system is consistentlinear systemconsistent if its solution set + is non-empty (that is, there exists a solution for the +system). Otherwise it is inconsistent.linear systeminconsistent +

+ + + + + +

+ All linear systems are one of the following: +

+
    +
  1. Consistent with one solution: + its solution set contains a single vector, e.g. + \setList{\left[\begin{array}{c}1\\2\\3\end{array}\right]} +
    2. +
  2. + Consistent with infinitely-many solutions: + its solution set contains infinitely many vectors, e.g. + + \setBuilder + { + \left[\begin{array}{c}1\\2-3a\\a\end{array}\right] + }{ + a\in\IR + } + +
    3. +
  3. Inconsistent: + its solution set is the empty set, denoted by either \{\} or \emptyset. +
    4. +
+ + + + + +

+All inconsistent linear systems contain a logical contradiction. + Find a contradiction in this system to show that its solution set + is the empty set. +

+ + + -x_1+2x_2 &= 5 + + + 2x_1-4x_2 &= 6 + + + + + + + +

+ Consider the following consistent linear system. +

+ + + -x_1+2x_2 &= -3 + + + 2x_1-4x_2 &= 6 + + + + +

+ Find three different solutions + for this system. +

+ + +

+ Let x_2=a where a is an arbitrary real number, then find an + expression for x_1 in terms of a. Use this to write + the solution set + + \setBuilder + { + \left[\begin{array}{c} + \unknown \\ + a + \end{array}\right] + }{ + a \in \IR + } + + for the linear system. +

+ + + + + +

+ Consider the following linear system. +

+ + + x_1 &\,+\,& 2x_2 &\, \,& &\,-\,& x_4 &\,=\,& 3 + + + &\, \,& &\, \,& x_3 &\,+\,& 4x_4 &\,=\,& -2 + + +

+ Describe the solution set + + \setBuilder + { + \left[\begin{array}{c} + \unknown \\ + a \\ + \unknown \\ + b + \end{array}\right] + }{ + a,b \in \IR + } + + to the linear system + by setting x_2=a and x_4=b, and then solving for x_1 and + x_3. +

+ + + + + +

+ Solving linear systems of two variables by graphing or substitution is + reasonable for two-variable systems, but these simple techniques + won't usually cut it for equations with + more than two variables or more than two equations. For example, + + + -2x_1 &\,-\,& 4x_2 &\,+\,& x_3 &\,-\,& 4x_4 &\,=\,& -8 + + + x_1 &\,+\,& 2x_2 &\,+\,& 2x_3 &\,+\,& 12x_4 &\,=\,& -1 + + + x_1 &\,+\,& 2x_2 &\,+\,& x_3 &\,+\,& 8x_4 &\,=\,& 1 + + + has the exact same solution set as the system in the previous + activity, but we'll want to learn new techniques + to compute these solutions efficiently. +

+ + + + + +

+ The only important information in a linear system are its coefficients and + constants. +

+ +

+ Original linear system: + + + x_1 + 3x_3 &\,=\,& 3 + + + 3x_1 - 2x_2 + 4x_3 &\,=\,& 0 + + + -x_2 + x_3 &\,=\,& -2 + + +

+

+ Verbose standard form: + + + 1x_1 &\,+\,& 0x_2 &\,+\,& 3x_3 &\,=\,& 3 + + + 3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0 + + + 0x_1 &\,-\,& 1x_2 &\,+\,& 1x_3 &\,=\,& -2 + + +

+

+ Coefficients/constants: + + + 1 & & 0 &\,\,& 3 &\,|\,& 3 + + + 3 &\, \,& -2 &\,\,& 4 &\,|\,& 0 + + + 0 &\, \,& -1 &\,\,& 1 &\,|\,& -2 + + +

+ + + + + + +

+ A system of m linear equations with n variables is often represented + by writing its coefficients and constants in an augmented matrix.augmented matrix +

+ + + a_{11}x_1 &\,+\,& a_{12}x_2 &\,+\,& \dots &\,+\,& a_{1n}x_n &\,=\,& b_1 + + + a_{21}x_1 &\,+\,& a_{22}x_2 &\,+\,& \dots &\,+\,& a_{2n}x_n &\,=\,& b_2 + + + \vdots& &\vdots& && &\vdots&&\vdots + + + a_{m1}x_1 &\,+\,& a_{m2}x_2 &\,+\,& \dots &\,+\,& a_{mn}x_n &\,=\,& b_m + + + + \left[\begin{array}{cccc|c} + a_{11} & a_{12} & \cdots & a_{1n} & b_1\\ + a_{21} & a_{22} & \cdots & a_{2n} & b_2\\ + \vdots & \vdots & \ddots & \vdots & \vdots\\ + a_{m1} & a_{m2} & \cdots & a_{mn} & b_m + \end{array}\right] + + + + + + +

+ The corresponding augmented matrix for this system is obtained by + simply writing the coefficients and constants in matrix form. +

+ +

+Linear system: + + +x_1 && &\,+\,& 3x_3 &\,=\,& 3 + + +3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0 + + +&\,-\,& x_2 &\,+\,& x_3 &\,=\,& -2 + + +

+

+Augmented matrix: + + \left[\begin{array}{ccc|c} + 1 & 0 & 3 & 3 \\ + 3 & -2 & 4 & 0 \\ + 0 & -1 & 1 & -2 + \end{array}\right] + +

+ +

+ Vector equation: + x_1 \left[\begin{array}{c} 1 \\ 3 \\ 0 \end{array}\right]+ x_2 \left[\begin{array}{c} 0 \\ -2 \\ -1 \end{array}\right] + x_3 \left[\begin{array}{c} 3 \\ 4 \\1 \end{array}\right] = \left[\begin{array}{c} 3 \\ 0 \\ -2 \end{array}\right] +

+ + + + + + Videos +
+ + Video: Converting between systems, vector equations, and augmented matrices +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + +

Choose a value for the real constant k such that the following system has one, many, or no solutions. In each case, write the solution set. +

+

+ Consider the linear system: + + + x_1 - x_2 &\,=\,& 1 + + + 3x_1 - 3x_2 &\,=\,& k + + +

+ + + + +

+

+ Consider the linear system: + + + ax_1 + bx_2 &\,=\,& j + + + cx_1 + dx_2 &\,=\,& k + + +

+ Assume j and k are arbitrary real numbers. +
    +
  • Choose values for a,b,c, and d, such that ad-bc = 0. Show that this system is inconsistent.
    • +
  • Prove that, if ad-bc \neq 0, the system is consistent with exactly one solution. +
+

+ + + + +

Given a set S, we can define a relation between two arbitrary elements a,b \in S. If the two elements are related, we denote this a \sim b. + +

+

+ Any relation on a set S that satisfies the properties below is an equivalence relationequivalence relation. +

    +
  • Reflexive: For any a \in S, a \sim a
    • +
  • Symmetric: For a,b \in S, if a\sim b, then b \sim a
    • + +
  • Transitive: for any a,b,c \in S, a \sim b \mbox{ and } b \sim c \mbox{ implies } a\sim c
    • +

+ +

For each of the following relations, show that it is or is not an equivalence relation. +

    +
  • For a,b, \in \mathbb{R}, a \sim b if an only if a \leq b.
    • +
  • For a,b, \in \mathbb{R}, a \sim b if an only if |a|=|b|.
    • +

+ + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/01-LE/02.ptx b/linear-algebra/source/01-LE/02.ptx new file mode 100644 index 00000000..f0097da6 --- /dev/null +++ b/linear-algebra/source/01-LE/02.ptx @@ -0,0 +1,782 @@ + +
+ Row Reduction of Matrices (LE2) + + + +Class Activities + + +

+ Two systems of linear equations (and their corresponding augmented + matrices) are said to be equivalentequivalent matrices if they have the same + solution set. +

+

+ For example, both of these systems share the same solution set + \setList{ \left[\begin{array}{c} 1 \\ 1\end{array}\right] }. +

+ +

+ + + 3x_1 &\,-\,& 2x_2 &\,=\,& 1 + + + x_1 &\,+\,& 4x_2 &\,=\,& 5 + + +

+

+ + + 3x_1 &\,-\,& 2x_2 &\,=\,& 1 + + + 4x_1 &\,+\,& 2x_2 &\,=\,& 6 + + +

+ +

+ Therefore these augmented matrices are equivalent + (even though they're not equal), + which we denote with \sim: +

+ + \left[\begin{array}{cc|c} + 3 & -2 & 1\\ + 1 & 4 & 5\\ + \end{array}\right] + \not= + \left[\begin{array}{cc|c} + 3 & -2 & 1\\ + 4 & 2 & 6\\ + \end{array}\right] + + + \left[\begin{array}{cc|c} + 3 & -2 & 1\\ + 1 & 4 & 5\\ + \end{array}\right] + \sim + \left[\begin{array}{cc|c} + 3 & -2 & 1\\ + 4 & 2 & 6\\ + \end{array}\right] + + + + + + +

+Consider whether these matrix manipulations (A) must keep or +(B) could change +the solution set for the corresponding linear system. +

+ + + +

+Swapping two rows, for example: +

+ +

+ + \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 4 & 5 & 6 \\ 1 & 2 & 3 \end{array}\right] + +

+

+ + x +2y \amp= 3 \amp 4x + 5y \amp= 6 + 4x + 5y \amp= 6 \amp x +2y \amp= 3 + +

+ + + + + +

+Swapping two columns, for example: +

+ +

+ + \left[\begin{array}{cc|c} 1 & 2 & 3 + \\ 4 & 5 & 6 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 2 & 1 & 3 + \\ 5 & 4 & 6 \end{array}\right] + +

+

+ + x +2y \amp= 3 \amp 2x + y \amp= 3 + 4x + 5y \amp= 6 \amp 5x +4y \amp= 6 + +

+ + + + + +

+Add a constant to every term of a row, for example: +

+ +

+ + \left[\begin{array}{cc|c} 1 & 2 & 3 + \\ 4 & 5 & 6 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 1+6 & 2+6 & 3+6 + \\ 4 & 5 & 6 \end{array}\right] + +

+

+ + x +2y \amp= 3 \amp 7x + 8y \amp= 9 + 4x + 5y \amp= 6 \amp 4x +5y \amp= 6 + +

+ + + + + +

+Multiply a row by a nonzero constant, for example: +

+ +

+ + \left[\begin{array}{cc|c} 1 & 2 & 3 + \\ 4 & 5 & 6 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 3 & 6 & 9 + \\ 4 & 5 & 6 \end{array}\right] + +

+

+ + x +2y \amp= 3 \amp 3x + 6y \amp= 9 + 4x + 5y \amp= 6 \amp 4x +5y \amp= 6 + +

+ + + + + +

+Add a constant multiple of one row to another row, for example: +

+ +

+ + \left[\begin{array}{cc|c} 1 & 2 & 3 + \\ 4 & 5 & 6 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 1 & 2 & 3 + \\ 4+3 & 5+6 & 6+9 \end{array}\right] + +

+

+ + x +2y \amp= 3 \amp \unknown x +\unknown y \amp= \unknown + 4x + 5y \amp= 6 \amp \unknown x +\unknown y \amp= \unknown + +

+ + + + + +

+Replace a column with zeros, for example: +

+ +

+ + \left[\begin{array}{cc|c} 1 & 2 & 3 + \\ 4 & 5 & 6 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 1 & 0 & 3 + \\ 4 & 0 & 6 \end{array}\right] + +

+

+ + x +2y \amp= 3 \amp \unknown x +\unknown y \amp= \unknown + 4x + 5y \amp= 6 \amp \unknown x +\unknown y \amp= \unknown + +

+ + + + + +

+Replace a row with zeros, for example: +

+ +

+ + \left[\begin{array}{cc|c} 1 & 2 & 3 + \\ 4 & 5 & 6 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 1 & 2 & 3 + \\ 0 & 0 & 0 \end{array}\right] + +

+

+ + x +2y \amp= 3 \amp \unknown x +\unknown y \amp= \unknown + 4x + 5y \amp= 6 \amp \unknown x +\unknown y \amp= \unknown + +

+ + + + + + + + +

+ The following three row operationsrow operations produce equivalent + augmented matrices. +

    +

  1. + Swap two rows, for example, R_1\leftrightarrow R_2: + + \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 4 & 5 & 6 \\ 1 & 2 & 3 \end{array}\right] + +

    +
    2. +
  2. +

    Multiply a row by a nonzero constant, for example, 2R_1\rightarrow R_1: + + \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 2(1) & 2(2) & 2(3) \\ 4 & 5 & 6 \end{array}\right] + +

    +
    3. +

  3. Add a constant multiple of one row to another row, + for example, R_2-4R_1\rightarrow R_2: + + \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right] + \sim + \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 4-4(1) & 5-4(2) & 6-4(3) \end{array}\right] + +

    +
    4. +
+

+

+Observe that we will use the following +notation: (Combination of old rows) \rightarrow (New row). +

+ + + + + +

+Each of the following linear systems has the same solution set. +

+ +

A) + + + x &\,+\,& 2y &\,+\,& z &\,=\,& 3 + + + -x &\,-\,& y &\,+\,& z &\,=\,& 1 + + + 2x &\,+\,& 5y &\,+\,& 3z &\,=\,& 7 + + +

+

B) + + + 2x &\,+\,& 5y &\,+\,& 3z &\,=\,& 7 + + + -x &\,-\,& y &\,+\,& z &\,=\,& 1 + + + x &\,+\,& 2y &\,+\,& z &\,=\,& 3 + + +

+

C) + + + x & & &\,-\,& z &\,=\,& 1 + + + & & y &\,+\,& 2z &\,=\,& 4 + + + & & y &\,+\,& z &\,=\,& 1 + + +

+ + +

D) + + + x &\,+\,& 2y &\,+\,& z &\,=\,& 3 + + + & & y &\,+\,& 2z &\,=\,& 4 + + + 2x &\,+\,& 5y &\,+\,& 3z &\,=\,& 7 + + +

+

E) + + + x & & &\,-\,& z &\,=\,& 1 + + + & & y &\,+\,& 2z &\,=\,& 4 + + + & & & & z &\,=\,& 3 + + +

+

F) + + + x &\,+\,& 2y &\,+\,& z &\,=\,& 3 + + + & & y &\,+\,& 2z &\,=\,& 4 + + + & & y &\,+\,& z &\,=\,& 1 + + +

+ +

+Sort these six equivalent linear systems from +most complicated to simplest (in your opinion). +

+ + + + + + +

+Here we've written the sorted linear systems from +as augmented matrices. +

+ + +\left[\begin{array}{ccc|c} + 2 & 5 & 3 & 7 \\ + -1 & -1 & 1 & 1 \\ + 1 & 2 & 1 & 3 +\end{array}\right] & \sim & + +\left[\begin{array}{ccc|c} +\markedPivot{1} & 2 & 1 & 3 \\ +-1 & -1 & 1 & 1 \\ +2 & 5 & 3 & 7 +\end{array}\right] & \sim & + +\left[\begin{array}{ccc|c} +\markedPivot{1} & 2 & 1 & 3 \\ +0 & 1 & 2 & 4 \\ +2 & 5 & 3 & 7 +\end{array}\right]\sim + + +\sim \left[\begin{array}{ccc|c} +\markedPivot{1} & 2 & 1 & 3 \\ +0 & \markedPivot{1} & 2 & 4 \\ +0 & 1 & 1 & 1 +\end{array}\right] & \sim & + +\left[\begin{array}{ccc|c} +\markedPivot{1} & 0 & -1 & 1 \\ +0 & \markedPivot{1} & 2 & 4 \\ +0 & 1 & 1 & 1 +\end{array}\right] & \sim & + +\left[\begin{array}{ccc|c} +\markedPivot{1} & 0 & -1 & 1 \\ +0 & \markedPivot{1} & 2 & 4 \\ +0 & 0 & -1 & -3 \end{array}\right] + + +

+Assign the following row operations to each step used to manipulate each +matrix to the next: + +

R_3-1R_2\to R_3

+

R_2+1R_1\to R_2

+

R_1\leftrightarrow R_3

+ + +

R_3-2R_1\to R_3

+

R_1-2R_3\to R_1

+ +

+ + + + + +

+A matrix is in reduced row echelon form (RREF)Reduced row echelon form if +

+
    +
  1. The leftmost nonzero term of each row is 1. + We call these terms pivots.pivot +
    2. +
  2. Each pivot is to the right of every higher pivot. +
    3. +
  3. Each term that is either above or below a pivot is 0. +
    4. +
  4. All zero rows (rows whose terms are all 0) are at the bottom of the matrix. +
    5. +
+

+Every matrix has a unique reduced row echelon form. If A is a matrix, we write \RREF(A) for the reduced row echelon form of that matrix. +

+ + + + + +

+Recall that a matrix is in reduced row echelon form (RREF) if +

+
    +
  1. The leftmost nonzero term of each row is 1. + We call these terms pivots. +
    2. +
  2. Each pivot is to the right of every higher pivot. +
    3. +
  3. Each term that is either above or below a pivot is 0. +
    4. +
  4. All zero rows (rows whose terms are all 0) are at the bottom of the matrix. +
    5. +
+

+For each matrix, mark the leading terms, and label it as RREF or not RREF. +For the ones not in RREF, determine which rule is violated and how it might be fixed. +

+ +

A=\left[\begin{array}{ccc|c} 1 & 0 & 0 & 3 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right]

+

B=\left[\begin{array}{ccc|c} 1 & 0 & 4 & 3 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right]

+

C=\left[\begin{array}{ccc|c} 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array}\right]

+ + + + + + +

+Recall that a matrix is in reduced row echelon form (RREF) if +

+
    +
  1. The leftmost nonzero term of each row is 1. + We call these terms pivots. +
    2. +
  2. Each pivot is to the right of every higher pivot. +
    3. +
  3. Each term that is either above or below a pivot is 0. +
    4. +
  4. All zero rows (rows whose terms are all 0) are at the bottom of the matrix. +
    5. +
+

+For each matrix, mark the leading terms, and label it as RREF or not RREF. +For the ones not in RREF, determine which rule is violated and how it might be fixed. +

+ +

D=\left[\begin{array}{ccc|c} 1 & 0 & 2 & -3 \\ 0 & 3 & 3 & -3 \\ 0 & 0 & 0 & 0 \end{array}\right]

+

E=\left[\begin{array}{ccc|c} 0 & 1 & 0 & 7 \\ 1 & 0 & 0 & 4 \\ 0 & 0 & 0 & 0 \end{array}\right]

+

F=\left[\begin{array}{ccc|c} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 0 \end{array}\right]

+ + + + + + +

+In practice, if we simply need to convert a matrix into reduced row echelon form, +we use technology to do so. +

+

+However, it is also important to understand the Gauss-Jordan eliminationGauss-Jordan elimination algorithm +that a computer or calculator uses to convert a matrix (augmented or not) into reduced row echelon form. +Understanding this algorithm will help us better understand how to interpret the results +in many applications we use it for in . +

+ + + + + +

+Consider the matrix +\left[\begin{array}{cccc}2 & 6 & -1 & 6 \\ 1 & 3 & -1 & 2 \\ -1 & -3 & 2 & 0 \end{array}\right]. +Which row operation is the best choice for the first move in converting to RREF? +

+
    +
  1. Add row 3 to row 2 (R_2+R_3 \rightarrow R_2) +
    2. +
  2. Add row 2 to row 3 (R_3+R_2 \rightarrow R_3) +
    3. +
  3. Swap row 1 to row 2 (R_1 \leftrightarrow R_2) +
    4. +
  4. Add -2 row 2 to row 1 (R_1-2R_2 \rightarrow R_1) +
    5. +
+ + + + + +

+Consider the matrix +\left[\begin{array}{cccc} \markedPivot{1} & 3 & -1 & 2 \\ 2 & 6 & -1 & 6 \\ -1 & -3 & 2 & 0 \end{array}\right]. +Which row operation is the best choice for the next move in converting to RREF? +

+
    +
  1. Add row 1 to row 3 (R_3+R_1 \rightarrow R_3) +
    2. +
  2. Add -2 row 1 to row 2 (R_2-2R_1 \rightarrow R_2) +
    3. +
  3. Add 2 row 2 to row 3 (R_3+2R_2 \rightarrow R_3) +
    4. +
  4. Add 2 row 3 to row 2 (R_2+2R_3 \rightarrow R_2) +
    5. +
+ + + + + +

+Consider the matrix +\left[\begin{array}{cccc}\markedPivot{1} & 3 & -1 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right]. +Which row operation is the best choice for the next move in converting to RREF? +

+
    +
  1. Add row 1 to row 2 (R_2+R_1 \rightarrow R_2) +
    2. +
  2. Add -1 row 3 to row 2 (R_2-R_3 \rightarrow R_2) +
    3. +
  3. Add -1 row 2 to row 3 (R_3-R_2 \rightarrow R_3) +
    4. +
  4. Add row 2 to row 1 (R_1+R_2 \rightarrow R_1) +
    5. +
+ + + + +

+The steps for the Gauss-Jordan elimination algorithm may be summarized +as follows: +

    +
  1. +

    +Ignoring any rows that already have marked pivots, identify the leftmost column +with a nonzero entry. +

    +
    2. +
  2. +

    +Use row operations to obtain a pivot of value 1 in the topmost row +that does not already have a marked pivot. +

    +
    3. +
  3. +

    +Mark this pivot, then use row operations to change all values above and below the +marked pivot to 0. +

    +
    4. +
  4. +

    +Repeat these steps until the matrix is in RREF. +

    +
    5. +
+

+

+In particular, once a pivot is marked, it should remain in the same position. +This will keep you from undoing your progress towards an RREF matrix. +

+ + + + +

+Complete the following RREF calculation (multiple row operations may be needed +for certain steps): + + A= + \left[\begin{array}{cccc}2 & 3 & 2 & 3 \\ -2 & 1 & 6 & 1 \\ -1 & -3 & -4 & 1 \end{array}\right] + \sim \left[\begin{array}{cccc}\markedPivot{1} & \unknown & \unknown & \unknown \\ -2 & 1 & 6 & 1 \\ -1 & -3 & -4 & 1 \end{array}\right] + \sim \left[\begin{array}{cccc}\markedPivot{1} & \unknown & \unknown & \unknown \\ 0 & \unknown & \unknown & \unknown \\ 0 & \unknown & \unknown & \unknown \end{array}\right] + + + \sim \left[\begin{array}{cccc}\markedPivot{1} & \unknown & \unknown & \unknown \\ 0 & \markedPivot{1} & \unknown & \unknown \\ 0 & \unknown & \unknown & \unknown \end{array}\right] + \sim \left[\begin{array}{cccc}\markedPivot{1} & 0 & \unknown & \unknown \\ 0 & \markedPivot{1} & \unknown & \unknown \\ 0 & 0 & \unknown & \unknown \end{array}\right] + \sim \cdots + \sim \left[\begin{array}{cccc}\markedPivot{1} & 0 & -2 & 0 \\ 0 & \markedPivot{1} & 2 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] + +

+ + + + + +

+Consider the matrix +A=\left[\begin{array}{cccc} 2 & 4 & 2 & -4 \\ -2 & -4 & 1 & 1 \\ 3 & 6 & -1 & -4 \end{array}\right]. +

+

+Compute \RREF(A). +

+ + + + + +

+Consider the non-augmented and augmented matrices +A=\left[\begin{array}{cccc} 2 & 4 & 2 & -4 \\ -2 & -4 & 1 & 1 \\ 3 & 6 & -1 & -4 \end{array}\right]\hspace{2em} +B=\left[\begin{array}{ccc|c} 2 & 4 & 2 & -4 \\ -2 & -4 & 1 & 1 \\ 3 & 6 & -1 & -4 \end{array}\right]. +

+

+Can \RREF(A) be used to find \RREF(B)? +

    +
  1. Yes, \RREF(A) and \RREF(B) are exactly the same.
    2. +
  2. Yes, \RREF(A) may be slightly modified to find \RREF(B).
    3. +
  3. No, a new calculuation is required.
    4. +
+

+ + + + + +

+Free browser-based technologies for mathematical computation +are available online. +

+
    +

  • + Go to . +

    +
    • +
  • + In the dropdown on the right, you can select a number of different languages. + Select "Octave" for the Matlab-compatible syntax used by this text. +
    • +
  • +

    Type rref([1,3,2;2,5,7]) and then press the Evaluate button to compute the \RREF of + \left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 7 \end{array}\right]. +

    +
    • +
+ + + + + +

+In the HTML version of this text, code cells are often embedded for your convenience +when RREFs need to be computed. +

+

+Try this out to compute +\RREF\left[\begin{array}{cc|c} 2 & 3 & 1 \\ 3 & 0 & 6 \end{array}\right]. +

+ + + + +rref([2,3,1;3,0,6]) + + + + + + + + Videos +
+ + Video: Row reduction +
+ + + + + +

Exercises available at .

+ + + + Mathematical Writing Explorations + + +

Prove that Gauss-Jordan Elimination preserves the solution set of a system of linear equations in + n variables. Make sure your proof includes each of the following. Just because I've used + bullet points here does not mean you should use bullet points in your proof. +

    +

  • Write an arbitrary system of linear equations in n variables. Your notation should be unambiguous.

    • +

  • Label an element of your solution set. You won't know what it is exactly, so you'll have to use a variable. Remember what it means (by definition!) to be in the solution set.

    • +

  • Describe the three operations used in Gauss-Jordan Elimination.

    • +

  • Consider all three operations in Gauss-Jordan Elimination. After each one is used, show that the element of the solution set you picked still satisfies the definition.

    • +
+

+ + + + + +

+ Let M_{2,2} indicate the set of all 2 \times 2 matrices with real entries. Show that equivalence of matrices as defined in this section is an equivalence relation, as in exploration

+ + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/01-LE/03.ptx b/linear-algebra/source/01-LE/03.ptx new file mode 100644 index 00000000..15ef4fa5 --- /dev/null +++ b/linear-algebra/source/01-LE/03.ptx @@ -0,0 +1,443 @@ + +
+ Counting Solutions for Linear Systems (LE3) + + + + + + Warm Up + + + +

+ Without referring to your Activity Book, which of the four criteria for a matrix to be in Reduced Row Echelon Form (RREF) can you recall? +

+ + + + +

+ Which, if any, of the following matrices are in RREF? You may refer to the Activity Book now for criteria that you may have forgotten. +

+ +

P=\left[\begin{array}{ccc|c} 1 & 0 & \frac{2}{3} & -3 \\ 0 & 3 & 3 & -\frac{3}{5} \\ 0 & 0 & 0 & 0 \end{array}\right]

+

Q=\left[\begin{array}{ccc|c} 0 & 1 & 0 & 7 \\ 1 & 0 & 0 & 4 \\ 0 & 0 & 0 & 0 \end{array}\right]

+

R=\left[\begin{array}{ccc|c} 1 & 0 & \frac{1}{2} & 4 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 0 \end{array}\right]

+ + + + + +Class Activities + + + +

+We will frequently need to know the reduced row echelon form of matrices +during the remainder of this course, so unless you're told otherwise, +feel free to use technology (see ) to compute RREFs efficiently. +

+ + + + + +

+Consider the following system of equations. +

+ + + 3x_1 &\,-\,& 2x_2 &\,+\,& 13x_3 &\,=\,& 6 + + + 2x_1 &\,-\,& 2x_2 &\,+\,& 10x_3 &\,=\,& 2 + + + -x_1 &\,+\,& 3x_2 &\,-\,& 6x_3 &\,=\,& 11 + + . + + + +

+Convert this to an augmented matrix and use technology to compute its +reduced row echelon form: + + \RREF + \left[\begin{array}{ccc|c} + \unknown&\unknown&\unknown&\unknown\\ + \unknown&\unknown&\unknown&\unknown\\ + \unknown&\unknown&\unknown&\unknown\\ + \end{array}\right] += + \left[\begin{array}{ccc|c} + \unknown&\unknown&\unknown&\unknown\\ + \unknown&\unknown&\unknown&\unknown\\ + \unknown&\unknown&\unknown&\unknown\\ + \end{array}\right] + +

+ + +

+Use the \RREF matrix to write a linear system equivalent +to the original system. +

+ + +

+How many solutions must this system have? +

+
    +
  1. +

    Zero

    +
    2. +
  2. +

    Only one

    +
    3. +
  3. +

    Infinitely-many

    +
    4. +
+ + + + +rref([3,-2,13,6;2,-2,10,2;-1,3,-6,11]) + + + + + + +

+Consider the vector equation +

+ + x_1 \left[\begin{array}{c} 3 \\ 2\\ -1 \end{array}\right] + +x_2 \left[\begin{array}{c}-2 \\ -2 \\ 0 \end{array}\right] + +x_3\left[\begin{array}{c} 13 \\ 10 \\ -3 \end{array}\right] + =\left[\begin{array}{c} 6 \\ 2 \\ 1 \end{array}\right] + + + +

+Convert this to an augmented matrix and use technology to compute its +reduced row echelon form: +

+ + \RREF + \left[\begin{array}{ccc|c} + \unknown&\unknown&\unknown&\unknown\\ + \unknown&\unknown&\unknown&\unknown\\ + \unknown&\unknown&\unknown&\unknown\\ + \end{array}\right] += + \left[\begin{array}{ccc|c} + \unknown&\unknown&\unknown&\unknown\\ + \unknown&\unknown&\unknown&\unknown\\ + \unknown&\unknown&\unknown&\unknown\\ + \end{array}\right] + + + +

+Use the \RREF matrix to write a linear system equivalent +to the original system. +

+ + +

+How many solutions must this system have? +

+
    +
  1. +

    Zero

    +
    2. +
  2. +

    Only one

    +
    3. +
  3. +

    Infinitely-many

    +
    4. +
+ + + + +rref([3,-2,13,6;2,-2,10,2;-1,0,-3,1]) + + + + +

+What contradictory equations besides 0=1 may be obtained from +the RREF of an augmented matrix? +

+
    +

  1. x=0 is an obtainable contradiction

    2. +

  2. x=y is an obtainable contradiction

    3. +

  3. 0=17 is an obtainable contradiction

    4. +

  4. 0=1 is the only obtainable contradiction

    5. +
+ + + + + +

+ Consider the following linear system. +

+ + + x_1 &+ 2x_2 &+ 3x_3 &= 1 + + 2x_1 &+ 4x_2 &+ 8x_3 &= 0 + + + + +

+ Find its corresponding augmented matrix A and + find \RREF(A). +

+ + +

+Use the \RREF matrix to write a linear system equivalent +to the original system. +

+ + +

+How many solutions must this system have? +

+
    +
  1. +

    Zero

    +
    2. +
  2. +

    One

    +
    3. +
  3. +

    Infinitely-many

    +
    4. +
+ + + + + + + + +

+We will see in that the intuition established here generalizes: a consistent system with more variables +than equations (ignoring 0=0) will always have infinitely many solutions. +

+ + + + + +

+By finding \RREF(A) from a linear system's corresponding augmented matrix A, +we can immediately tell how many solutions the system has. +

+
    +
  • +

    +If the linear system given by \RREF(A) includes the contradiction +0=1, that is, the row +\left[\begin{array}{ccc|c}0&\cdots&0&1\end{array}\right], +then the system is inconsistent, which means it has zero solutions +and its solution set is written as \emptyset or \{\}. +

    +
    • +
  • +

    +If the linear system given by \RREF(A) sets each variable of the system +to a single value; that is, x_1=s_1, x_2=s_2, and so on; +then the system is consistent with exactly one solution +\left[\begin{array}{c}s_1\\s_2\\\vdots\end{array}\right], +and its solution set is +\setList{ \left[\begin{array}{c}s_1\\s_2\\\vdots\end{array}\right] }. +

    +
    • +
  • +

    +Otherwise, the system must have more variables than non-trivial +equations (equations other than 0=0). This means it is +consistent with infinitely-many +different solutions. We'll learn how to find such solution sets in +. +

    +
    • +
+ + + + + +

+For each vector equation, write an explanation for +whether each solution set +has no solutions, one solution, or infinitely-many solutions. +If the set is finite, describe it using set notation. +

+ + +

+ + x_{1} \left[\begin{array}{c} +1 \\ +-1 \\ +1 +\end{array}\right] + x_{2} \left[\begin{array}{c} +4 \\ +-3 \\ +1 +\end{array}\right] + x_{3} \left[\begin{array}{c} +7 \\ +-6 \\ +4 +\end{array}\right] = \left[\begin{array}{c} +10 \\ +-6 \\ +4 +\end{array}\right] + +

+ + +

+ + x_{1} \left[\begin{array}{c} +-2 \\ +-1 \\ +-2 +\end{array}\right] + x_{2} \left[\begin{array}{c} +3 \\ +1 \\ +1 +\end{array}\right] + x_{3} \left[\begin{array}{c} +-2 \\ +-2 \\ +-5 +\end{array}\right] = \left[\begin{array}{c} +1 \\ +4 \\ +13 +\end{array}\right] + +

+ + +

+ + x_{1} \left[\begin{array}{c} +-1 \\ +-2 \\ +1 +\end{array}\right] + x_{2} \left[\begin{array}{c} +-5 \\ +-5 \\ +4 +\end{array}\right] + x_{3} \left[\begin{array}{c} +-7 \\ +-9 \\ +6 +\end{array}\right] = \left[\begin{array}{c} +3 \\ +1 \\ +-2 +\end{array}\right] + +

+ + + + + + + + + Cool Down + + +

+ In , we stated, but did not prove the assertion that + all linear systems are one of the following: +

    +
  1. Consistent with one solution: + its solution set contains a single vector, e.g. + \setList{\left[\begin{array}{c}1\\2\\3\end{array}\right]} +
    2. +
  2. + Consistent with infinitely-many solutions: + its solution set contains infinitely many vectors, e.g. + + \setBuilder + { + \left[\begin{array}{c}1\\2-3a\\a\end{array}\right] + }{ + a\in\IR + } + +
    3. +
  3. Inconsistent: + its solution set is the empty set, denoted by either \{\} or \emptyset. +
    4. +
+

+ + +

+ Explain why this fact is a consequence of above. +

+ + + + + Videos +
+ + Video: Finding the number of solutions for a system +
+ + + + + +

Exercises available at .

+ + + + + Mathematical Writing Explorations + + +

A system of equations with all constants equal to 0 is called homogeneous. These are addressed in detail in section +

    +
  • Choose three systems of equations from this chapter that you have already solved. Replace the constants with 0 to make the systems homogeneous. Solve the homogeneous systems and make a conjecture about the relationship between the earlier solutions you found and the associated homogeneous systems. +
    • +
  • Prove or disprove. A system of linear equations is homogeneous if an only if it has the the zero vector as a solution.

+ + + + + Sample Problem and Solution +

+ Sample problem . +

+ + + +
diff --git a/linear-algebra/source/01-LE/04.ptx b/linear-algebra/source/01-LE/04.ptx new file mode 100644 index 00000000..ffd1c120 --- /dev/null +++ b/linear-algebra/source/01-LE/04.ptx @@ -0,0 +1,471 @@ + +
+ Linear Systems with Infinitely-Many Solutions (LE4) + + + + + + Warm Up + + +

+ Write down any three linear systems and determine if they are consistent, have a single solution, or have infinitely many solutions. +

+ + + + +Class Activities + + +

+Consider this simplified linear system found to be equivalent to the system +from : +

+ + + x_1 &+ 2x_2 & &= 4 + + + & &\phantom{+}x_3 &= -1 + + +

+Earlier, we determined this system has infinitely-many solutions. +

+ + +

+Let x_1=a and write the solution set in the form + + \setBuilder + { + \left[\begin{array}{c} a \\ \unknown \\ \unknown \end{array}\right] + }{ + a \in \IR + } +. +

+ + +

+Let x_2=b and write the solution set in the form + + \setBuilder + { + \left[\begin{array}{c} \unknown \\ b \\ \unknown \end{array}\right] + }{ + b \in \IR + } +. +

+ + +

+Which of these was easier? What features of the RREF matrix + + \left[\begin{array}{ccc|c} + \markedPivot{1} & 2 & 0 & 4 \\ + 0 & 0 & \markedPivot{1} & -1 + \end{array}\right] + caused this? +

+ + + + + +

+Recall that the pivots of a matrix in \RREF form are the leading +1s in each non-zero row. +

+

+The pivot columns in an augmented matrix correspond to the +bound variablesbound variables in the system of equations (x_1,x_3 below). +The remaining variables are called free variablesfree variables (x_2 below). +

+ + + \left[\begin{array}{ccc|c} + \markedPivot{1} & 2 & 0 & 4 \\ + 0 & 0 & \markedPivot{1} & -1 + \end{array}\right] + +

+To efficiently solve a system in RREF form, assign letters to the free +variables, and then solve for the bound variables. +

+ + + + + +

+Find the solution set for the system + + +2x_1&\,-\,&2x_2&\,-\,&6x_3&\,+\,&x_4&\,-\,&x_5&\,=\,&3 + + +-x_1&\,+\,&x_2&\,+\,&3x_3&\,-\,&x_4&\,+\,&2x_5 &\,=\,& -3 + + +x_1&\,-\,&2x_2&\,-\,&x_3&\,+\,&x_4&\,+\,&x_5 &\,=\,& 2 + + +by doing the following. +

+ + +

+Row-reduce its augmented matrix. +

+ + +

+Assign letters to the free variables (given by the non-pivot columns): +

\unknown = a \unknown = b

+

+ + +

+Solve for the bound variables (given by the pivot columns) to show that +\unknown = 1+5a+2b\unknown = 1+2a+3b\unknown=3+3b +

+ + + +

+Replace x_1 through x_5 with the appropriate expressions of a,b +in the following set-builder notation. +

+

+ + \setBuilder + { + \left[\begin{array}{c} + \hspace{2em}x_1\hspace{2em} \\ + \hspace{2em}x_2\hspace{2em} \\ + \hspace{2em}x_3\hspace{2em} \\ + \hspace{2em}x_4\hspace{2em} \\ + \hspace{2em}x_5\hspace{2em} + \end{array}\right] + }{ + a,b\in \IR + } + +

+ + + + + + + + + + + +

+Don't forget to correctly express the solution set of a linear system. +Systems with zero or one solutions may be written by listing their elements, + while systems with infinitely-many solutions may be written using + set-builder notation. +

+
    +

  • Inconsistent: \emptyset or \{\} +

    • (not 0 or + + \left[\begin{array}{c}0\\0\\0\end{array}\right] + )
    +

    +
    • +

  • Consistent with one solution: e.g. + \setList{ \left[\begin{array}{c}1\\2\\3\end{array}\right] } +

    • (not just + \left[\begin{array}{c}1\\2\\3\end{array}\right])
    +

    +
    • +

  • + Consistent with infinitely-many solutions: e.g. + + \setBuilder + { + \left[\begin{array}{c}1\\2-3a\\a\end{array}\right] + }{ + a\in\IR + } + +

    • (not just + + \left[\begin{array}{c}1\\2-3a\\a\end{array}\right] + + )
    +

    +
    • +
+ + + + + +

+ Consider the following + system of linear equations. +x_{1} \left[\begin{array}{c} +1 \\ +0 \\ +1 +\end{array}\right] + x_{2} \left[\begin{array}{c} +0 \\ +1 \\ +-1 +\end{array}\right] + x_{3} \left[\begin{array}{c} +-1 \\ +5 \\ +-5 +\end{array}\right] + x_{4} \left[\begin{array}{c} +-3 \\ +13 \\ +-13 +\end{array}\right] = \left[\begin{array}{c} +-3 \\ +12 \\ +-12 +\end{array}\right]. +

+ + +

+Explain how to find a simpler system or vector equation +that has the same solution set. +

+ + +

+Explain how to describe this solution set using set notation. +

+ + + + + +

+Consider the following +system of linear equations. + + \begin{matrix} + x_{1} & & & - & 2 \, x_{3} & = & -3 \\ + 5 \, x_{1} & + & x_{2} & - & 7 \, x_{3} & = & -18 \\ + 5 \, x_{1} & - & x_{2} & - & 13 \, x_{3} & = & -12 \\ + x_{1} & + & 3 \, x_{2} & + & 7 \, x_{3} & = & -12 \\ + \end{matrix} + + +

+ +

+Explain how to find a simpler system or vector equation +that has the same solution set. +

+ + +

+Explain how to describe this solution set using set notation. +

+ + + + + + + + + + + Cool Down + + +

+ Consider the following linear system, its augmented matrix A, and \RREF(A): +\begin{matrix} x_{1} & - & x_{2} & + & x_{3} & = & 4 \\ & & x_{2} & - & 2 \, x_{3} & = & -1 \\ & & x_{2} & - & 2 \, x_{3} & = & -3 \\ x_{1} & + & 2 \, x_{2} & - & 5 \, x_{3} & = & 0 \\ \end{matrix} +A=\left[\begin{array}{ccc|c} + 1 & -1 & 1 & 4 \\ + 0 & 1 & -2 & -1 \\ + 0 & 1 & -2 & -3 \\ + 1 & 2 & -5 & 0 + \end{array}\right],\ \RREF(A)=\left[\begin{array}{ccc|c} + 1 & 0 & -1 & 0 \\ + 0 & 1 & -2 & 0 \\ + 0 & 0 & 0 & 1 \\ + 0 & 0 & 0 & 0 + \end{array}\right]. + All of the following statements are not accurate or otherwise incorrect; identify what is problematic about the statements and correct them. +

+ + + +

+ The matrix A is inconsistent. +

+ + + + +

+ The linear system has two bound variables and one free variable. +

+ + + + +

+ The solution set to the given linear system is \{\emptyset\}. +

+ + + + + + +

+ Consider the following linear system, its augmented matrix B, and \RREF(B): +\begin{matrix} 2 \, x_{1} & - & 2 \, x_{2} & - & 8 \, x_{3} & + & 3 \, x_{4} & - & 9 \, x_{5} & = & -17 \\ -x_{1} & & & + & x_{3} & - & x_{4} & + & 2 \, x_{5} & = & 6 \\ 2 \, x_{1} & - & x_{2} & - & 5 \, x_{3} & + & x_{4} & - & 5 \, x_{5} & = & -10 \\ -x_{1} & + & 3 \, x_{2} & + & 10 \, x_{3} & & & + & 7 \, x_{5} & = & 6 \\ \end{matrix} +B=\left[\begin{array}{ccccc|c} + 2 & -2 & -8 & 3 & -9 & -17\\ + -1 & 0 & 1 & -1 & 2 & 6\\ + 2 & -1 & -5 & 1 & -5 & -10\\ + -1 & 3 & 10 & 0 & 7 & 6\\ + \end{array}\right] + + \RREF(B)=\left[\begin{array}{ccccc|c} + 1 & 0 & -1 & 0 & -1 & -3\\ + 0 & 1 & 3 & 0 & 2 & 1\\ + 0 & 0 & 0 & 1 & -1 & -3\\ + 0 & 0 & 0 & 0 & 0 & 0\\ + \end{array}\right]. + All of the following statements are not accurate or otherwise incorrect; identify what is problematic about the statements and correct them. +

+ + + +

+ The matrix B is consistent with infinitely many solutions. +

+ + + + +

+ The solution set is given by + \left[\begin{array}{c} a + b - 3 \\ -3 \, a - 2 \, b + 1 \\ a \\ b - 3 \\ b \end{array}\right]. +

+ + + + +

+ The variables x_3,x_5 are free. Setting them equal to a,b respectively and solving for the bound variables, the solution set to the linear system is given by + \left\{ \left[\begin{array}{c} a + b - 3 \\ -3 \, a - 2 \, b + 1 \\ b - 3 \end{array}\right] \,\middle|\, a,b \in\mathbb R \right\}. +

+ + + + + + + Videos +
+ + Video: Solving a system of linear equations with infinitely-many solutions +
+ + + + + +

Exercises available at .

+ + + + Mathematical Writing Explorations + + +

Construct a system of 3 equations in 3 variables having: +

  • 0 free variables
    • +
  • 1 free variable
    • +
  • 2 free variables
    • +
+ In each case, solve the system you have created. Conjecture a relationship between the number of free variables and the type of solution set that can be obtained from a given system. +

+ + + + +

For each of the following, decide if it's true or false. If you think it's true, can we construct a proof? If you think it's false, can we find a counterexample? +

    +
  • If the coefficient matrix of a system of linear equations has a pivot in the rightmost column, then the system is inconsistent.
    • +
  • If a system of equations has two equations and four unknowns, then it must be consistent.
    • +
  • If a system of equations having four equations and three unknowns is consistent, then the solution is unique.
    • +
  • Suppose that a linear system has four equations and four unknowns and that the coefficient matrix has four pivots. Then the linear system is consistent and has a unique solution.
    • +
  • Suppose that a linear system has five equations and three unknowns and that the coefficient matrix has a pivot in every column. Then the linear system is consistent and has a unique solution.

 + + + + Sample Problem and Solution +

+ Sample problem . +

+ + + + +
diff --git a/linear-algebra/source/01-LE/main.ptx b/linear-algebra/source/01-LE/main.ptx new file mode 100644 index 00000000..08e295ca --- /dev/null +++ b/linear-algebra/source/01-LE/main.ptx @@ -0,0 +1,10 @@ + + + Systems of Linear Equations (LE) + + + + + + + diff --git a/linear-algebra/source/01-LE/outcomes/01.ptx b/linear-algebra/source/01-LE/outcomes/01.ptx new file mode 100644 index 00000000..0d7cbb79 --- /dev/null +++ b/linear-algebra/source/01-LE/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Translate back and forth between a system of linear equations, a vector equation, and the corresponding augmented matrix. +

\ No newline at end of file diff --git a/linear-algebra/source/01-LE/outcomes/02.ptx b/linear-algebra/source/01-LE/outcomes/02.ptx new file mode 100644 index 00000000..c61c055b --- /dev/null +++ b/linear-algebra/source/01-LE/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Explain why a matrix isn’t in reduced row echelon form, and put a matrix in reduced row echelon form. +

\ No newline at end of file diff --git a/linear-algebra/source/01-LE/outcomes/03.ptx b/linear-algebra/source/01-LE/outcomes/03.ptx new file mode 100644 index 00000000..2da0af43 --- /dev/null +++ b/linear-algebra/source/01-LE/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Determine the number of solutions for a system of linear equations or a vector equation. +

\ No newline at end of file diff --git a/linear-algebra/source/01-LE/outcomes/04.ptx b/linear-algebra/source/01-LE/outcomes/04.ptx new file mode 100644 index 00000000..b3df3175 --- /dev/null +++ b/linear-algebra/source/01-LE/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Compute the solution set for a system of linear equations or a vector equation with infinitely many solutions. +

diff --git a/linear-algebra/source/01-LE/outcomes/main.ptx b/linear-algebra/source/01-LE/outcomes/main.ptx new file mode 100644 index 00000000..60798603 --- /dev/null +++ b/linear-algebra/source/01-LE/outcomes/main.ptx @@ -0,0 +1,23 @@ + +> + + +

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
+ \ No newline at end of file diff --git a/linear-algebra/source/01-LE/outcomes/question.ptx b/linear-algebra/source/01-LE/outcomes/question.ptx new file mode 100644 index 00000000..f55fc4b0 --- /dev/null +++ b/linear-algebra/source/01-LE/outcomes/question.ptx @@ -0,0 +1,4 @@ + +

+How can we solve systems of linear equations? +

diff --git a/linear-algebra/source/01-LE/readiness.ptx b/linear-algebra/source/01-LE/readiness.ptx new file mode 100644 index 00000000..9b406fcd --- /dev/null +++ b/linear-algebra/source/01-LE/readiness.ptx @@ -0,0 +1,39 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Determine if a system to a two-variable system of linear equations + will have zero, one, or infinitely-many solutions by graphing.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    2. +
  2. +

    + Find the unique solution to a two-variable system of linear equations + by back-substitution.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    3. +
  3. +

    Describe sets using set-builder notation, and check if an element is a member of a set described by set-builder notation.

    +
      +
    • +

      Review: YouTube

      +
      • +
    +
    4. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/01-LE/samples/01.ptx b/linear-algebra/source/01-LE/samples/01.ptx new file mode 100644 index 00000000..2cacebc8 --- /dev/null +++ b/linear-algebra/source/01-LE/samples/01.ptx @@ -0,0 +1,44 @@ + + + LE1 + +

+Consider the scalar system of equations + + +3x_1 &\,+\,& 2x_2 &\,\,& &\,+\,&x_4 &= 1 +-x_1 &\,-\,& 4x_2 &\,+\,&x_3&\,-\,&7x_4 &= 0 + &\,\,& x_2 &\,-\,&x_3 &\,\,& &= -2 + +

+
    +
  1. +Rewrite this system as a vector equation. +
    2. +
  2. +Write an augmented matrix corresponding to this system. +
    3. +
+ + +
    +

  1. + x_1\left[\begin{array}{c} 3 \\ -1 \\ 0 \end{array}{}\right] + + x_2 \left[\begin{array}{c}2 \\ -4 \\ 1 \end{array}{}\right]+ + x_3 \left[\begin{array}{c} 1 \\ 1 \\ -1 \end{array}{}\right] + + x_4 \left[\begin{array}{c} 1 \\ -7 \\ 0 \end{array}{}\right] + = \left[\begin{array}{c} 1 \\ 0 \\ -2 \end{array}{}\right]

    +
    2. +
  2. +

    + +\left[\begin{array}{cccc|c} +3 & 2 & 0 & 1 & 1 \\ +-1 & -4 & 1 & -7 & 0 \\ +0 & 1 & -1 & 0 & -2 +\end{array}\right] +

    +
    3. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/01-LE/samples/02.ptx b/linear-algebra/source/01-LE/samples/02.ptx new file mode 100644 index 00000000..7fbe4c0c --- /dev/null +++ b/linear-algebra/source/01-LE/samples/02.ptx @@ -0,0 +1,152 @@ + + +LE2 + +
    +
  1. +

    +For each of the following matrices, explain why it is not in reduced row +echelon form. + + A = \left[\begin{array}{ccc} +-4 & 0 & 4 \\ +0 & 1 & -2 \\ +0 & 0 & 0 \\ +0 & 0 & 0 \\ +0 & 0 & 0 +\end{array}\right] \hspace{2em} + B = \left[\begin{array}{ccc} +0 & 1 & 2 \\ +1 & 0 & -3 \\ +0 & 0 & 0 \\ +0 & 0 & 0 \\ +0 & 0 & 0 +\end{array}\right] \hspace{2em} + C = \left[\begin{array}{ccc} +1 & -4 & 4 \\ +0 & 1 & 0 \\ +0 & 0 & 0 \\ +0 & 0 & 0 \\ +0 & 0 & 0 +\end{array}\right] \hspace{2em} + \hspace{2em} + +

    +
    2. +
  2. +

    +Show step-by-step why +\RREF + \left[\begin{array}{cccc} + 0 & 3 & 1 & 2 \\ + 1 & 2 & -1 & -3 \\ + 2 & 4 & -1 & -1 + \end{array}\right] + = \left[\begin{array}{cccc} + 1 & 0 & 0 & 4 \\ + 0 & {1} & 0 & -1 \\ + 0 & 0 & {1} & 5 + \end{array}\right]. + +

    +
    3. +
+ + +
    +

  1. +

      +
    • +

      +A=\left[\begin{array}{ccc} +-4 & 0 & 4 \\ +0 & 1 & -2 \\ +0 & 0 & 0 \\ +0 & 0 & 0 \\ +0 & 0 & 0 +\end{array}\right] is not in reduced row echelon form because +the pivots are not all 1. +

      +
      • +
    • +

      +B=\left[\begin{array}{ccc} +0 & 1 & 2 \\ +1 & 0 & -3 \\ +0 & 0 & 0 \\ +0 & 0 & 0 \\ +0 & 0 & 0 +\end{array}\right] is not in reduced row echelon form because +the pivots are not descending to the right. +

      +
      • +
    • +

      +C=\left[\begin{array}{ccc} +1 & -4 & 4 \\ +0 & 1 & 0 \\ +0 & 0 & 0 \\ +0 & 0 & 0 \\ +0 & 0 & 0 +\end{array}\right] is not in reduced row echelon form because +not every entry above and below each pivot is zero. +

      +
      • +

    +
    2. +

  2. + + + \left[\begin{array}{cccc} + 0 & 3 & 1 & 2 \\ + 1 & 2 & -1 & -3 \\ + 2 & 4 & -1 & -1 + \end{array}\right] + &\sim& + \left[\begin{array}{cccc} + \markedPivot{1} & 2 & -1 & -3 \\ + 0 & 3 & 1 & 2 \\ + 2 & 4 & -1 & -1 + \end{array}\right] + &\hspace{0.2in} \text{Swap Rows 1 and 2}& + &\sim& + \left[\begin{array}{cccc} + \markedPivot{1} & 2 & -1 & -3 \\ + 0 & 3 & 1 & 2 \\ + 0 & 0 & 1 & 5 + \end{array}\right] + &\hspace{0.2in} \text{Add } -2 \text{ Row 1 to Row 3}& + &\sim& + \left[\begin{array}{cccc} + \markedPivot{1} & 2 & -1 & -3 \\ + 0 & \markedPivot{1} & \frac{1}{3} & \frac{2}{3} \\ + 0 & 0 & 1 & 5 + \end{array}\right] + &\hspace{0.2in} \text{Multiply Row 3 by } \frac{1}{3}& + &\sim& + \left[\begin{array}{cccc} + \markedPivot{1} & 0 & -\frac{5}{3} & -\frac{13}{3} \\ + 0 & \markedPivot{1} & \frac{1}{3} & \frac{2}{3} \\ + 0 & 0 & \markedPivot{1} & 5 + \end{array}\right] + &\hspace{0.2in} \text{Add } -2 \text{ Row 2 to Row 1}& + &\sim& + \left[\begin{array}{cccc} + \markedPivot{1} & 0 & -\frac{5}{3} & -\frac{13}{3} \\ + 0 & \markedPivot{1} & 0 & -1 \\ + 0 & 0 & \markedPivot{1} & 5 + \end{array}\right] + &\hspace{0.2in} \text{Add } -\frac{1}{3} \text{ Row 3 to Row 2}& + &\sim& + \left[\begin{array}{cccc} + \markedPivot{1} & 0 & 0 & 4 \\ + 0 & \markedPivot{1} & 0 & -1 \\ + 0 & 0 & \markedPivot{1} & 5 + \end{array}\right] + &\hspace{0.2in} \text{Add } \frac{5}{3} \text{ Row 3 to Row 1}& + +

    3. + +
+ + diff --git a/linear-algebra/source/01-LE/samples/03.ptx b/linear-algebra/source/01-LE/samples/03.ptx new file mode 100644 index 00000000..2f223901 --- /dev/null +++ b/linear-algebra/source/01-LE/samples/03.ptx @@ -0,0 +1,159 @@ + +LE3 + +

+Consider each of the following systems of linear equations +or vector equations. +

+
    +
  1. +

    + + \begin{matrix} + -2 \, x_{1} & + & x_{2} & + & x_{3} & = & -2 \\ + -2 \, x_{1} & - & 3 \, x_{2} & - & 3 \, x_{3} & = & 0 \\ + 3 \, x_{1} & + & x_{2} & + & x_{3} & = & 3 \\ + \end{matrix} + +

    +
    2. +
  2. +

    + + x_{1} \left[\begin{array}{c} +-5 \\ +3 \\ +-1 +\end{array}\right] + x_{2} \left[\begin{array}{c} +3 \\ +-2 \\ +2 +\end{array}\right] + x_{3} \left[\begin{array}{c} +14 \\ +-9 \\ +7 +\end{array}\right] = \left[\begin{array}{c} +1 \\ +0 \\ +-4 +\end{array}\right] + +

    +
    3. +
  3. +

    + + x_{1} \left[\begin{array}{c} +0 \\ +-1 \\ +-1 +\end{array}\right] + x_{2} \left[\begin{array}{c} +1 \\ +-4 \\ +-4 +\end{array}\right] + x_{3} \left[\begin{array}{c} +2 \\ +-4 \\ +-3 +\end{array}\right] = \left[\begin{array}{c} +-5 \\ +11 \\ +8 +\end{array}\right] + +

    +
    4. +
+ + + +
    +
  1. +

    + \mathrm{RREF}\left[\begin{array}{ccc|c} +-2 & 1 & 1 & -2 \\ +-2 & -3 & -3 & 0 \\ +3 & 1 & 1 & 3 +\end{array}\right]=\left[\begin{array}{ccc|c} +1 & 0 & 0 & 0 \\ +0 & 1 & 1 & 0 \\ +0 & 0 & 0 & 1 +\end{array}\right] + This matrix corresponds to the simpler system + + \begin{matrix} + x_{1} & & & & & = & 0 \\ + & & x_{2} & + & x_{3} & = & 0 \\ + & & & & 0 & = & 1 \\ + \end{matrix} + + The third equation 0=1 indicates that the system has no solutions. + The solution set is \emptyset. +

    +
    2. +
  2. +

    + \mathrm{RREF}\left[\begin{array}{ccc|c} +-5 & 3 & 14 & 1 \\ +3 & -2 & -9 & 0 \\ +-1 & 2 & 7 & -4 +\end{array}\right]=\left[\begin{array}{ccc|c} +1 & 0 & -1 & -2 \\ +0 & 1 & 3 & -3 \\ +0 & 0 & 0 & 0 +\end{array}\right] + This matrix corresponds to the simpler system + + \begin{matrix} + x_{1} & & & - & x_3 & = & -2 \\ + & & x_{2} & + & 3\,x_{3} & = & -3 \\ + & & & & 0 & = & 0 \\ + \end{matrix}. + + Since there are three variables and two nontrivial equations, the solution set has infinitely-many solutions. +

    +
    3. +
  3. +

    + \mathrm{RREF}\left[\begin{array}{ccc|c} +0 & 1 & 2 & -5 \\ +-1 & -4 & -4 & 11 \\ +-1 & -4 & -3 & 8 +\end{array}\right]=\left[\begin{array}{ccc|c} +1 & 0 & 0 & -3 \\ +0 & 1 & 0 & 1 \\ +0 & 0 & 1 & -3 +\end{array}\right] + This matrix corresponds to the simpler system + + \begin{matrix} + x_{1} & & & & & = & -3 \\ + & & x_{2} & & & = & 1 \\ + & & & & x_{3} & = & -3 \\ + \end{matrix}. + + This system has one solution. + The solution set is \left\{ \left[\begin{array}{c} +-3 \\ +1 \\ +-3 +\end{array}\right] \right\}. +

    +
    4. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/01-LE/samples/04.ptx b/linear-algebra/source/01-LE/samples/04.ptx new file mode 100644 index 00000000..dfbb51a6 --- /dev/null +++ b/linear-algebra/source/01-LE/samples/04.ptx @@ -0,0 +1,82 @@ + +LE4 + +

+Consider the following +vector equation. + + + x_{1} \left[\begin{array}{c} +-3 \\ +0 \\ +4 +\end{array}\right] + x_{2} \left[\begin{array}{c} +-3 \\ +0 \\ +4 +\end{array}\right] + x_{3} \left[\begin{array}{c} +0 \\ +1 \\ +0 +\end{array}\right] + x_{4} \left[\begin{array}{c} +-4 \\ +-5 \\ +5 +\end{array}\right] = \left[\begin{array}{c} +-11 \\ +-9 \\ +14 +\end{array}\right] + +

+
    +
  1. +

    +Explain how to find a simpler system or vector equation +that has the same solution set. +

    +
    2. +
  2. +

    +Explain how to describe this solution set using set notation. +

    +
    3. +
+ + + +

First, we compute + \mathrm{RREF}\left[\begin{array}{cccc|c} +-3 & -3 & 0 & -4 & -11 \\ +0 & 0 & 1 & -5 & -9 \\ +4 & 4 & 0 & 5 & 14 +\end{array}\right]=\left[\begin{array}{cccc|c} +1 & 1 & 0 & 0 & 1 \\ +0 & 0 & 1 & 0 & 1 \\ +0 & 0 & 0 & 1 & 2 +\end{array}\right]. + This corresponds to the simpler system + + \begin{matrix} + x_{1} & + & x_2 & & & & & = & 1 \\ + & & & & x_3 & & & = & 1 \\ + & & & & & & x_4 & = & 2 \\ + \end{matrix}. + + Since the second column is a non-pivot column, we let x_2=a. Making this substitution + and then solving for x_1, x_3, and x_4 produces the system + \begin{matrix} + x_1 &=& 1-a \\ + x_2 &=& a \\ + x_3 &=& 1 \\ + x_4 &=& 2 \\ + \end{matrix} + Thus, the solution set is \left\{ \left[\begin{array}{c} +-a + 1 \\ +a \\ +1 \\ +2 +\end{array}\right] \,\middle|\, a \in\mathbb R \right\} . +

+ + \ No newline at end of file diff --git a/linear-algebra/source/02-EV/01.ptx b/linear-algebra/source/02-EV/01.ptx new file mode 100644 index 00000000..27a9f5e2 --- /dev/null +++ b/linear-algebra/source/02-EV/01.ptx @@ -0,0 +1,536 @@ + +
+ Linear Combinations (EV1) + + + + + + +Warm Up + + +

+Discuss which of the vectors +\vec{u}=\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right] and +\vec{v}=\left[\begin{array}{c} 0 \\ 3 \\ -1 \end{array}\right] +is a solution to the given vector equation: + +x_1\left[\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right]+ +x_2\left[\begin{array}{c} 2 \\ -1 \\ 0 \end{array}\right]+ +x_3\left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right]= +\left[\begin{array}{c} -1 \\ 1 \\ 5 \end{array}\right] + +

+ + + + + + + +Class Activities + +

+We've been working with Euclidean vector spacesEuclideanvector space +of the form + + \IR^n=\setBuilder{\left[\begin{array}{c}x_1\\x_2\\\vdots\\x_n\end{array}\right]}{x_1,x_2,\dots,x_n\in\IR} +. +There are other kinds of vector spaces as well (e.g. polynomials, matrices), which we will investigate in +. But understanding +the structure of Euclidean vectors on their own will be beneficial, even when we turn our attention +to other kinds of vectors. +

+ + +

+Likewise, when we multiply a vector by a real number, as in +-3 \left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]=\left[\begin{array}{c} -3 \\ 3 \\ -6 \end{array}\right], +we refer to this real number as a scalar. +

+ + + + + +

+ A linear combination linear combinationof a set of vectors + \{\vec v_1,\vec v_2,\dots,\vec v_m\} is given by + c_1\vec v_1+c_2\vec v_2+\dots+c_m\vec v_m for any choice of + scalar multiples c_1,c_2,\dots,c_m. +

+

+ For example, we can say \left[\begin{array}{c}3 \\0 \\ 5\end{array}\right] + is a linear combination of the vectors \left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right] + and \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] since + + \left[\begin{array}{c} 3 \\ 0 \\ 5 \end{array}\right] = + 2 \left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right] + + 1\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] + . +

+ + + + + +

+ The spanspan of a set of vectors is the collection of all linear + combinations of that set: + + \vspan\{\vec v_1,\vec v_2,\dots,\vec v_m\} = + \setBuilder{c_1\vec v_1+c_2\vec v_2+\dots+c_m\vec v_m}{ + c_i\in\IR} + . +

+

+ For example: + + \vspan\setList + { + \left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right], + \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] + } = \setBuilder + { + a\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]+ + b\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] + }{ + a,b\in\IR + } + . +

+ + + + + +

+ Consider \vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right]\right\}. +

+ + +

+ Sketch the four Euclidean vectors + 1\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}1\\2\end{array}\right],\hspace{1em} + 3\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}3\\6\end{array}\right],\hspace{1em} + 0\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right],\hspace{1em} + -2\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}-2\\-4\end{array}\right] + in the xy plane by placing a dot at the (x,y) coordinate associated with each vector. +

+ + +

+ Sketch a representation of all the vectors belonging to + + \vspan\setList{\left[\begin{array}{c}1\\2\end{array}\right]} + = + \setBuilder{a\left[\begin{array}{c}1\\2\end{array}\right]}{a\in\IR} + + in the xy plane by plotting their (x,y) coordinates as dots. + What best describes this sketch? +

    +
  1. A line
    2. +
  2. A plane
    3. +
  3. A parabola
    4. +
  4. A circle
    5. +
+

+ + + + + +

+It is important to remember that +\{\vec v_1,\vec v_2,\dots,\vec v_m\}\not=\vspan\{\vec v_1,\vec v_2,\dots,\vec v_m\}. +

+

+For example, + +\setList +{ +\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right], +\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] +} + +is a set containing exactly two vectors, while + +\vspan\setList +{ +\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right], +\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] +} = \setBuilder +{ +a\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]+ +b\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] +}{ +a,b\in\IR +} + +is a set containing infinitely-many vectors. +

+ + + + + + + +

+ Consider \vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right], + \left[\begin{array}{c}-1\\1\end{array}\right]\right\}. +

+ + +

+ Sketch the following five Euclidean vectors in the xy plane. + + 1\left[\begin{array}{c}1\\2\end{array}\right]+ + 0\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown\hspace{2em} + 0\left[\begin{array}{c}1\\2\end{array}\right]+ + 1\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown\hspace{2em} + 1\left[\begin{array}{c}1\\2\end{array}\right]+ + 1\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown + + + -2\left[\begin{array}{c}1\\2\end{array}\right]+ + 1\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown\hspace{2em} + -1\left[\begin{array}{c}1\\2\end{array}\right]+ + -2\left[\begin{array}{c}-1\\1\end{array}\right]=\unknown + +

+ + +

+ Sketch a representation of all the vectors belonging to + \vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right], + \left[\begin{array}{c}-1\\1\end{array}\right]\right\}= + \setBuilder{a\left[\begin{array}{c}1\\2\end{array}\right]+ + b\left[\begin{array}{c}-1\\1\end{array}\right]}{a, b \in \IR} + in the xy plane. What best describes this sketch? +

    +
  1. A line
    2. +
  2. A plane
    3. +
  3. A parabola
    4. +
  4. A circle
    5. +
+

+ + + + + +

+ Sketch a representation of all the vectors belonging to + \vspan\left\{\left[\begin{array}{c}6\\-4\end{array}\right], + \left[\begin{array}{c}-3\\2\end{array}\right]\right\} + in the xy plane. What best describes this sketch? +

    +
  1. A line
    2. +
  2. A plane
    3. +
  3. A parabola
    4. +
  4. A cube
    5. +
+

+ + + + + + +

+Consider the following questions to discover whether a Euclidean +vector belongs to a span. +

+ + + +

+The Euclidean vector +\left[\begin{array}{c}-1\\-6\\1\end{array}\right] belongs to +\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], +\left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\} exactly when +there exists a solution to which of these vector equations? +

+

+

    +
  1. +x_1\left[\begin{array}{c}-1\\-6\\1\end{array}\right]+ +x_2\left[\begin{array}{c}1\\0\\-3\end{array}\right] +=\left[\begin{array}{c}-1\\-3\\2\end{array}\right] +
    2. +
  2. +x_1\left[\begin{array}{c}1\\0\\-3\end{array}\right]+ +x_2\left[\begin{array}{c}-1\\-3\\2\end{array}\right] +=\left[\begin{array}{c}-1\\-6\\1\end{array}\right] +
    3. +
  3. +x_1\left[\begin{array}{c}-1\\-3\\2\end{array}\right]+ +x_2\left[\begin{array}{c}-1\\-6\\1\end{array}\right]+ +x_3\left[\begin{array}{c}1\\0\\-3\end{array}\right]=0 +
    4. +
+

+ + + +

+ Use technology to find \RREF of the corresponding augmented matrix, + and then use that matrix to find the solution set of the vector equation. +

+ + + +

+ Given this solution set, does + \left[\begin{array}{c}-1\\-6\\1\end{array}\right] belong to + \vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], + \left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}? +

+ + + + + + +

+The following are all equivalent statements: +

+
    +
  • +

    + The vector \vec{b} belongs to \vspan\{\vec v_1,\dots,\vec v_n\}. +

    +
    • +
  • +

    + The vector \vec{b} is a linear combination of the vectors \vec v_1,\dots,\vec v_n. +

    +
    • +
  • +

    + The vector equation x_1 \vec{v}_1+\cdots+x_n \vec{v}_n=\vec{b} is consistent. +

    +
    • +
  • +

    + The linear system corresponding to + \left[\vec v_1\,\dots\,\vec v_n \,|\, \vec b\right] + is consistent. +

    +
    • +
  • +

    + \RREF\left[\vec v_1\,\dots\,\vec v_n \,|\, \vec b\right] + doesn't have a row [0\,\cdots\,0\,|\,1] + representing the contradiction 0=1. +

    +
    • +
+ + + + +

Consider this claim about a vector equation:

+

\left[\begin{array}{c} -6 \\ 2 \\ -6 \end{array}\right]is a linear combination of the vectors \left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \text{ and } \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right].

+ + + +

Write a statement involving the solutions of a vector equation that's equivalent to this claim.

+ + + + +

Explain why the statement you wrote is true.

+ + + + +

Since your statement was true, use the solution set to describe a linear combination of \left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \text{ and } \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right] that equals \left[\begin{array}{c} -6 \\ 2 \\ -6 \end{array}\right].

+ + + + + + + + + +

Consider this claim about a vector equation:

+

\left[\begin{array}{c} -5 \\ -1 \\ -7 \end{array}\right] belongs to + \vspan\left\{\left[\begin{array}{c} 1 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 6 \end{array}\right] , + \left[\begin{array}{c} 2 \\ 0 \\ 4 \end{array}\right] , \left[\begin{array}{c} -4 \\ 1 \\ -5 \end{array}\right]\right\}.

+ + + +

Write a statement involving the solutions of a vector equation that's equivalent to this claim.

+ + + + +

Explain why the statement you wrote is false, to conclude that the vector does not belong to the span.

+ + + + + + + + + + + + + + + + +Cool Down + + +

+Before next class, find some time to do the following: +

+ + +

+Without referring to your activity book, write down the definition of a linear combination of vectors. +

+ + +

+Let \vec{u}=\left[\begin{array}{c} 1 \\ 2 \\0 \end{array}\right] and +\vec{v}=\left[\begin{array}{c} -1 \\ 3 \\ 0\end{array}\right]. +Write down an example +\vec{w_1}=\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown\end{array}\right] +of a linear combination of \vec{u},\vec{v}. Then write down an example +\vec{w_2}=\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown\end{array}\right] +that is not a linear combination of \vec{u},\vec{v}. +

+ + +

+Draw a rough sketch of the vectors \vec{u}=\left[\begin{array}{c} 1 \\ 2 \\0 \end{array}\right], +\vec{v}=\left[\begin{array}{c} -1 \\ 3 \\ 0\end{array}\right], +\vec{w_1}=\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown\end{array}\right], and +\vec{w_2}=\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown\end{array}\right] +in \IR^3. +

+ + + + + + Videos +
+ + Video: Linear combinations +
+ + + + + +

Exercises available at .

+ + +Mathematical Writing Explorations + + +

Suppose S = \{\vec{v_1},\ldots, \vec{v_n}\} is a set of vectors. Show that \vec{v_0} is a linear combination of members of S, if an only if there are a set of scalars \{c_0,c_1,\ldots, c_n\} such that \vec{z} = c_0\vec{v_0} + \cdots + c_n\vec{v_n}. + +We can do this in a few parts. I've used bullets here to indicate all that needs to be done. This is an "if and only if" proof, so it needs two parts. +

  • First, assume that \vec{0} = c_0\vec{v_0} + \cdots + c_n\vec{v_n} has a solution, with c_0 \neq 0. Show that \vec{v_0} is a linear combination of elements of S.
    • +
  • Next, assume that \vec{v_0} is a linear combination of elements of S. Can you find the appropriate \{c_0,c_1,\ldots, c_n\} to make the equation \vec{z} = c_0\vec{v_0} + \cdots + c_n\vec{v_n} true?
    • +
  • In either of your proofs above, does the case when \vec{v_0} = \vec{z} change your thinking? Explain why or why not.
    • +
+

+ + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/02-EV/02.ptx b/linear-algebra/source/02-EV/02.ptx new file mode 100644 index 00000000..5e87aa57 --- /dev/null +++ b/linear-algebra/source/02-EV/02.ptx @@ -0,0 +1,463 @@ + +
+ Spanning Sets (EV2) + + + + + + Warm Up + + +

+ Given a set of ingredients and a meal, a recipe is a list of amounts of each ingredient required to prepare the given meal. +

+ + + +

+ Use the words vector and linear combination to create a new statement that is analogous to one above. +

+ + + + +

+ Building on your analogy, what role might the word span play? +

+ + + + + +Class Activities + + +

+Any single non-zero vector/number x in \IR^1 spans \IR^1, +since \IR^1=\setBuilder{cx}{c\in\IR}. +

+
+ + +\begin{tikzpicture} + \draw[<->] (-3,0) -- (3,0); + \draw[thick,->,blue] (0,0) -- (2,0) node[above] {x}; + \draw (0,-0.2) -- (0,0.2) node[above] {0}; +\end{tikzpicture} + + + An \IR^1 vector +
+ + + + + +

+ How many vectors are required to span \IR^2? + Sketch a drawing in the xy plane to support your answer. +

+
+ + +\begin{tikzpicture} + \draw[<->] (-4,0) -- (4,0); + \draw[<->] (0,-4) -- (0,4); +\end{tikzpicture} + + + The xy plane \IR^2 +
+
    +
  1. +

    + 1 +

    +
    2. +
  2. +

    + 2 +

    +
    3. +
  3. +

    + 3 +

    +
    4. +
  4. +

    + 4 +

    +
    5. +
  5. +

    + Infinitely Many +

    +
    6. +
+ + + + + +

+ How many vectors are required to span \IR^3? +

+
+ + + \begin{tikzpicture}[x={(210:0.8cm)}, y={(0:1cm)}, z={(90:1cm)}] + \draw[->] (0,0,0) -- (6,0,0); + \draw[->] (0,0,0) -- (0,6,0); + \draw[->] (0,0,0) -- (0,0,6); + \end{tikzpicture} + + + \IR^3 space +
+
    +
  1. +

    + 1 +

    +
    2. +
  2. +

    + 2 +

    +
    3. +
  3. +

    + 3 +

    +
    4. +
  4. +

    + 4 +

    +
    5. +
  5. +

    + Infinitely Many +

    +
    6. +
+ + + + + +

+ At least n vectors are required to span \IR^n. +

+
+ + + \begin{tikzpicture}[scale=0.5] + \draw[<->] (-4,0) -- (4,0); + \draw[<->] (0,-4) -- (0,4); + \draw[blue!50] (2,-4) -- (-2,4); + \draw[thick,blue,->] (0,0) -- (1,-2); + \end{tikzpicture} + \begin{tikzpicture}[x={(210:0.8cm)}, y={(0:1cm)}, z={(90:1cm)},scale=0.5] + \draw[->] (0,0,0) -- (6,0,0); + \draw[->] (0,0,0) -- (0,6,0); + \draw[->] (0,0,0) -- (0,0,6); + \draw[fill=purple!20,fill opacity=0.5] + (-2,-2,2) -- (6,-2,-2) -- (2,2,-2) -- (-6,2,2) -- (-2,-2,2); + \draw[thick,blue,->] (0,0,0) -- (1,-1,0); + \draw[thick,red,->] (0,0,0) -- (-2,0,1); + \end{tikzpicture} + + + Failed attempts to span \IR^n by <n vectors +
+ + + + + +

+ Consider the question: Does every vector in \IR^3 belong to + \vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right], + \left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}? +

+ + + +

+ Determine if \left[\begin{array}{c} 7 \\ -3 \\ -2 \end{array}\right] belongs to + \vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right], + \left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}. +

+ + + + +

+ Determine if \left[\begin{array}{c} 2 \\ 5 \\ 7 \end{array}\right] belongs to + \vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right], + \left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}. +

+ + + + +

+An arbitrary vector \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] belongs to +\vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right], +\left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\} provided the equation +x_1\left[\begin{array}{c}1\\-1\\0\end{array}\right]+ + x_2\left[\begin{array}{c}-2\\0\\1\end{array}\right]+ + x_3\left[\begin{array}{c}-2\\-2\\2\end{array}\right]=\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] has... +

    +
  1. no solutions.
    2. +
  2. exactly one solution.
    3. +
  3. at least one solution.
    4. +
  4. infinitely-many solutions.
    5. +
+

+ + + + +

+We're guaranteed at least one solution if the RREF of the corresponding augmented matrix has no contradictions; +likewise, we have no solutions if the RREF corresponds to the contradiction 0=1. Given +\left[\begin{array}{ccc|c}1&-2&-2&\unknown\\-1&0&-2&\unknown\\0&1&2&\unknown\end{array}\right]\sim +\left[\begin{array}{ccc|c}1&0&2&\unknown\\0&1&2&\unknown\\0&0&0&\unknown\end{array}\right] +we may conclude that the set does not span all of \IR^3 because... +

    +
  1. the row [0\,1\,2\,|\,\unknown] prevents a contradiction.
    2. +
  2. the row [0\,1\,2\,|\,\unknown] allows a contradiction.
    3. +
  3. the row [0\,0\,0\,|\,\unknown] prevents a contradiction.
    4. +
  4. the row [0\,0\,0\,|\,\unknown] allows a contradiction.
    5. +
+

+ + + + + + +

+ The set \{\vec v_1,\dots,\vec v_m\} spans all of \IR^n + exactly when the vector equation + x_1 \vec{v}_1 + \cdots + x_m\vec{v}_m = \vec{w} + is consistent for every vector \vec{w}. +

+

+ Likewise, the set \{\vec v_1,\dots,\vec v_m\} fails to span all of \IR^n + exactly when the vector equation + x_1 \vec{v}_1 + \cdots + x_m\vec{v}_m = \vec{w} + is inconsistent for some vector \vec{w}. +

+

+ Note these two possibilities are decided based on whether or not + \RREF[\vec v_1\,\dots\,\vec v_m] has either all pivot rows, or at + least one non-pivot row (a row of zeroes): + \left[\begin{array}{ccc|c}1&-2&-2\\-1&0&-2\\0&1&2\end{array}\right]\sim + \left[\begin{array}{ccc|c}1&0&2\\0&1&2\\0&0&0\end{array}\right]. +

+ + + + + +

+ Consider the set of vectors S=\left\{ + \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], + \left[\begin{array}{c}1\\-4\\3\\0\end{array}\right], + \left[\begin{array}{c}1\\7\\-3\\-1\end{array}\right], + \left[\begin{array}{c}0\\3\\5\\7\end{array}\right], + \left[\begin{array}{c}3\\13\\7\\16\end{array}\right] + \right\} + and the question + Does + \IR^4=\vspan S? +

+ + +

+ Rewrite this question in terms of the solutions to a vector equation. +

+ + +

+ Answer your new question, and use this to answer the original question. +

+ + + + + + + + + + + +

+Let \vec{v}_1, \vec{v}_2, \vec{v}_3 \in \IR^7 be three Euclidean vectors, +and suppose \vec{w} is another vector with +\vec{w} \in \vspan \left\{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\}. +What can you conclude about + \vspan \left\{ \vec{w}, \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} ? +

+
    +
  1. \vspan \left\{ \vec{w}, \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} is larger than \vspan \left\{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} . +
    2. +
  2. \vspan \left\{ \vec{w}, \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} is the same as \vspan \left\{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} . +
    3. +
  3. \vspan \left\{ \vec{w}, \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} is smaller than \vspan \left\{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} . +
    4. +
+ + + + + + Cool Down + + +

+ One of our important results in this lesson is , which states that a set of n vectors is required to span \IR^n. + While we developed some geometric intuition for why this true, we did not prove it in class. + Before coming to class next time, follow the steps outlined below to convince yourself of this fact using the concepts we learned in this lesson. +

+ + + +

+ Let \{\vec{v}_1,\dots, \vec{v}_m\} be a set of vectors living in \IR^n and assume that m <n. + How many rows and how many columns will the matrix [\vec{v}_1\cdots \vec{v}_m] have? +

+ + + + +

+ Given no additional information about the vectors \vec{v}_1,\dots, \vec{v}_m, what is the maximum possible number of pivots in \RREF[\vec v_1\,\dots\,\vec v_m]? +

+ + + + +

+ Conclude that our given set of vector cannot span all of \IR^n. +

+ + + + + + + Videos +
+ + Video: Determining if a set spans a Euclidean space +
+ + + + + +

Exercises available at .

+ + + + Mathematical Writing Explorations + + + Construct each of the following, or show that it is impossible: +
    +
  • A set of 2 vectors that spans \mathbb{R}^3
    • +
  • A set of 3 vectors that spans \mathbb{R}^3
    • +
  • A set of 3 vectors that does not span \mathbb{R}^3
    • +
  • A set of 4 vectors that spans \mathbb{R}^3
    • +
+ For any of the sets you constructed that did span the required vector space, are any of the vectors a linear combination of the others in your set? + + + + + Based on these results, generalize this a conjecture about how a set of n-1, n and n+1 vectors would or would not span \mathbb{R}^n. + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + + +
diff --git a/linear-algebra/source/02-EV/03.ptx b/linear-algebra/source/02-EV/03.ptx new file mode 100644 index 00000000..7e4bd2fa --- /dev/null +++ b/linear-algebra/source/02-EV/03.ptx @@ -0,0 +1,938 @@ + +
+ Subspaces (EV3) + + + + + + Warm Up + + +

+ Consider the linear equation x+2y+z=0. +

+ + + +

+ Verify that both \vec{v}=\left[\begin{array}{c}1\\-1\\1\end{array}\right] and \vec{w}=\left[\begin{array}{c}1\\0\\-1\end{array}\right] are solutions. +

+ + + + +

+ Is the vector 2\vec{v}-3\vec{w} also a solution? +

+ + + + + + +Class Activities + + +

+ Recall that if S=\left\{\vec{v}_1,\dots, \vec{v}_n\right\} is subset of vectors in \IR^n, then \vspan(S) is the set of all linear combinations of vectors in S. + In EV2 (), we learned how to decide whether \vspan(S) was equal to all of \IR^n or something strictly smaller. +

+ + + +

+ Let S denote a set of vectors in \IR^n and suppose that \vec{u},\vec{v}\in\vspan(S), + c\in\IR and that \vec{w}\in\IR^n. Which of the following vectors might + not belong to \vspan(S)? +

    +
  1. \vec{0}
    2. +
  2. \vec{u}+\vec{w}
    3. +
  3. \vec{u}+\vec{v}
    4. +
  4. c\vec{u}
    5. +
+

+ + + + + +

+ A homogeneoushomogeneous system of linear equations is one of the form: + + + a_{11}x_1 &\,+\,& a_{12}x_2 &\,+\,& \dots &\,+\,& a_{1n}x_n &\,=\,& 0 + + + a_{21}x_1 &\,+\,& a_{22}x_2 &\,+\,& \dots &\,+\,& a_{2n}x_n &\,=\,& 0 + + + \vdots& &\vdots& && &\vdots&&\vdots + + + a_{m1}x_1 &\,+\,& a_{m2}x_2 &\,+\,& \dots &\,+\,& a_{mn}x_n &\,=\,& 0 + + +

+

+ This system is equivalent to the vector equation: + x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0} + and the augmented matrix: + + \left[\begin{array}{cccc|c} + a_{11} & a_{12} & \cdots & a_{1n} & 0\\ + a_{21} & a_{22} & \cdots & a_{2n} & 0\\ + \vdots & \vdots & \ddots & \vdots & \vdots\\ + a_{m1} & a_{m2} & \cdots & a_{mn} & 0 + \end{array}\right] + +

+ + + + + +

+ Consider the homogeneous vector equation x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0}. +

+ + +

+ Is this equation consistent? +

+
    +
  1. +

    + no. +

    +
    2. +
  2. +

    + yes. +

    +
    3. +
  3. +

    + more information is required. +

    +
    4. +
+ + +

+ Note that if \left[\begin{array}{c} a_1 \\ \vdots \\ a_n \end{array}\right] and + \left[\begin{array}{c} b_1 \\ \vdots \\ b_n \end{array}\right] are both solutions, + we know that + a_1 \vec{v}_1+\cdots+a_n \vec{v}_n = \vec{0} + \text{ and } + b_1 \vec{v}_1+\cdots+b_n \vec{v}_n = \vec{0} . + Therefore by adding these equations: + (a_1 + b_1) \vec{v}_1+\cdots+(a_n+b_n) \vec{v}_n = \vec{0}, +

+

+ we may conclude that the vector \left[\begin{array}{c} a_1+ b_1 \\ \vdots \\ a_n+b_n \end{array}\right] + is... +

+
    +
  1. +

    + another solution. +

    +
    2. +
  2. +

    + not a solution. +

    +
    3. +
  3. +

    + is equal to \vec{0}. +

    +
    4. +
+ + +

+ Similarly, if c \in \IR, then since multiplying by c yields: + (ca_1)\vec{v}_1+\cdots+(ca_n)\vec{v}_n=\vec{0}, + + we may conclude that the vector \left[\begin{array}{c} ca_1 \\ \vdots \\ ca_n \end{array}\right] is... +

+
    +
  1. +

    + another solution. +

    +
    2. +
  2. +

    + not a solution. +

    +
    3. +
  3. +

    + is equal to \vec{0}. +

    +
    4. +
  4. +

    + The empty set. +

    +
    5. +
+ + + + +

+ If S is any set of vectors in \IR^n, then the set \vspan(S) has the following properties: +

    +
  • +

    + the set \vspan(S) is non-empty. +

    +
    • +
  • +

    + the set \vspan(S) is closed under addition: for any \vec{u},\vec{v}\in \vspan(S), the sum \vec{u}+\vec{v} is also in \vspan(S). +

    +
    • +
  • +

    + the set \vspan(S) is closed under scalar multiplication: for any \vec{u}\in\vspan(S) and scalar c\in\IR, the product c\vec{u} is also in \vspan(S). +

    +
    • +
+ Likewise, if W is the solution set to a homogenous vector equation, it too satisfies: +
    +
  • +

    + the set W is non-empty. +

    +
    • +
  • +

    + the set W is closed under addition: for any \vec{u},\vec{v}\in W, the sum \vec{u}+\vec{v} is also in W. +

    +
    • +
  • +

    + the set \vspan(S) is closed under scalar multiplication: for any \vec{u}\in W and scalar c\in\IR, the product c\vec{u} is also in W. +

    +
    • +
+

+ + + + +

+ A subset W of a vector space is called a subspacesubspace + provided that it satisfies the following properties: +

    +
  • +

    + the subset is non-empty. +

    +
    • +
  • +

    + the subset is closed under additionvector spaceclosed under addition: for any \vec{u},\vec{v} \in W, the sum \vec{u}+\vec{v} is also in W. +

    +
    • +
  • +

    + the subset is closed under scalar multiplicationvector spaceclosed under scalar multiplication: for any \vec{u} \in W and scalar c \in \IR, the product c\vec{u} is also in W. +

    +
    • +
+

+ + + + +

+Note the similarities between a planar subspace spanned by two non-colinear vectors in \IR^3, +and the Euclidean plane \IR^2. While they are not the same thing (and shouldn't be referred to +interchangably), algebraists call such similar spaces isomorphicisomorphic; +we'll learn what this means more carefully in a later chapter. +

+
+ A planar subset of \IR^3 compared with the plane \IR^2. + + A planar subset of \IR^3 compared with the plane \IR^2. + + \begin{tikzpicture}[scale=0.5] + \draw[->] (0,0,0) -- (6,0,0); + \draw[->] (0,0,0) -- (0,6,0); + \draw[->] (0,0,0) -- (0,0,6); + \draw[fill=purple!20,fill opacity=0.5] + (-2,-2,2) -- (6,-2,-2) -- (2,2,-2) -- (-6,2,2) -- (-2,-2,2); + \draw[thick,blue,->] (0,0,0) -- (1,-1,0); + \draw[thick,red,->] (0,0,0) -- (-2,0,1); + \begin{scope}[shift={(12,2)}] + \draw (-4,0) -- (4,0); + \draw (0,-4) -- (0,4); + \draw[fill=purple!20,fill opacity=0.5] + (-3.8,-3.8) -- (-3.8,3.8) -- (3.8,3.8) -- (3.8,-3.8) -- (-3.8,-3.8); + \draw[very thick,blue,->] (0,0) -- (1,0); + \draw[very thick,red,->] (0,0) -- (0,1); + \end{scope} + \end{tikzpicture} + + +
+ + + + + + +

+Let W=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}. +

+ + + +

+ Is W the empty set? +

+ + + +

+ Let's assume that \vec{v}=\left[\begin{array}{c} x \\ y \\ z \end{array}\right] and + \vec{w} = \left[\begin{array}{c} a \\ b \\ c \end{array}\right] are in W. + What are we allowed to assume? +

    +
  1. +

    + x+2y+z=0. +

    +
    2. +
  2. +

    + a+2b+c=0. +

    +
    3. +
  3. +

    + Both of these. +

    +
    4. +
  4. +

    + Neither of these. +

    +
    5. +
+

+ + +

+ Which equation must be verified to show that + \vec v+\vec w = \left[\begin{array}{c} x+a \\ y+b \\ z+c \end{array}\right] + also belongs to W? +

    +
  1. +

    + (x+a)+2(y+b)+(z+c)=0. +

    +
    2. +
  2. +

    + x+a+2y+b+z+c=0. +

    +
    3. +
  3. +

    + x+2y+z=a+2b+c. +

    +
    4. +
+

+ + +

+Use the assumptions from (a) to verify the equation from (b). +

+ + +

+ Is W is a subspace of \IR^3? +

    +
  1. +

    + Yes +

    +
    2. +
  2. +

    + No +

    +
    3. +
  3. +

    + Not enough information +

    +
    4. +
+

+ + +

+ Show that k\vec v=\left[\begin{array}{c}kx\\ky\\kz\end{array}\right] + also belongs to W for any k\in\IR by verifying + (kx)+2(ky)+(kz)=0 under these assumptions. +

+ + +

+ Is W is a subspace of \IR^3? +

    +
  1. +

    + Yes +

    +
    2. +
  2. +

    + No +

    +
    3. +
  3. +

    + Not enough information +

    +
    4. +
+

+ + + + + +

+Let W=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=4}. +

+ + + +

+ Is W the empty set? +

+ + + +

+Which of these statements is valid? +

    +
  1. +

    +\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W, and +\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\in W, so W is a subspace. +

    +
    2. +
  2. +

    +\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W, and +\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\in W, so W is not a subspace. +

    +
    3. +
  3. +

    +\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W, but +\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\not\in W, so W is a subspace. +

    +
    4. +
  4. +

    +\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W, but +\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\not\in W, so W is not a subspace. +

    +
    5. +
+

+ + +

+Which of these statements is valid? +

    +
  1. +

    +\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W, and +\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\in W, so W is a subspace. +

    +
    2. +
  2. +

    +\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W, and +\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\in W, so W is not a subspace. +

    +
    3. +
  3. +

    +\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W, but +\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\not\in W, so W is a subspace. +

    +
    4. +
  4. +

    +\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W, but +\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\not\in W, so W is not a subspace. +

    +
    5. +
+

+ + + + +

+In summary, any one of the following is enough to +prove that a nonempty subset W is not a subspace: +

    +
  • +

    +Find specific values for \vec u,\vec v\in W +such that \vec u+\vec v\not\in W. +

    +
    • +
  • +

    +Find specific values for c\in\IR,\vec v\in W +such that c\vec v\not\in W. +

    +
    • +
  • +

    +Show that \vec 0\not\in W. +

    +
    • +
+

+

+If you cannot do any of these, then W can be proven to be a subspace +by doing all of the following: +

    +
  1. +

    + Show that W is non-empty. +

    +
    2. +
  2. +

    +For all \vec v,\vec w\in W (not just specific values), +\vec u+\vec v\in W. +

    +
    3. +
  3. +

    +For all \vec v\in W and c\in \IR (not just specific values), +c\vec v\in W. +

    +
    4. +
+

+ + + + +

+ Consider these subsets of \IR^3: + + R= + \setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{y=z+1} + \hspace{2em} + S= + \setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{y=|z|} + \hspace{2em} + T= + \setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{z=xy} + . +

+ + +

+ Show R isn't a subspace by showing that \vec 0\not\in R. +

+ + +

+ Show S isn't a subspace by finding two vectors \vec u,\vec v\in S + such that \vec u+\vec v\not\in S. +

+ + +

+ Show T isn't a subspace by finding a vector \vec v\in T + such that 2\vec v\not\in T. +

+ + + + + +

Consider the following two sets of Euclidean vectors: U=\left\{ \left[\begin{array}{c} x \\ y \end{array}\right] \middle|\,7 \, x + 4 \, y = 0\right\} \hspace{2em} W=\left\{ \left[\begin{array}{c} x \\ y \end{array}\right] \middle|\,3 \, x y^{2} = 0\right\}

+

Explain why one of these sets is a subspace of \mathbb{R}^2 and one is not.

+ + + + + + +

+ Consider the following attempted proof that + U=\left\{ \left[\begin{array}{c} x \\ y \end{array}\right] \middle| x+y=xy\right\} + is closed under scalar multiplication. +

+
+

+Let \left[\begin{array}{c} x \\ y \end{array}\right]\in U, so we know that x+y=xy. +We want to show k\left[\begin{array}{c} x \\ y \end{array}\right]=\left[\begin{array}{c} kx \\ ky \end{array}\right]\in U, +that is, (kx)+(ky)=(kx)(ky). This is verified by the following calculation: + + +(kx)+(ky)&=(kx)(ky) + + +k(x+y)&=k^2xy + + +0[k(x+y)]&=0[k^2xy] + + +0&=0 + + +

+
+

+ Is this reasoning valid? +

    +
  1. +

    + Yes +

    +
    2. +
  2. +

    + No +

    +
    3. +
+

+ + + + + +

+Proofs of an equality \mathrm{LEFT}=\mathrm{RIGHT} should generally be of one of these forms: +

    +
  1. +

    + Using a chain of equalities: + + + \mathrm{LEFT} &= \cdots + + + &= \cdots + + + &= \cdots + + + &= \mathrm{RIGHT} + + + Alternatively: + + + \mathrm{LEFT} &= \cdots & \mathrm{RIGHT} &=\cdots + + + &= \cdots & &= \cdots + + + &= \cdots & &= \cdots + + + &= \mathrm{SAME}& &= \mathrm{SAME} + + +

    +
    2. +
  2. +

    +When the assumption \mathrm{THIS}=\mathrm{THAT} is already known or assumed to be true : + + + && \mathrm{THIS} &= \mathrm{THAT} + + + & \Rightarrow& \cdots &= \cdots + + + & \Rightarrow& \cdots &= \cdots + + + & \Rightarrow& \mathrm{LEFT} &= \mathrm{RIGHT} + + +

    +
    3. +
+

+ + + +

+The following proof is invalid. + + + && \mathrm{LEFT} &= \mathrm{RIGHT} + + + & \Rightarrow& \cdots &= \cdots + + + & \Rightarrow& \cdots &= \cdots + + + & \Rightarrow& 0 &= 0 + + + & \Rightarrow& \mathrm{ANYTHING} &= \mathrm{ANYTHING} + + +Basically, you cannot prove something is true by assuming it's true, +and it's not helpful to prove to someone that zero equals itself +(they probably already know that). +

+ + + + + + + + + + + Cool Down + + Recall that in we used the words vector, linear combination, and span to make an anology with recipes, ingredients, and meals. + In this analogy, a recipe was defined to be a list of amounts of each ingredient to build a particular meal. + + + + +

+ Given the set of ingredients S=\{\textrm{flour}, \textrm{yeast}, \textrm{salt}, \textrm{water}, \textrm{sugar}, \textrm{milk}\}, how should we think of the subspace \vspan(S)? +

+ + + + +

+ What is one meal that lives in the subspace \vspan(S)? +

+ + + + +

+ What is one meal that does not live in the subspace \vspan(S)? +

+ + + + + + +

+ Let W=\left\{\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]\middle|x+y=3z+2w\right\}. + The set W is a subspace. Below are two attempted proofs of the fact that W is closed under vector addition. + Both of them are invalid; explain why. +

+ + + +

+ Let \vec{u}=\left[\begin{array}{c}1\\4\\1\\1\end{array}\right],\vec{v}=\left[\begin{array}{c}2\\-1\\1\\-1\end{array}\right]. + Then both \vec{u},\vec{w} are elements of W. Their sum is + \vec{w}=\left[\begin{array}{c}3\\3\\2\\0\end{array}\right] + and since 3+3=3\cdot (2)+2\cdot (0), + it follows that \vec{w} is also in W and so W is closed under vector addition. +

+ + + + +

+ If \left[\begin{array}{c}x\\y\\z\\w\end{array}\right],\left[\begin{array}{c}a\\b\\c\\d\end{array}\right] are in W, we need to show that \left[\begin{array}{c}x+a\\y+b\\z+c\\w+d\end{array}\right] is also in W. + To be in W, we need + (x+a)+(y+b)=3(z+c)+2(w+d). + Well, if (x+a)+(y+b)=3(z+c)+2(w+d), + then we know that x+y-3z-2w+a+b-3c-2d=0 + by moving everything over to the left hand side. + Since we are assumming that x+y-3z-2w=0 and a+b-3c-2d=0, it follows that + 0=0, which is true, which proves that vector addition is closed. +

+ + + + + + + + + Videos +
+ + Video: Showing that a subset of a vector space is a subspace +
+
+ + Video: Showing that a subset of a vector space is not a subspace +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + + A square matrix M is symmetricsymmetric matrix if, for each index i,j, the entries m_{ij} = m_{ji}. That is, the matrix is itself when reflected over the diagonal from upper left to lower right. + +Prove that the set of n \times n symmetric matrices is a subspace of M_{n \times n}. + s + + + + The space of all real-valued function of one real variable is a vector space. First, define \oplus and \odot for this vector space. Check that you have closure (both kinds!) and show what the zero vector is under your chosen addition. + +Decide if each of the following is a subspace. If so, prove it. If not, provide the counterexample. +
    +
  • The set of even functions, \{f:\mathbb{R} \rightarrow \mathbb{R}: f(-x) = f(x) \mbox{ for all } x\}.
    • +
  • The set of odd functions, \{f:\mathbb{R} \rightarrow \mathbb{R}: f(-x) = -f(x) \mbox{ for all } x\}.
    • +
+ + + +Give an example of each of these, or explain why it's not possible that such a thing would exist. +
  • A nonempty subset of M_{2 \times 2} that is not a subspace.
    • +
  • A set of two vectors in \mathbb{R}^2 that is not a spanning set.
    • +
+ + + +Let V be a vector space and S = \{\vec{v}_1,\vec{v}_2,\ldots,\vec{v}_n\} a subset of V. Show that the span of S is a subspace. + +Is it possible that there is a subset of V containing fewer vectors than S, but whose span contains all of the vectors in the span of S? + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
\ No newline at end of file diff --git a/linear-algebra/source/02-EV/04.ptx b/linear-algebra/source/02-EV/04.ptx new file mode 100644 index 00000000..93d37e0d --- /dev/null +++ b/linear-algebra/source/02-EV/04.ptx @@ -0,0 +1,565 @@ + +
+ Linear Independence (EV4) + + + + + + + Warm Up + + +

+ Consider the vector equation + x_1\left[\begin{array}{c}1\\1\\1\end{array}\right]+x_2\left[\begin{array}{c}2\\0\\-1\end{array}\right]+x_3\left[\begin{array}{c}-1\\2\\0\end{array}\right]=\left[\begin{array}{c}-1\\7\\6\end{array}\right]. +

+ + + +

+ Decide which of \left[\begin{array}{c}3\\-1\\2\end{array}\right] or \left[\begin{array}{c}1\\1\\1\end{array}\right] is a solution vector. +

+ + + + +

+ Consider now the following vector equation: + y_1\left[\begin{array}{c}1\\1\\1\end{array}\right]+y_2\left[\begin{array}{c}2\\0\\-1\end{array}\right]+y_3\left[\begin{array}{c}-1\\2\\0\end{array}\right]+y_4\left[\begin{array}{c}-1\\7\\6\end{array}\right]=\vec{0}. + How is this vector equation related to the original one? +

+ + + + +

+ Use the solution vector you found in part (a) to construct a solution vector to this new equation. +

+ + + + + +Class Activities + + +

+ Consider the two sets + + S=\left\{ + \left[\begin{array}{c}2\\3\\1\end{array}\right], + \left[\begin{array}{c}1\\1\\4\end{array}\right] + \right\} \hspace{3em} + T=\left\{ + \left[\begin{array}{c}2\\3\\1\end{array}\right], + \left[\begin{array}{c}1\\1\\4\end{array}\right], + \left[\begin{array}{c}-1\\0\\-11\end{array}\right] + \right\} + . + Which of the following is true? +

+
    +
  1. \vspan S is bigger than \vspan T. +
    2. +
  2. \vspan S and \vspan T are the same size. +
    3. +
  3. \vspan S is smaller than \vspan T. +
    4. +
+ + + + + +

+ We say that a set of vectors is linearly dependentlinearly dependent if one vector + in the set belongs to the span of the others. Otherwise, we say the set + is linearly independent.linearly independent +

+
+ + + \begin{tikzpicture}[x={(210:0.8cm)}, y={(0:1cm)}, z={(90:1cm)}] + \draw[->] (0,0,0) -- (6,0,0); + \draw[->] (0,0,0) -- (0,6,0); + \draw[->] (0,0,0) -- (0,0,6); + \draw[fill=purple!20,fill opacity=0.5] + (-2,-2,2) -- (6,-2,-2) -- (2,2,-2) -- (-6,2,2) -- (-2,-2,2); + \draw[thick,blue,->] (0,0,0) -- (1,-1,0); + \draw[thick,red,->] (0,0,0) -- (-2,0,1); + \draw[thick,purple,->] (0,0,0) -- (1,1,-1); + \end{tikzpicture} + + + A linearly dependent set of three vectors +
+

+ You can think of linearly dependent sets as containing a redundant vector, + in the sense that you can drop a vector out without reducing the span of the set. In the above image, all three vectors lay in the same planar subspace, + but only two vectors are needed to span the plane, so the set is + linearly dependent. +

+ + + + +

Consider the following three vectors in \IR^3: + \vec v_1=\left[\begin{array}{c}-2 \\ 0 \\ 0\end{array}\right], + \vec v_2=\left[\begin{array}{c}1 \\ 3 \\ 0\end{array}\right], + \text{ and } + \vec v_3=\left[\begin{array}{c}-2 \\ 5 \\ 4\end{array}\right] + . +

+ + +

+ Let \vec w = 3\vec v_1 - \vec v_2 - 5 \vec v_3 = \left[\begin{array}{c}\unknown \\ \unknown \\ \unknown\end{array}\right]. + The set \{\vec v_1,\vec v_2,\vec v_3,\vec w\} is... +

    +

  1. linearly dependent: at least one vector is a linear combination of others

    2. +

  2. linearly independent: no vector is a linear combination of others

    3. +
+

+ + +

+ Find \RREF \left[\begin{array}{cccc} + \vec v_1 & \vec v_2 & \vec v_3 & \vec w \\ + \end{array}\right]= + \RREF \left[\begin{array}{cccc} + -2 & 1 &-2 & \unknown \\ + 0 & 3 & 5 & \unknown \\ + 0 &0 &4 & \unknown + \end{array}\right]= \unknown . +

+

+ What does this tell you about solution set for the vector equation + x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec w=\vec{0}? +

    +
  1. +

    + It is inconsistent. +

    +
    2. +
  2. +

    + It is consistent with one solution. +

    +
    3. +
  3. +

    + It is consistent with infinitely many solutions. +

    +
    4. +
+

+ + +

+ Which of these might explain the connection? +

    +
  1. +

    +A pivot column establishes linear independence and creates a +contradiction. +

    +
    2. +
  2. +

    +A non-pivot column both describes a linear combination and +reveals the number of solutions. +

    +
    3. +
  3. +

    +A pivot row describes the bound variables and prevents +a contradiction. +

    +
    4. +
  4. +

    +A non-pivot row prevents contradictions and makes the vector +equation solvable. +

    +
    5. +
+

+ + + + + +

+ For any vector space, + the set \{\vec v_1,\dots\vec v_n\} is linearly dependent if and only + if the vector equation x_1\vec v_1+ x_2 \vec v_2+\dots+x_n\vec v_n=\vec{0} is consistent with + infinitely many solutions. +

+

+ Likewise, the set of vectors + \{\vec v_1,\dots\vec v_n\} is linearly independent + if and only the vector equation x_1\vec v_1+ x_2 \vec v_2 + \cdots + x_n\vec v_n = \vec{0} + has exactly one solution: \left[\begin{array}{c} + x_1 \\ \vdots \\ x_n + \end{array}\right]=\left[\begin{array}{c} + 0 \\ \vdots \\ 0 + \end{array}\right]. +

+ + + + + +

+ Find + \RREF\left[\begin{array}{ccccc|c} + 2&2&3&-1&4&0\\ + 3&0&13&10&3&0\\ + 0&0&7&7&0&0\\ + -1&3&16&14&1&0 + \end{array}\right] + + and mark the part of the matrix that demonstrates that + S=\left\{ + \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], + \left[\begin{array}{c}2\\0\\0\\3\end{array}\right], + \left[\begin{array}{c}3\\13\\7\\16\end{array}\right], + \left[\begin{array}{c}-1\\10\\7\\14\end{array}\right], + \left[\begin{array}{c}4\\3\\0\\1\end{array}\right] + \right\} + + is linearly dependent (the part that shows its linear system has + infinitely many solutions). +

+ + + + + + + + +

+ Compare the following results: +

+
    +
  • +

    + A set of \IR^m vectors + \{\vec v_1,\dots\vec v_n\} is linearly independent if and only + if \RREF\left[\begin{array}{ccc}\vec v_1&\dots&\vec v_n\end{array}\right] + has all pivot columns. +

    +
    • +
  • +

    + A set of \IR^m vectors + \{\vec v_1,\dots\vec v_n\} is linearly dependent if and only + if \RREF\left[\begin{array}{ccc}\vec v_1&\dots&\vec v_n\end{array}\right] + has at least one non-pivot column. +

    +
    • +
  • +

    + A set of \IR^m vectors + \{\vec v_1,\dots\vec v_n\} spans \IR^m if and only + if \RREF\left[\begin{array}{ccc}\vec v_1&\dots&\vec v_n\end{array}\right] + has all pivot rows. +

    +
    • +
  • +

    + A set of \IR^m vectors + \{\vec v_1,\dots\vec v_n\} fails to span \IR^m if and only + if \RREF\left[\begin{array}{ccc}\vec v_1&\dots&\vec v_n\end{array}\right] + has at least one non-pivot row. +

    +
    • +
+ + + + + + +

Write a statement involving the solutions of a vector equation that's equivalent to each claim:

+ + + +

+ The set of vectors \left\{ \left[\begin{array}{c} + 1 \\ + -1 \\ + 0 \\ + -1 + \end{array}\right] , \left[\begin{array}{c} + 5 \\ + 5 \\ + 3 \\ + 1 + \end{array}\right] , \left[\begin{array}{c} + 9 \\ + 11 \\ + 6 \\ + 3 + \end{array}\right] \right\} is linearly independent. +

+ + + + +

+ The set of vectors \left\{ \left[\begin{array}{c} + 1 \\ + -1 \\ + 0 \\ + -1 + \end{array}\right] , \left[\begin{array}{c} + 5 \\ + 5 \\ + 3 \\ + 1 + \end{array}\right] , \left[\begin{array}{c} + 9 \\ + 11 \\ + 6 \\ + 3 + \end{array}\right] \right\} is linearly dependent. +

+ + + + + +

Explain how to determine which of these statements is true.

+ + + + + + + + + + + +

+What is the largest number of \IR^4 vectors that can form a linearly independent set? +

+
    +
  1. +

    + 3 +

    +
    2. +
  2. +

    + 4 +

    +
    3. +
  3. +

    + 5 +

    +
    4. +
  4. +

    + You can have infinitely many vectors and still be linearly independent. +

    +
    5. +
+ + + + + + + + +

+ Is it possible for the set of Euclidean vectors \{\vec v_1, \vec v_2,\ldots, \vec v_n, \vec 0\} + to be linearly independent? +

    +
  1. +

    + Yes +

    +
    2. +
  2. +

    + No +

    +
    3. +
+

+ + + + + Cool Down + + Recall that in we used the words vector, linear combination, and span to make an anology with recipes, ingredients, and meals. + In this analogy, a recipe was defined to be a list of amounts of each ingredient to build a particular meal. + + + +

+ Consider the statement: The set of vectors \left\{\vec{v}_1,\vec{v}_2,\vec{v}_3\right\} is linearly independent because the vector \vec{v}_3 is a linear combination of \vec{v_1} and \vec{v}_2. + Construct an analogous statement involving ingredients, meals, and recipes, using the terms linearly independent and linear combination. +

+ + + + +

+ The following exercises are designed to help develop your geometric intution around linear dependence. +

+ + + +

+ Draw sketches that depict the following: +

    +
  • +

    + Three linearly independent vectors in \IR^3. +

    +
    • +
  • +

    + Three linearly dependent vectors in \IR^3. +

    +
    • +
+

+ + + + +

+ If you have three linearly dependent vectors, is it necessarily the case that one of the vectors is a multiple of the other? +

+ + + + + + + Videos +
+ + Video: Linear independence +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + + Prove the result of , by showing that, given a set S = \{\vec{v}_1,\vec{v}_2,\ldots,\vec{v}_n\} of vectors, S is linearly independent iff the equation x_1\vec{v}_1 + x_2\vec{v}_2 + \ldots\ + x_n\vec{v}_n = \vec{0} is only true when x_1 = x_2 = \cdots = x_n = 0. + + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/02-EV/05.ptx b/linear-algebra/source/02-EV/05.ptx new file mode 100644 index 00000000..a240549c --- /dev/null +++ b/linear-algebra/source/02-EV/05.ptx @@ -0,0 +1,585 @@ + +
+ Identifying a Basis (EV5) + + + + + + + Warm Up + + Recall that in we used the words vector, linear combination, and span to make an anology with recipes, ingredients, and meals. + In this analogy, a recipe was defined to be a list of amounts of each ingredient to build a particular meal. + + + +

+ Consider the following set of ingredients: + S=\left\{\textrm{tomato}, \textrm{olive oil}, \textrm{dough}, \textrm{cheese}, \textrm{pizza sauce}, \textrm{garlic}\right\} +

+ + + +

+ Does "pizza" live inside of \vspan(S)? +

+ + + + +

+ Identify which ingredients in S make the set linearly dependent. +

+ + + + +

+ Can you think of a subset S' of S that is linearly independent and for which "pizza" is still in \vspan{S'}? +

+ + + + + +Class Activities + + + +

+ Consider the set of vectors + S=\left\{ + \left[\begin{array}{c} 3 \\ -2 \\ -1 \\ 0 \end{array} \right], + \left[\begin{array}{c} 2 \\ 4 \\ 1 \\ 1 \end{array} \right], + \left[\begin{array}{c} 0 \\ -16 \\ -5 \\ -3 \end{array} \right], + \left[\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array} \right], + \left[\begin{array}{c} 3 \\ 3 \\ 0 \\ 1 \end{array} \right] + \right\} + . +

+ + + +

+ Express the vector \left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right] as a linear combination + of the vectors in S, i.e. find scalars such that + + \left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right] = + \unknown \left[\begin{array}{c} 3 \\ -2 \\ -1 \\ 0 \end{array} \right] + + \unknown \left[\begin{array}{c} 2 \\ 4 \\ 1 \\ 1 \end{array} \right] + + \unknown \left[\begin{array}{c} 0 \\ -16 \\ -5 \\ -3 \end{array} \right] + + \unknown \left[\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array} \right] + + \unknown \left[\begin{array}{c} 3 \\ 3 \\ 0 \\ 1 \end{array} \right] + . +

+ + + + +

+ Find a different way to express the vector \left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right] + as a linear combination of the vectors in S. +

+ + + + +

+ Consider another vector \left[\begin{array}{c} 8 \\ 6 \\ 7 \\ 5 \end{array} \right]. + Without computing the RREF of another matrix, + how many ways can this vector be written as a linear combination of the vectors in S? +

    +

  1. Zero.

    2. +

  2. One.

    3. +

  3. Infinitely-many.

    4. +

  4. Computing a new matrix RREF is necessary.

    5. +
+

+ + + + + + + + + +

+ Let's review some of the terminology we've been dealing with... +

+ + +

+ If every vector in a vector space can be constructed as one or more linear combinations of vectors in a set S, we can say... +

    +

  1. the set S spans the vector space.

    2. +

  2. the set S fails to span the vector space.

    3. +

  3. the set S is linearly independent.

    4. +

  4. the set S is linearly dependent.

    5. +
+

+ + +

+ If the zero vector \vec 0 can be constructed as a unique linear combination of vectors in a set S + (the combination multiplying every vector by the scalar value 0), we can say... +

    +

  1. the set S spans the vector space.

    2. +

  2. the set S fails to span the vector space.

    3. +

  3. the set S is linearly independent.

    4. +

  4. the set S is linearly dependent.

    5. +
+

+ + +

+ If every vector of a vector space can either be constructed as a unique linear combination of vectors in a set S, + or not at all, we can say... +

    +

  1. the set S spans the vector space.

    2. +

  2. the set S fails to span the vector space.

    3. +

  3. the set S is linearly independent.

    4. +

  4. the set S is linearly dependent.

    5. +
+

+ + + + + +

+ A basis basis of a vector space V is a set of vectors S contained in V for which +

+
    +
  1. +

    + Every vector in the vector space can be expressed as a linear combination of the vectors in S. +

    +
    2. +
  2. +

    + For each vector \vec{v} in the vector space, there is only one way to write it as a linear combination + of the vectors in S. +

    +
    3. +
+ +

These two properties may be expressed more succintly as the statement "Every vector in V can be expressed + uniquely as a linear combination of the vectors in S". +

+ + + + +

In terms of a vector equation, a set S=\left\{\vec{v}_1,\ldots,\vec{v}_n\right\} is a basis of a vector space if the vector equation + x_1 \vec{v_1}+\cdots+x_n\vec{v_n}=\vec{w} + has a unique solution for every vector \vec{w} in the vector space.

+

Put another way, a basis may be thought of as a minimal set of building blocks that can be used to construct + any other vector of the vector space.

+ + + +

+ Let S be a basis () for a vector space. Then... +

    +

  1. the set S must both span the vector space and be linearly independent.

    2. +

  2. the set S must span the vector space but could be linearly dependent.

    3. +

  3. the set S must be linearly independent but could fail to span the vector space.

    4. +

  4. the set S could fail to span the vector space and could be linearly dependent.

    5. +
+

+ + + +

+The vectors + + +\hat i &= (1,0,0)=\left[\begin{array}{c}1 \\ 0 \\ 0 \\ \end{array}\right] & +\hat j &= (0,1,0)=\left[\begin{array}{c}0 \\ 1 \\ 0 \end{array}\right] & +\hat k &=(0,0,1)= \left[\begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] + + +form a basis \{\hat i,\hat j,\hat k\} used frequently in +multivariable calculus. +

+

+Find the unique linear combination of these vectors + + \unknown\hat i+\unknown\hat j+\unknown\hat k + +that equals the vector + + (3,-2,4)=\left[\begin{array}{c}3 \\ -2 \\ 4\end{array}\right] + + in xyz space. +

+ + + + +

+ The standard basisbasisstandard of \IR^n is the set \{\vec{e}_1, \ldots, \vec{e}_n\} where + + + \vec{e}_1 &= \left[\begin{array}{c}1 \\ 0 \\ 0 \\ \vdots \\ 0 \\ 0 \end{array}\right] & + \vec{e}_2 &= \left[\begin{array}{c}0 \\ 1 \\ 0 \\ \vdots \\ 0 \\ 0 \end{array}\right] & + \cdots & & + \vec{e}_n = \left[\begin{array}{c}0 \\ 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{array}\right] + + . +

+

+In particular, the standard basis for \mathbb R^3 is \{\vec e_1,\vec e_2,\vec e_3\}=\{\hat i,\hat j,\hat k\}. +

+ + + + + +

+ Take the RREF of an appropriate matrix to determine + if each of the following sets is a basis for \IR^4. +

+ + +

+ + \left\{ + \left[\begin{array}{c}1\\0\\0\\0\end{array}\right], + \left[\begin{array}{c}0\\1\\0\\0\end{array}\right], + \left[\begin{array}{c}0\\0\\1\\0\end{array}\right], + \left[\begin{array}{c}0\\0\\0\\1\end{array}\right] + \right\} + +

    +
  1. +

    +A basis, because it both spans \IR^4 and is linearly independent. +

    +
    2. +
  2. +

    +Not a basis, because while it spans \IR^4, it is linearly dependent. +

    +
    3. +
  3. +

    +Not a basis, because while it is linearly independent, it fails to span \IR^4. +

    +
    4. +
  4. +

    +Not a basis, because not only does it fail to span \IR^4, it's also linearly dependent. +

    +
    5. +
+

+ + +

+ + \left\{ + \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], + \left[\begin{array}{c}2\\0\\0\\3\end{array}\right], + \left[\begin{array}{c}4\\3\\0\\2\end{array}\right], + \left[\begin{array}{c}-3\\0\\1\\3\end{array}\right] + \right\} + +

    +
  1. +

    +A basis, because it both spans \IR^4 and is linearly independent. +

    +
    2. +
  2. +

    +Not a basis, because while it spans \IR^4, it is linearly dependent. +

    +
    3. +
  3. +

    +Not a basis, because while it is linearly independent, it fails to span \IR^4. +

    +
    4. +
  4. +

    +Not a basis, because not only does it fail to span \IR^4, it's also linearly dependent. +

    +
    5. +
+

+ + +

+ + \left\{ + \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], + \left[\begin{array}{c}2\\0\\0\\3\end{array}\right], + \left[\begin{array}{c}3\\13\\7\\16\end{array}\right], + \left[\begin{array}{c}-1\\10\\7\\14\end{array}\right], + \left[\begin{array}{c}4\\3\\0\\2\end{array}\right] + \right\} + +

    +
  1. +

    +A basis, because it both spans \IR^4 and is linearly independent. +

    +
    2. +
  2. +

    +Not a basis, because while it spans \IR^4, it is linearly dependent. +

    +
    3. +
  3. +

    +Not a basis, because while it is linearly independent, it fails to span \IR^4. +

    +
    4. +
  4. +

    +Not a basis, because not only does it fail to span \IR^4, it's also linearly dependent. +

    +
    5. +
+

+ + +

+ + \left\{ + \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], + \left[\begin{array}{c}4\\3\\0\\2\end{array}\right], + \left[\begin{array}{c}-3\\0\\1\\3\end{array}\right], + \left[\begin{array}{c}3\\6\\1\\5\end{array}\right] + \right\} + +

    +
  1. +

    +A basis, because it both spans \IR^4 and is linearly independent. +

    +
    2. +
  2. +

    +Not a basis, because while it spans \IR^4, it is linearly dependent. +

    +
    3. +
  3. +

    +Not a basis, because while it is linearly independent, it fails to span \IR^4. +

    +
    4. +
  4. +

    +Not a basis, because not only does it fail to span \IR^4, it's also linearly dependent. +

    +
    5. +
+

+ + +

+ + \left\{ + \left[\begin{array}{c}5\\3\\0\\-1\end{array}\right], + \left[\begin{array}{c}-2\\1\\0\\3\end{array}\right], + \left[\begin{array}{c}4\\5\\1\\3\end{array}\right] + \right\} + +

    +
  1. +

    +A basis, because it both spans \IR^4 and is linearly independent. +

    +
    2. +
  2. +

    +Not a basis, because while it spans \IR^4, it is linearly dependent. +

    +
    3. +
  3. +

    +Not a basis, because while it is linearly independent, it fails to span \IR^4. +

    +
    4. +
  4. +

    +Not a basis, because not only does it fail to span \IR^4, it's also linearly dependent. +

    +
    5. +
+

+ + + + + + + + +

+ If \{\vec v_1,\vec v_2,\vec v_3,\vec v_4\} is a basis for + \IR^4, that means \RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4] + has a pivot in every row (because it spans), and has a pivot in every column + (because it's linearly independent). +

+

+ What is \RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4]? + + \RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4] + = + \left[\begin{array}{cccc} + \unknown & \unknown & \unknown & \unknown \\ + \unknown & \unknown & \unknown & \unknown \\ + \unknown & \unknown & \unknown & \unknown \\ + \unknown & \unknown & \unknown & \unknown \\ + \end{array}\right] + +

+ + + + + +

+ The set \{\vec v_1,\dots,\vec v_m\} is a basis for \IR^n if and + only if m=n and + \RREF[\vec v_1\,\dots\,\vec v_n]= + \left[\begin{array}{cccc} + 1&0&\dots&0\\ + 0&1&\dots&0\\ + \vdots&\vdots&\ddots&\vdots\\ + 0&0&\dots&1 + \end{array}\right] + . +

+

+ That is, a basis for \IR^n must have exactly n vectors and + its square matrix must row-reduce to the so-called identity matrixidentity matrix + containing all zeros except for a downward diagonal of ones. + (We will learn where the identity matrix gets its name in a later module.) +

+ + + + + + + Cool Down + + +

+ Let S denote a set of vectors in \IR^n. Without referring to your Activity Book, write down: +

+ + + +

+ The definition of what it means for S to be linearly independent. +

+ + + + +

+ The definition of what it means for S to span \IR^n. +

+ + + + +

+ The definition of what it means for S to be a basis for \IR^n. +

+ + + + + +

+ You are going on a trip and need to pack. Let S denote the set of items that you are packing in your suitcase. +

+ + + +

+ Give an example of such a set of items S that you would say "spans" everything you need, but is linearly dependent. +

+ + + + +

+ Give an example of such a set of items S that is linearly independent, but does not "span" everything you need. +

+ + + + +

+ Give an example of such a set S that you might reasonably consider to be a "basis" for what you need? +

+ + + + + + + + Videos +
+ + Video: Verifying that a set of vectors is a basis of a vector space +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + +
    +
  • What is a basis for M_{2,2}?
    • +
  • What about M_{3,3}?
    • +
  • Could we write each of these in a way that looks like the standard basis vectors in \mathbb{R}^m for some m? Make a conjecture about the relationship between these spaces of matrices and standard Eulidean space.
    • +
+ + + + +Recall our earlier definition of symmetric matrices. Find a basis for each of the following: +
    +
  • The space of 2 \times 2 symmetric matrices.
    • +
  • The space of 3 \times 3 symmetric matrices.
    • +
  • The space of n \times n symmetric matrices.
    • +
+ + + + +Must a basis for the space P_2, the space of all quadratic polynomials, contain a polynomial of each degree less than or equal to 2? Generalize your result to polynomials of arbitrary degree. + + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/02-EV/06.ptx b/linear-algebra/source/02-EV/06.ptx new file mode 100644 index 00000000..713ff0ce --- /dev/null +++ b/linear-algebra/source/02-EV/06.ptx @@ -0,0 +1,483 @@ + +
+ Subspace Basis and Dimension (EV6) + + + + + + Warm Up + + +

+ Consider the set S of vectors in \IR^4 given by + S=\left\{\left[\begin{array}{c}2\\3\\0\\1\end{array}\right],\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right]\right\} +

+ + + +

+ Is the set S linearly independent or linearly dependent? +

+ + + + +

+ How would you describe the subspace \vspan{S} geometrically? +

+ + + + +

+ What do the spaces \vspan{S} and \IR^2 have in common? In what ways do they differ? +

+ + + + + +Class Activities + +

+Recall from section that a subspace of a vector space is +the result of spanning a set of vectors from that vector space. +

+

+Recall also that a linearly dependent set contains redundant vectors. For example, +only two of the three vectors in are needed to span +the planar subspace. +

+ + + + +

+ Consider the subspace of \IR^4 given by W=\vspan\left\{ + \left[\begin{array}{c}2\\3\\0\\1\end{array}\right], + \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], + \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], + \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right] + \right\} + . +

+ + +

+ Mark the column of \RREF\left[\begin{array}{cccc} + 2&2&2&1\\ + 3&0&-3&5\\ + 0&1&2&-1\\ + 1&-1&-3&0 + \end{array}\right] that shows that W's spanning set + is linearly dependent. +

+ + +

+What would be the result of removing the vector that gave us this column? +

    +
  1. The set still spans W, and remains linearly dependent.
    2. +
  2. The set still spans W, but is now also linearly independent.
    3. +
  3. The set no longer spans W, and remains linearly dependent.
    4. +
  4. The set no longer spans W, but is now linearly independent.
    5. +
+

+ + + + +rref([2,2,2,1; 3,0,-3,5; 0,1,2,-1; 1,-1,-3,0]) + + + + +

+ Let W be a subspace of a vector space. A basis for + W is a linearly independent set of vectors that spans W + (but not necessarily the entire vector space). +

+ + + +

+ So given a set S=\{\vec v_1,\dots,\vec v_m\}, + to compute a basis for the subspace \vspan S, + simply remove the vectors corresponding to the non-pivot columns of + \RREF[\vec v_1\,\dots\,\vec v_m]. + For example, since + + \RREF + \left[\begin{array}{cccc} + 1 & 2 & 0 & 1 \\ + 2 & 4 & -2 & 2 \\ + 3 & 6 & -2 & 1 \\ + \end{array}\right] + = + \left[\begin{array}{cccc} + \markedPivot{1} & 2 & 0 & 1 \\ + 0 & 0 & \markedPivot{1} & 1 \\ + 0 & 0 & 0 & 0 + \end{array}\right] + + the subspace + + W=\vspan\setList{ + \left[\begin{array}{c}1\\2\\3\end{array}\right], + \left[\begin{array}{c}2\\4\\6\end{array}\right], + \left[\begin{array}{c}0\\-2\\-2\end{array}\right], + \left[\begin{array}{c}1\\2\\1\end{array}\right] + } + + has + + \setList{ + \left[\begin{array}{c}1\\2\\3\end{array}\right], + \left[\begin{array}{c}0\\-2\\-2\end{array}\right] + } + + as a basis. +

+ + + + + + + + + + + + +

+ Find a basis for \vspan S where + +S=\left\{ +\left[\begin{array}{c}2\\3\\0\\1\end{array}\right], +\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], +\left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], +\left[\begin{array}{c}1\\5\\-1\\0\end{array}\right] +\right\} + . +

+ + +

+ Find a basis for \vspan T where + +T=\left\{ +\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], +\left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], +\left[\begin{array}{c}1\\5\\-1\\0\end{array}\right], +\left[\begin{array}{c}2\\3\\0\\1\end{array}\right] +\right\} + . +

+ + + + + + + +

+ Even though we found different bases for them, + \vspan S and \vspan T are exactly the same subspace of \IR^4, + since + + S=\left\{ + \left[\begin{array}{c}2\\3\\0\\1\end{array}\right], + \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], + \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], + \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right] + \right\} + = + \left\{ + \left[\begin{array}{c}2\\0\\1\\-1\end{array}\right], + \left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right], + \left[\begin{array}{c}1\\5\\-1\\0\end{array}\right], + \left[\begin{array}{c}2\\3\\0\\1\end{array}\right] + \right\}=T + . +

+

+Thus the basis for a subspace is not unique in general. +

+ + + + + +

+ Any non-trivial real vector space has infinitely-many different bases, but all + the bases for a given vector space are exactly the same size. +

+

+ For example, + + \setList{\vec e_1,\vec e_2,\vec e_3} + \text{ and } + \setList{ + \left[\begin{array}{c}1\\0\\0\end{array}\right], + \left[\begin{array}{c}0\\1\\0\end{array}\right], + \left[\begin{array}{c}1\\1\\1\end{array}\right] + } + \text{ and } + \setList{ + \left[\begin{array}{c}1\\0\\-3\end{array}\right], + \left[\begin{array}{c}2\\-2\\1\end{array}\right], + \left[\begin{array}{c}3\\-2\\5\end{array}\right] + } + + are all valid bases for \IR^3, and they all contain three vectors. +

+ + + + + +

+ The dimension of a vector space or subspace is equal to the size + of any basis for the vector space. +

+

+ As you'd expect, \IR^n has dimension n. + For example, \IR^3 has dimension 3 because any basis for \IR^3 + such as + + \setList{\vec e_1,\vec e_2,\vec e_3} + \text{ and } + \setList{ + \left[\begin{array}{c}1\\0\\0\end{array}\right], + \left[\begin{array}{c}0\\1\\0\end{array}\right], + \left[\begin{array}{c}1\\1\\1\end{array}\right] + } + \text{ and } + \setList{ + \left[\begin{array}{c}1\\0\\-3\end{array}\right], + \left[\begin{array}{c}2\\-2\\1\end{array}\right], + \left[\begin{array}{c}3\\-2\\5\end{array}\right] + } + + contains exactly three vectors. +

+ + + + + +

Consider the following subspace W of \mathbb R^4: W=\mathrm{span}\,\left\{ \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ 0 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} -3 \\ 1 \\ -5 \\ 5 \end{array}\right] , \left[\begin{array}{c} 12 \\ -3 \\ 15 \\ -18 \end{array}\right] \right\}.

+ + + +

Explain and demonstrate how to find a basis of W.

+ + + + +

Explain and demonstrate how to find the dimension of W.

+ + + + + + + +

The dimension of a subspace may be found by doing what with an appropriate RREF matrix? +

    +
  1. Count the rows.
    2. +
  2. Count the non-pivot columns.
    3. +
  3. Count the pivots.
    4. +
  4. Add the number of pivot rows and pivot columns.
    5. +
+

+ + + + + + Cool Down + + +

+ In , we found a basis for the subspace + W=\vspan\setList{ + \left[\begin{array}{c}1\\2\\3\end{array}\right], + \left[\begin{array}{c}2\\4\\6\end{array}\right], + \left[\begin{array}{c}0\\-2\\-2\end{array}\right], + \left[\begin{array}{c}1\\2\\1\end{array}\right] + }. + To do so, we use the results of the calculation: + + \RREF + \left[\begin{array}{cccc} + 1 & 2 & 0 & 1 \\ + 2 & 4 & -2 & 2 \\ + 3 & 6 & -2 & 1 \\ + \end{array}\right] + = + \left[\begin{array}{cccc} + \markedPivot{1} & 2 & 0 & 1 \\ + 0 & 0 & \markedPivot{1} & 1 \\ + 0 & 0 & 0 & 0 + \end{array}\right] + + to conclude that the set \setList{ + \left[\begin{array}{c}1\\2\\3\end{array}\right], + \left[\begin{array}{c}0\\-2\\-2\end{array}\right] + }, the set of vectors corresponding to the pivot columns of the RREF, is a basis for W. +

+ + + +

+ Explain why neither of the vectors + \left[\begin{array}{c}1\\0\\0\end{array}\right], + \left[\begin{array}{c}0\\1\\0\end{array}\right] are elements of W. +

+ + + + +

+ Explain why this shows that, in general, when we calculate a basis for W=\vspan\{\vec{v}_1,\dots, \vec{v}_n\}, the pivot columns of \RREF[\vec{v}_1\dots \vec{v}_n] themselves do not form a basis for W. +

+ + + + + + Videos +
+ + Video: Finding a basis of a subspace and computing the dimension of a subspace +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + + Prove each of the following statements is true. +
    +
  • If \{\vec{b}_1, \vec{b}_2,\ldots, \vec{b}_m\} and \{\vec{c}_1,\vec{c}_2,\ldots,\vec{c}_n\} are each a basis for a vector space V, then m=n.
    • +
  • If \{\vec{v}_1,\vec{v}_2\ldots, \vec{v}_n\} is linearly independent, then so is \{\vec{v}_1,\vec{v}_1 + \vec{v}_2, \ldots, \vec{v}_1 + \vec{v}_2 + \cdots + \vec{v}_n\}.
    • +
  • Let V be a vector space of dimension n, and \vec{v} \in V. Then there exists a basis for V which contains \vec{v}.
    • +
+ + + + + Suppose we have the set of all function f:S \rightarrow \mathbb{R}. We claim that this is a vector space under the usual operation of function addition and scalar multiplication. What is the dimension of this space for each choice of S below: + +
    +
  • S = \{1\}
    • +
  • S = \{1,2\}
    • +
  • S = \{1,2,\ldots ,n\}
    • +
  • S = \mathbb{R}
    • +
+ + + + Suppose you have the vector space V = \left\{\left(\begin{array}{c}x\\y\\z\end{array}\right)\in \mathbb{R}^3: x + y + z = 1\right\} with the operations \left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right) \oplus \left(\begin{array}{c}x_2\\y_2\\z_2\end{array}\right) = \left(\begin{array}{c}x_1 + x_2 - 1\\y_1 + y_2\\z_1+z_2\end{array}\right) \mbox{ and } \alpha\odot\left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right) = \left(\begin{array}{c}\alpha x_1 - \alpha +1\\\alpha y_1\\\alpha z_1\end{array}\right). Find a basis for V and determine it's dimension. + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/02-EV/07.ptx b/linear-algebra/source/02-EV/07.ptx new file mode 100644 index 00000000..884dfde4 --- /dev/null +++ b/linear-algebra/source/02-EV/07.ptx @@ -0,0 +1,421 @@ + +
+ Homogeneous Linear Systems (EV7) + + + + + + Warmup + +

+ Recall from that a homogeneoushomogeneous system of linear equations is one of the form: + + + a_{11}x_1 &\,+\,& a_{12}x_2 &\,+\,& \dots &\,+\,& a_{1n}x_n &\,=\,& 0 + + + a_{21}x_1 &\,+\,& a_{22}x_2 &\,+\,& \dots &\,+\,& a_{2n}x_n &\,=\,& 0 + + + \vdots& &\vdots& && &\vdots&&\vdots + + + a_{m1}x_1 &\,+\,& a_{m2}x_2 &\,+\,& \dots &\,+\,& a_{mn}x_n &\,=\,& 0 + + +

+

+ This system is equivalent to the vector equation: + x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0} + and the augmented matrix: + + \left[\begin{array}{cccc|c} + a_{11} & a_{12} & \cdots & a_{1n} & 0\\ + a_{21} & a_{22} & \cdots & a_{2n} & 0\\ + \vdots & \vdots & \ddots & \vdots & \vdots\\ + a_{m1} & a_{m2} & \cdots & a_{mn} & 0 + \end{array}\right]. + +

+ + + + +

+ In , we observed that if + x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0} + is a homogenous vector equation, then: +

    +
  • +

    + The zero vector \vec{0} is a solution; +

    +
    • +
  • +

    + The sum of any two solutions is again a solution; +

    +
    • +
  • +

    + Multiplying a solution by a scalar produces another solution. +

    +
    • +
+

+ +

+ Based on this recollection, which of the following best describes the solution set to the homogenous equation? +

    +
  1. +

    + A basis for \IR^n. +

    +
    2. +
  2. +

    + A subspace of \IR^n. +

    +
    3. +
  3. +

    + All of \IR^n. +

    +
    4. +
  4. +

    + The empty set. +

    +
    5. +
+

+ + + +Class Activities + + + +

+Consider the homogeneous system of equations + + +x_1&\,+\,&2x_2&\,\,& &\,+\,& x_4 &=& 0 + + +2x_1&\,+\,&4x_2&\,-\,&x_3 &\,-\,&2 x_4 &=& 0 + + +3x_1&\,+\,&6x_2&\,-\,&x_3 &\,-\,& x_4 &=& 0 + + +

+ + +

+Find its solution set (a subspace of \IR^4). +

+ + +

+Rewrite this solution space in the form \setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right] }{a,b \in \IR}. +

+ + +

+Rewrite this solution space in the form \vspan\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\}. +

+ + +

+Which of these choices best describes the set of two vectors +\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\} +used in this span? +

    +
  1. +

    + The set is linearly dependent. +

    +
    2. +
  2. +

    + The set is linearly independent. +

    +
    3. +
  3. +

    + The set spans all of \IR^4. +

    +
    4. +
  4. +

    + The set fails to span the solution space. +

    +
    5. +
+

+ + + + + + + + +

+ The coefficients of the free variables in the solution space of a linear system + always yield linearly independent vectors that span the solution space. +

+

+ Thus if + + \setBuilder{ + a \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right] + + b \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right] + }{ + a,b \in \IR + } = \vspan\left\{ \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right], + \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right] \right\} + + is the solution space for a homogeneous system, then + + \setList{ + \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right], + \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right] + } + + is a basis for the solution space. +

+ + + + + + + +

+Consider the homogeneous system of equations + + + 2x_1&\,+\,&4x_2&\,+\,& 2x_3&\,-\,&4x_4 &=& 0 + + +-2x_1&\,-\,&4x_2&\,+\,&x_3 &\,+\,& x_4 &=& 0 + + +3x_1&\,+\,&6x_2&\,-\,&x_3 &\,-\,&4 x_4 &=& 0 + + +

+

+Find a basis for its solution space. +

+ + + + + + + + +

+Consider the homogeneous vector equation + + x_1 \left[\begin{array}{c} 2 \\ -2 \\ 3 \end{array}\right]+ + x_2 \left[\begin{array}{c} 4 \\ -4 \\ 6 \end{array}\right]+ + x_3 \left[\begin{array}{c} 2 \\ 1 \\ -1 \end{array}\right]+ + x_4 \left[\begin{array}{c} -4 \\ 1 \\ -4 \end{array}\right]= + \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right] + +

+

+Find a basis for its solution space. +

+ + + + + + + + +

+Consider the homogeneous system of equations + + +x_1&\,-\,&3x_2&\,+\,& 2x_3 &=& 0 + + +2x_1&\,+\,&6x_2&\,+\,&4x_3 &=& 0 + + +x_1&\,+\,&6x_2&\,-\,&4x_3 &=& 0 + + +

+ + + +

+Find its solution space. +

+ + + + +

+Which of these is the best choice of basis for this +solution space? +

    +
  1. \{\}
    2. +
  2. \{\vec 0\}
    3. +
  3. The basis does not exist
    4. +
+

+ + + + + + + + + +

+To create a computer-animated film, an animator first models a scene +as a subset of \mathbb R^3. Then to transform this three-dimensional +visual data for display on a two-dimensional movie screen or television set, +the computer could apply a linear tranformation that maps visual information +at the point (x,y,z)\in\mathbb R^3 onto the pixel located at +(x+y,y-z)\in\mathbb R^2. +

+ + + +

+What homoegeneous linear system describes the positions (x,y,z) +within the original scene that would be aligned with the +pixel (0,0) on the screen? +

+ + + + +

+Solve this system to describe these locations. +

+ + + + + + + + + + + Cool Down + + +

+ Let S=\setList{ + \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right], + \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right], + \left[\begin{array}{c} 1 \\ 0 \\ -2 \\ 3 \end{array}\right] + } and A=\left[\begin{array}{ccc} + -2 & -1 &1\\ + 1 & 0 &0\\ + 0 & -4 &-2\\ + 0 & 1 &3 + \end{array}\right]; note that + \RREF(A)=\left[\begin{array}{ccc} + 1 & 0 &0\\ + 0 & 1 &0\\ + 0 & 0 &1\\ + 0 & 0 &0 + \end{array}\right]. + The following statements are all invalid for at least one reason. Determine what makes them invalid and, suggest alternative valid statements that the author may have meant instead. +

+ + + +

+ The matrix A is linearly independent because \RREF(A) has a pivot in each column. +

+ + + + +

+ The matrix A does not span \IR^4 because \RREF(A) has a row of zeroes. +

+ + + + +

+ The set of vectors S spans. +

+ + + + +

+ The set of vectors S is a basis. +

+ + + + + + + + Videos +
+ + Video: Polynomial and matrix calculations +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + +

An n \times n matrix M is non-singularnon-singular if the associated homogeneous system with coefficient matrix M is consistent with one solution. Assume the matrices in the writing explorations in this section are all non-singular. +

+
    +
  • Prove that the reduced row echelon form of M is the identity matrix. +
    • +
  • +Prove that, for any column vector \vec{b} = \left[\begin{array}{c}b_1\\b_2\\ \vdots \\b_n \end{array}\right], the system of equations given by \left[\begin{array}{c|c}M & \vec{b}\end{array} + \right] has a unique solution. +
    • + +
  • + Prove that the columns of M form a basis for \mathbb{R}^n.
    • + +
  • + Prove that the rank of M is n.
    • +
+ + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/02-EV/main.ptx b/linear-algebra/source/02-EV/main.ptx new file mode 100644 index 00000000..c730df71 --- /dev/null +++ b/linear-algebra/source/02-EV/main.ptx @@ -0,0 +1,13 @@ + + + Euclidean Vectors (EV) + + + + + + + + + + diff --git a/linear-algebra/source/02-EV/outcomes/01.ptx b/linear-algebra/source/02-EV/outcomes/01.ptx new file mode 100644 index 00000000..96f669c0 --- /dev/null +++ b/linear-algebra/source/02-EV/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Determine if a Euclidean vector can be written as a linear combination of a given set of Euclidean vectors by solving an appropriate vector equation. +

\ No newline at end of file diff --git a/linear-algebra/source/02-EV/outcomes/02.ptx b/linear-algebra/source/02-EV/outcomes/02.ptx new file mode 100644 index 00000000..68ed3c8b --- /dev/null +++ b/linear-algebra/source/02-EV/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Determine if a set of Euclidean vectors spans \IR^n by solving appropriate vector equations. +

\ No newline at end of file diff --git a/linear-algebra/source/02-EV/outcomes/03.ptx b/linear-algebra/source/02-EV/outcomes/03.ptx new file mode 100644 index 00000000..ef44b3d0 --- /dev/null +++ b/linear-algebra/source/02-EV/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Determine if a subset of \IR^n is a subspace or not. +

\ No newline at end of file diff --git a/linear-algebra/source/02-EV/outcomes/04.ptx b/linear-algebra/source/02-EV/outcomes/04.ptx new file mode 100644 index 00000000..2a0a2b29 --- /dev/null +++ b/linear-algebra/source/02-EV/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Determine if a set of Euclidean vectors is linearly dependent or independent by solving an appropriate vector equation. +

\ No newline at end of file diff --git a/linear-algebra/source/02-EV/outcomes/05.ptx b/linear-algebra/source/02-EV/outcomes/05.ptx new file mode 100644 index 00000000..32f3af87 --- /dev/null +++ b/linear-algebra/source/02-EV/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+Explain why a set of Euclidean vectors is or is not a basis of \IR^n. +

\ No newline at end of file diff --git a/linear-algebra/source/02-EV/outcomes/06.ptx b/linear-algebra/source/02-EV/outcomes/06.ptx new file mode 100644 index 00000000..adf90cda --- /dev/null +++ b/linear-algebra/source/02-EV/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+Compute a basis for the subspace spanned by a given set of Euclidean vectors, and determine the dimension of the subspace. +

\ No newline at end of file diff --git a/linear-algebra/source/02-EV/outcomes/07.ptx b/linear-algebra/source/02-EV/outcomes/07.ptx new file mode 100644 index 00000000..15345309 --- /dev/null +++ b/linear-algebra/source/02-EV/outcomes/07.ptx @@ -0,0 +1,4 @@ + +

+Find a basis for the solution set of a homogeneous system of equations. +

\ No newline at end of file diff --git a/linear-algebra/source/02-EV/outcomes/main.ptx b/linear-algebra/source/02-EV/outcomes/main.ptx new file mode 100644 index 00000000..2590299b --- /dev/null +++ b/linear-algebra/source/02-EV/outcomes/main.ptx @@ -0,0 +1,32 @@ + +> + + +

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
  7. + +
    8. +
+ \ No newline at end of file diff --git a/linear-algebra/source/02-EV/outcomes/question.ptx b/linear-algebra/source/02-EV/outcomes/question.ptx new file mode 100644 index 00000000..f5cbc8d8 --- /dev/null +++ b/linear-algebra/source/02-EV/outcomes/question.ptx @@ -0,0 +1,4 @@ + +

+What is a space of Euclidean vectors? +

diff --git a/linear-algebra/source/02-EV/readiness.ptx b/linear-algebra/source/02-EV/readiness.ptx new file mode 100644 index 00000000..d9b4199a --- /dev/null +++ b/linear-algebra/source/02-EV/readiness.ptx @@ -0,0 +1,43 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Use set builder notation to describe sets of vectors.

    +
      +
    • +

      Review: YouTube

      +
      • +
    +
    2. +
  2. +

    Add Euclidean vectors and multiply Euclidean vectors by scalars.

    +
      +
    • +

      Review: Khan Academy (1) + (2)

      +
      • +
    +
    3. +
  3. +

    Perform basic manipulations of augmented matrices and linear + systems.

    +
      +
    • +

      + Review: + , + , + +

      +
      • +
    +
    4. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/02-EV/samples/01.ptx b/linear-algebra/source/02-EV/samples/01.ptx new file mode 100644 index 00000000..853859bc --- /dev/null +++ b/linear-algebra/source/02-EV/samples/01.ptx @@ -0,0 +1,214 @@ + +EV1 + +
    +
  1. +

    +Write a statement involving the solutions of a vector equation +that's equivalent to each claim below. +

    +
      +
    • +

      +\left[\begin{array}{c} +-13 \\ +3 \\ +-13 +\end{array}\right]is a linear combination of the vectors \left[\begin{array}{c} +1 \\ +0 \\ +1 +\end{array}\right] , \left[\begin{array}{c} +2 \\ +0 \\ +2 +\end{array}\right] , \left[\begin{array}{c} +3 \\ +0 \\ +3 +\end{array}\right] , \text{ and } \left[\begin{array}{c} +-5 \\ +1 \\ +-5 +\end{array}\right]. +

      +
      • +
    • +

      +\left[\begin{array}{c} +-13 \\ +3 \\ +-15 +\end{array}\right]is a linear combination of the vectors \left[\begin{array}{c} +1 \\ +0 \\ +1 +\end{array}\right] , \left[\begin{array}{c} +2 \\ +0 \\ +2 +\end{array}\right] , \left[\begin{array}{c} +3 \\ +0 \\ +3 +\end{array}\right] , \text{ and } \left[\begin{array}{c} +-5 \\ +1 \\ +-5 +\end{array}\right]. +

      +
      • +
    +
    2. +
  2. +

    +Use these statements to determine if each vector +is or is not a linear combination. +If it is, give an example of such a linear combination. +

    +
    3. +
+ + + + + \ No newline at end of file diff --git a/linear-algebra/source/02-EV/samples/02.ptx b/linear-algebra/source/02-EV/samples/02.ptx new file mode 100644 index 00000000..691f3853 --- /dev/null +++ b/linear-algebra/source/02-EV/samples/02.ptx @@ -0,0 +1,151 @@ + +EV2 + +
    +
  1. +

    +Write a statement involving the solutions of a vector equation that's equivalent to each +claim below. +

    +
      +
    • +

      +The set of vectors \left\{ \left[\begin{array}{c} +1 \\ +-1 \\ +2 \\ +0 +\end{array}\right] , \left[\begin{array}{c} +3 \\ +-2 \\ +3 \\ +3 +\end{array}\right] , \left[\begin{array}{c} +10 \\ +-7 \\ +11 \\ +9 +\end{array}\right] , \left[\begin{array}{c} +-6 \\ +3 \\ +-3 \\ +-9 +\end{array}\right] \right\} spans \mathbb R^4. +

      +
      • +
    • +

      +The set of vectors \left\{ \left[\begin{array}{c} +1 \\ +-1 \\ +2 \\ +0 +\end{array}\right] , \left[\begin{array}{c} +3 \\ +-2 \\ +3 \\ +3 +\end{array}\right] , \left[\begin{array}{c} +10 \\ +-7 \\ +11 \\ +9 +\end{array}\right] , \left[\begin{array}{c} +-6 \\ +3 \\ +-3 \\ +-9 +\end{array}\right] \right\} does not span \mathbb R^4. +

      +
      • +
    +
    2. +
  2. +

    +Explain how to determine which of these statements is true. +

    +
    3. +
+ + +

The set of vectors \left\{ \left[\begin{array}{c} +1 \\ +-1 \\ +2 \\ +0 +\end{array}\right] , \left[\begin{array}{c} +3 \\ +-2 \\ +3 \\ +3 +\end{array}\right] , \left[\begin{array}{c} +10 \\ +-7 \\ +11 \\ +9 +\end{array}\right] , \left[\begin{array}{c} +-6 \\ +3 \\ +-3 \\ +-9 +\end{array}\right] \right\} spans \mathbb{R}^4 exactly when the vector equation + x_1 \left[\begin{array}{c} +1 \\ +-1 \\ +2 \\ +0 +\end{array}\right] +x_2 \left[\begin{array}{c} +3 \\ +-2 \\ +3 \\ +3 +\end{array}\right] +x_3 \left[\begin{array}{c} +10 \\ +-7 \\ +11 \\ +9 +\end{array}\right] +x_4 \left[\begin{array}{c} +-6 \\ +3 \\ +-3 \\ +-9 +\end{array}\right] =\vec{v} has a solution for all \vec{v} \in \mathbb{R}^4. If there is some vector \vec{v} \in \mathbb{R}^4 for which this vector equation has no solution, then the set does not span \mathbb{R}^4. To answer this, we compute + +\mathrm{RREF}\, \left[\begin{array}{cccc} +1 & 3 & 10 & -6 \\ +-1 & -2 & -7 & 3 \\ +2 & 3 & 11 & -3 \\ +0 & 3 & 9 & -9 +\end{array}\right] = \left[\begin{array}{cccc} +1 & 0 & 1 & 3 \\ +0 & 1 & 3 & -3 \\ +0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 +\end{array}\right] + . + We see that for some \vec{v} \in \mathbb{R}^4, this vector equation will not have a solution, so the set of vectors \left\{ \left[\begin{array}{c} +1 \\ +-1 \\ +2 \\ +0 +\end{array}\right] , \left[\begin{array}{c} +3 \\ +-2 \\ +3 \\ +3 +\end{array}\right] , \left[\begin{array}{c} +10 \\ +-7 \\ +11 \\ +9 +\end{array}\right] , \left[\begin{array}{c} +-6 \\ +3 \\ +-3 \\ +-9 +\end{array}\right] \right\} + does not span +\mathbb{R}^4. +

+ + diff --git a/linear-algebra/source/02-EV/samples/03.ptx b/linear-algebra/source/02-EV/samples/03.ptx new file mode 100644 index 00000000..f9b49e08 --- /dev/null +++ b/linear-algebra/source/02-EV/samples/03.ptx @@ -0,0 +1,129 @@ + + +EV3 + +

+Consider the following two sets of Euclidean vectors. + + W = \setBuilder{\left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] }{x+y=3z+2w} +\hspace{3em} + U = \setBuilder{\left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right]}{x+y=3z+w^2} + +Explain why one of these sets is a subspace of \IR^3, and +why the other is not. +

+ + +

+To show that W is a subspace, first note that it is nonempty as +\left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \in W, since 0+0=3(0)+3(0). Then let + + \vec v=\left[\begin{array}{c} x_1 \\y_1 \\ z_1 \\ w_1 \end{array}\right]\in W + and + + \vec w=\left[\begin{array}{c} x_2 \\y_2 \\ z_2 \\ w_2 \end{array}\right] \in W +, +so we know that x_1+y_1=3z_1+2w_1 and x_2+y_2=3z_2+2w_2. +

+

+Consider + +\left[\begin{array}{c} x_1 \\y_1 \\ z_1 \\ w_1\end{array}\right] ++\left[\begin{array}{c} x_2 \\y_2 \\ z_2 \\ w_2 \end{array}\right] +=\left[\begin{array}{c} x_1+x_2 \\y_1+y_2 \\ z_1+z_2 \\w_1+w_2 \end{array}\right] +. +To see if \vec{v}+\vec{w} \in W, we need to check if (x_1+x_2)+(y_1+y_2) = 3(z_1+z_2)+2(w_1+w_2). +We compute + + + (x_1+x_2)+(y_1+y_2) &= (x_1+y_1)+(x_2+y_2) &\text{by regrouping} + + + &= (3z_1+2w_1)+(3z_2+2w_2) & \text{since \vec{v},\vec{w} \in W} + + + &=3(z_1+z_2)+2(w_1+w_2) & \text{by regrouping.} + + +Thus \vec v+\vec w\in W, so W is closed under vector addition. +

+

+Now consider + +c\vec v +=\left[\begin{array}{c} cx_1 \\cy_1 \\ cz_1 \\ cw_1 \end{array}\right] +. +Similarly, to check that c\vec{v} \in W, we need to check if cx_1+cy_1=3(cz_1)+2(cw_1), so we compute + + +cx_1+cy_1 & = c(x_1+y_1) &\text{by factoring} + + +&=c(3z_1+2w_1) &\text{since \vec{v} \in W} + + +&=3(cz_1)+2(cw_1) &\text{by regrouping} + + +and we see that c\vec v\in W, so W is closed under scalar +multiplication. Therefore W is a subspace of \IR^3. +

+

+Now, to show U is not a subspace, we will show that it is not closed under vector addition. +

+ + + \ No newline at end of file diff --git a/linear-algebra/source/02-EV/samples/04.ptx b/linear-algebra/source/02-EV/samples/04.ptx new file mode 100644 index 00000000..2b2442f9 --- /dev/null +++ b/linear-algebra/source/02-EV/samples/04.ptx @@ -0,0 +1,131 @@ + + +EV4 + +
    +
  1. +

    +Write a statement involving the solutions of a vector equation that's equivalent to each +claim below. +

    +
      +
    • +

      +The set of vectors \left\{ \left[\begin{array}{c} +1 \\ +3 \\ +4 \\ +-4 +\end{array}\right] , \left[\begin{array}{c} +-1 \\ +-3 \\ +-4 \\ +4 +\end{array}\right] , \left[\begin{array}{c} +0 \\ +1 \\ +3 \\ +-3 +\end{array}\right] \right\} is linearly independent. +

      +
      • +
    • +

      +The set of vectors \left\{ \left[\begin{array}{c} +1 \\ +3 \\ +4 \\ +-4 +\end{array}\right] , \left[\begin{array}{c} +-1 \\ +-3 \\ +-4 \\ +4 +\end{array}\right] , \left[\begin{array}{c} +0 \\ +1 \\ +3 \\ +-3 +\end{array}\right] \right\} is linearly dependent. +

      +
      • +
    +
    2. +
  2. +

    +Explain how to determine which of these statements is true. +

    +
    3. +
+ + +

The set of vectors \left\{ \left[\begin{array}{c} +1 \\ +3 \\ +4 \\ +-4 +\end{array}\right] , \left[\begin{array}{c} +-1 \\ +-3 \\ +-4 \\ +4 +\end{array}\right] , \left[\begin{array}{c} +0 \\ +1 \\ +3 \\ +-3 +\end{array}\right] \right\} is linearly independent exactly when the vector equation + x_1 \left[\begin{array}{c} +1 \\ +3 \\ +4 \\ +-4 +\end{array}\right] +x_2 \left[\begin{array}{c} +-1 \\ +-3 \\ +-4 \\ +4 +\end{array}\right] +x_3 \left[\begin{array}{c} +0 \\ +1 \\ +3 \\ +-3 +\end{array}\right] =\left[\begin{array}{c}0 \\ 0 \\ 0 \\ 0 \end{array}\right] has no non-trivial (i.e. nonzero) solutions. + The set is linearly dependent when there exists a nontrivial (i.e. nonzero) solution. We compute + +\mathrm{RREF}\, \left[\begin{array}{ccc} +1 & -1 & 0 \\ +3 & -3 & 1 \\ +4 & -4 & 3 \\ +-4 & 4 & -3 +\end{array}\right] = \left[\begin{array}{ccc} +1 & -1 & 0 \\ +0 & 0 & 1 \\ +0 & 0 & 0 \\ +0 & 0 & 0 +\end{array}\right] + . +

+

+ Thus, this vector equation has a solution set \left\{ \left[\begin{array}{c}a \\ a \\ 0 \end{array}\right]\ \middle|\ a \in \mathbb{R}\right\}. + Since there are nontrivial solutions, we conclude that the set of vectors \left\{ \left[\begin{array}{c} +1 \\ +3 \\ +4 \\ +-4 +\end{array}\right] , \left[\begin{array}{c} +-1 \\ +-3 \\ +-4 \\ +4 +\end{array}\right] , \left[\begin{array}{c} +0 \\ +1 \\ +3 \\ +-3 +\end{array}\right] \right\} is linearly + +dependent. +

+ + \ No newline at end of file diff --git a/linear-algebra/source/02-EV/samples/05.ptx b/linear-algebra/source/02-EV/samples/05.ptx new file mode 100644 index 00000000..a988c1be --- /dev/null +++ b/linear-algebra/source/02-EV/samples/05.ptx @@ -0,0 +1,138 @@ + +EV5 + +
    +
  1. +

    +Write a statement involving spanning and independence properties +that's equivalent to each claim below. +

    +
      +
    • +

      +The set of vectors \left\{ \left[\begin{array}{c} +1 \\ +3 \\ +4 \\ +-4 +\end{array}\right] , \left[\begin{array}{c} +0 \\ +1 \\ +3 \\ +-3 +\end{array}\right] , \left[\begin{array}{c} +3 \\ +11 \\ +18 \\ +-18 +\end{array}\right] , \left[\begin{array}{c} +-2 \\ +-7 \\ +-11 \\ +11 +\end{array}\right] \right\} is a basis +of \mathbb{R}^4. +

      +
      • +
    • +

      +The set of vectors \left\{ \left[\begin{array}{c} +1 \\ +3 \\ +4 \\ +-4 +\end{array}\right] , \left[\begin{array}{c} +0 \\ +1 \\ +3 \\ +-3 +\end{array}\right] , \left[\begin{array}{c} +3 \\ +11 \\ +18 \\ +-18 +\end{array}\right] , \left[\begin{array}{c} +-2 \\ +-7 \\ +-11 \\ +11 +\end{array}\right] \right\} is not a basis +of \mathbb{R}^4. +

      +
      • +
    +
    2. +
  2. +

    +Explain how to determine which of these statements is true. +

    +
    3. +
+ + +

The set of vectors \left\{ \left[\begin{array}{c} +1 \\ +3 \\ +4 \\ +-4 +\end{array}\right] , \left[\begin{array}{c} +0 \\ +1 \\ +3 \\ +-3 +\end{array}\right] , \left[\begin{array}{c} +3 \\ +11 \\ +18 \\ +-18 +\end{array}\right] , \left[\begin{array}{c} +-2 \\ +-7 \\ +-11 \\ +11 +\end{array}\right] \right\} is a basis +of \mathbb{R}^4 exactly when it is linearly independent and the set spans \mathbb{R}^4. + If it is either linearly dependent, or the set does not span \mathbb{R}^4, then the set is not a basis. +

+

To answer this, we compute + +\mathrm{RREF}\, \left[\begin{array}{cccc} +1 & 0 & 3 & -2 \\ +3 & 1 & 11 & -7 \\ +4 & 3 & 18 & -11 \\ +-4 & -3 & -18 & 11 +\end{array}\right] = \left[\begin{array}{cccc} +1 & 0 & 3 & -2 \\ +0 & 1 & 2 & -1 \\ +0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 +\end{array}\right] + . +

+

We see that this set of vectors is linearly dependent, so therefore +the set of vectors \left\{ \left[\begin{array}{c} +1 \\ +3 \\ +4 \\ +-4 +\end{array}\right] , \left[\begin{array}{c} +0 \\ +1 \\ +3 \\ +-3 +\end{array}\right] , \left[\begin{array}{c} +3 \\ +11 \\ +18 \\ +-18 +\end{array}\right] , \left[\begin{array}{c} +-2 \\ +-7 \\ +-11 \\ +11 +\end{array}\right] \right\} is + +not a basis. +

+ + diff --git a/linear-algebra/source/02-EV/samples/06.ptx b/linear-algebra/source/02-EV/samples/06.ptx new file mode 100644 index 00000000..cf3ccee6 --- /dev/null +++ b/linear-algebra/source/02-EV/samples/06.ptx @@ -0,0 +1,68 @@ + + +EV6 + +

+Consider the subspace + + W = \vspan \left\{ + \left[\begin{array}{c} 1 \\ -3 \\ -1 \\ 2 \end{array}\right] , + \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ -2 \end{array}\right] , + \left[\begin{array}{c} 3 \\ -6 \\ -1 \\ 2 \end{array}\right] , + \left[\begin{array}{c} 1 \\ 6 \\ 1 \\ -1 \end{array}\right] , + \left[\begin{array}{c} 2 \\ 3 \\ 0 \\ 1 \end{array}\right] + \right\} . + +

+
    +
  1. +

    +Explain how to find a basis of W. +

    +
    2. +
  2. +

    +Explain how to find the dimension of W. +

    +
    3. +
+ + +
    +
  1. +

    +Observe that + \RREF + \left[\begin{array}{ccccc} + 1 & 1 & 3 & 1 & 2 \\ + -3 & 0 & -6 & 6 & 3 \\ + -1 & 1 & -1 & 1 & 0 \\ + 2 & -2 & 2 & -1 & 1 + \end{array}\right] = + \left[\begin{array}{ccccc} + 1 & 0 & 2 & 0 & 1 \\ + 0 & 1 & 1 & 0 & 0 \\ + 0 & 0 & 0 & 1 & 1 \\ + 0 & 0 & 0 & 0 & 0 + \end{array}\right] + +If we remove the vectors yielding non-pivot columns, the resulting +set will span the same vectors while being linearly independent. +Therefore +\left\{ + \left[\begin{array}{c} 1 \\ -3 \\ -1 \\ 2 \end{array}\right] , + \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ -2 \end{array}\right] , + \left[\begin{array}{c} 1 \\ 6 \\ 1 \\ -1 \end{array}\right] + \right\} + +is a basis of W. +

    +
    2. +
  2. +

    +Since this (and thus every other) basis has three vectors in it, the dimension of W is 3. +

    +
    3. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/02-EV/samples/07.ptx b/linear-algebra/source/02-EV/samples/07.ptx new file mode 100644 index 00000000..61cf7eda --- /dev/null +++ b/linear-algebra/source/02-EV/samples/07.ptx @@ -0,0 +1,91 @@ + + +EV7 + +

+Consider the homogeneous system of equations + + +x_1 &\,+\,& x_2 &\,+\,& 3x_3 &\,+\,& x_4 &\,+\,& 2x_5 &=& 0 + + +-3x_1 &\,\,& &\,-\,& 6x_3 &\,+\,&6 x_4 &\,+\,& 3x_5 &=& 0 + + +-x_1 &\,+\,& x_2 &\,-\,& x_3 &\,+\,& x_4 &\,\,& &=& 0 + + +2x_1 &\,-\,& 2x_2 &\,+\,& 2x_3 &\,-\,& x_4 &\,+\,& x_5 &=& 0 + + +

+
    +
  1. +

    +Find the solution space of the system. +

    +
    2. +
  2. +

    +Find a basis of the solution space. +

    +
    3. +
+ + +
    +
  1. +

    +Observe that + \RREF + \left[\begin{array}{ccccc|c} + 1 & 1 & 3 & 1 & 2 & 0\\ + -3 & 0 & -6 & 6 & 3 & 0\\ + -1 & 1 & -1 & 1 & 0 & 0\\ + 2 & -2 & 2 & -1 & 1& 0 + \end{array}\right] = + \left[\begin{array}{ccccc|c} + 1 & 0 & 2 & 0 & 1 &0\\ + 0 & 1 & 1 & 0 & 0 &0\\ + 0 & 0 & 0 & 1 & 1 &0\\ + 0 & 0 & 0 & 0 & 0&0 + \end{array}\right] + +

    +

    +Letting x_3=a and x_5=b +(since those correspond to the non-pivot columns), +this is equivalent to the system + + + +x_1 &\,\,& &\,+\,& 2x_3 &\,\,& &\,+\,& x_5 &=& 0 + + + &\,\,& x_2 &\,+\,& x_3 &\,\,& &\,\,& &=& 0 + + + &\,\,& &\,\,& x_3 &\,\,& &\,\,& &=& a + + + &\,\,& &\,\,& &\,\,& x_4 &\,+\,& x_5 &=& 0 + + + &\,\,& &\,\,& &\,\,& &\,\,& x_5 &=& b + + +Thus, the solution set is + \setBuilder{\left[\begin{array}{c} -2a-b \\ -a \\ a \\ -b \\ b \end{array}\right]}{a,b \in \IR} . +

    +
    2. +
  2. +

    +Since we can write \left[\begin{array}{c} -2a-b \\ -a \\ a \\ -b \\ b \end{array}\right] = +a \left[\begin{array}{c} -2 \\ -1 \\ 1 \\ 0 \\ 0 \end{array}\right] + b \left[\begin{array}{c} -1 \\ 0 \\ 0 \\ -1 \\ 1 \end{array}\right], +a basis for the solution space is + \left \{ \left[\begin{array}{c} -2 \\ -1 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ 0 \\ 0 \\ -1 \\ 1 \end{array}\right] \right\}. +

    +
    3. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/03-AT/01.ptx b/linear-algebra/source/03-AT/01.ptx new file mode 100644 index 00000000..ad9aba0c --- /dev/null +++ b/linear-algebra/source/03-AT/01.ptx @@ -0,0 +1,775 @@ + +
+ Linear Transformations (AT1) + + + + + + Warm Up + + + +

+ What is our definition for a set S of vectors to be linearly independent? +

+ + + + +

+ What specific calculation would you perform to test is a set S of Euclidean vectors is linearly independent? +

+ + + + + + +

+ What is our definition for a set S of vectors in \IR^n to span \IR^n ? +

+ + + + +

+ What specific calculation would you perform to test is a set S of Euclidean vectors spans all of \IR^n ? +

+ + + + +Class Activities + + + +

+A linear transformation linear transformation(also called a linear map) +is a map between vector spaces that preserves the vector space operations. +More precisely, if V and W are vector spaces, a map +T:V\rightarrow W is called a linear transformation if +

    +
  1. +

    + T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w}) + for any \vec{v},\vec{w} \in V, and +

    +
    2. +
  2. +

    + T(c\vec{v}) = cT(\vec{v}) + for any c \in \IR, and \vec{v} \in V. +

    +
    3. +
+In other words, a map is linear when vector space operations +can be applied before or after the transformation without affecting the result. +

+ + + + + +

+Given a linear transformation T:V\to W, +V is called the domain of T and +W is called the co-domain of T. +

+
+ A linear transformation with a domain of \IR^3 and a co-domain of \IR^2 + + +\begin{tikzpicture}[x=0.2in,y=0.2in] + \begin{scope}[shift={(0,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (2,1) + node[anchor=south west] {\(\vec v\)}; + \node[anchor=west] at (-1,-1) {domain \(\IR^3\)}; + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \node[anchor=south] at (5,4) {Linear transformation \(T:\IR^3\to\IR^2\)}; + \begin{scope}[shift={(9,0.5)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,red] (0,0) -- (-1.5,1) + node[anchor=south east] {\(T(\vec v)\)}; + \node[anchor=west] at (0,-1) {co-domain \(\IR^2\)}; + \end{scope} +\end{tikzpicture} + + +
+ + + + +

+One example of a linear transformation \IR^3\to\IR^2 +is the projection of three-dimesional data onto a two-dimensional screen, +as is necessary for computer animiation in film or video games. +

+
+ A projection of a 3D teapot onto a 2D screen + +
+ + + + +

+Let T : \IR^3 \rightarrow \IR^2 be given by + +T\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) += +\left[\begin{array}{c} x-z \\ 3y \end{array}\right]. + +

+ + +

+Compute the result of adding vectors before a T transformation: + +T\left( + \left[\begin{array}{c} x \\ y \\ z \end{array}\right] + + \left[\begin{array}{c} u \\ v \\ w \end{array}\right] +\right) += +T\left( + \left[\begin{array}{c} x+u \\ y+v \\ z+w \end{array}\right] +\right) + +

    +
  1. +

    + \left[\begin{array}{c} x-u+z-w \\ 3y-3v \end{array}\right] +

    +
    2. +
  2. +

    + \left[\begin{array}{c} x+u-z-w \\ 3y+3v \end{array}\right] +

    +
    3. +
  3. +

    + \left[\begin{array}{c} x+u \\ 3y+3v \\ z+w \end{array}\right] +

    +
    4. +
  4. +

    + \left[\begin{array}{c} x-u \\ 3y-3v \\ z-w \end{array}\right] +

    +
    5. +
+

+ + +

+Compute the result of adding vectors after a T transformation: + +T\left( + \left[\begin{array}{c} x \\ y \\ z \end{array}\right] +\right) + T\left( + \left[\begin{array}{c} u \\ v \\ w \end{array}\right] +\right) += +\left[\begin{array}{c} x-z \\ 3y \end{array}\right] + +\left[\begin{array}{c} u-w \\ 3v \end{array}\right] + +

    +
  1. +

    + \left[\begin{array}{c} x-u+z-w \\ 3y-3v \end{array}\right] +

    +
    2. +
  2. +

    + \left[\begin{array}{c} x+u-z-w \\ 3y+3v \end{array}\right] +

    +
    3. +
  3. +

    + \left[\begin{array}{c} x+u \\ 3y+3v \\ z+w \end{array}\right] +

    +
    4. +
  4. +

    + \left[\begin{array}{c} x-u \\ 3y-3v \\ z-w \end{array}\right] +

    +
    5. +
+

+ + +

+Is T a linear transformation? +

    +
  1. +

    + Yes. +

    +
    2. +
  2. +

    + No. +

    +
    3. +
  3. +

    + More work is necessary to know. +

    +
    4. +
+

+ + +

+ Compute the result of scalar multiplcation before a T transformation: + +T\left(c\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) += +T\left(\left[\begin{array}{c} cx \\ cy \\ cz \end{array}\right] \right) + +

    +
  1. +

    + \left[\begin{array}{c} cx-cz\\ 3cy \end{array}\right] +

    +
    2. +
  2. +

    + \left[\begin{array}{c} cx+cz \\ -3cy \end{array}\right] +

    +
    3. +
  3. +

    + \left[\begin{array}{c} x+c \\ 3y+c \\ z+c \end{array}\right] +

    +
    4. +
  4. +

    + \left[\begin{array}{c} x-c \\ 3y-c \\ z-c \end{array}\right] +

    +
    5. +
+

+ + +

+ Compute the result of scalar multiplcation after a T transformation: + + cT\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) += + c\left[\begin{array}{c} x-z \\ 3y \end{array}\right] + +

    +
  1. +

    + \left[\begin{array}{c} cx-cz\\ 3cy \end{array}\right] +

    +
    2. +
  2. +

    + \left[\begin{array}{c} cx+cz \\ -3cy \end{array}\right] +

    +
    3. +
  3. +

    + \left[\begin{array}{c} x+c \\ 3y+c \\ z+c \end{array}\right] +

    +
    4. +
  4. +

    + \left[\begin{array}{c} x-c \\ 3y-c \\ z-c \end{array}\right] +

    +
    5. +
+

+ + +

+Is T a linear transformation? +

    +
  1. +

    + Yes. +

    +
    2. +
  2. +

    + No. +

    +
    3. +
  3. +

    + More work is necessary to know. +

    +
    4. +
+

+ + + + + +

+Let S : \IR^2 \rightarrow \IR^4 be given by + + S\left(\left[\begin{array}{c} x \\ y \end{array}\right] \right) += + \left[\begin{array}{c} x+y \\ x^2 \\ y+3 \\ y-2^x \end{array}\right] + +

+ + +

+Compute + + S\left( + \left[\begin{array}{c} 0 \\ 1 \end{array}\right] + + \left[\begin{array}{c} 2 \\ 3 \end{array}\right] + \right) += + S\left( + \left[\begin{array}{c} 2 \\ 4 \end{array}\right] + \right) + +

    +
  1. +

    + \left[\begin{array}{c} 6 \\ 4 \\ 7 \\ 0 \end{array}\right] +

    +
    2. +
  2. +

    + \left[\begin{array}{c} -3 \\ 0 \\ 1 \\ 5 \end{array}\right] +

    +
    3. +
  3. +

    + \left[\begin{array}{c} -3 \\ -1 \\ 7 \\ 5 \end{array}\right] +

    +
    4. +
  4. +

    + \left[\begin{array}{c} 6 \\ 4 \\ 10 \\ -1 \end{array}\right] +

    +
    5. +
+

+ + +

+Compute + + S\left( + \left[\begin{array}{c} 0 \\ 1 \end{array}\right] + \right) + S\left( + \left[\begin{array}{c} 2 \\ 3 \end{array}\right] + \right) += + \left[\begin{array}{c} 0+1 \\ 0^2 \\ 1+3 \\ 1-2^0 \end{array}\right] + + \left[\begin{array}{c} 2+3 \\ 2^2 \\ 3+3 \\ 3-2^2 \end{array}\right] + +

    +
  1. +

    + \left[\begin{array}{c} 6 \\ 4 \\ 7 \\ 0 \end{array}\right] +

    +
    2. +
  2. +

    + \left[\begin{array}{c} -3 \\ 0 \\ 1 \\ 5 \end{array}\right] +

    +
    3. +
  3. +

    + \left[\begin{array}{c} -3 \\ -1 \\ 7 \\ 5 \end{array}\right] +

    +
    4. +
  4. +

    + \left[\begin{array}{c} 6 \\ 4 \\ 10 \\ -1 \end{array}\right] +

    +
    5. +
+

+ + +

+Is T a linear transformation? +

    +
  1. +

    + Yes. +

    +
    2. +
  2. +

    + No. +

    +
    3. +
  3. +

    + More work is necessary to know. +

    +
    4. +
+

+ + + + + + + + + + + + +

+Fill in the \unknowns, assuming T:\mathbb R^3\to\mathbb R^3 is linear: + + T\left(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\right) + = T\left(\unknown \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\right) + = \unknown T\left(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\right) + = \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \end{array}\right] + +

+ + + + + + +

+In summary, any one of the following is enough to +prove that T:V\to W is not a linear transformation: +

    +
  • +

    +Find specific values for \vec v,\vec w\in V +such that T(\vec v+\vec w)\not=T(\vec v)+T(\vec w). +

    +
    • +
  • +

    +Find specific values for \vec v\in V and c\in \IR +such that T(c\vec v)\not=cT(\vec v). +

    +
    • +
  • +

    +Show T(\vec 0)\not=\vec 0. +

    +
    • +
+

+

+If you cannot do any of these, then T can be proven to be a +linear transformation by doing both of the following: +

    +
  1. +

    +For all \vec v,\vec w\in V (not just specific values), +T(\vec v+\vec w)=T(\vec v)+T(\vec w). +

    +
    2. +
  2. +

    +For all \vec v\in V and c\in \IR (not just specific values), +T(c\vec v)=cT(\vec v). +

    +
    3. +
+

+

+(Note the similarities between this process and showing that a subset of a vector +space is or is not a subspace: .) +

+ + + + + + +

Consider the following maps of Euclidean vectors P:\mathbb R^3\rightarrow\mathbb R^3 and + Q:\mathbb R^3\rightarrow\mathbb R^3 defined by + P\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)= + \left[\begin{array}{c} -2 \, x - 3 \, y - 3 \, z \\ 3 \, x + 4 \, y + 4 \, z \\ 3 \, x + 4 \, y + 5 \, z \end{array}\right] + \hspace{1em} \text{and} \hspace{1em} Q\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)= + \left[\begin{array}{c} x - 4 \, y + 9 \, z \\ y - 2 \, z \\ 8 \, y^{2} - 3 \, x z \end{array}\right]. + Which do you suspect? +

    +
  1. +

    + P is linear, but Q is not. +

    +
    2. +
  2. +

    + Q is linear, but P is not. +

    +
    3. +
  3. +

    + Both maps are linear. +

    +
    4. +
  4. +

    + Neither map is linear. +

    +
    5. +
+

+ + + + +

Consider the following map of Euclidean vectors S:\mathbb R^2\rightarrow\mathbb R^2 S\left( \left[\begin{array}{c} x \\ y \end{array}\right]\right)= \left[\begin{array}{c} x + 2 \, y \\ 9 \, x y \end{array}\right]. Prove that S is not a linear transformation.

+ + + + + +

Consider the following map of Euclidean vectors T:\mathbb R^2\rightarrow\mathbb R^2 T\left( \left[\begin{array}{c} x \\ y \end{array}\right] \right)= \left[\begin{array}{c} 8 \, x - 6 \, y \\ 6 \, x - 4 \, y \end{array}\right]. Prove that T is a linear transformation.

+ + + + + + + + + + + Cool Down + +

+ Let f(x)=x^3-1. + Then, f\colon\IR\to\IR is a function with domain and codomain equal to \IR. + Is f(x) is a linear transformation? +

+ + +

+ Consider two vectors \vec{u},\vec{v} and their sum \vec{u}+\vec{v}. +

+ + +

+ Is it the case that rotating \vec{u}+\vec{v} about the origin by \frac{\pi}{2}=90^\circ is the same as first rotating each of \vec{u},\vec{v} and then adding them together? +

+ + + + +

+ Is it the case that rotating 5\vec{u} about the origin by \frac{\pi}{2}=90^\circ is the same as first rotating \vec{u} by \frac{\pi}{2}=90^\circ and then scaling by 5? +

+ + + + +

+ Based on this, do you suspect that the transformation R\colon\IR^2\to\IR^2 given by rotating vectors about the origin through an angle of \frac{\pi}{2}=90^\circ is linear? Do you think there is anything special about the angle \frac{\pi}{2}=90^\circ? +

+ + + + + In , we made an analogy between vectors and linear combinations with ingredients and recipes. + Let us think of cooking as a transformation of ingredients. + In this analogy, would it be appropriate for us to consider "cooking" to be a linear transformation or not? + Describe your reasoning. + + + + + Videos +
+ + Video: Showing a transformation is linear +
+
+ + Video: Showing a transformation is not linear +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + +

If V,W are vectors spaces, with associated zero vectors \vec{0}_V and \vec{0}_W, and T:V \rightarrow W is a linear transformation, does T(\vec{0}_V) = \vec{0}_W? Prove this is true, or find a counterexample.

 + + + + +

Assume f: V \rightarrow W is a linear transformation between vector spaces. Let \vec{v} \in V with additive inverse \vec{v}^{-1}. Prove that f(\vec{v}^{-1}) = [f(\vec{v})]^{-1}.

+ + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/03-AT/02.ptx b/linear-algebra/source/03-AT/02.ptx new file mode 100644 index 00000000..ec74905c --- /dev/null +++ b/linear-algebra/source/03-AT/02.ptx @@ -0,0 +1,546 @@ + +
+ Standard Matrices (AT2) + + + + + + Warm Up + + + +

+Recall that a linear map T:V\rightarrow W +satisfies +

+
    +
  1. +

    + T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w}) + for any \vec{v},\vec{w} \in V. +

    +
    2. +
  2. +

    + T(c\vec{v}) = cT(\vec{v}) + for any c \in \IR,\vec{v} \in V. +

    +
    3. +
+

+In other words, a map is linear when vector space operations +can be applied before or after the transformation without affecting the result. +

+ + + + +

+ Can you recall the following? +

+ + + +

+ Given a transformation, what do the terms domain and codomain mean? +

+ + + + +

+ What does the notation T\colon V\to W mean? +

+ + + + + +Class Activities + + +

+Suppose T: \IR^3 \rightarrow \IR^2 is a linear map, and you know + + T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right) += + \left[\begin{array}{c} 2 \\ 1 \end{array}\right] + +and + + T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right) += + \left[\begin{array}{c} -3 \\ 2 \end{array}\right] +. +What is T\left(\left[\begin{array}{c} 3 \\ 0 \\ 0 \end{array}\right]\right)? +

+
    +
  1. \left[\begin{array}{c} 6 \\ 3\end{array}\right] +
    2. +
  2. \left[\begin{array}{c} -9 \\ 6 \end{array}\right] +
    3. +
  3. \left[\begin{array}{c} -4 \\ -2 \end{array}\right] +
    4. +
  4. \left[\begin{array}{c} 6 \\ -4 \end{array}\right] +
    5. +
+ + + + + +

+Suppose T: \IR^3 \rightarrow \IR^2 is a linear map, and you know + + T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right) += + \left[\begin{array}{c} 2 \\ 1 \end{array}\right] + +and + + T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right) += + \left[\begin{array}{c} -3 \\ 2 \end{array}\right] +. +What is T\left(\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right]\right)? +

+
    +
  1. +

    + \left[\begin{array}{c} 2 \\ 1\end{array}\right] +

    +
    2. +
  2. +

    + \left[\begin{array}{c} 3 \\ -1 \end{array}\right] +

    +
    3. +
  3. +

    + \left[\begin{array}{c} -1 \\ 3 \end{array}\right] +

    +
    4. +
  4. +

    + \left[\begin{array}{c} 5 \\ -8 \end{array}\right] +

    +
    5. +
+ + + + + +

+Suppose T: \IR^3 \rightarrow \IR^2 is a linear map, and you know + + T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right) += + \left[\begin{array}{c} 2 \\ 1 \end{array}\right] + +and + + T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right) += + \left[\begin{array}{c} -3 \\ 2 \end{array}\right] +. +What is T\left(\left[\begin{array}{c} -2 \\ 0 \\ -3 \end{array}\right]\right)? +

+
    +
  1. +

    + \left[\begin{array}{c} 2 \\ 1\end{array}\right] +

    +
    2. +
  2. +

    + \left[\begin{array}{c} 3 \\ -1 \end{array}\right] +

    +
    3. +
  3. +

    + \left[\begin{array}{c} -1 \\ 3 \end{array}\right] +

    +
    4. +
  4. +

    + \left[\begin{array}{c} 5 \\ -8 \end{array}\right] +

    +
    5. +
+ + + + + +

+Suppose T: \IR^3 \rightarrow \IR^2 is a linear map, and you know + + T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right) += + \left[\begin{array}{c} 2 \\ 1 \end{array}\right] + +and + + T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right) += + \left[\begin{array}{c} -3 \\ 2 \end{array}\right] +. +What piece of information would help you compute T\left(\left[\begin{array}{c}0\\4\\-1\end{array}\right]\right)? +

+
    +
  1. +

    + The value of T\left(\left[\begin{array}{c} 0\\-4\\0\end{array}\right]\right). +

    +
    2. +
  2. +

    + The value of T\left(\left[\begin{array}{c} 0\\1\\0\end{array}\right]\right). +

    +
    3. +
  3. +

    + The value of T\left(\left[\begin{array}{c} 1\\1\\1\end{array}\right]\right). +

    +
    4. +
  4. +

    + Any of the above. +

    +
    5. +
+ + + + + +

+Consider any basis \{\vec b_1,\dots,\vec b_n\} for V. Since every +vector \vec v can be written as a linear combination of +basis vectors, \vec v = x_1\vec b_1+\dots+ x_n\vec b_n, we may compute +T(\vec v) as follows: +

+ + T(\vec v)=T(x_1\vec b_1+\dots+ x_n\vec b_n)= + x_1T(\vec b_1)+\dots+x_nT(\vec b_n) +. +

+Therefore any linear transformation T:V \rightarrow W can be defined +by just describing the values of T(\vec b_i). +

+

+Put another way, the images of the basis vectors completely determinedetermine the transformation T. +

+ + + + + +

+Since a linear transformation T:\IR^n\to\IR^m is determined by its action on +the standard basis \{\vec e_1,\dots,\vec e_n\}, it is convenient to +store this information in an m\times n matrix, called the standard matrixstandard matrix of T, given by +[T(\vec e_1) \,\cdots\, T(\vec e_n)]. +

+

+For example, +let T: \IR^3 \rightarrow \IR^2 be the linear map determined by +the following values for T applied to the standard basis of \IR^3. +

+\scriptsize + T\left(\vec e_1 \right) += + T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right] \right) += + \left[\begin{array}{c} 3 \\ 2\end{array}\right] +\hspace{2em} + T\left(\vec e_2 \right) += + T\left(\left[\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right] \right) += + \left[\begin{array}{c} -1 \\ 4\end{array}\right] +\hspace{2em} + T\left(\vec e_3 \right) += + T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right) += + \left[\begin{array}{c} 5 \\ 0\end{array}\right] + +

+Then the standard matrix corresponding to T is + + \left[\begin{array}{ccc}T(\vec e_1) & T(\vec e_2) & T(\vec e_3)\end{array}\right] += + \left[\begin{array}{ccc}3 & -1 & 5 \\ 2 & 4 & 0 \end{array}\right] +. +

+ + + + + +

+ Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by + + T\left(\vec e_1 \right) + = + \left[\begin{array}{c} 0 \\ 3 \\ -2\end{array}\right] + \hspace{2em} + T\left(\vec e_2 \right) + = + \left[\begin{array}{c} -3 \\ 0 \\ 1\end{array}\right] + \hspace{2em} + T\left(\vec e_3 \right) + = + \left[\begin{array}{c} 4 \\ -2 \\ 1\end{array}\right] + \hspace{2em} + T\left(\vec e_4 \right) + = + \left[\begin{array}{c} 2 \\ 0 \\ 0\end{array}\right] + +Write the standard matrix [T(\vec e_1) \,\cdots\, T(\vec e_n)] for T. +

+ + + + + +

+ Let T: \IR^3 \rightarrow \IR^2 be the linear transformation given by +T\left(\left[\begin{array}{c} x\\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} x+3z \\ 2x-y-4z \end{array}\right] +

+ + +

+Compute T(\vec e_1), T(\vec e_2), and T(\vec e_3). +

+ + +

+Find the standard matrix for T. +

+ + + + + +

+ Because every linear map T:\IR^m\to\IR^n has a linear combination + of the variables in each component, and thus + T(\vec e_i) yields exactly the coefficients of x_i, + the standard matrix for T is simply an array of + the coefficients of the x_i: + + T\left(\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]\right) + = + \left[\begin{array}{c} + ax+by+cz+dw \\ + ex+fy+gz+hw + \end{array}\right] + \hspace{2em} + A + = + \left[\begin{array}{cccc} + a & b & c & d \\ + e & f & g & h + \end{array}\right] + +

+

+Since the formula for a linear transformation T and +its standard matrix A may both be used to compute the transformation +of a vector \vec x, we will often write +T(\vec x) and A\vec x interchangably: + + T\left(\left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right]\right) + = + \left[\begin{array}{c} + ax_1+bx_2+cx_3+dx_4 \\ + ex_1+fx_2+gx_3+hx_4 + \end{array}\right] + = + A\left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] + = + \left[\begin{array}{cccc} + a & b & c & d \\ + e & f & g & h + \end{array}\right] + \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] + +

+ + + + + +

+ Let T: \IR^3 \rightarrow \IR^3 be the linear transformation given by the standard matrix + + \left[\begin{array}{ccc} 3 & -2 & -1 \\ 4 & 5 & 2 \\ 0 & -2 & 1 \end{array}\right] + . +

+ + +

+Compute T\left(\left[\begin{array}{c} 1\\ 2 \\ 3 \end{array}\right] \right) . +

+ + +

+Compute T\left(\left[\begin{array}{c} x\\ y \\ z \end{array}\right] \right) . +

+ + + + + +

Compute the following linear transformations of vectors given their + standard matrices.

 + +

+ T_1\left(\left[\begin{array}{c}1\\2\end{array}\right]\right) + \text{ for the standard matrix } + A_1=\left[\begin{array}{cc}4&3\\0&-1\\1&1\\3&0\end{array}\right] +

+

+ T_2\left(\left[\begin{array}{c}1\\1\\0\\-3\end{array}\right]\right) + \text{ for the standard matrix } + A_2=\left[\begin{array}{cccc}4&3&0&-1\\1&1&3&0\end{array}\right] +

+

+ T_3\left(\left[\begin{array}{c}0\\-2\\0\end{array}\right]\right) + \text{ for the standard matrix } + A_3=\left[\begin{array}{ccc}4&3&0\\0&-1&3\\5&1&1\\3&0&0\end{array}\right] +

+ + + + + + Cool Down + + +

+ Consider the linear transformation R\colon\IR^2\to\IR^2 given by rotating vectors about the origin through an angle of \frac{\pi}{4}=45^\circ. +

+ + + +

+ If \vec{e}_1,\vec{e}_2 are the standard basis vectors of \IR^2, calculate R(\vec{e}_1),R(\vec{e}_2). +

+ + + + +

+ What is the standard matrix representing R? +

+ + + + + + +

+ Consider the linear transformation S\colon\IR^2\to\IR^2 given by reflecting vectors across the line x_1=x_2. +

+ + + +

+ If \vec{e}_1,\vec{e}_2 are the standard basis vectors of \IR^2, calculate S(\vec{e}_1),S(\vec{e}_2). +

+ + + + +

+ What is the standard matrix representing S? +

+ + + + + + + + Videos +
+ + Video: Using the standard matrix to compute the image of a vector +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations +

We can represent images in the plane \mathbb{R}^2 using vectors, and manipulate those images with linear transformations. We introduce some notation in these explorations that is needed for their completion, but is not essential to the rest of the text. These have a geometric flair to them, and can be understood by thinking of geometric transformations in terms of standard matrices.

+ +

Given two vectors \vec{v} = \left[\begin{array}{c}v_1\\v_2\\ \vdots \\ v_n\end{array}\right] and \vec{w} = \left[\begin{array}{c}w_1 \\ w_2\\ \vdots \\ w_n\end{array}\right], we define the dot product dot product as + \vec{v}\cdot \vec{w} = v_1w_1 + v_2w_2 + \cdots v_nw_n. +

+ + + For each of the following properties, determine if it is held by the dot product. Either provide a proof it the property holds, or provide a counter-example if it does not. + +
    +
  • Distributive over addition (e.g., (\vec{u} + \vec{v})\cdot \vec{w} = \vec{u}\cdot\vec{w} + \vec{v}\cdot\vec{w})?
    • +
  • Associative?
    • +
  • Commutative?
    • +
+ + + + Given the properties you proved in the last exploration, could the dot product take the place of \oplus as a vector space operation on \mathbb{R}^n? + +Is the dot product a linear operator? That is, given vectors \vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n, and k,m \in \mathbb{R}, is it true that +\vec{u} \cdot (k\vec{v} + m\vec{w}) = k(\vec{u} \cdot \vec{v}) + m(\vec{u}\cdot\vec{w}). Prove or provide a counter-example. + + + + +

Assume \vec{v} = \left[\begin{array}{c}v_1\\v_2\\ \vdots \\ v_n\end{array}\right] and define the length of a vector by |\vec{v}| = \left(\sum_{i=1}^n v_i^2 \right)^{1/2}. Prove that |\vec{u}| = |\vec{v}| if an only if \vec{u} + \vec{v} and \vec{u} - \vec{v} are perpendicular. You may use the fact (try and prove it!) that two vectors are perpendicular if and only if their dot product is zero. +

+ + + + + +
    +
  • A dilation is given by by mapping a vector \vec{v} = \left[\begin{array}{c}x\\y\end{array}\right] to some scalar multiple of \vec{v}.
    • + +
  • A rotation is given by \vec{v} \mapsto \left[\begin{array}{c} \cos(\theta)x - \sin(\theta)y\\ \cos(\theta)y + \sin(\theta)x\end{array}\right].
    • + +
  • A reflection of \vec{v} over a line l can be found by first finding a vector \vec{l} = \left[\begin{array}{c} l_x\\l_y\end{array}\right] along l, then \vec{v} \mapsto 2\frac{\vec{l}\cdot\vec{v}}{\vec{l}\cdot\vec{l}}\vec{l} - \vec{v}.
    • +
+Represent each of the following transformations with respect to the standard basis in \mathbb{R}^2. +
    +
  • Rotation through an angle \theta.
    • +
  • Reflection over a line l passing through the origin.
    • +
  • Dilation by some scalar s.
    • +
+Prove that each transformation is linear, and that your matrix representations are correct. + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/03-AT/03.ptx b/linear-algebra/source/03-AT/03.ptx new file mode 100644 index 00000000..215f64f2 --- /dev/null +++ b/linear-algebra/source/03-AT/03.ptx @@ -0,0 +1,724 @@ + +
+ Image and Kernel (AT3) + + + + + + + Warm Up + + +

+ Consider the matrix A=\left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right]. +

+ + + +

+ The matrix A is the standard matrix of a linear transformation T. + What is the domain and the codomain of the transformation T? +

+ + + + +

+ Describe how T transforms the standard basis vectors of the domain that you found above. +

+ + + + + +Class Activities + + + +

+Let T: \IR^2 \rightarrow \IR^3 be given by + + T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right] + \hspace{3em} + \text{with standard matrix } + \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] + +Which of these subspaces of \IR^2 describes +the set of all vectors that transform into \vec 0? +

+
    +
  1. +

    + \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR} +

    +
    2. +
  2. +

    + \setBuilder{\left[\begin{array}{c}a \\ 0\end{array}\right]}{a\in\IR} +

    +
    3. +
  3. +

    + \setList{\left[\begin{array}{c}0\\0\end{array}\right]} +

    +
    4. +
  4. +

    + \setBuilder{\left[\begin{array}{c}a \\ b\end{array}\right]}{a,b\in\IR} +

    +
    5. +
+ + + + + + +

+Let T: V \rightarrow W be a linear transformation, and let \vec{z} be the additive +identity (the zero vector) of W. The kernelkernel of T +is an important subspace of V defined by + +\ker T = \left\{ \vec{v} \in V\ \big|\ T(\vec{v})=\vec{z}\right\} + +

+
+ + + \begin{tikzpicture}[x=0.15in,y=0.15in] + \begin{scope}[shift={(0,0)}] + \draw (0,0) node[anchor=north west] {\(\ker T\)} -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,latex-latex,blue] (-3,-3) -- (3,3); + \end{scope} + \draw[dashed,-latex] (3.5,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,0.5)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \fill[blue] (0,0) circle (0.2) + node[anchor=south east] {\(\vec{0}\)}; + \end{scope} + \end{tikzpicture} + + + The kernel of a linear transformation +
+ + + + + + +

+Let T: \IR^3 \rightarrow \IR^2 be given by + + T\left(\left[\begin{array}{c}x \\ y\\z \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y \end{array}\right] + \hspace{3em} + \text{with standard matrix } + \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] + +Which of these subspaces of \IR^3 describes \ker T, +the set of all vectors that transform into \vec 0? +

+
    +
  1. +

    + \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR} +

    +
    2. +
  2. +

    + \setBuilder{\left[\begin{array}{c}a \\ a\\ 0\end{array}\right]}{a\in\IR} +

    +
    3. +
  3. +

    + \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]} +

    +
    4. +
  4. +

    + \setBuilder{\left[\begin{array}{c}a \\ b\\c\end{array}\right]}{a,b,c\in\IR} +

    +
    5. +
+ + + + + +

+Let T: \IR^3 \rightarrow \IR^2 be the linear transformation given by the +standard matrix +T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{c} 3x+4y-z \\ x+2y+z \end{array}\right] +

+ + +

+Set + + T\left(\left[\begin{array}{c}x\\y\\z\end{array}\right]\right) + = + \left[\begin{array}{c}0\\0\end{array}\right] + to find a linear system of equations whose solution set is the kernel. +

+ + +

+Use \RREF(A) to solve this homogeneous system of equations and find a basis +for the kernel of T. +

+ + + + + + + +

+Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by + T\left(\left[\begin{array}{c} x \\ y \\ z \\ w \end{array}\right] \right) = +\left[\begin{array}{c} 2x+4y+2z-4w \\ -2x-4y+z+w \\ 3x+6y-z-4w\end{array}\right]. +

+

+Find a basis for the kernel of T. +

+ + + + + + + + + +

+Let T: \IR^2 \rightarrow \IR^3 be given by + + T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right] + \hspace{3em} + \text{with standard matrix } + \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] + +Which of these subspaces of \IR^3 describes the set of all vectors that are the result of using T to transform +\IR^2 vectors? +

+
    +
  1. +

    + \setBuilder{\left[\begin{array}{c}0 \\ 0\\ a\end{array}\right]}{a\in\IR} +

    +
    2. +
  2. +

    + \setBuilder{\left[\begin{array}{c}a \\ b\\ 0\end{array}\right]}{a,b\in\IR} +

    +
    3. +
  3. +

    + \setList{\left[\begin{array}{c}0\\0\\0\end{array}\right]} +

    +
    4. +
  4. +

    + \setBuilder{\left[\begin{array}{c}a \\ b\\c\end{array}\right]}{a,b,c\in\IR} +

    +
    5. +
+ + + + +

+Let T: V \rightarrow W be a linear transformation. +The image of T is an important subspace of W defined by + +\Im T = \left\{ \vec{w} \in W\ \big|\ \text{there is some }\vec v\in V \text{ with } T(\vec{v})=\vec{w}\right\} + +

+

+In the examples below, the left example's image is all of \IR^2, but the +right example's image is a planar subspace of \IR^3. +

+
+ + +\begin{tikzpicture}[x=0.15in,y=0.15in] + \begin{scope}[shift={(0,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (1,2); + \draw[thick,-latex,blue] (0,0) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (1,-1); + \draw[thick,-latex,blue] (0,0) -- (-2,3); + \draw[thick,-latex,blue] (0,0) -- (-3,-2); + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (0.5,2); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-1.5,1); + \draw[thick,-latex,blue] (0,0) -- (0,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2,-2); + \fill[color=blue, opacity=0.5] (-2,-2) rectangle (2,2); + \end{scope} +\end{tikzpicture} +\hspace{3em} +\begin{tikzpicture}[x=0.15in,y=0.15in] + \begin{scope}[shift={(0,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (0.5,2); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-1.5,1); + \draw[thick,-latex,blue] (0,0) -- (0,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2,-2); + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (0.5,1.5); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-2.5,1); + \draw[thick,-latex,blue] (0,0) -- (-0.5,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2.5,-0.5); + \fill[color=blue, opacity=0.5] (-2,-2) -- (3,-1) -- (2,2) -- (-3,1) -- (-2,-2); + \end{scope} +\end{tikzpicture} + + + The image of a linear transformation +
+ + + + +

+Let T: \IR^3 \rightarrow \IR^2 be given by + + T\left(\left[\begin{array}{c}x \\ y\\z \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y \end{array}\right] + \hspace{3em} + \text{with standard matrix } + \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] + +Which of these subspaces of \IR^2 describes \Im T, +the set of all vectors that are the result of using T to transform +\IR^3 vectors? +

+
    +
  1. +

    + \setBuilder{\left[\begin{array}{c}a \\ a\end{array}\right]}{a\in\IR} +

    +
    2. +
  2. +

    + \setBuilder{\left[\begin{array}{c}a \\ 0\end{array}\right]}{a\in\IR} +

    +
    3. +
  3. +

    + \setList{\left[\begin{array}{c}0\\0\end{array}\right]} +

    +
    4. +
  4. +

    + \setBuilder{\left[\begin{array}{c}a \\ b\end{array}\right]}{a,b\in\IR} +

    +
    5. +
+ + + + + +

Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by the + standard matrix + + A + = + \left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right] + = + \left[\begin{array}{cccc}T(\vec e_1)&T(\vec e_2)&T(\vec e_3)&T(\vec e_4)\end{array}\right] + . +

+

+ Consider the question: Which vectors \vec{w} in \IR^3 belong to + \Im T? +

+ + + +

+ Determine if \left[\begin{array}{c} 12 \\ 3 \\ 3 \end{array}\right] belongs to + \Im T. +

+ + + + +

+ Determine if \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] belongs to + \Im T. +

+ + + + +

+An arbitrary vector \left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right] belongs to +\Im T provided the equation +x_1 T(\vec{e}_1)+x_2 T(\vec{e}_2)+x_3T(\vec{e}_3)+x_4T(\vec{e}_4)=\vec{w} has... +

    +
  1. no solutions.
    2. +
  2. exactly one solution.
    3. +
  3. at least one solution.
    4. +
  4. infinitely-many solutions.
    5. +
+

+ + + + +

+ Based on this, how do \Im T and \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} relate to each other? +

    +
  1. The set \Im T contains \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} but is not equal to it.
    2. +
  2. The set \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} contains \Im T but is not equal to it.
    3. +
  3. The set \Im T and \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} are equal to each other.
    4. +
  4. There is no relation between these two sets.
    5. +
+

+ + + + + + +

+Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by the +standard matrix + + A + = + \left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right] +. +

+

+Since the set + + \setList{ + \left[\begin{array}{c}3\\-1\\2\end{array}\right], + \left[\begin{array}{c}4\\1\\1\end{array}\right], + \left[\begin{array}{c}7\\0\\3\end{array}\right], + \left[\begin{array}{c}1\\2\\-1\end{array}\right] + } + +spans \Im T, we can obtain a basis for \Im T by finding + + \RREF A + = + \left[\begin{array}{cccc} 1 & 0 & 1 & -1\\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] + +and only using the vectors corresponding to pivot columns: + + \setList{ + \left[\begin{array}{c}3\\-1\\2\end{array}\right], + \left[\begin{array}{c}4\\1\\1\end{array}\right] + } + +

+ + + + + +

+Let T:\IR^n\to\IR^m be a linear transformation with standard matrix A. +

+
    +
  • +

    + The kernel of T is the solution set of the homogeneous system given +by the augmented matrix \left[\begin{array}{c|c}A&\vec 0\end{array}\right]. +Use the coefficients of its free variables to get a basis for the kernel (as in ). +

    +
    • +
  • +

    + The image of T is the span of the columns of A. Remove +the vectors creating non-pivot columns in \RREF A to get a basis +for the image (as in ). +

    +
    • +
+ + + + + + +

+Let T: \IR^3 \rightarrow \IR^4 be the linear transformation given by the +standard matrix + + A + = + \left[\begin{array}{ccc} 1 & -3 & 2\\ 2 & -6 & 0 \\ 0 & 0 & 1 \\ -1 & 3 & 1 \end{array}\right] +. +

+

+Find a basis for the kernel and a basis for the image of T. +

+ + + + + + + +

+Let T: \IR^n \rightarrow \IR^m be a linear transformation with standard matrix A. +Which of the following is equal to the dimension of the kernel of T? +

+
    +
  1. +

    + The number of pivot columns +

    +
    2. +
  2. +

    + The number of non-pivot columns +

    +
    3. +
  3. +

    + The number of pivot rows +

    +
    4. +
  4. +

    + The number of non-pivot rows +

    +
    5. +
+ + + + + +

+Let T: \IR^n \rightarrow \IR^m be a linear transformation with standard matrix A. +Which of the following is equal to the dimension of the image of T? +

+
    +
  1. +

    + The number of pivot columns +

    +
    2. +
  2. +

    + The number of non-pivot columns +

    +
    3. +
  3. +

    + The number of pivot rows +

    +
    4. +
  4. +

    + The number of non-pivot rows +

    +
    5. +
+ + + + + +

+Combining these with the observation that the number of columns is the dimension of the domain of T, we have the rank-nullity theorem: +

+
+

+The dimension of the domain of T equals \dim(\ker T)+\dim(\Im T). +

+
+

+The dimension of the image is called the rank of T (or A) and the dimension of the kernel is called the nullity. +

+ + + + +

Let T:\mathbb{R}^4 \to \mathbb{R}^3 be the linear transformation given by T\left( \left[\begin{array}{c} x \\ y \\ z \\ {w} \end{array}\right] \right) = \left[\begin{array}{c} x - y + 5 \, z + 3 \, {w} \\ -x - 4 \, z - 2 \, {w} \\ y - 2 \, z - {w} \end{array}\right].

+ + + +

Explain and demonstrate how to find the image of T and a basis for that image.

+ + + + + +

Explain and demonstrate how to find the kernel of T and a basis for that kernel.

+ + + + + +

Explain and demonstrate how to find the rank and nullity of T, and why the rank-nullity theorem holds for T.

+ + + + + + + + + + + Cool Down + + +

+ In this section, we've introduced two important subspaces that are associated with a linear transformation T\colon V\to W, namely: \Im T, the image of T, and \ker T, the kernel of T. + The following sequence is designed to help you internalize these definitions. + Try to complete them without referring to your Activity Book, and then check your answers. +

+ + + +

+ One of \ker T and \Im T is a subspace of the domain and the other is a subspace of the codomain. + Which is which? +

+ + + + +

+ Write down the precise definitions of these subspaces. +

+ + + + +

+ How would you describe these definitions to a layperson? +

+ + + + +

+ What picture, or other study strategy would be helpful to you in conceptualizing how these defintions fit together? +

+ + + + + + + Videos +
+ + Video: The kernel and image of a linear transformation +
+
+ + Video: Finding a basis of the image of a linear transformation +
+
+ + Video: Finding a basis of the kernel of a linear transformation +
+
+ + Video: The rank-nullity theorem +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + +

+ Assume f:V \rightarrow W is a linear map. +Let \{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\} be a set of vectors in V, and set \vec{w_i} = f(\vec{v_i}). +

    +
  • If the set \{\vec{w_1},\vec{w_2},\ldots,\vec{w_n}\} is linearly independent, must the set \{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\} also be linearly independent?
    • +
  • If the set \{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\} is linearly independent, must the set \{\vec{w_1},\vec{w_2},\ldots,\vec{w_n}\} also be linearly independent?
    • +
  • If the set \{\vec{w_1},\vec{w_2},\ldots,\vec{w_n}\} spans W, must the set \{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\} also span V?
    • +
  • If the set \{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\} spans V, must the set \{\vec{w_1},\vec{w_2},\ldots,\vec{w_n}\} also span W?
    • +
  • In light of this, is the image of the basis of a vector space always a basis for the codomain?
    • +

+ + + + + Prove the Rank-Nullity Theorem. Use the steps below to help you. +
    +
  • The theorem states that, given a linear map h:V \rightarrow W, with V and W vector spaces, the rank of h, plus the nullity of h, equals the dimension of the domain V. Assume that the dimension of V is n.
    • +
  • For simplicity, denote the rank of h by \mathcal{R}(h), and the nullity by \mathcal{N}(h).
    • +
  • Recall that \mathcal{R}(h) is the dimension of the range space of h. State the precise definition.
    • +
  • Recall that \mathcal{N}(h) is the dimension of the null space of h. State the precise definition.
    • +
  • Begin with a basis for the null space, denoted B_N = \{\vec{\beta_1}, \vec{\beta_2}, \ldots, \vec{\beta_k}\}. Show how this can be extended to a basis B_V for V, with +B_V = \{\vec{\beta_1}, \vec{\beta_2}, \ldots, \vec{\beta_k}, \vec{\beta_{k+1}}, \vec{\beta_{k+2}}, \ldots, \vec{\beta_n}\}. In this portion, you should assume k \leq n, and construct additional vectors which are not linear combinations of vectors in B_N. Prove that you can always do this until you have n total linearly independent vectors.
    • +
  • Show that B_R = \{h(\vec{\beta_{k+1}}), h(\vec{\beta_{k+2}}), \ldots, h(\vec{\beta_n})\} is a basis for the range space. Start by showing that it is linearly independent, and be sure you prove that each element of the range space can be written as a linear combination of B_R.
    • +
  • Show that B_R spans the range space.
    • +
  • State your conclusion.
    • +
+ + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/03-AT/04.ptx b/linear-algebra/source/03-AT/04.ptx new file mode 100644 index 00000000..5f5f7e9d --- /dev/null +++ b/linear-algebra/source/03-AT/04.ptx @@ -0,0 +1,1066 @@ + +
+ Injective and Surjective Linear Maps (AT4) + + + + + + Warm Up + + +

+ Consider the linear transformation T\colon\IR^4\to\IR^3 that is represented by the standard matrix A=\left[\begin{array}{cccc} 3 & 4 & 7 & 1\\ -1 & 1 & 0 & 2 \\ 2 & 1 & 3 & -1 \end{array}\right]. + Which of the following processes helps us compute a basis for \Im T and which helps us compute a basis for \ker T? +

+
    +
  1. +

    + Compute \RREF(A) and consider the set of columns of A that correspond to columns in \RREF(A) with pivots. +

    +
    2. +
  2. +

    + Calculate a basis for the solution space to the homogenous system of equations for which A is the coefficient matrix. +

    +
    3. +
+ + + +Class Activities + + + +

+Let T: V \rightarrow W be a linear transformation. +T is called injective or one-to-one if T does not map two +distinct vectors to the same place. More precisely, T is injective if +T(\vec{v}) \neq T(\vec{w}) whenever \vec{v} \neq \vec{w}. +

+
+ An injective transformation and a non-injective transformation + + +\begin{tikzpicture}[x=0.15in,y=0.15in] + \begin{scope}[shift={(0,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (-1.5,1) + node[anchor=south east] {\(\vec v\)}; + \draw[thick,-latex,red] (0,0) -- (1.5,-2) + node[anchor=south west] {\(\vec w\)}; + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (2,1) + node[anchor=south west] {\(T(\vec v)\)}; + \draw[thick,-latex,red] (0,0) -- (1,2) + node[anchor=south west] {\(T(\vec w)\)}; + \end{scope} + \node[anchor=north] at (5,-1) {injective}; +\end{tikzpicture} +\hspace{3em} +\begin{tikzpicture}[x=0.15in,y=0.15in] + \begin{scope}[shift={(0,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (2,1) + node[anchor=south west] {\(\vec v\)}; + \draw[thick,-latex,red] (0,0) -- (1,2) + node[anchor=south west] {\(\vec w\)}; + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,purple] (0,0) -- (0.5,2) + node[anchor=south west] {\(T(\vec v)=T(\vec w)\)}; + \end{scope} + \node[anchor=north] at (5,-1) {not injective}; +\end{tikzpicture} + + +
+ + + + + +

+Let T: \IR^3 \rightarrow \IR^2 be given by + + T\left(\left[\begin{array}{c}x \\ y\\z \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y \end{array}\right] + \hspace{3em} + \text{with standard matrix } + \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] + +Is T injective? +

+
    +
  1. +

    + Yes, because T(\vec v)=T(\vec w) whenever \vec v=\vec w. +

    +
    2. +
  2. +

    + Yes, because T(\vec v)\not=T(\vec w) whenever \vec v\not=\vec w. +

    +
    3. +
  3. +

    + No, because + + T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right) + \not= + T\left(\left[\begin{array}{c}0\\0\\2\end{array}\right]\right) + . +

    +
    4. +
  4. +

    + No, because + + T\left(\left[\begin{array}{c}0\\0\\1\end{array}\right]\right) + = + T\left(\left[\begin{array}{c}0\\0\\2\end{array}\right]\right) + . +

    +
    5. +
+ + + + + +

+Let T: \IR^2 \rightarrow \IR^3 be given by + + T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right] + \hspace{3em} + \text{with standard matrix } + \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] + +Is T injective? +

+
    +
  1. +

    + Yes, because T(\vec v)=T(\vec w) whenever \vec v=\vec w. +

    +
    2. +
  2. +

    + Yes, because T(\vec v)\not=T(\vec w) whenever \vec v\not=\vec w. +

    +
    3. +
  3. +

    + No, because + + T\left(\left[\begin{array}{c}1\\2\end{array}\right]\right) + \not= + T\left(\left[\begin{array}{c}3\\4\end{array}\right]\right) + . +

    +
    4. +
  4. +

    + No, because + + T\left(\left[\begin{array}{c}1\\2\end{array}\right]\right) + = + T\left(\left[\begin{array}{c}3\\4\end{array}\right]\right) + . +

    +
    5. +
+ + + + + +

+Let T: V \rightarrow W be a linear transformation. +T is called surjective or onto if every element of W is mapped to by an element of V. More precisely, for every \vec{w} \in W, there is some \vec{v} \in V with T(\vec{v})=\vec{w}. +

+
+ + +\begin{tikzpicture}[x=0.15in,y=0.15in] + \begin{scope}[shift={(0,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (1,2); + \draw[thick,-latex,blue] (0,0) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (1,-1); + \draw[thick,-latex,blue] (0,0) -- (-2,3); + \draw[thick,-latex,blue] (0,0) -- (-3,-2); + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (0.5,2); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-1.5,1); + \draw[thick,-latex,blue] (0,0) -- (0,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2,-2); + \fill[color=blue, opacity=0.5] (-2,-2) rectangle (2,2); + \end{scope} + \node[anchor=north] at (5,-2) {surjective}; +\end{tikzpicture} +\hspace{3em} +\begin{tikzpicture}[x=0.15in,y=0.15in] + \begin{scope}[shift={(0,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (0.5,2); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-1.5,1); + \draw[thick,-latex,blue] (0,0) -- (0,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2,-2); + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (0.5,1.5); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-2.5,1); + \draw[thick,-latex,blue] (0,0) -- (-0.5,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2.5,-0.5); + \fill[color=blue, opacity=0.5] (-2,-2) -- (3,-1) -- (2,2) -- (-3,1) -- (-2,-2); + \end{scope} + \node[anchor=north] at (5,-2) {not surjective}; +\end{tikzpicture} + + + A surjective transformation and a non-surjective transformation +
+ + + + + +

+Let T: \IR^2 \rightarrow \IR^3 be given by + + T\left(\left[\begin{array}{c}x \\ y \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y \\ 0 \end{array}\right] + \hspace{3em} + \text{with standard matrix } + \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] + +Is T surjective? +

+
    +
  1. +

    + Yes, because for every \vec w=\left[\begin{array}{c}x\\y\\z\end{array}\right]\in\IR^3, +there exists \vec v=\left[\begin{array}{c}x\\y\end{array}\right]\in\IR^2 such that +T(\vec v)=\vec w. +

    +
    2. +
  2. +

    + No, because + + T\left(\left[\begin{array}{c}x\\y\end{array}\right]\right) + +can never equal + + \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] + . +

    +
    3. +
  3. +

    + No, because + + T\left(\left[\begin{array}{c}x\\y\end{array}\right]\right) + +can never equal + + \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right] + . +

    +
    4. +
+ + + + + +

+Let T: \IR^3 \rightarrow \IR^2 be given by + + T\left(\left[\begin{array}{c}x \\ y\\z \end{array}\right] \right) + = + \left[\begin{array}{c} x \\ y \end{array}\right] + \hspace{3em} + \text{with standard matrix } + \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] + +Is T surjective? +

+
    +
  1. +

    + Yes, because for every \vec w=\left[\begin{array}{c}x\\y\end{array}\right]\in\IR^2, +there exists \vec v=\left[\begin{array}{c}x\\y\\42\end{array}\right]\in\IR^3 such that +T(\vec v)=\vec w. +

    +
    2. +
  2. +

    + Yes, because for every \vec w=\left[\begin{array}{c}x\\y\end{array}\right]\in\IR^2, +there exists \vec v=\left[\begin{array}{c}0\\0\\z\end{array}\right]\in\IR^3 such that +T(\vec v)=\vec w. +

    +
    3. +
  3. +

    + No, because + + T\left(\left[\begin{array}{c}x\\y\\z\end{array}\right]\right) + +can never equal + + \left[\begin{array}{c} 3\\-2 \end{array}\right] + . +

    +
    4. +
+ + + + + + + +

+Let T: V \rightarrow W be a linear transformation where +\ker T contains multiple vectors. What can you conclude? +

+
    +
  1. +

    + T is injective +

    +
    2. +
  2. +

    + T is not injective +

    +
    3. +
  3. +

    + T is surjective +

    +
    4. +
  4. +

    + T is not surjective +

    +
    5. +
+ + + + + +

+A linear transformation T is injective if and only if \ker T = \{\vec{0}\}. +Put another way, an injective linear transformation may be +recognized by its trivial kernel. +

+
+ + +\begin{tikzpicture}[x=0.15in,y=0.15in] + \begin{scope}[shift={(0,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (-1.5,1) + node[anchor=south east] {\(\vec v\)}; + \draw[thick,-latex,red] (0,0) -- (1.5,-2) + node[anchor=south west] {\(\vec w\)}; + \fill[green!50!black] (0,0) circle (0.2) + node[anchor=south west] {\(\vec{0}\)}; + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (2,1) + node[anchor=south west] {\(T(\vec v)\)}; + \draw[thick,-latex,red] (0,0) -- (1,2) + node[anchor=south west] {\(T(\vec w)\)}; + \fill[green!50!black] (0,0) circle (0.2) + node[anchor=south east] {\(T(\vec{0})=\vec{0}\)}; + \end{scope} +\end{tikzpicture} + + + A linear transformation with trivial kernel, which is therefore injective +
+ + + + + +

+Let T: V \rightarrow \IR^3 be a linear transformation where +\Im T may be spanned by only two vectors. +What can you conclude? +

+
    +
  1. +

    + T is injective +

    +
    2. +
  2. +

    + T is not injective +

    +
    3. +
  3. +

    + T is surjective +

    +
    4. +
  4. +

    + T is not surjective +

    +
    5. +
+ + + + + +

+A linear transformation T:V \rightarrow W is surjective if and only if +\Im T = W. Put another way, a surjective linear transformation may be +recognized by its identical codomain and image. +

+
+ + +\begin{tikzpicture}[x=0.15in,y=0.15in] + \begin{scope}[shift={(0,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (1,2); + \draw[thick,-latex,blue] (0,0) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (1,-1); + \draw[thick,-latex,blue] (0,0) -- (-2,3); + \draw[thick,-latex,blue] (0,0) -- (-3,-2); + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (0.5,2); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-1.5,1); + \draw[thick,-latex,blue] (0,0) -- (0,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2,-2); + \fill[color=blue, opacity=0.5] (-2,-2) rectangle (2,2); + \end{scope} + \node[anchor=north] at (5,-2) {surjective, \(\Im T=\IR^2\)}; +\end{tikzpicture} +\hspace{3em} +\begin{tikzpicture}[x=0.15in,y=0.15in] + \begin{scope}[shift={(0,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (0.5,2); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-1.5,1); + \draw[thick,-latex,blue] (0,0) -- (0,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2,-2); + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (0.5,1.5); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-2.5,1); + \draw[thick,-latex,blue] (0,0) -- (-0.5,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2.5,-0.5); + \fill[color=blue, opacity=0.5] (-2,-2) -- (3,-1) -- (2,2) -- (-3,1) -- (-2,-2); + \end{scope} + \node[anchor=north] at (5,-2) {not surjective, \(\Im T\not=\IR^3\)}; +\end{tikzpicture} + + + A linear transformation with identical codomain and image, which is therefore surjective; and a linear transformation with an image smaller than the codomain \IR^3, which is therefore not surjective. +
+ + + + + +

+ A transformation that is both injective and surjective is said to be + bijective. +

+ + + + + +

+Let T: \IR^n \rightarrow \IR^m be a linear map with standard matrix A. +Determine whether each of the following statements +means T is (A) injective, (B) surjective, or +(C) bijective (both). +

    +

  1. The kernel of T is trivial, i.e. \ker T=\{\vec 0\}. +

    2. +

  2. The image of T equals its codomain, i.e. \Im T=\IR^m. +

    3. +

  3. For every \vec w\in \IR^m, the set \{\vec v\in \IR^n|T(\vec v)=\vec w\} contains + exactly one vector. +

    4. +
+

+ + + + + +

+Let T: \IR^n \rightarrow \IR^m be a linear map with standard matrix A. +Determine whether each of the following statements +means T is (A) injective, (B) surjective, or +(C) bijective (both). +

    +

  1. The columns of A span \IR^m. +

    2. +

  2. The columns of A form a basis for \IR^m. +

    3. +

  3. The columns of A are linearly independent. +

    4. +
+

+ + + + + +

+Let T: \IR^n \rightarrow \IR^m be a linear map with standard matrix A. +Determine whether each of the following statements +means T is (A) injective, (B) surjective, or +(C) bijective (both). +

    +

  1. \RREF(A) is the identity matrix. +

    2. +

  2. Every column of \RREF(A) has a pivot. +

    3. +

  3. Every row of \RREF(A) has a pivot. +

    4. +
+

+ + + + + +

+Let T: \IR^n \rightarrow \IR^m be a linear map with standard matrix A. +Determine whether each of the following statements +means T is (A) injective, (B) surjective, or +(C) bijective (both). +

    +

  1. The system of linear equations given by the augmented matrix + \left[\begin{array}{c|c}A & \vec{b} \end{array}\right] has a solution for all \vec{b} \in \IR^m. +

    2. +

  2. The system of linear equations given by the augmented matrix + \left[\begin{array}{c|c}A & \vec{b} \end{array}\right] has exactly one solution for all \vec{b} \in \IR^m. +

    3. +

  3. The system of linear equations given by the augmented matrix + \left[\begin{array}{c|c} A & \vec{0} \end{array}\right] has exactly one solution. +

    4. +
+

+ + + + +

+ The easiest way to determine if the linear map with standard matrix A + is injective is to see if \RREF(A) has a pivot in each column. +

+

+ The easiest way to determine if the linear map with standard matrix A + is surjective is to see if \RREF(A) has a pivot in each row. +

+ + + + +

+ What can you conclude about the linear map + T:\IR^2\to\IR^3 with standard matrix + \left[\begin{array}{cc} a & b \\ c & d \\ e & f \end{array}\right]? +

+
    +

  1. Its standard matrix has more columns than rows, so T is not + injective. +

    2. +

  2. Its standard matrix has more columns than rows, so T is + injective. +

    3. +

  3. Its standard matrix has more rows than columns, so T is not + surjective. +

    4. +

  4. Its standard matrix has more rows than columns, so T is + surjective. +

    5. +
+ + + + + +

+ What can you conclude about the linear map T:\IR^3\to\IR^2 with standard matrix + \left[\begin{array}{ccc} a & b & c \\ d & e & f \end{array}\right]? +

+
    +

  1. Its standard matrix has more columns than rows, so T is not + injective. +

    2. +

  2. Its standard matrix has more columns than rows, so T is + injective. +

    3. +

  3. Its standard matrix has more rows than columns, so T is not + surjective. +

    4. +

  4. Its standard matrix has more rows than columns, so T is + surjective. +

    5. +
+ + + + + + +

The following are true for any linear map T:V\to W:

+
    +

  • If \dim(V)>\dim(W), then T is not injective. +

    • +

  • If \dim(V)<\dim(W), then T is not surjective. +

    • +
+

+ Basically, a linear transformation cannot reduce dimension without collapsing + vectors into each other, and a linear transformation cannot + increase dimension from its domain to its image. +

+
+ + + \begin{tikzpicture}[x=0.12in,y=0.12in] + \begin{scope}[shift={(0,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (2,1) + node[anchor=south west] {\(\vec v\)}; + \draw[thick,-latex,red] (0,0) -- (1,2) + node[anchor=south west] {\(\vec w\)}; + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,purple] (0,0) -- (0.5,2) + node[anchor=south west] {\(T(\vec v)=T(\vec w)\)}; + \end{scope} + \node[anchor=north] at (5,-1) {not injective, \(3>2\)}; + \end{tikzpicture} + \hspace{3em} + \begin{tikzpicture}[x=0.12in,y=0.12in] + \begin{scope}[shift={(0,1)}] + \draw (-2,0) -- (2,0); + \draw (0,-2) -- (0,2); + \draw[thick,-latex,blue] (0,0) -- (0.5,2); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-1.5,1); + \draw[thick,-latex,blue] (0,0) -- (0,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2,-2); + \end{scope} + \draw[dashed,-latex] (3,3) to [bend left=30] (7,3); + \begin{scope}[shift={(9,0)}] + \draw (0,0) -- (3,0); + \draw (0,0) -- (0,3); + \draw (0,0) -- (-2,-1); + \draw[thick,-latex,blue] (0,0) -- (0.5,1.5); + \draw[thick,-latex,blue] (0,0) -- (2,1); + \draw[thick,-latex,blue] (0,0) -- (-2.5,1); + \draw[thick,-latex,blue] (0,0) -- (-0.5,-1.5); + \draw[thick,-latex,blue] (0,0) -- (2.5,-0.5); + \fill[color=blue, opacity=0.5] (-2,-2) -- (3,-1) -- (2,2) -- (-3,1) -- (-2,-2); + \end{scope} + \node[anchor=north] at (5,-2) {not surjective, \(2<3\)}; + \end{tikzpicture} + + + A linear transformation whose domain has a larger dimension than its codomain, and is therefore not injective; and a linear transformation whose domain has a smaller dimension than its codomain, and is therefore not surjective. +
+

+ But dimension arguments cannot be used to prove a map is + injective or surjective. +

+ + + + + + +

+Suppose T: \IR^n \rightarrow \IR^4 with standard matrix +A=\left[\begin{array}{cccc} + a_{11}&a_{12}&\cdots&a_{1n}\\ + a_{21}&a_{22}&\cdots&a_{2n}\\ + a_{31}&a_{32}&\cdots&a_{3n}\\ + a_{41}&a_{42}&\cdots&a_{4n}\\ +\end{array}\right] is bijective. +

+ + +

+How many pivot rows must \RREF A have? +

+ + +

+ How many pivot columns must \RREF A have? +

+ + +

+What is \RREF A? +

+ + + + + + +

+Let T: \IR^n \rightarrow \IR^n be a bijective linear map with +standard matrix A. Label each of the following as true or false. +

+
    +

  1. \RREF(A) is the identity matrix. +

    2. +

  2. The columns of A form a basis for \IR^n +

    3. +

  3. The system of linear equations given by the augmented matrix \left[\begin{array}{c|c} A & \vec{b} \end{array}\right] has exactly one solution +for each \vec b \in \IR^n. +

    4. +
+ + + + + +

+ The easiest way to show that the linear map with standard matrix A + is bijective is to show that \RREF(A) is the identity matrix. +

+ + + + + +

+Let T: \IR^3 \rightarrow \IR^3 be given by the standard matrix +A=\left[\begin{array}{ccc} 2&1&-1 \\ 4&1&1 \\ 6&2&1\end{array}\right]. +Which of the following must be true? +

+
    +

  1. T is neither injective nor surjective +

    2. +

  2. T is injective but not surjective +

    3. +

  3. T is surjective but not injective +

    4. +

  4. T is bijective. +

    5. +
+ + + + +rref([2,1,-1; 4,1,1; 6,2,1]) + + + + +

+Let T: \IR^3 \rightarrow \IR^3 be given by +T\left(\left[\begin{array}{ccc} x \\ y \\ z \end{array}\right] \right) = +\left[\begin{array}{c} 2x+y-z \\ 4x+y+z \\ 6x+2y\end{array}\right]. +Which of the following must be true? +

+
    +

  1. T is neither injective nor surjective +

    2. +

  2. T is injective but not surjective +

    3. +

  3. T is surjective but not injective +

    4. +

  4. T is bijective. +

    5. +
+ + + +rref([2,1,-1; 4,1,1; 6,2,0]) + + + + +

+Let T: \IR^2 \rightarrow \IR^3 be given by +T\left(\left[\begin{array}{c} x \\ y \end{array}\right] \right) = +\left[\begin{array}{c} 2x+3y \\ x-y \\ x+3y\end{array}\right]. +Which of the following must be true? +

+
    +

  1. T is neither injective nor surjective +

    2. +

  2. T is injective but not surjective +

    3. +

  3. T is surjective but not injective +

    4. +

  4. T is bijective. +

    5. +
+ + + +rref([2,3;1,-1;1,3]) + + + + +

+Let T: \IR^3 \rightarrow \IR^2 be given by +T\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = +\left[\begin{array}{c} 2x+y-z \\ 4x+y+z\end{array}\right]. +Which of the following must be true? +

+
    +

  1. T is neither injective nor surjective +

    2. +

  2. T is injective but not surjective +

    3. +

  3. T is surjective but not injective +

    4. +

  4. T is bijective. +

    5. +
+ + + +rref([2,1,-1;4,1,1]) + + + + + + Cool Down + + +

+ Let T\colon\IR^n\to\IR^m be a linear transformation with standard matrix A. + We reasoned during class that the following statements are logically equivalent: +

+
    +
  1. +

    + The columns of A are linearly independent. +

    +
    2. +
  2. +

    + \RREF(A) has a pivot in each column. +

    +
    3. +
  3. +

    + The transformation T is injective. +

    +
    4. +
  4. +

    + The system of equations given by [A|\vec{0}] has a unique solution. +

    +
    5. +
+ While they are all logically equivalent, they are different statements that offer varied perspectives on our growing conceptual knowledge of linear algebra. + + + +

+ If you are asked to decide if a transformation T is injective, which of the above statements do you think is the most useful? +

+ + + + +

+ Can you think of some situations in which translating between these four statements might be useful to you? +

+ + + + + + +

+ Let T\colon\IR^n\to\IR^m be a linear transformation with standard matrix A. + We reasoned during class that the following statements are logically equivalent: +

+
    +
  1. +

    + The columns of A span all of \IR^m. +

    +
    2. +
  2. +

    + \RREF(A) has a pivot in each row. +

    +
    3. +
  3. +

    + The transformation T is surjective. +

    +
    4. +
  4. +

    + The system of equations given by [A|\vec{b}] is always consistent. +

    +
    5. +
+ While they are all logically equivalent, they are different statements that offer varied perspectives on our growing conceptual knowledge of linear algebra. + + + +

+ If you are asked to decide if a transformation T is surjective, which of the above statements do you think is the most useful? +

+ + + + +

+ Can you think of some situations in which translating between these four statements might be useful to you? +

+ + + + + + + + Videos +
+ + Video: The kernel and image of a linear transformation +
+
+ + Video: Finding a basis of the image of a linear transformation +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + + Suppose that f:V \rightarrow W is a linear transformation between two vector spaces V and W. State carefully what conditions f must satisfy. Let \vec{0_V} and \vec{0_W} be the zero vectors in V and W respectively. +
    +
  • Prove that f is one-to-one if and only if f(\vec{0_V}) = \vec{0_W}, and that \vec{0_V} is the unique element of V which is mapped to \vec{0_W}. Remember that this needs to be done in both directions. First prove the if and only if statement, and then show the uniqueness.
    • + +
  • Do not use subtraction in your proof. The only vector space operation we have is addition, and a structure preserving function only preserves addition. If you are writing \vec{v} - \vec{v} = \vec{0}_V, what you really mean is that \vec{v} \oplus \vec{v}^{-1} = \vec{0}_V, where \vec{v}^{-1} is the additive inverse of \vec{v}.
+ + + +

Start with an n-dimensional vector space V. We can define the dual of V, denoted V^*, by +V^* = \{h:V \rightarrow \mathbb{R}: h \mbox{ is linear}\}. +Prove that V is isomorphic toV^*. Here are some things to think about as you work through this. +

    +
  • Start by assuming you have a basis for V. How many basis vectors should you have?
    • +
  • For each basis vector in V, define a function that returns 1 if it's given that basis vector, and returns 0 if it's given any other basis vector. For example, if \vec{b_i} and \vec{b_j} are each members of the basis for V, and you'll need a function f_i:V \rightarrow \{0,1\}, where f_i(b_i) = 1 and f_i(b_j)= 0 for all j \neq i.
    • +
  • How many of these functions will you need? Show that each of them is in V^*.
    • +
  • Show that the functions you found in the last part are a basis for V^*? To do this, take an arbitrary function h \in V^* and some vector \vec{v} \in V. Write \vec{v} in terms of the basis you chose earlier. How can you write h(\vec{v}), with respect to that basis? Pay attention to the fact that all functions in V^* are linear.
    • +
  • Now that you've got a basis for V and a basis for V^*, can you find an isomorphism?
    • +
+

+ + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/03-AT/05.ptx b/linear-algebra/source/03-AT/05.ptx new file mode 100644 index 00000000..f190c0d9 --- /dev/null +++ b/linear-algebra/source/03-AT/05.ptx @@ -0,0 +1,752 @@ + +
+ Vector Spaces (AT5) + + + + + + Warm Up + + + +

+ How would you describe a sandwich to someone who has never seen a sandwich before? +

+ + + + +

+ How would you describe to someone what a vector is? +

+ + + + + +Class Activities + +

+Consider the following applications of properties of the real numbers +\mathbb R: +

+
    +

  1. 1+(2+3)=(1+2)+3. +

    +
    2. +
  2. +

    + + 7+4=4+7. +

    +
    3. +
  3. +

    + There exists some \unknown + where 5+\unknown=5. +

    +
    4. +
  4. +

    + There exists some \unknown + where 9+\unknown=0. +

    +
    5. +
  5. +

    + \frac{1}{2}(1+7) + is the only number that is equally distant from + 1 and 7. +

    +
    6. +
+ + + + +

+Which of the following properites of \IR^2 Euclidean vectors is NOT true? +

+
    +

  1. \left[\begin{array}{c} x_1\\x_2\end{array}\right] + +\left(\left[\begin{array}{c} y_1\\y_2\end{array}\right] + +\left[\begin{array}{c} z_1\\z_2\end{array}\right]\right)= + \left(\left[\begin{array}{c} x_1\\x_2\end{array}\right] + +\left[\begin{array}{c} y_1\\y_2\end{array}\right]\right) + +\left[\begin{array}{c} z_1\\z_2\end{array}\right]. +

    +
    2. +
  2. +

    + + \left[\begin{array}{c}x_1\\x_2\end{array}\right] + + + \left[\begin{array}{c}y_1\\y_2\end{array}\right] + = + \left[\begin{array}{c}y_1\\y_2\end{array}\right] + + + \left[\begin{array}{c}x_1\\x_2\end{array}\right]. +

    +
    3. +
  3. +

    + There exists some \left[\begin{array}{c}\unknown\\\unknown\end{array}\right] + where \left[\begin{array}{c}x_1\\x_2\end{array}\right] + +\left[\begin{array}{c}\unknown\\\unknown\end{array}\right] + =\left[\begin{array}{c}x_1\\x_2\end{array}\right]. +

    +
    4. +
  4. +

    + There exists some \left[\begin{array}{c}\unknown\\\unknown\end{array}\right] + where \left[\begin{array}{c}x_1\\x_2\end{array}\right]+ + \left[\begin{array}{c}\unknown\\\unknown\end{array}\right]= + \left[\begin{array}{c}0\\0\end{array}\right]. +

    +
    5. +
  5. +

    + \displaystyle\frac{1}{2}\left(\left[\begin{array}{c}x_1\\x_2\end{array}\right] + + \left[\begin{array}{c}y_1\\y_2\end{array}\right] \right) + is the only vector whose endpoint is equally distant from the endpoints of + \left[\begin{array}{c}x_1\\x_2\end{array}\right] and + \left[\begin{array}{c}y_1\\y_2\end{array}\right]. +

    +
    6. +
+ + + +

+ Consider the following applications of properites of the real numbers + \mathbb R: +

+
    +
  1. +

    + 3(2(7))=(3\cdot 2)(7). +

    +
    2. +
  2. +

    + 1(19)=19. +

    +
    3. +
  3. +

    + There exists some \unknown + such that \unknown \cdot 4= + 9. +

    +
    4. +
  4. +

    + 3\cdot (2+8)=3\cdot 2+3\cdot 8. +

    +
    5. +
  5. +

    + (2+7)\cdot 4=2\cdot 4+7\cdot 4. +

    +
    6. +
+ + + +

+Which of the following properites of \IR^2 Euclidean vectors is NOT true? +

+
    +
  1. +

    + a\left(b\left[\begin{array}{c}x_1\\x_2\end{array}\right]\right)= + ab\left[\begin{array}{c}x_1\\x_2\end{array}\right]. +

    +
    2. +
  2. +

    + 1\left[\begin{array}{c}x_1\\x_2\end{array}\right]= + \left[\begin{array}{c}x_1\\x_2\end{array}\right]. +

    +
    3. +
  3. +

    + There exists some \unknown + such that \unknown\left[\begin{array}{c}x_1\\x_2\end{array}\right]= + \left[\begin{array}{c}y_1\\y_2\end{array}\right]. +

    +
    4. +
  4. +

    + a(\vec u+\vec v)=a\vec u+a\vec v. +

    +
    5. +
  5. +

    + (a+b)\vec v=a\vec v+b\vec v. +

    +
    6. +
+ + + + + +

+ Every Euclidean vector space \mathbb R^n satisfies the following properties, where + \vec u,\vec v,\vec w are Euclidean vectors and a,b are scalars. +

+
    +

  1. Vector addition is associative: \vec u + (\vec v + \vec w)= + (\vec u + \vec v) + \vec w. +

    +
    2. +
  2. +

    Vector addition is commutative: + \vec u + \vec v= + \vec v + \vec u. +

    +
    3. +
  3. +

    An additive identity exists: + There exists some \vec z + where \vec v + \vec z=\vec v. +

    +
    4. +
  4. +

    Additive inverses exist: + There exists some -\vec v + where \vec v + (-\vec v)=\vec z. +

    +
    5. +
  5. +

    Scalar multiplication is associative: + a (b \vec v)=(ab) \vec v. +

    +
    6. +
  6. +

    1 is a multiplicative identity: + 1 \vec v=\vec v. +

    +
    7. +
  7. +

    Scalar multiplication distributes over vector addition: + a (\vec u + \vec v)=(a \vec u) + (a \vec v). +

    +
    8. +
  8. +

    Scalar multiplication distributes over scalar addition: + (a+ b) \vec v=(a \vec v) + (b \vec v). +

    +
    9. +
+ + + + + + +

+ A vector spacevector space V is any set of mathematical objects, called vectors vector , and a set of numbers, called scalars scalar , with + associated addition \oplus and scalar multiplication \odot + operations that satisfy the following properties. + Let \vec u,\vec v,\vec w be vectors belonging to V, and let a,b be scalars. +

+ +
    +

  1. Vector addition is associative: \vec u\oplus (\vec v\oplus \vec w)= + (\vec u\oplus \vec v)\oplus \vec w. +

    +
    2. +
  2. +

    Vector addition is commutative: + \vec u\oplus \vec v= + \vec v\oplus \vec u. +

    +
    3. +
  3. +

    An additive identity exists: + There exists some \vec z + where \vec v\oplus \vec z=\vec v.additive identity +

    +
    4. +
  4. +

    Additive inverses exist: + There exists some -\vec v + where \vec v\oplus (-\vec v)=\vec z.additive inverse +

    +
    5. +
  5. +

    Scalar multiplication is associative: + a\odot(b\odot\vec v)=(ab)\odot\vec v. +

    +
    6. +
  6. +

    1 is a multiplicative identity: + 1\odot\vec v=\vec v. +

    +
    7. +
  7. +

    Scalar multiplication distributes over vector addition: + a\odot(\vec u\oplus \vec v)=(a\odot\vec u)\oplus(a\odot\vec v). +

    +
    8. +
  8. +

    Scalar multiplication distributes over scalar addition: + (a+ b)\odot\vec v=(a\odot\vec v)\oplus(b\odot \vec v). +

    +
    9. +
+ + + + + + + + + + +

+ Consider the + set \IC of complex numbers with the usual defintion for addition: + (a+b\mathbf i)\oplus(c+d\mathbf i)=(a+c)+(b+d)\mathbf i. +

+

Let + \vec u=a+b\mathbf{i}, \vec v=c+d\mathbf{i}, and \vec w=e+f\mathbf{i}. Then + + + \vec u\oplus(\vec v \oplus \vec w) + &= + (a+b\mathbf{i})\oplus((c+d\mathbf{i})\oplus(e+f\mathbf{i})) + + + &= + (a+b\mathbf{i})\oplus((c+e)+(d+f)\mathbf{i}) + + + &=(a+c+e)+(b+d+f)\mathbf{i} + + + + + (\vec u\oplus\vec v) \oplus \vec w + &= + ((a+b\mathbf{i})\oplus(c+d\mathbf{i}))\oplus(e+f\mathbf{i}) + + + &=((a+c)+(b+d)\mathbf{i})\oplus(e+f\mathbf{i}) + + + &=(a+c+e)+(b+d+f)\mathbf{i} + + +

+

+This proves that complex addition is associative: +\vec u\oplus(\vec v \oplus \vec w) = (\vec u\oplus\vec v) \oplus \vec w. +The seven other vector space properties may also be verified, so \IC +is an example of a vector space. +

+ + + + +

+ The following sets are just a few examples of vector spaces, with the usual/natural + operations for addition and scalar multiplication. +

+
    +
  • +

    + \IR^n: Euclidean vectors with n components. +

    +
    • +
  • +

    + \IC: Complex numbers. +

    +
    • +
  • +

    + M_{m,n}: Matrices of real numbers with m rows and + n columns. +

    +
    • +
  • +

    + \P_n: Polynomials of degree n or less. +

    +
    • +
  • +

    + \P: Polynomials of any degree. +

    +
    • +
  • +

    + C(\IR): Real-valued continuous functions. +

    +
    • +
+ + + + + +

+ Consider the set V=\setBuilder{(x,y)}{y=2^x}. +

+

Which of the following vectors is not in V?

+
    +
  1. +

    + (0, 0) +

    +
    2. +
  2. +

    + (1, 2) +

    +
    3. +
  3. +

    + (2, 4) +

    +
    4. +
  4. +

    + (3, 8) +

    +
    5. +
+ + + + +

+ Consider the set V=\setBuilder{(x,y)}{y=2^x} with + the operation \oplus defined by + + (x_1,y_1)\oplus (x_2,y_2)=(x_1+x_2,y_1y_2) + . +

+

+Let \vec u, \vec v be in V with +\vec u=(1, 2) and \vec v=(2, 4). +Using the operations defined for V, which of the +following is \vec u\oplus\vec v?

+
    +
  1. +

    + (2, 6) +

    +
    2. +
  2. +

    + (2, 8) +

    +
    3. +
  3. +

    + (3, 6) +

    +
    4. +
  4. +

    + (3, 8) +

    +
    5. +
+ + + + +

+Consider the set V=\setBuilder{(x,y)}{y=2^x} with +operations \oplus,\odot defined by + + (x_1,y_1)\oplus (x_2,y_2)=(x_1+x_2,y_1y_2) + \hspace{3em} + c\odot (x,y)=(cx,y^c) + . +

+

+Let a=2, b=-3 be scalars and +\vec u=(1,2) \in V. +

+ + +

+Verify that +(a+b)\odot \vec u=\left(-1,\frac{1}{2}\right). +

+ + +

+Compute the value of + +\left(a\odot \vec u\right)\oplus \left(b\odot \vec u\right) + . +

+ + + + +

+Consider the set V=\setBuilder{(x,y)}{y=2^x} with +operations \oplus,\odot defined by + + (x_1,y_1)\oplus (x_2,y_2)=(x_1+x_2,y_1y_2) + \hspace{3em} + c\odot (x,y)=(cx,y^c) + . +

+

+Let a, b be unspecified +scalars in \mathbb R and \vec u = (x,y) be an +unspecified vector in V. +

+ + +

+Show that both sides of the equation + +(a+b)\odot (x,y)= +\left(a\odot (x,y)\right)\oplus \left(b\odot (x,y)\right) + +simplify to the expression (ax+bx,y^ay^b). +

+ + + +

+Show that V contains an additive identity element +\vec{z}=(\unknown,\unknown) satisfying +(x,y)\oplus(\unknown,\unknown)=(x,y) +for all (x,y)\in V. +

+

+That is, pick appropriate values for +\vec{z}=(\unknown,\unknown) and then +simplify (x,y)\oplus(\unknown,\unknown) +into just (x,y). +

+ + +

+ Is V a vector space? +

    +

  1. Yes

    2. +

  2. No

    3. +

  3. More work is required

    4. +
+

+ + + + + + +

+ It turns out V=\setBuilder{(x,y)}{y=2^x} with +operations \oplus,\odot defined by + + (x_1,y_1)\oplus (x_2,y_2)=(x_1+x_2,y_1y_2) + \hspace{3em} + c\odot (x,y)=(cx,y^c) + +satisifes all eight properties from . +

+

+ Thus, V is a vector space. +

+ + + + + + +

+ Let V=\setBuilder{(x,y)}{x,y\in\IR} + have operations defined by + + (x_1,y_1)\oplus (x_2,y_2)=(x_1+y_1+x_2+y_2,x_1^2+x_2^2) + + + c\odot (x,y)=(x^c,y+c-1) + . +

+ + +

+ Show that 1 is the scalar multiplication identity element + by simplifying 1\odot(x,y) to (x,y). +

+ + +

+ Show that V does not have an additive identity element + \vec z=(z,w) by showing that + (0,-1)\oplus(z,w)\not=(0,-1) no matter what the values of z,w + are. +

+ + +

+ Is V a vector space? +

    +

  1. Yes

    2. +

  2. No

    3. +

  3. More work is required

    4. +
+

+ + + + + +

+ Let V=\setBuilder{(x,y)}{x,y\in\IR} have operations defined by + + (x_1,y_1)\oplus (x_2,y_2)=(x_1+x_2,y_1+3y_2) + \hspace{3em} + c\odot (x,y)=(cx,cy) + . +

+ + +

+Show that scalar multiplication distributes over vector addition, i.e. + c \odot \left( (x_1,y_1) \oplus (x_2,y_2) \right) = c\odot (x_1,y_1) \oplus c\odot (x_2,y_2) +for all c\in \IR,\, (x_1,y_1),(x_2,y_2) \in V. +

+ + +

+ Show that vector addition is not associative, i.e. + (x_1,y_1) \oplus \left((x_2,y_2) \oplus (x_3,y_3)\right) \neq \left((x_1,y_1)\oplus (x_2,y_2)\right) \oplus (x_3,y_3) + for some vectors (x_1,y_1), (x_2,y_2), (x_3,y_3) \in V. +

+ + +

+ Is V a vector space? +

    +

  1. Yes

    2. +

  2. No

    3. +

  3. More work is required

    4. +
+

+ + + + + + Cooldown + + + +

+ What are some objects that are important to you personally, academically, or otherwise that appear vector-like to you? + What makes them feel vector-like? Which axiom for vector spaces does not hold for these objects, if any. +

+ + + + +

+ Our vector space axioms have eight properties. While these eight properties are enough to capture vectors, the objects that we study in the real-world often have additional structures not captured by these axioms. + What are some structures that you have encountered in other classes, or in previous experiences, that are not captured by these eight axioms? +

+ + + + + + + Videos +
+ + Video: Verifying that a vector space property holds +
+
+ + Video: Showing something is not a vector space +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + +

+

    +
  • Show that \mathbb{R}^+, the set of positive real numbers, is a vector space, but where x\oplus y really means the product (so 2 \oplus 3 = 6), and where scalar multiplication \alpha\odot x really means x^\alpha. Yes, you really do need to check all of the properties, but this is the only time I'll make you do so. Remember, examples aren't proofs, so you should start with arbitrary elements of \mathbb R^+ for your vectors. Make sure you're careful about telling the reader what \alpha means.
    • +
  • Prove that the additive identity \vec{z} in an arbitrary vector space is unique.
    • +
  • Prove that additive inverses are unique. Assume you have a vector space V and some \vec{v} \in V. Further, assume \vec{w_1},\vec{w_2} \in V with \vec{v} \oplus \vec{w_1} = \vec{v} \oplus \vec{w_2} = \vec{z}. Prove that \vec{w_1} = \vec{w_2}.
    • +

+ + + + +

Consider the vector space of polynomials, \P_n. Suppose further that n= ab, where a \mbox{ and } b are each positive integers. Conjecture a relationship between M_{a,b} and \P_n. We will investigate this further in section +

+ + + Sample Problem and Solution +

+ Sample problem . +

+ + + + +
diff --git a/linear-algebra/source/03-AT/06.ptx b/linear-algebra/source/03-AT/06.ptx new file mode 100644 index 00000000..53b63ed6 --- /dev/null +++ b/linear-algebra/source/03-AT/06.ptx @@ -0,0 +1,444 @@ + +
+ Polynomial and Matrix Spaces (AT6) + + + + + + Warm Up + + +

+ Consider the following vector equation and statements about it: + x_1\vec{v}_1+x_2\vec{v}_2+\cdots+x_n\vec{v}_n=\vec{w} +

    +
  1. +

    + The above vector equation is consistent for every choice of \vec{w}. +

    +
    2. +
  2. +

    + When the right hand is equal to \vec{0}, the equation has a unique solution. +

    +
    3. +
  3. +

    + The given equation always has a unique solution, no matter what \vec{w} is. +

    +
    4. +
+

+ +

+ Which, if any, of these statements make sense if we no longer assume that the vectors \vec{v}_1,\dots, \vec{v}_n are Euclidean vectors, but rather elements of a vector space? +

+ + + +Class Activities + +

+Nearly every term we've defined for Euclidean vector spaces \mathbb R^n +was actually defined for all kinds of vector spaces: +

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+ + + + +

+Let V be a vector space with the basis \{\vec v_1,\vec v_2,\vec v_3\}. +Which of these completes the following definition for a bijective linear map +T:V\to\mathbb R^3? +T(\vec v)=T(a\vec v_1+b\vec v_2+c\vec v_3)=\left[\begin{array}{c} + \unknown\\\unknown\\\unknown + \end{array}\right] +

    +

  1. \left[\begin{array}{c} + 0\\ 0\\ 0 + \end{array}\right]

    2. +

  2. \left[\begin{array}{c} + a+b+c\\ 0\\ 0 + \end{array}\right]

    3. +

  3. \left[\begin{array}{c} + a\\ b\\ c + \end{array}\right]

    4. +
+

+ + + + + +

+ Every vector space with finite dimension, that is, every + vector space V with a basis of the form + \{\vec v_1,\vec v_2,\dots,\vec v_n\} has a linear bijection T + with Euclidean space \IR^n that simply swaps its basis with + the standard basis \{\vec e_1,\vec e_2,\dots,\vec e_n\} for \IR^n: + + T(c_1\vec v_1+c_2\vec v_2+\dots+c_n\vec v_n) + = + c_1\vec e_1+c_2\vec e_2+\dots+c_n\vec e_n + = + \left[\begin{array}{c} + c_1\\c_2\\\vdots\\c_n + \end{array}\right] + + This transformation (in fact, any linear bijection between vector spaces) + is called an + isomorphismisomorphism, and V is said to be + isomorphicisomorphic to \IR^n. +

+

Note, in particular, that every vector space of dimension n is isomorphic to \IR^n.

+ + + + + + + +

+The matrix space M_{2,2}=\left\{\left[\begin{array}{cc} + a&b\\c&d +\end{array}\right]\middle| a,b,c,d\in\IR\right\} has the basis + +\left\{ +\left[\begin{array}{cc} + 1&0\\0&0 +\end{array}\right], +\left[\begin{array}{cc} + 0&1\\0&0 +\end{array}\right], +\left[\begin{array}{cc} + 0&0\\1&0 +\end{array}\right], +\left[\begin{array}{cc} + 0&0\\0&1 +\end{array}\right] +\right\} +. +

+ + + +

What is the dimension of M_{2,2}? +

    +
  1. 2
    2. +
  2. 3
    3. +
  3. 4
    4. +
  4. 5
    5. +
+

+ + + + +

+ Which Euclidean space is M_{2,2} isomorphic to? +

    +
  1. \IR^2
    2. +
  2. \IR^3
    3. +
  3. \IR^4
    4. +
  4. \IR^5
    5. +
+

+ + + + +

+ Describe an isomorphism T:M_{2,2}\to\IR^{\unknown}: + + T\left(\left[\begin{array}{cc} + a&b\\c&d + \end{array}\right]\right)=\left[\begin{array}{c} + \unknown\\\\\vdots\\\\\unknown + \end{array}\right] + +

+ + + + + + +

+The polynomial space \P^4=\left\{a+bx+cx^2+dx^3+ex^4\middle| a,b,c,d,e\in\IR\right\} has the basis + +\left\{1,x,x^2,x^3,x^4\right\} +. +

+ + + +

What is the dimension of \P^4? +

    +
  1. 2
    2. +
  2. 3
    3. +
  3. 4
    4. +
  4. 5
    5. +
+

+ + + + +

+ Which Euclidean space is \P^4 isomorphic to? +

    +
  1. \IR^2
    2. +
  2. \IR^3
    3. +
  3. \IR^4
    4. +
  4. \IR^5
    5. +
+

+ + + + +

+ Describe an isomorphism T:\P^4\to\IR^{\unknown}: + + T\left(a+bx+cx^2+dx^3+ex^4\right)=\left[\begin{array}{c} + \unknown\\\\\vdots\\\\\unknown + \end{array}\right] + +

+ + + + + +

+Since any finite-dimensional vector space is isomorphic to a Euclidean +space \IR^n, one approach to answering questions about such spaces +is to answer the corresponding question about \IR^n. +

+ + + + + +

+Consider how to construct the polynomial x^3+x^2+5x+1 +as a linear combination of polynomials from the set +\left\{ x^{3} - 2 \, x^{2} + x + 2 , 2 \, x^{2} - 1 , +-x^{3} + 3 \, x^{2} + 3 \, x - 2 , x^{3} - 6 \, x^{2} + 9 \, x + 5 \right\}. +

+ + + +

Describe the vector space involved in this problem, and an isomorphic Euclidean space and relevant Eucldean vectors that can be used to solve this problem.

+ + + + +

Show how to construct an appropriate Euclidean vector from an approriate set of Euclidean vectors.

+ + + + +

Use this result to answer the original question.

+ + + + + + +

+ The space of polynomials \P (of any degree) + has the basis \{1,x,x^2,x^3,\dots\}, + so it is a natural example of an infinite-dimensional vector space. +

+

+ Since \P and other infinite-dimensional vector spaces cannot be treated as + an isomorphic finite-dimensional Euclidean space \IR^n, vectors in + such vector spaces cannot be studied by converting them into Euclidean vectors. + Fortunately, most of the examples we will be + interested in for this course will be finite-dimensional. +

+ + + + + + + Cool Down + + +

+ Let A=\left[\begin{array}{ccc} + -2 & -1 &1\\ + 1 & 0 &0\\ + 0 & -4 &-2\\ + 0 & 1 &3 + \end{array}\right] and let T\colon\IR^3\to\IR^4 denote the corresponding linear transformation. + Note that + \RREF(A)=\left[\begin{array}{ccc} + 1 & 0 &0\\ + 0 & 1 &0\\ + 0 & 0 &1\\ + 0 & 0 &0 + \end{array}\right]. + The following statements are all invalid for at least one reason. Determine what makes them invalid and, suggest alternative valid statements that the author may have meant instead. +

+ + + +

+ The matrix A is injective because \RREF(A) has a pivot in each column. +

+ + + + +

+ The matrix A does not span \IR^4 because \RREF(A) has a row of zeroes. +

+ + + + +

+ The transformation T does not span \IR^4. +

+ + + + +

+ The transformation T is linearly independent. +

+ + + + + + + Videos +
+ + Video: Polynomial and matrix calculations +
+ + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + + Given a matrix M +
    +
  • the span of the set of all columns is the column spacecolumn space
    • + +
  • the span of the set of all rows is the row spacerow space
    • +
  • the rankrank of a matrix is the dimension of the column space.
    • +
+ + + +Calculate the rank of these matrices. +
    +
  • \left[\begin{array}{ccc}2 & 1&3\\1&-1&2\\1&0&3\end{array}\right]
    • +
  • \left[\begin{array}{cccc}1&-1&2&3\\3&-3&6&3\\-2&2&4&5\end{array}\right]
    • +
  • \left[\begin{array}{ccc}1&3&2\\5&1&1\\6&4&3\end{array}\right]
    • +
  • \left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right]
    • +
+ + + + + + +Calculate a basis for the row space and a basis for the column space of the matrix + \left[\begin{array}{cccc}2&0&3&4\\0&1&1&-1\\3&1&0&2\\10&-4&-1&-1\end{array}\right]. + + + + +If you are given the values of a,b, and c, what value of d will cause the matrix \left[\begin{array}{cc}a&b\\c&d\end{array}\right] to have rank 1? + + + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/03-AT/main.ptx b/linear-algebra/source/03-AT/main.ptx new file mode 100644 index 00000000..a7c32f70 --- /dev/null +++ b/linear-algebra/source/03-AT/main.ptx @@ -0,0 +1,12 @@ + + + Algebraic Properties of Linear Maps (AT) + + + + + + + + + diff --git a/linear-algebra/source/03-AT/outcomes/01.ptx b/linear-algebra/source/03-AT/outcomes/01.ptx new file mode 100644 index 00000000..55ba2b78 --- /dev/null +++ b/linear-algebra/source/03-AT/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Determine if a map between Euclidean vector spaces is linear or not. +

\ No newline at end of file diff --git a/linear-algebra/source/03-AT/outcomes/02.ptx b/linear-algebra/source/03-AT/outcomes/02.ptx new file mode 100644 index 00000000..58016b93 --- /dev/null +++ b/linear-algebra/source/03-AT/outcomes/02.ptx @@ -0,0 +1,6 @@ + +

+Translate back and forth between a + linear transformation of Euclidean spaces and its standard matrix, and + perform related computations. +

\ No newline at end of file diff --git a/linear-algebra/source/03-AT/outcomes/03.ptx b/linear-algebra/source/03-AT/outcomes/03.ptx new file mode 100644 index 00000000..795e8953 --- /dev/null +++ b/linear-algebra/source/03-AT/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Compute a basis for the kernel and a basis for the image of a linear map, and verify that the rank-nullity theorem holds for a given linear map. +

\ No newline at end of file diff --git a/linear-algebra/source/03-AT/outcomes/04.ptx b/linear-algebra/source/03-AT/outcomes/04.ptx new file mode 100644 index 00000000..4a18a13e --- /dev/null +++ b/linear-algebra/source/03-AT/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Determine if a given linear map is injective and/or surjective. +

\ No newline at end of file diff --git a/linear-algebra/source/03-AT/outcomes/05.ptx b/linear-algebra/source/03-AT/outcomes/05.ptx new file mode 100644 index 00000000..bacd63a3 --- /dev/null +++ b/linear-algebra/source/03-AT/outcomes/05.ptx @@ -0,0 +1,5 @@ + +

+Explain why a given set with defined addition and scalar multiplication does satisfy a +given vector space property, but nonetheless isn't a vector space. +

\ No newline at end of file diff --git a/linear-algebra/source/03-AT/outcomes/06.ptx b/linear-algebra/source/03-AT/outcomes/06.ptx new file mode 100644 index 00000000..d378cca2 --- /dev/null +++ b/linear-algebra/source/03-AT/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+Answer questions about vector spaces of polynomials or matrices. +

\ No newline at end of file diff --git a/linear-algebra/source/03-AT/outcomes/main.ptx b/linear-algebra/source/03-AT/outcomes/main.ptx new file mode 100644 index 00000000..baf2a08c --- /dev/null +++ b/linear-algebra/source/03-AT/outcomes/main.ptx @@ -0,0 +1,29 @@ + + + + +

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
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    6. +
  6. + +
    7. +
+ \ No newline at end of file diff --git a/linear-algebra/source/03-AT/outcomes/question.ptx b/linear-algebra/source/03-AT/outcomes/question.ptx new file mode 100644 index 00000000..b02e22d2 --- /dev/null +++ b/linear-algebra/source/03-AT/outcomes/question.ptx @@ -0,0 +1,4 @@ + +

+How can we understand linear maps algebraically? +

diff --git a/linear-algebra/source/03-AT/readiness.ptx b/linear-algebra/source/03-AT/readiness.ptx new file mode 100644 index 00000000..87924280 --- /dev/null +++ b/linear-algebra/source/03-AT/readiness.ptx @@ -0,0 +1,48 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    State the definition of a spanning set, and determine if a set of Euclidean vectors spans \IR^n.

    +
      +
    • +

      Review:

      +
      • +
    +
    2. +
  2. +

    State the definition of linear independence, and determine if a set of Euclidean vectors is linearly dependent or independent.

    +
      +
    • +

      Review:

      +
      • +
    +
    3. +
  3. +

    + State the definition of a basis, and determine if a set of Euclidean vectors is a basis. +

    +
      +
    • +

      Review: , +

      +
      • +
    +
    4. +
  4. +

    Find a basis of the solution space to a homogeneous system +of linear equations.

    +
      +
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      Review:

      +
      • +
    +
    5. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/03-AT/samples/01.ptx b/linear-algebra/source/03-AT/samples/01.ptx new file mode 100644 index 00000000..ad0badc7 --- /dev/null +++ b/linear-algebra/source/03-AT/samples/01.ptx @@ -0,0 +1,128 @@ + + +AT1 + +

Answer the following questions about transformations.

+
    +
  1. +

    Consider the following maps of Euclidean vectors P:\mathbb R^3\rightarrow\mathbb R^3 and + Q:\mathbb R^3\rightarrow\mathbb R^3 defined by + P\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)= + \left[\begin{array}{c} 3 \, x - y + z \\ 2 \, x - 2 \, y + 4 \, z \\ -2 \, x - 2 \, y - 3 \, z \end{array}\right] + \hspace{1em} \text{and} \hspace{1em} Q\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right)= + \left[\begin{array}{c} y - 2 \, z \\ -3 \, x - 4 \, y + 12 \, z \\ 5 \, x y + 3 \, z \end{array}\right]. + Without writing a proof, explain why only one of these maps is likely to be a linear transformation. +

    +
    2. +
  2. +

    Consider the following map of Euclidean vectors S:\mathbb R^2\rightarrow\mathbb R^2 + S\left( \left[\begin{array}{c} x \\ y \end{array}\right] \right)=\left[\begin{array}{c} x + 2 \, y \\ -3 \, x y \end{array}\right]. + Prove that S is not a linear transformation. +

    +
    3. +
  3. +

    Consider the following map of Euclidean vectors T:\mathbb R^2\rightarrow\mathbb R^2 + T\left( \left[\begin{array}{c} x \\ y \end{array}\right] \right)=\left[\begin{array}{c} -4 \, x - 5 \, y \\ 2 \, x - 4 \, y \end{array}\right]. + Prove that T is a linear transformation.

    +
    4. +
+ + +
    +
  1. +

    A linear map between Euclidean spaces must consist of linear polynomials in each component. + All three components of P are linear so P is likely to be linear; however, the + third component of Q contains the nonlinear term xy, so Q is unlikely + to be linear. +

    +
    2. +
  2. +

    We need to show either that S fails to preserve either vector addition or + that S fails to preserve scalar multiplication. +

    +

    + For example, for a scalar c \in \IR and a vector \left[\begin{array}{c}x \\y \end{array}\right] \in \IR^2, + we can compute + + S\left(c\left[\begin{array}{c} x \\ y \end{array} \right]\right) = + S\left(\left[\begin{array}{c} cx \\ cy \end{array} \right]\right) = + \left[\begin{array}{c}cx+2cy \\ -3c^2xy \end{array} \right] + + whereas + + cS\left(\left[\begin{array}{c} x \\ y \end{array} \right]\right) = + c\left[\begin{array}{c}x+2y \\ -3xy \end{array} \right] = + \left[\begin{array}{c}cx+2cy \\ -3cxy \end{array} \right]. + + Since -3c^2xy \neq -3cxy, we see that S\left(c\left[\begin{array}{c} x \\ y \end{array} \right]\right) \neq cS\left(\left[\begin{array}{c} x \\ y \end{array} \right]\right), + so S fails to preserve scalar multiplication and cannot be a linear transformation. +

    +

    + Alternatively, we could instead take two vectors \left[\begin{array}{c}x_1 \\y_1 \end{array}\right], \left[\begin{array}{c}x_2 \\y_2 \end{array}\right] \in \IR^2 + and compute + + S \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right] + \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) = + S \left( \left[\begin{array}{c}x_1 +x_2 \\y_1 + y_2 \end{array}\right] \right) = + \left[\begin{array}{c} (x_1+x_2)+2(y_1+y_2) \\ -3(x_1+x_2)(y_1+y_2) \end{array} \right] + + whereas + + S \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right]\right) +S\left( \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) = + \left[\begin{array}{c} x_1+2y_1 \\ -3x_1y_1 \end{array}\right] + + \left[\begin{array}{c} x_2+2y_2 \\ -3x_2y_2 \end{array}\right] = + \left[\begin{array}{c} x_1+2y_1+x_2+2y_2 \\ -3x_1y_1-3x_2y_2 \end{array} \right] + + Since -3(x_1+x_2)(y_1+y_2) \neq -3x_1y_1-3x_2y_2 , we see that S \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right] + \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) \neq + S \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right]\right) +S\left( \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) + , + so S fails to preserve addition and cannot be a linear transformation. +

    +
    3. +
  3. +

    We need to show that T preserves both vector addition and + that T preserves scalar multiplication. +

    +

    + First, let us take two vectors \left[\begin{array}{c}x_1 \\y_1 \end{array}\right], \left[\begin{array}{c}x_2 \\y_2 \end{array}\right] \in \IR^2 + and compute + + T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right] + \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) = + T \left( \left[\begin{array}{c}x_1 +x_2 \\y_1 + y_2 \end{array}\right] \right) = + \left[\begin{array}{c} -4(x_1+x_2)-5(y_1+y_2) \\ 2(x_1+x_2)-4(y_1+y_2)\end{array} \right] + + and + + T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right]\right) +T\left( \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) = + \left[\begin{array}{c} -4x_1-5y_1 \\ 2x_1-4y_1 \end{array}\right] + + \left[\begin{array}{c} -4x_2-5y_2 \\ 2x_2-4y_2 \end{array}\right] = + \left[\begin{array}{c} -4x_1-5y_1-4x_2-5y_2 \\ 2x_1-4y_1+2x_2-4y_2\end{array} \right] + + So we see that T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right] + \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) = + T \left( \left[\begin{array}{c}x_1 \\y_1 \end{array}\right]\right) +T\left( \left[\begin{array}{c}x_2 \\y_2 \end{array}\right]\right) + , + so T preserves addition. +

    +

    + Now, take a scalar c \in \IR and a vector \left[\begin{array}{c}x \\y \end{array}\right] \in \IR^2, + and compute + + T\left(c\left[\begin{array}{c} x \\ y \end{array} \right]\right) = + T\left(\left[\begin{array}{c} cx \\ cy \end{array} \right]\right) = + \left[\begin{array}{c} -4cx-5cy \\ 2cx-4cy\end{array} \right] + + and + + cT\left(\left[\begin{array}{c} x \\ y \end{array} \right]\right) = + c\left[\begin{array}{c}-4x-5y \\ 2x-4y \end{array} \right] = + \left[\begin{array}{c}-4cx-5cy \\ 2cx-4cy\end{array} \right]. + + We see that T\left(c\left[\begin{array}{c} x \\ y \end{array} \right]\right) = cT\left(\left[\begin{array}{c} x \\ y \end{array} \right]\right), + so T preserves scalar multiplication. +

    +

    + Since T preserves both addition and scalar multiplication, we have proven that T is a linear transformation. +

    +
    4. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/03-AT/samples/02.ptx b/linear-algebra/source/03-AT/samples/02.ptx new file mode 100644 index 00000000..d46e90d4 --- /dev/null +++ b/linear-algebra/source/03-AT/samples/02.ptx @@ -0,0 +1,51 @@ + +AT2 + +
    +
  1. +

    +Find the standard matrix for the linear transformation T: \IR^3\rightarrow \IR^4 given by +T\left(\left[\begin{array}{c} x \\ y \\ z \\ \end{array}\right] \right) = \left[\begin{array}{c} -x+y \\ -x+3y-z \\ 7x+y+3z \\ 0 \end{array}\right]. +

    +
    2. +
  2. +

    +Let S: \IR^4 \rightarrow \IR^3 be the linear transformation given by the standard matrix +\left[\begin{array}{cccc} 2 & 3 & 4 & 1 \\ 0 & 1 & -1 & -1 \\ 3 & -2 & -2 & 4 \end{array}\right]. +Compute S\left( \left[\begin{array}{c} -2 \\ 1 \\ 3 \\ 2\end{array}\right] \right) . +

    +
    3. +
+ + +
    +
  1. +

    +Since + + +T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]\right) &= \left[\begin{array}{c} -1 \\ -1 \\ 7 \\0\end{array}\right] + + +T\left(\left[\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right]\right) &= \left[\begin{array}{c} 1 \\ 3 \\ 1 \\0 \end{array}\right] + + +T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right]\right) &= \left[\begin{array}{c} 0 \\ -1 \\ 3 \\ 0 \end{array}\right], + + +the standard matrix for T is \left[\begin{array}{ccc} -1 & 1 & 0 \\ -1 & 3 & -1 \\ 7 & 1 & 3 \\ 0 & 0 & 0 \end{array}\right] . +

    +
    2. +
  2. +

    + +S\left(\left[\begin{array}{c} -2 \\ 1 \\ 3 \\ 2 \end{array}\right] \right) = -2S(\vec{e}_1)+S(\vec{e}_2)+3S(\vec{e}_3)+2S(\vec{e}_4) + + += -2 \left[\begin{array}{c} 2 \\ 0 \\ 3 \end{array}\right] + \left[\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right] + 3 \left[\begin{array}{c} 4 \\ -1 \\ -2 \end{array}\right]+2\left[\begin{array}{c} 1 \\ -1 \\ 4 \end{array}\right] += \left[\begin{array}{c} 13 \\ -4 \\ -6\end{array}\right]. +

    +
    3. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/03-AT/samples/03.ptx b/linear-algebra/source/03-AT/samples/03.ptx new file mode 100644 index 00000000..18b3d144 --- /dev/null +++ b/linear-algebra/source/03-AT/samples/03.ptx @@ -0,0 +1,100 @@ + + +AT3 + +

+Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by + + T\left(\left[\begin{array}{c}x\\y\\z\\w\end{array}\right] \right) = + \left[\begin{array}{c} + x+3y+2z-3w \\ + 2x+4y+6z-10w \\ + x+6y-z+3w + \end{array}\right] + +

+
    +
  1. +

    +Explain how to find the image of T and the kernel of T. +

    +
    2. +
  2. +

    +Explain how to find a basis of the image of T and a basis of the kernel of T. +

    +
    3. +
  3. +

    +Explain how to find the rank and nullity of T, and why the rank-nullity theorem holds for T. +

    +
    4. +
+ + +
    +
  1. +

    + To find the image we compute + \Im(T) = T\left(\vspan\left\{\vec{e}_1,\vec{e}_2,\vec{e}_3,\vec{e}_4\right\}\right) + += \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} + + += \vspan\left\{\left[\begin{array}{c}1 \\ 2 \\ 1 \end{array}\right], \left[\begin{array}{c} 3 \\ 4 \\ 6 \end{array}\right], \left[\begin{array}{c} 2 \\ 6 \\ -1 \end{array}\right], \left[\begin{array}{c} -3 \\ -10 \\ 3 \end{array}\right]\right\}. +

    +
    2. +
  2. +

    +The kernel is the solution set of the corresponding homogeneous system of equations, i.e. + + +x &+& 3y &+& 2z &-& 3w &=& 0 + + +2x &+& 4y &+& 6z &-& 10w &=& 0 + + +x &+& 6y &-& z &+& 3w &=& 0 . + + +So we compute +\RREF\left[\begin{array}{cccc|c} 1 & 3 & 2 & -3 & 0 \\ 2 & 4 & 6 & -10 &0 \\ 1 & 6 & -1 & 3 & 0 \end{array}\right] += \left[\begin{array}{cccc|c} 1 & 0 & 5 & -9 & 0 \\ 0 & 1 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]. +Then, letting z=a and w=b we have + \ker T = \setBuilder{\left[\begin{array}{c}-5a+9b \\ a-2b \\ a \\ b \end{array}\right]}{a,b \in \IR}. +

    +
    3. +
  3. +

    +Since +\Im(T) = \vspan\left\{\left[\begin{array}{c}1 \\ 2 \\ 1 \end{array}\right], \left[\begin{array}{c} 3 \\ 4 \\ 6 \end{array}\right], +\left[\begin{array}{c} 2 \\ 6 \\ -1 \end{array}\right], \left[\begin{array}{c} -3 \\ -10 \\ 3 \end{array}\right]\right\}, +we simply need to find a linearly independent subset of these four spanning vectors. So we compute +\RREF \left[\begin{array}{cccc}1 & 3 & 2 & -3 \\ 2 & 4 & 6 & -10 \\ 1 & 6 & -1 & 3 \end{array}\right] += \left[\begin{array}{cccc} 1 & 0 & 5 & -9 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 0\end{array}\right]. +Since the first two columns are pivot columns, they form a linearly independent spanning set, so a basis for \Im T +is \setList{\left[\begin{array}{c}1\\2\\1 \end{array}\right], \left[\begin{array}{c}3\\4\\6 \end{array}\right]}. +

    +

    +To find a basis for the kernel, note that + \ker T = \setBuilder{\left[\begin{array}{c}-5a+9b \\ a-2b \\ a \\ b \end{array}\right]}{a,b \in \IR} + += \setBuilder{a \left[\begin{array}{c}-5 \\ 1 \\ 1 \\ 0 \end{array}\right]+b \left[\begin{array}{c} 9 \\ -2 \\ 0 \\ 1 \end{array}\right]}{a,b \in \IR} + += \vspan\left\{ \left[\begin{array}{c} -5 \\ 1 \\ 1 \\ 0 \end{array}\right], \left[\begin{array}{c} 9 \\ -2 \\ 0 \\ 1 \end{array}\right]\right\}. +so a basis for the kernel is +\setList{\left[\begin{array}{c}-5 \\ 1 \\ 1 \\ 0 \end{array}\right], +\left[\begin{array}{c}9 \\ -2 \\ 0 \\ 1 \end{array}\right]}. +

    +
    4. +
  4. +

    +The dimension of the image (the rank) is 2, the dimension of the kernel (the nullity) is 2, +and the dimension of the domain of T is 4, so we see 2+2=4, which verifies that the sum +of the rank and nullity of T is the dimension of the domain of T. +

    +
    5. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/03-AT/samples/04.ptx b/linear-algebra/source/03-AT/samples/04.ptx new file mode 100644 index 00000000..1cc1d46f --- /dev/null +++ b/linear-algebra/source/03-AT/samples/04.ptx @@ -0,0 +1,38 @@ + + +AT4 + +

+Let T: \IR^4 \rightarrow \IR^3 be the linear transformation given by the standard matrix \left[\begin{array}{cccc} 1 & 3 & 2 & -3 \\ 2 & 4 & 6 & -10 \\ 1 & 6 & -1 & 3 \end{array}\right]. +

+
    +
  1. +

    +Explain why T is or is not injective. +

    +
    2. +
  2. +

    +Explain why T is or is not surjective. +

    +
    3. +
+ + +

+Compute \RREF\left[\begin{array}{cccc}1 & 3 & 2 & -3 \\ 2 & 4 & 6 & -10 \\ 1 & 6 & -1 & 3 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 5 & -9 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 0\end{array}\right]. +

+
    +
  1. +

    +Note that the third and fourth columns are non-pivot columns, which means \ker T contains infinitely many vectors, so T is not injective. +

    +
    2. +
  2. +

    +Since there are only two pivots, the image (i.e. the span of the columns) is a 2-dimensional subspace (and thus does not equal \IR^3), so T is not surjective. +

    +
    3. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/03-AT/samples/05.ptx b/linear-algebra/source/03-AT/samples/05.ptx new file mode 100644 index 00000000..7ae0ed60 --- /dev/null +++ b/linear-algebra/source/03-AT/samples/05.ptx @@ -0,0 +1,139 @@ + + + +AT5 + +

+Let V be the set of all pairs of numbers (x,y) of real numbers together with the following operations: +

+ + +(x_1,y_1) \oplus (x_2,y_2) &= (2x_1+2x_2,2y_1+2y_2) + + +c\odot (x,y) &= (cx,c^2y) + + +
    +
  1. +

    +Show that scalar multiplication distributes over vector addition: +

    +c\odot \left((x_1,y_1) \oplus (x_2,y_2) \right) = c \odot (x_1,y_1) \oplus c \odot (x_2,y_2) +
    2. +
  2. +

    +Explain why V nonetheless is not a vector space. +

    +
    3. +
+ + +
    +
  1. +

    +We compute both sides: +

    + + + c \odot \left((x_1,y_1) \oplus (x_2,y_2) \right) + &= c \odot (2x_1+2x_2,2y_1+2y_2) + + + &= (c(2x_1+2x_2),c^2(2y_1+2y_2)) + + + &= (2cx_1+2cx_2,2c^2y_1+2c^2y_2) + + +

    +and +

    + + + c\odot (x_1,y_1) \oplus c \odot (x_2,y_2) + &= (cx_1,c^2y_1) \oplus (cx_2,c^2y_2) + + + &= (2cx_1+2cx_2,2c^2y_1+2c^2y_2) + + +

    +Since these are the same, we have shown that the property holds. +

    +
    2. +
  2. +

    +To show V is not a vector space, we must show that it fails one of the 8 defining properties of vector spaces. +We will show that scalar multiplication does not distribute over scalar addition, i.e., there are values +such that + (c+d)\odot(x,y) \neq c \odot(x,y) \oplus d\odot(x,y) +

    +
      +
    • +

      +(Solution method 1) +First, we compute +

      + + +(c+d)\odot(x,y) &= ((c+d)x,(c+d)^2y) + + +&= ( (c+d)x, (c^2+2cd+d^2)y). + + +

      +Then we compute +

      + + +c\odot (x,y) \oplus d\odot(x,y) &= (cx,c^2y) \oplus (dx,d^2y) + + +&= ( 2cx+2dx, 2c^2y+2d^2y). + + +

      +Since (c+d)x\not=2cx+2dy when c,d,x,y=1, the property fails to hold. +

      +
      • +
    • +

      +(Solution method 2) When we let c,d,x,y=1, we may simplify both sides as follows. +

      + + +(c+d)\odot(x,y) &= 2\odot(1,1) + +&= (2\cdot1,2^2\cdot1) + +&=(2,4) + + + + +c\odot (x,y) \oplus d\odot(x,y) &= 1\odot(1,1)\oplus 1\odot(1,1) + + +&= (1\cdot1,1^2\cdot1)\oplus(1\cdot1,1^2\cdot1) + + +&= (1,1)\oplus(1,1) + + +&= (2\cdot1+2\cdot1,2\cdot1+2\cdot1) + + +&= (4,4) + + +

      +Since these ordered pairs are different, the property fails to hold. +

      +
      • +
    +
    3. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/03-AT/samples/06.ptx b/linear-algebra/source/03-AT/samples/06.ptx new file mode 100644 index 00000000..9232c4d3 --- /dev/null +++ b/linear-algebra/source/03-AT/samples/06.ptx @@ -0,0 +1,77 @@ + +AT6 + +
    +
  1. +

    +Given the set +\left\{ x^{3} - 2 \, x^{2} + x + 2 , 2 \, x^{2} - 1 , -x^{3} + 3 \, x^{2} + 3 \, x - 2 , x^{3} - 6 \, x^{2} + 9 \, x + 5 \right\} +write a statement involving the solutions to a + +polynomial +equation that's equivalent to each claim below. +

    +
      +
    • +

      +The set of + +polynomials +is linearly independent. +

      +
      • +
    • +

      +The set of + +polynomials +is linearly dependent. +

      +
      • +
    +
    2. +
  2. +

    +Explain how to determine which of these statements is true. +

    +
    3. +
+ + +

The set of polynomials \left\{ x^{3} - 2 \, x^{2} + x + 2 , 2 \, x^{2} - 1 , -x^{3} + 3 \, x^{2} + 3 \, x - 2 , x^{3} - 6 \, x^{2} + 9 \, x + 5 \right\} + is linearly independent exactly when the polynomial equation + y_1\left( x^{3} - 2 \, x^{2} + x + 2 \right)+y_2\left( 2 \, x^{2} - 1 \right)+y_3\left( -x^{3} + 3 \, x^{2} + 3 \, x - 2 \right)+y_4\left( x^{3} - 6 \, x^{2} + 9 \, x + 5\right)=0 + has no nontrivial (i.e. nonzero) solutions. The set is linearly dependent when this equation has a nontrivial (i.e. nonzero) solution.

+

To solve this equation, we distribute and then collect coefficients to obtain + \left(y_1-y_3+y_4\right)x^3+\left(-2y_1+2y_2+3y_3-6y_4\right)x^2+\left(y_1+3y_3+9y_4\right)x+\left(2y_1-y_2-2y_3+5y_4\right)=0. + These polynomials are equal precisely when their coefficients are equal, leading to the system + \begin{matrix} + y_1 & & &-&y_3 & +&y_4 & = & 0 \\ + -2 y_1 & + & 2y_2 &+&3y_3 & -&6y_4 & = & 0 \\ + y_1 & + & &+&3y_3 & +&9y_4 & = & 0 \\ + 2 y_1 & - & y_2 &-&2y_3 & +&5y_4 & = & 0 + \end{matrix} + . +

+

+ To solve this, we compute + +\mathrm{RREF}\, \left[\begin{array}{cccc|c} +1 & 0 & -1 & 1 & 0\\ + -2 & 2 & 3 & -6 & 0\\ + 1 & 0 & 3 & 9 & 0\\ + 2 & -1 & -2 & 5 & 0 +\end{array}\right] = \left[\begin{array}{cccc|c} +1 & 0 & 0 & 3 & 0\\ +0 & 1 & 0 & -3 & 0\\ +0 & 0 & 1 & 2 & 0\\ +0 & 0 & 0 & 0 & 0 +\end{array}\right] + +

+

+The system has (infintely many) nontrivial solutions, so we that the set of polynomials + is linearly dependent. +

+ + \ No newline at end of file diff --git a/linear-algebra/source/04-MX/01.ptx b/linear-algebra/source/04-MX/01.ptx new file mode 100644 index 00000000..513f555d --- /dev/null +++ b/linear-algebra/source/04-MX/01.ptx @@ -0,0 +1,355 @@ + +
+ Matrices and Multiplication (MX1) + + + + +Class Activities + + +

+If T: \IR^n \rightarrow \IR^m and S: \IR^m \rightarrow \IR^k are linear maps, +then the composition map S\circ T +computed as (S \circ T)(\vec{v})=S(T(\vec{v})) +is a linear map from \IR^n \rightarrow \IR^k. +

+
+ + +\begin{tikzcd}[ampersand replacement=\&] +\IR^n \arrow[rr, bend right, "S\circ T"'] \arrow[r,"T"] \& \IR^m \arrow[r,"S"] \&\IR^k +\end{tikzcd} + + + The composition of two linear maps. +
+ + + + + + +

+Let T: \IR^3 \rightarrow \IR^2 be defined by +the 2\times 3 standard matrix B and +S: \IR^2 \rightarrow \IR^4 be defined by the +4\times 2 standard matrix A: +B=\left[\begin{array}{ccc} 2 & 1 & -3 \\ 5 & -3 & 4 \end{array}\right] + \hspace{2em} +A=\left[\begin{array}{cc} 1 & 2 \\ 0 & 1 \\ 3 & 5 \\ -1 & -2 \end{array}\right]. +

+ + +

+What are the domain and codomain of the composition map S \circ T? +

+
    +

  1. The domain is \IR ^3 and the codomain is \IR^2 +

    2. +

  2. The domain is \IR ^2 and the codomain is \IR^4 +

    3. +

  3. The domain is \IR ^3 and the codomain is \IR^4 +

    4. +

  4. The domain is \IR ^4 and the codomain is \IR^3 +

    5. +
+ + +

+ What size will the standard matrix of S \circ T be? +

+
    +
  1. 4 \text{ (rows)} \times 3 \text{ (columns)} +
    2. +
  2. 3 \text{ (rows)} \times 4 \text{ (columns)} +
    3. +
  3. 3 \text{ (rows)} \times 2 \text{ (columns)} +
    4. +
  4. 2 \text{ (rows)} \times 4 \text{ (columns)} +
    5. +
+ + +

+ Compute + + (S \circ T)(\vec{e}_1) + = + S(T(\vec{e}_1)) + = + S\left(\left[\begin{array}{c} 2 \\ 5\end{array}\right]\right) + = + \left[\begin{array}{c}\unknown\\\unknown\\\unknown\\\unknown\end{array}\right]. + +

+ + +

+ Compute + + (S \circ T)(\vec{e}_2) + . +

+ + +

+ Compute + + (S \circ T)(\vec{e}_3) + . +

+ + +

+ Use (S \circ T)(\vec{e}_1),(S \circ T)(\vec{e}_2),(S \circ T)(\vec{e}_3) + to write the standard matrix for S \circ T. +

+ + + + + +

+We define the product AB of a m \times n matrix A and a +n \times k +matrix B to be the m \times k standard matrix of the composition map of the +two corresponding linear functions. +

+

+For the previous activity, +T was a map \IR^3 \rightarrow \IR^2, and +S was a map \IR^2 \rightarrow \IR^4, so S \circ T gave a map \IR^3 \rightarrow \IR^4 with a +4\times 3 standard matrix: + + AB + = + \left[\begin{array}{cc} 1 & 2 \\ 0 & 1 \\ 3 & 5 \\ -1 & -2 \end{array}\right] + \left[\begin{array}{ccc} 2 & 1 & -3 \\ 5 & -3 & 4 \end{array}\right] + + + = + \left[ + (S \circ T)(\vec{e}_1) \hspace{1em} + (S\circ T)(\vec{e}_2) \hspace{1em} + (S \circ T)(\vec{e}_3) + \right] + = + \left[\begin{array}{ccc} + 12 & -5 & 5 \\ + 5 & -3 & 4 \\ + 31 & -12 & 11 \\ + -12 & 5 & -5 + \end{array}\right] +. +

+ + + + + + + +

+Let S: \IR^3 \rightarrow \IR^2 be given by the matrix +A=\left[\begin{array}{ccc} -4 & -2 & 3 \\ 0 & 1 & 1 \end{array}\right] +and T: \IR^2 \rightarrow \IR^3 be given by the matrix +B=\left[\begin{array}{cc} 2 & 3 \\ 1 & -1 \\ 0 & -1 \end{array}\right]. +

+ + +

+Write the dimensions (rows \times columns) +for A, B, AB, and BA. +

+ + +

+Find the standard matrix AB of S \circ T. +

+ + +

+Find the standard matrix BA of T \circ S. +

+ + + + + +

+Consider the following three matrices. +

+ + A = \left[\begin{array}{ccc}1&0&-3\\3&2&1\end{array}\right] + \hspace{2em} + B = \left[\begin{array}{ccccc}2&2&1&0&1\\1&1&1&-1&0\\0&0&3&2&1\\-1&5&7&2&1\end{array}\right] + \hspace{2em} + C = \left[\begin{array}{cc}2&2\\0&-1\\3&1\\4&0\end{array}\right] + + + +

+Find the domain and codomain of each of the three linear maps corresponding to A, B, and C. +

+ + +

+Only one of the matrix products +AB,AC,BA,BC,CA,CB can actually be computed. +Compute it. +

+ + + + + +

+Let B=\left[\begin{array}{ccc} 3 & -4 & 0 \\ 2 & 0 & -1 \\ 0 & -3 & 3 \end{array}\right], +and let A=\left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right]. +

+ + +

+ Compute the product BA by hand. +

+ + +

+ Check your work using technology. Using Octave: +

+ +B = [3 -4 0 ; 2 0 -1 ; 0 -3 3] +A = [2 7 -1 ; 0 3 2 ; 1 1 -1] +B*A + + + + + +B = [3 -4 0 ; 2 0 -1 ; 0 -3 3] +A = [2 7 -1 ; 0 3 2 ; 1 1 -1] +B*A + + + + +

Of the following three matrices, only two may be multiplied. + + A=\left[\begin{array}{cccc} +-1 & 3 & -2 & -3 \\ +1 & -4 & 2 & 3 +\end{array}\right] \hspace{1em} B=\left[\begin{array}{ccc} +1 & -6 & -1 \\ +0 & 1 & 0 +\end{array}\right] \hspace{1em} C=\left[\begin{array}{ccc} +1 & -1 & -1 \\ +0 & 1 & -2 \\ +-2 & 4 & -1 \\ +-2 & 3 & -1 +\end{array}\right] + + Explain which two can be multiplied and why. Then show how to find their product.

+ + + + +

Let + +T\left(\left[\begin{array}{c}x\\y \end{array}\right]\right)= +\left[\begin{array}{c} x+2y \\ y \\ 3x +5y \\ -x-2y \end{array}\right] + In we adopted the notation + +T\left(\left[\begin{array}{c}x\\y \end{array}\right]\right)= +\left[\begin{array}{c} x+2y \\ y \\ 3x +5y \\ -x-2y \end{array}\right]= +A +\left[\begin{array}{c}x\\y \end{array}\right] = +\left[\begin{array}{cc} 1 & 2 \\ 0 & 1 \\ 3 & 5 \\ -1 & -2 \end{array}\right] +\left[\begin{array}{c}x\\y \end{array}\right] + . + Verify that + \left[\begin{array}{cc} 1 & 2 \\ 0 & 1 \\ 3 & 5 \\ -1 & -2 \end{array}\right] + \left[\begin{array}{c}x\\y \end{array}\right] = + \left[\begin{array}{c} x+2y \\ y \\ 3x +5y \\ -x-2y \end{array}\right] + +in terms of matrix multiplication.

+ + + + + + + Videos +
+ + Video: Multiplying matrices +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + + Construct 3 matrices, A,B,\mbox{ and } C, such that +
    +
  • AB:\mathbb{R}^4\rightarrow\mathbb{R}^2
    • +
  • BC:\mathbb{R}^2\rightarrow\mathbb{R}^3
    • +
  • CA:\mathbb{R}^3\rightarrow\mathbb{R}^4
    • . +
  • ABC:\mathbb{R}^2\rightarrow\mathbb{R}^2
    • +
+ + + + Construct 3 examples of matrix multiplication, with all matrix dimensions at least 2. +
    +
  • Where A and B are not square, but AB is square.
    • +
  • Where AB = BA.
    • +
  • Where AB \neq BA.
    • +
+ + + + +Use the included map in this problem. +
+ Adjacency map, showing roads between 5 cities + + A map with 5 dots. A is connected to B, B is connected to C, C is connected to D and E, and D and E are connected to each other + +\tikz{\node[shape=circle,draw=black] (A) at (0,0) {A}; + \node[shape=circle,draw=black] (B) at (0,2) {B}; + \node[shape=circle,draw=black] (C) at (0,4) {C}; + \node[shape=circle,draw=black] (D) at (-2,6) {D}; + \node[shape=circle,draw=black] (E) at (2,6) {E}; + \draw (A)--(B)--(C)--(D)--(E)--(C)} + + +
+ +
    +
  • An adjacency matrix for this map is a matrix that has the number of roads from city i to city j in the (i,j) entry of the matrix. A road is a path of length exactly 1. All (i,i)entries are 0. Write the adjacency matrix for this map, with the cities in alphabetical order.
    • + +
  • What does the square of this matrix tell you about the map? The cube? The n-th power?
    • +
+ + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/04-MX/02.ptx b/linear-algebra/source/04-MX/02.ptx new file mode 100644 index 00000000..e01c6ee0 --- /dev/null +++ b/linear-algebra/source/04-MX/02.ptx @@ -0,0 +1,318 @@ + +
+ The Inverse of a Matrix (MX2) + + + + + +Class Activities + + +

+ Let A=\left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right]. + Find a 3 \times 3 matrix B such that BA=A, that is, + + \left[\begin{array}{ccc} \unknown & \unknown & \unknown \\ + \unknown & \unknown & \unknown + \\ \unknown & \unknown & \unknown \end{array}\right] + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] + = + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] + + Check your guess using technology. +

+ + + + + + + +

+ The identity matrix I_n (or just I when n is obvious from context) is + the n \times n matrix + I_n = \left[\begin{array}{cccc} 1 & 0 & \cdots & 0 \\ 0 & 1 & \ddots & \vdots \\ + \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & 1 \end{array}\right]. + It has a 1 on each diagonal element and a 0 in every other position. +

+ + + + + +

+ For any square matrix A, IA=AI=A: + + \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] + = + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] + \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] + = + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] + +

+ + + + +

+Let T: \IR^n \rightarrow \IR^m be a linear map with standard matrix A. +Sort the following items into three groups of statements: a group that means +T is injective, a group that means T is surjective, +and a group that means T is bijective. +

+
    +

  1. T(\vec x)=\vec b has a solution for all \vec b\in\IR^m +

    2. +

  2. T(\vec x)=\vec b has a unique solution for all \vec b\in\IR^m +

    3. +

  3. T(\vec x)=\vec 0 has a unique solution. +

    4. +

  4. The columns of A span \IR^m +

    5. +

  5. The columns of A are linearly independent +

    6. +

  6. The columns of A are a basis of \IR^m +

    7. +

  7. Every column of \RREF(A) has a pivot +

    8. +

  8. Every row of \RREF(A) has a pivot +

    9. +

  9. m=n and \RREF(A)=I +

    10. +
+ + + + + +

+Let T: \IR^n \rightarrow \IR^n be a linear bijection with standard matrix +A. +

+

+By item (B) from +we may define an inverse mapinverse map +T^{-1} : \IR^n \rightarrow \IR^n +that defines T^{-1}(\vec b) as the unique solution \vec x satisfying +T(\vec x)=\vec b, that is, T(T^{-1}(\vec b))=\vec b. +

+

+Furthermore, let +A^{-1}=[T^{-1}(\vec e_1)\hspace{1em}\cdots\hspace{1em}T^{-1}(\vec e_n)] +be the standard matrix for T^{-1}. We call A^{-1} the +inverse matrixinverse matrix of A, +and we also say that A is an invertibleinvertible +matrix. +

+ + + + + + +

+Let T: \IR^3 \rightarrow \IR^3 be the linear bijection given by the standard matrix +A=\left[\begin{array}{ccc} 2 & -1 & -6 \\ 2 & 1 & 3 \\ 1 & 1 & 4 \end{array}\right]. +

+ + +

+To find \vec x = T^{-1}(\vec{e}_1), we need to find the unique +solution for T(\vec x)=\vec e_1. Which of these linear systems +can be used to find this solution? +

    +
  1. +\begin{array}{cccc} +2x_1 & -1x_2 & -6x_3 & =x_1 \\ +2x_1 & +1x_2 & +3x_3 & =0 \\ +1x_1 & +1x_2 & +4x_3 & =0 +\end{array} +
    2. +
  2. +\begin{array}{cccc} +2x_1 & -1x_2 & -6x_3 & =x_1 \\ +2x_1 & +1x_2 & +3x_3 & =x_2 \\ +1x_1 & +1x_2 & +4x_3 & =x_3 +\end{array} +
    3. +
  3. +\begin{array}{cccc} +2x_1 & -1x_2 & -6x_3 & =1 \\ +2x_1 & +1x_2 & +3x_3 & =0 \\ +1x_1 & +1x_2 & +4x_3 & =0 +\end{array} +
    4. +
  4. +\begin{array}{cccc} +2x_1 & -1x_2 & -6x_3 & =1 \\ +2x_1 & +1x_2 & +3x_3 & =1 \\ +1x_1 & +1x_2 & +4x_3 & =1 +\end{array} +
    5. +
+

+ + +

+Use that system to find the solution \vec x=T^{-1}(\vec{e}_1) +for T(\vec x)=\vec{e}_1. +

+ + +

+Similarly, solve T(\vec x)=\vec{e}_2 to find T^{-1}(\vec{e}_2), +and solve T(\vec x)=\vec{e}_3 to find T^{-1}(\vec{e}_3). +

+ + +

+Use these to write +A^{-1}= [T^{-1}(\vec e_1)\hspace{1em} +T^{-1}(\vec e_2)\hspace{1em}T^{-1}(\vec e_3)] +, the standard matrix for T^{-1}. +

+ + + + + + + + + +

+Find the inverse A^{-1} of the matrix +A=\left[\begin{array}{cccc} 0 & 0 & 0 & -1 \\ 1 & 0 & -1 & -4 \\ 1 & 1 & 0 & -4 \\ 1 & -1 & -1 & 2 \end{array}\right] +by computing how it transforms each of the standard +basis vectors for \mathbb R^4: +T^{-1}(\vec e_1), T^{-1}(\vec e_2), T^{-1}(\vec e_3), +and T^{-1}(\vec e_4). +

+ + + + + + + + +

+Is the matrix \left[\begin{array}{ccc} 2 & 3 & 1 \\ -1 & -4 & 2 \\ 0 & -5 & 5 \end{array}\right] invertible? +

    +
  1. Yes, because its transformation is a bijection.
    2. +
  2. Yes, because its transformation is not a bijection.
    3. +
  3. No, because its transformation is a bijection.
    4. +
  4. No, because its transformation is not a bijection.
    5. +
+

+ + + + + + + +

+ An n\times n matrix A is invertible if and only if \RREF(A) = I_n. +

+ + + + + +

+ Let T:\IR^2\to\IR^2 be the bijective linear map defined by + T\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)=\left[\begin{array}{c} 2x -3y \\ -3x + 5y\end{array}\right], + with the inverse map + T^{-1}\left(\left[\begin{array}{c}x\\y\end{array}\right]\right)=\left[\begin{array}{c} 5x+ 3y \\ 3x + 2y\end{array}\right]. +

+ + +

+ Compute (T^{-1}\circ T)\left(\left[\begin{array}{c}-2\\1\end{array}\right]\right). +

+ + +

+ If A is the standard matrix for T and A^{-1} is the + standard matrix for T^{-1}, find the 2\times 2 matrix + A^{-1}A=\left[\begin{array}{ccc}\unknown&\unknown\\\unknown&\unknown\end{array}\right]. +

+ + + + + +

+ T^{-1}\circ T=T\circ T^{-1} is the identity map for any bijective + linear transformation T. Therefore + A^{-1}A=AA^{-1} equals the identity matrix I for any invertible matrix + A. +

+ + + + + + + Videos +
+ + Video: Invertible matrices +
+
+ + Video: Finding the inverse of a matrix +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + + Assume A is an n \times n matrix. Prove the following are equivalent. Some of these results you have proven previously. +
    +
  • A row reduces to the identity matrix.
    • +
  • For any choice of \vec{b} \in \mathbb{R}^n, the system of equations represented by the augmented matrix [A|\vec{b}] has a unique solution.
    • +
  • The columns of A are a linearly independent set.
    • +
  • The columns of A form a basis for \mathbb{R}^n.
    • +
  • The rank of A is n.
    • +
  • The nullity of A is 0.
    • +
  • A is invertible.
    • +
  • The linear transformation T with standard matrix A is injective and surjective. Such a map is called an isomorphism.
    • +
+ + + + + +
  • Assume T is a square matrix, and T^4 is the zero matrix. Prove that (I - T)^{-1} = I + T + T^2 + T^3. You will need to first prove a lemma that matrix multiplication distributes over matrix addition.
    • +
  • Generalize your result to the case where T^n is the zero matrix.
    • +
+ + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/04-MX/03.ptx b/linear-algebra/source/04-MX/03.ptx new file mode 100644 index 00000000..c0774caf --- /dev/null +++ b/linear-algebra/source/04-MX/03.ptx @@ -0,0 +1,259 @@ + +
+ Solving Systems with Matrix Inverses (MX3) + + + + + + +Class Activities + + +

+Consider the following linear system with a unique solution: + +\begin{matrix} +3x_{1} & - & 2x_{2} & - & 2x_{3} & - & 4x_{4} & = & -7 \\ +2x_{1} & - & x_{2} & - & x_{3} & - & x_{4} & = & -1 \\ +-x_{1} & & & + & x_{3} & & & = & -1 \\ + & - & x_{2} & & & - & 2x_{4} & = & -5 \\ +\end{matrix} + +

+ + +

+Suppose we let + +T\left(\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]\right)= +\left[\begin{matrix} +3x_{1} & - & 2x_{2} & - & 2x_{3} & - & 4x_{4} \\ +2x_{1} & - & x_{2} & - & x_{3} & - & x_{4} \\ +-x_{1} & & & + & x_{3} & & \\ + & - & x_{2} & & & - & 2x_{4} \\ +\end{matrix}\right] +. +Which of these choices would help us solve the given system? +

    +
  1. +Compute +T\left( +\left[\begin{matrix} +-7 \\ + -1 \\ + -1 \\ + -5 \\ +\end{matrix}\right]\right) +
    2. +
  2. +Find \left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right] where +T\left(\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right]\right)= +\left[\begin{matrix} +-7 \\ + -1 \\ + -1 \\ + -5 \\ +\end{matrix}\right] +
    3. +
+

+ + +

+How can we express this in terms of matrix multiplication? +

+
    +
  1. + +\left[\begin{matrix} +3 & -2 & -2 & - 4 \\ +2 & -1 & -1 & -1 \\ +-1 & 0 & 1 & 0 \\ +0 & -1 & 0 & -2 \\ +\end{matrix}\right] +\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right] += +\left[\begin{matrix} +-7 \\ + -1 \\ + -1 \\ + -5 \\ +\end{matrix}\right] + +
    2. +
  2. + +\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right] += +\left[\begin{matrix} +3 & -2 & -2 & - 4 \\ +2 & -1 & -1 & -1 \\ +-1 & 0 & 1 & 0 \\ +0 & -1 & 0 & -2 \\ +\end{matrix}\right] +\left[\begin{matrix} +-7 \\ + -1 \\ + -1 \\ + -5 \\ +\end{matrix}\right] + +
    3. +
  3. + +\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right] +\left[\begin{matrix} +3 & -2 & -2 & - 4 \\ +2 & -1 & -1 & -1 \\ +-1 & 0 & 1 & 0 \\ +0 & -1 & 0 & -2 \\ +\end{matrix}\right] += +\left[\begin{matrix} +-7 \\ + -1 \\ + -1 \\ + -5 \\ +\end{matrix}\right] + +
    4. +
  4. + +\left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right] += +\left[\begin{matrix} +-7 \\ + -1 \\ + -1 \\ + -5 \\ +\end{matrix}\right] +\left[\begin{matrix} +3 & -2 & -2 & - 4 \\ +2 & -1 & -1 & -1 \\ +-1 & 0 & 1 & 0 \\ +0 & -1 & 0 & -2 \\ +\end{matrix}\right] + +
    5. +
+ + +

+How could a matrix equation of the form A\vec x=\vec b +be solved for \vec x? +

    +
  1. Multiply: (\RREF A)(A\vec x)=(\RREF A)\vec b
    2. +
  2. Add: (\RREF A) + A\vec x=(\RREF A)+\vec b
    3. +
  3. Multiply: (A^{-1})(A\vec x)=(A^{-1})\vec b
    4. +
  4. Add: (A^{-1}) + A\vec x=(A^{-1})+\vec b
    5. +
+

+ + +

+Find + \left[\begin{matrix}x_1\\x_2\\x_3\\x_4\end{matrix}\right] + using the method you chose in (c). +

+ + + + +

+The linear system described by the augmented matrix +[A \mid \vec b] has exactly the same solution set as +the matrix equation A\vec x=\vec b. +

+

+When A is invertible, then we have both +[A \mid \vec b]\sim[I \mid \vec x] and +\vec x=A^{-1}\vec b, which can be seen as + + && A\vec x&=\vec b + &\Rightarrow & A^{-1}A\vec x&=A^{-1}\vec b + &\Rightarrow &\vec x&=A^{-1}\vec b + + +

+ + + + +

Consider the vector equation x_{1} \left[\begin{array}{c} 1 \\ 2 \\ -2 \end{array}\right] + x_{2} \left[\begin{array}{c} -2 \\ -3 \\ 3 \end{array}\right] + x_{3} \left[\begin{array}{c} 1 \\ 4 \\ -3 \end{array}\right] = \left[\begin{array}{c} -3 \\ 5 \\ -1 \end{array}\right] with a unique solution.

+ + + +

Explain and demonstrate how this problem can be restated using matrix multiplication.

+ + + + +

Use the properties of matrix multiplication to find the unique solution.

+ + + + + + + Videos +

+Video coming soon to +this YouTube playlist. +

+ + + + + +

Exercises available at .

+ + + + Mathematical Writing Explorations + + + Use row reduction to find the inverse of the following general matrix. Give conditions on which this inverse exists. + \left[\begin{array}{ccc}1 & b & c \\ d & e & f \\ g & h & i \end{array}\right] + + + + + Assume that H is invertible, and that HG is the zero matrix. Prove that G must be the zero matrix. Would this still be true if H were not invertible? + + + + + If H is invertible and r \in \mathbb{R}, what is the inverse of rH? + + + + + If H and G are invertible, is H^{-1} + G^{-1} = (H+G)^{-1}? + + + + + Prove that if A, P, and Q are invertible with PAQ = I, then A^{-1} = QP. + + + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/04-MX/04.ptx b/linear-algebra/source/04-MX/04.ptx new file mode 100644 index 00000000..00cebaa6 --- /dev/null +++ b/linear-algebra/source/04-MX/04.ptx @@ -0,0 +1,354 @@ + +
+ Row Operations as Matrix Multiplication (MX4) + + + + + +Class Activities + + + +

+Tweaking the identity matrix slightly allows us to write row operations +in terms of matrix multiplication. +

+ + +

+Which of these tweaks of the identity matrix +yields a matrix that doubles the third row of A when +left-multiplying? (2R_3\to R_3) + + \left[\begin{array}{ccc} \unknown & \unknown & \unknown \\ \unknown & \unknown & \unknown \\ \unknown & \unknown & \unknown \end{array}\right] + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] += + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 2 & 2 & -2 \end{array}\right] + +

    +
  1. + +\left[\begin{array}{ccc} +2 & 0 & 0 \\ +0 & 1 & 0 \\ +0 & 0 & 1 +\end{array}\right] + +
    2. +
  2. + +\left[\begin{array}{ccc} +1 & 0 & 0 \\ +0 & 2 & 0 \\ +0 & 0 & 1 +\end{array}\right] + +
    3. +
  3. + +\left[\begin{array}{ccc} +1 & 0 & 0 \\ +0 & 1 & 0 \\ +0 & 0 & 2 +\end{array}\right] + +
    4. +
  4. + +\left[\begin{array}{ccc} +2 & 0 & 0 \\ +0 & 2 & 0 \\ +0 & 0 & 2 +\end{array}\right] + +
    5. +
+

+ + +

+Which of these tweaks of the identity matrix +yields a matrix that swaps the first and third rows of A +when left-multiplying? (R_1\leftrightarrow R_3) + + \left[\begin{array}{ccc} \unknown & \unknown & \unknown \\ \unknown & \unknown & \unknown \\ \unknown & \unknown & \unknown \end{array}\right] + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] + = + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 1 & 1 & -1 \\ 0 & 3 & 2 \end{array}\right] + +

    +
  1. + +\left[\begin{array}{ccc} +1 & 0 & 0 \\ +0 & 0 & 1 \\ +0 & 1 & 0 \\ +\end{array}\right] + +
    2. +
  2. + +\left[\begin{array}{ccc} +0 & 1 & 0 \\ +0 & 0 & 1 \\ +1 & 0 & 0 \\ +\end{array}\right] + +
    3. +
  3. + +\left[\begin{array}{ccc} +0 & 0 & 1 \\ +0 & 1 & 0 \\ +1 & 0 & 0 \\ +\end{array}\right] + +
    4. +
  4. + +\left[\begin{array}{ccc} +0 & 1 & 0 \\ +1 & 0 & 0 \\ +0 & 0 & 1 +\end{array}\right] + +
    5. +
+

+ + +

+Which of these tweaks of the identity matrix +yields a matrix that adds 5 times the third row of A to +the first row when left-multiplying? (R_1+5R_3\to R_1) + + \left[\begin{array}{ccc} \unknown & \unknown & \unknown \\ \unknown & \unknown & \unknown \\ \unknown & \unknown & \unknown \end{array}\right] + \left[\begin{array}{ccc} 2 & 7 & -1 \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] += + \left[\begin{array}{ccc} 2+5(1) & 7+5(1) & -1+5(-1) \\ 0 & 3 & 2 \\ 1 & 1 & -1 \end{array}\right] + +

    +
  1. + +\left[\begin{array}{ccc} +1 & 0 & 1 \\ +0 & 1 & 0 \\ +0 & 0 & 5 +\end{array}\right] + +
    2. +
  2. + +\left[\begin{array}{ccc} +1 & 0 & 5 \\ +0 & 1 & 0 \\ +0 & 0 & 1 +\end{array}\right] + +
    3. +
  3. + +\left[\begin{array}{ccc} +5 & 5 & 5 \\ +0 & 1 & 0 \\ +0 & 0 & 1 +\end{array}\right] + +
    4. +
  4. + +\left[\begin{array}{ccc} +1 & 0 & 5 \\ +0 & 1 & 0 \\ +0 & 0 & 5 +\end{array}\right] + +
    5. +
+

+ + + + + + + +

+If R is the result of applying a row operation to I, then +RA is the result of applying the same row operation to A. +

+
    +
  • +

    + Scaling a row: R= + \left[\begin{array}{ccc} + c & 0 & 0 \\ + 0 & 1 & 0 \\ + 0 & 0 & 1 + \end{array}\right] + +

    +
    • +
  • +

    + Swapping rows: R= + \left[\begin{array}{ccc} + 0 & 1 & 0 \\ + 1 & 0 & 0 \\ + 0 & 0 & 1 + \end{array}\right] + +

    +
    • +
  • +

    + Adding a row multiple to another row: R= + \left[\begin{array}{ccc} + 1 & 0 & c \\ + 0 & 1 & 0 \\ + 0 & 0 & 1 + \end{array}\right] + +

    +
    • +
+

+Such matrices can be chained together to emulate multiple row operations. +In particular, +\RREF(A)=R_k\dots R_2R_1A +for some sequence of matrices R_1,R_2,\dots,R_k. +

+ + + + + +

+What would happen if you right-multiplied by +the tweaked identity matrix rather than left-multiplied? +

    +
  1. The manipulated rows would be reversed.
    2. +
  2. Columns would be manipulated instead of rows.
    3. +
  3. The entries of the resulting matrix would be rotated 180 degrees.
    4. +
+

+ + + + + +

+Consider the two row operations +R_2\leftrightarrow R_3 and R_1+R_2\to R_1 +applied as follows to show A\sim B: +

+ + +A + = +\left[\begin{array}{ccc} +-1&4&5\\ +0&3&-1\\ +1&2&3\\ +\end{array}\right] + &\sim +\left[\begin{array}{ccc} +-1&4&5\\ +1&2&3\\ +0&3&-1\\ +\end{array}\right] + + + &\sim +\left[\begin{array}{ccc} +-1+1&4+2&5+3\\ +1&2&3\\ +0&3&-1\\ +\end{array}\right] + = +\left[\begin{array}{ccc} +0&6&8\\ +1&2&3\\ +0&3&-1\\ +\end{array}\right] + = +B + + +

+Express these row operations as matrix multiplication +by expressing B as the product of two matrices and A: + +B = +\left[\begin{array}{ccc} +\unknown&\unknown&\unknown\\ +\unknown&\unknown&\unknown\\ +\unknown&\unknown&\unknown +\end{array}\right] +\left[\begin{array}{ccc} +\unknown&\unknown&\unknown\\ +\unknown&\unknown&\unknown\\ +\unknown&\unknown&\unknown +\end{array}\right] +A + +Check your work using technology. +

+ + + + + + + +

Let A be any 4 \times 4 matrix.

+ + + +

Give a 4 \times 4 matrix M that may be used to perform the row operation -5 R_2 \to R_2.

+ + + + +

Give a 4 \times 4 matrix Y that may be used to perform the row operation R_2 \leftrightarrow R_3.

+ + + + +

Use matrix multiplication to describe the matrix obtained by applying -5 R_2 \to R_2 and then R_2 \leftrightarrow R_3 to A (note the order).

+ + + + + + + + + Videos +
+ + Video: Row operations as matrix multiplication +
+ + + + + +

Exercises available at .

+ + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/04-MX/main.ptx b/linear-algebra/source/04-MX/main.ptx new file mode 100644 index 00000000..94a84e8b --- /dev/null +++ b/linear-algebra/source/04-MX/main.ptx @@ -0,0 +1,10 @@ + + + Matrices (MX) + + + + + + + diff --git a/linear-algebra/source/04-MX/outcomes/01.ptx b/linear-algebra/source/04-MX/outcomes/01.ptx new file mode 100644 index 00000000..7d1174bf --- /dev/null +++ b/linear-algebra/source/04-MX/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Multiply matrices. +

\ No newline at end of file diff --git a/linear-algebra/source/04-MX/outcomes/02.ptx b/linear-algebra/source/04-MX/outcomes/02.ptx new file mode 100644 index 00000000..1a5f036b --- /dev/null +++ b/linear-algebra/source/04-MX/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+ Determine if a matrix is invertible, and if so, compute its inverse. +

\ No newline at end of file diff --git a/linear-algebra/source/04-MX/outcomes/03.ptx b/linear-algebra/source/04-MX/outcomes/03.ptx new file mode 100644 index 00000000..bbc24574 --- /dev/null +++ b/linear-algebra/source/04-MX/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+ Invert an appropriate matrix to solve a system of linear equations. +

\ No newline at end of file diff --git a/linear-algebra/source/04-MX/outcomes/04.ptx b/linear-algebra/source/04-MX/outcomes/04.ptx new file mode 100644 index 00000000..ccfeb834 --- /dev/null +++ b/linear-algebra/source/04-MX/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Express row operations through matrix multiplication. +

\ No newline at end of file diff --git a/linear-algebra/source/04-MX/outcomes/main.ptx b/linear-algebra/source/04-MX/outcomes/main.ptx new file mode 100644 index 00000000..60798603 --- /dev/null +++ b/linear-algebra/source/04-MX/outcomes/main.ptx @@ -0,0 +1,23 @@ + +> + + +

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
+ \ No newline at end of file diff --git a/linear-algebra/source/04-MX/outcomes/question.ptx b/linear-algebra/source/04-MX/outcomes/question.ptx new file mode 100644 index 00000000..95dd95e3 --- /dev/null +++ b/linear-algebra/source/04-MX/outcomes/question.ptx @@ -0,0 +1,4 @@ + +

+What algebraic structure do matrices have? +

diff --git a/linear-algebra/source/04-MX/readiness.ptx b/linear-algebra/source/04-MX/readiness.ptx new file mode 100644 index 00000000..9deee7ce --- /dev/null +++ b/linear-algebra/source/04-MX/readiness.ptx @@ -0,0 +1,52 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Compose functions of real numbers.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    2. +
  2. +

    Identify the domain and codomain of linear transformations.

    +
      +
    • +

      Review: YouTube

      +
      • +
    +
    3. +
  3. +

    Find the matrix corresponding to a linear transformation and compute the image of a vector given a standard matrix.

    +
      +
    • +

      Review:

      +
      • +
    +
    4. +
  4. +

    Determine if a linear transformation is injective and/or surjective.

    +
      +
    • +

      Review:

      +
      • +
    +
    5. +
  5. +

    Interpret the ideas of injectivity and surjectivity in multiple ways.

    +
      +
    • +

      Review: YouTube

      +
      • +
    +
    6. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/04-MX/samples/01.ptx b/linear-algebra/source/04-MX/samples/01.ptx new file mode 100644 index 00000000..c10d0119 --- /dev/null +++ b/linear-algebra/source/04-MX/samples/01.ptx @@ -0,0 +1,40 @@ + + +MX1 + +

+Of the following three matrices, only two may be multiplied. +

+ + +A &= \left[\begin{array}{cc} 1 & -3 \\ 0 & 1 \end{array}\right] & +B&= \left[\begin{array}{ccc} 4 & 1 & 2 \end{array}\right] & +C&= \left[\begin{array}{ccc} 0 & 1 & 3 \\ 1 & -2 & 5 \end{array}\right] + + +

+Explain which two may be multiplied and why. Then show how to find their product. +

+ + +

+AC is the only one that can be computed, since C corresponds to a linear transformation \mathbb{R}^3 \rightarrow \mathbb{R}^2 and A corresponds to a linear transfromation \mathbb{R}^2 \rightarrow \mathbb{R}^2. Thus the composition AC corresponds to a linear transformation \mathbb{R}^3 \rightarrow \mathbb{R}^2 with a 2\times 3 standard matrix. +We compute +

+ + +AC\left( \vec{e}_1 \right) &= A \left( \left[\begin{array}{c} 0 \\ 1 \end{array}\right] \right) = 0 \left[\begin{array}{c} 1 \\ 0 \end{array}\right] + 1\left[\begin{array}{c} -3 \\ 1 \end{array}\right] = \left[\begin{array}{c} -3 \\ 1 \end{array}\right] \\ + + +AC\left( \vec{e}_2 \right) &= A \left( \left[\begin{array}{c} 1 \\ -2 \end{array}\right] \right) = 1 \left[\begin{array}{c} 1 \\ 0 \end{array}\right] -2\left[\begin{array}{c} -3 \\ 1 \end{array}\right] = \left[\begin{array}{c} 7 \\ -2 \end{array}\right] \\ + + +AC\left( \vec{e}_3 \right) &= A \left( \left[\begin{array}{c} 3 \\ 5 \end{array}\right] \right) = 3 \left[\begin{array}{c} 1 \\ 0 \end{array}\right] + 5\left[\begin{array}{c} -3 \\ 1 \end{array}\right] = \left[\begin{array}{c} -12 \\ 5 \end{array}\right] \\ + + . +

+Thus +AC = \left[\begin{array}{ccc} -3 & 7 & -12 \\ 1 & -2 & 5 \end{array}\right]. +

+ + \ No newline at end of file diff --git a/linear-algebra/source/04-MX/samples/02.ptx b/linear-algebra/source/04-MX/samples/02.ptx new file mode 100644 index 00000000..93be5551 --- /dev/null +++ b/linear-algebra/source/04-MX/samples/02.ptx @@ -0,0 +1,106 @@ + + + + +MX2 + +

+Explain why each of the following matrices is or is not invertible by +disussing its corresponding linear transformation. +If the matrix is invertible, explain how to find its inverse. + +\hspace{2em} +D = \left[\begin{array}{cccc} +-1 & 1 & 0 & 2 \\ +-2 & 5 & 5 & -4 \\ +2 & -3 & -2 & 0 \\ +4 & -4 & -3 & 5 +\end{array}\right] +\hspace{2em} +N = \left[\begin{array}{cccc} +-3 & 9 & 1 & -11 \\ +3 & -9 & -2 & 13 \\ +3 & -9 & -3 & 15 \\ +-4 & 12 & 2 & -16 +\end{array}\right] +\hspace{2em} + +

+ + +

We compute +\mathrm{RREF}\left(D\right)=\left[\begin{array}{cccc} +1 & 0 & 0 & 0 \\ +0 & 1 & 0 & 0 \\ +0 & 0 & 1 & 0 \\ +0 & 0 & 0 & 1 +\end{array}\right]. +We see D is bijective, and therefore invertible. To compute the inverse, we solve D\vec{x}=\vec{e}_1 by computing +\mathrm{RREF}\,\left[\begin{array}{cccc|c} +-1 & 1 & 0 & 2 & 1\\ +-2 & 5 & 5 & -4 & 0 \\ +2 & -3 & -2 & 0 & 0\\ +4 & -4 & -3 & 5 & 0 +\end{array}\right]=\left[\begin{array}{cccc|c} +1 & 0 & 0 & 0 & 21 \\ +0 & 1 & 0 & 0 & 38\\ +0 & 0 & 1 & 0 & -36\\ +0 & 0 & 0 & 1 & -8 +\end{array}\right]. +

+Similarly, we solve D\vec{x}=\vec{e}_2 by computing +\mathrm{RREF}\,\left[\begin{array}{cccc|c} +-1 & 1 & 0 & 2 & 0\\ +-2 & 5 & 5 & -4 & 1 \\ +2 & -3 & -2 & 0 & 0\\ +4 & -4 & -3 & 5 & 0 +\end{array}\right]=\left[\begin{array}{cccc|c} +1 & 0 & 0 & 0 & 8 \\ +0 & 1 & 0 & 0 & 14\\ +0 & 0 & 1 & 0 & -13\\ +0 & 0 & 0 & 1 & -3 +\end{array}\right]. +

+ Similarly, we solve D\vec{x}=\vec{e}_3 by computing +\mathrm{RREF}\,\left[\begin{array}{cccc|c} +-1 & 1 & 0 & 2 & 0\\ +-2 & 5 & 5 & -4 & 0 \\ +2 & -3 & -2 & 0 & 1\\ +4 & -4 & -3 & 5 & 0 +\end{array}\right]=\left[\begin{array}{cccc|c} +1 & 0 & 0 & 0 & 23 \\ +0 & 1 & 0 & 0 & 41\\ +0 & 0 & 1 & 0 & -39\\ +0 & 0 & 0 & 1 & -9 +\end{array}\right]. +

Similarly, we solve D\vec{x}=\vec{e}_4 by computing +\mathrm{RREF}\,\left[\begin{array}{cccc|c} +-1 & 1 & 0 & 2 & 0\\ +-2 & 5 & 5 & -4 & 0 \\ +2 & -3 & -2 & 0 & 0\\ +4 & -4 & -3 & 5 & 1 +\end{array}\right]=\left[\begin{array}{cccc|c} +1 & 0 & 0 & 0 & -2 \\ +0 & 1 & 0 & 0 & -4\\ +0 & 0 & 1 & 0 & 4\\ +0 & 0 & 0 & 1 & 1 +\end{array}\right]. +

+Combining these, we obtain D^{-1}=\left[\begin{array}{cccc} +21 & 8 & 23 & -2 \\ +38 & 14 & 41 & -4 \\ +-36 & -13 & -39 & 4 \\ +-8 & -3 & -9 & 1 +\end{array}\right]. +

+

We compute +\mathrm{RREF}\left(N\right)=\left[\begin{array}{cccc} +1 & -3 & 0 & 3 \\ +0 & 0 & 1 & -2 \\ +0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 +\end{array}\right]. +We see N is not bijective and thus is not invertible. +

+ + \ No newline at end of file diff --git a/linear-algebra/source/04-MX/samples/03.ptx b/linear-algebra/source/04-MX/samples/03.ptx new file mode 100644 index 00000000..e15f112e --- /dev/null +++ b/linear-algebra/source/04-MX/samples/03.ptx @@ -0,0 +1,26 @@ + + +MX3 + + Use a matrix inverse to solve the following matrix-vector equation. + \left[\begin{array}{ccc} + 1& 2& 1\\ + 0& 0& 2\\ + 1& 1& 1\\ + \end{array}\right] \vec{v} = \left[\begin{array}{c}4\\ -2 \\ 2 \end{array}\right] + + + Using the techniques from section , and letting M = \left[\begin{array}{ccc} + 1& 2& 1\\ + 0& 0& 2\\ + 1& 1& 1\\ + \end{array}\right], we find M^{-1} = \left[\begin{array}{ccc} + -1& -1/2& 2\\ + 1& 0& -1\\ + 0& 1/2& 0\\ + \end{array}\right]. + Our equation can be written as M\vec{v} = \left[\begin{array}{c}4\\ -2 \\ 2 \end{array}\right], and may therefore be solved via + \vec{v} = I\vec{v} = M^{-1}M\vec{v} = M^{-1}\left[\begin{array}{c}4\\ -2 \\ 2 \end{array}\right] = + \left[\begin{array}{c}1\\ 2 \\ -1 \end{array}\right] + + \ No newline at end of file diff --git a/linear-algebra/source/04-MX/samples/04.ptx b/linear-algebra/source/04-MX/samples/04.ptx new file mode 100644 index 00000000..88a6142c --- /dev/null +++ b/linear-algebra/source/04-MX/samples/04.ptx @@ -0,0 +1,48 @@ + +MX4 + +

+Let A be a 4\times4 matrix. +

+
    +
  1. +

    +Give a 4\times 4 matrix P that may be used to perform the row + operation {R_3} \to R_3+4 \, {R_1} . +

    +
    2. +
  2. +

    +Give a 4\times 4 matrix Q that may be used to perform the row + operation {R_1} \to -4 \, {R_1}. +

    +
    3. +
  3. +

    +Use matrix multiplication to describe the matrix obtained by applying + {R_3} \to 4 \, {R_1} + {R_3} and then {R_1} \to -4 \, {R_1} + to A (note the order). +

    +
    4. +
+ + +
    +
  1. +

    +P=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 4 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] +

    +
    2. +
  2. +

    +Q=\left[\begin{array}{cccc} -4 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right] +

    +
    3. +
  3. +

    +QPA +

    +
    4. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/05-GT/01.ptx b/linear-algebra/source/05-GT/01.ptx new file mode 100644 index 00000000..319c0165 --- /dev/null +++ b/linear-algebra/source/05-GT/01.ptx @@ -0,0 +1,1131 @@ + +
+ Row Operations and Determinants (GT1) + + + + +Class Activities + + + + + +

+The image in illustrates how the linear transformation +T : \IR^2 \rightarrow \IR^2 given by the +standard matrix A = \left[\begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array}\right] +transforms the unit square. +

+
+ + +\begin{tikzpicture} +\fill[red!50!white] (0,0) rectangle (1,1);` +\draw[thin,gray,<->] (-4,0)-- (4,0); +\draw[thin,gray,<->] (0,-4)-- (0,4); +\draw[thick,blue,->] (0,0) -- node[below] {\(A \vec{e}_1= \left[\begin{array}{c}2 \\ 0 \end{array}\right]\)}++ (2,0); +\draw[thick,blue,->] (0,0) -- node[left] {\(A \vec{e}_2 = \left[\begin{array}{c} 0 \\ 3 \end{array}\right]\)}++(0,3); +\draw[thick,red,->] (0,0) -- ++(1,0); +\draw[thick,red,->] (0,0) -- ++(0,1); +\draw[blue,dashed] (2,0) -- (2,3) -- (0,3); +\draw[red,dashed] (1,0) -- (1,1) -- (0,1); +\end{tikzpicture} + + + Transformation of the unit square by the matrix A. +
+ +

+What are the lengths of A\vec e_1 and A\vec e_2? +

+

+What is the area of the transformed unit square? +

+ + + + + +

+The image below illustrates how the linear transformation +S : \IR^2 \rightarrow \IR^2 given by the +standard matrix B = \left[\begin{array}{cc} 2 & 3 \\ 0 & 4 \end{array}\right] +transforms the unit square. +

+
+ + +\begin{tikzpicture} +\fill[red!50!white] (0,0) rectangle (1,1); +\draw[thin,gray,<->] (-4,0)-- (4,0); +\draw[thin,gray,<->] (0,-4)-- (0,4); +\draw[thick,blue,->] (0,0) -- node[below] {\(B \vec{e}_1= \left[\begin{array}{c}2 \\ 0 \end{array}\right]\)}++ (2,0); +\draw[thick,blue,->] (0,0) -- ++(3,4) node[above] {\(B \vec{e}_2 = \left[\begin{array}{c} 3 \\ 4 \end{array}\right]\)}; +\draw[thick,red,->] (0,0) -- ++(1,0); +\draw[thick,red,->] (0,0) -- ++(0,1); +\draw[blue,dashed] (2,0) -- (5,4) -- (3,4); +\draw[red,dashed] (1,0) -- (1,1) -- (0,1); +\end{tikzpicture} + + + Transformation of the unit square by the matrix B +
+ + +

+What are the lengths of B\vec e_1 and B\vec e_2? +

+ + +

+What is the area of the transformed unit square? +

+ + + + + +

+ It is possible to find two nonparallel vectors that are scaled but not rotated by + the linear map given by B. +

+ B\vec e_1=\left[\begin{array}{cc} 2 & 3 \\ 0 & 4 \end{array}\right]\left[\begin{array}{c}1\\0\end{array}\right] + =\left[\begin{array}{c}2\\0\end{array}\right]=2\vec e_1 + + + B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right] + = + \left[\begin{array}{cc} 2 & 3 \\ 0 & 4 \end{array}\right]\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right] + = + \left[\begin{array}{c}3\\2\end{array}\right] + = + 4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right] + +
+ + + \begin{tikzpicture} + \fill[red!50!white] (0,0) -- (1,0) -- (1.75,0.5) -- (0.75,0.5) -- (0,0); + \draw[thin,gray,<->] (-4,0)-- (4,0); + \draw[thin,gray,<->] (0,-4)-- (0,4); + \draw[thick,blue,->] (0,0) -- node[below] {\(B\left[\begin{array}{c}1\\0\end{array}\right]=2\left[\begin{array}{c}1\\0\end{array}\right]\)}++ (2,0); + \draw[thick,blue,->] (0,0) -- ++(3,2) node[above] {\(B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]=4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]\)}; + \draw[thick,red,->] (0,0) -- (1,0); + \draw[thick,red,->] (0,0) -- (0.75,0.5); + \draw[red,dashed] (1,0) -- (1.75,0.5) -- (0.75,0.5); + \draw[blue,dashed] (2,0) -- (5,2) -- (3,2); + \end{tikzpicture} + + + Certain vectors are stretched out without being rotated. +
+

+ The process for finding such vectors will be covered later in this chapter. +

+ + + + + + +

+ Notice that while a linear map can transform vectors in various ways, + linear maps always transform parallelograms into parallelograms, + and these areas are always transformed by the same factor: in the case of + B=\left[\begin{array}{cc} 2 & 3 \\ 0 & 4 \end{array}\right], + this factor is 8. +

+
+ + +\begin{tikzpicture}[scale=0.7] +\fill[red!50!white] (0,0) rectangle (1,1); +\draw[thin,gray,<->] (-4,0)-- (4,0); +\draw[thin,gray,<->] (0,-4)-- (0,4); +\draw[thick,blue,->] (0,0) -- node[below] {\(B \vec{e}_1= \left[\begin{array}{c}2 \\ 0 \end{array}\right]\)}++ (2,0); +\draw[thick,blue,->] (0,0) -- ++(3,4) node[above] {\(B \vec{e}_2 = \left[\begin{array}{c} 3 \\ 4 \end{array}\right]\)}; +\draw[thick,red,->] (0,0) -- ++(1,0); +\draw[thick,red,->] (0,0) -- ++(0,1); +\draw[blue,dashed] (2,0) -- (5,4) -- (3,4); +\draw[red,dashed] (1,0) -- (1,1) -- (0,1); +\end{tikzpicture} + \begin{tikzpicture}[scale=0.7] + \fill[red!50!white] (0,0) -- (1,0) -- (1.75,0.5) -- (0.75,0.5) -- (0,0); + \draw[thin,gray,<->] (-4,0)-- (4,0); + \draw[thin,gray,<->] (0,-4)-- (0,4); + \draw[thick,blue,->] (0,0) -- node[below] {\(B\left[\begin{array}{c}1\\0\end{array}\right]=2\left[\begin{array}{c}1\\0\end{array}\right]\)}++ (2,0); + \draw[thick,blue,->] (0,0) -- ++(3,2) node[above] {\(B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]=4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]\)}; + \draw[thick,red,->] (0,0) -- (1,0); + \draw[thick,red,->] (0,0) -- (0.75,0.5); + \draw[red,dashed] (1,0) -- (1.75,0.5) -- (0.75,0.5); + \draw[blue,dashed] (2,0) -- (5,2) -- (3,2); + \end{tikzpicture} + + + A linear map transforming parallelograms into parallelograms. +
+

+Since this change in area is always the same for a given linear map, +it will be equal to the value of the transformed unit square (which +begins with area 1). +

+ + + + + +

+We will define the determinant of a square matrix B, +or \det(B) for short, to be the factor by which B scales areas. +In order to figure out how to compute it, we first figure out the properties it must satisfy. +

+
+ + +\begin{tikzpicture}[scale=0.7] +\fill[red!50!white] (0,0) rectangle (1,1); +\draw[thin,gray,<->] (-4,0)-- (4,0); +\draw[thin,gray,<->] (0,-4)-- (0,4); +\draw[thick,blue,->] (0,0) -- node[below] {\(B \vec{e}_1= \left[\begin{array}{c}2 \\ 0 \end{array}\right]\)}++ (2,0); +\draw[thick,blue,->] (0,0) -- ++(3,4) node[above] {\(B \vec{e}_2 = \left[\begin{array}{c} 3 \\ 4 \end{array}\right]\)}; +\draw[thick,red,->] (0,0) -- ++(1,0); +\draw[thick,red,->] (0,0) -- ++(0,1); +\draw[blue,dashed] (2,0) -- (5,4) -- (3,4); +\draw[red,dashed] (1,0) -- (1,1) -- (0,1); +\end{tikzpicture} + \begin{tikzpicture}[scale=0.7] + \fill[red!50!white] (0,0) -- (1,0) -- (1.75,0.5) -- (0.75,0.5) -- (0,0); + \draw[thin,gray,<->] (-4,0)-- (4,0); + \draw[thin,gray,<->] (0,-4)-- (0,4); + \draw[thick,blue,->] (0,0) -- node[below] {\(B\left[\begin{array}{c}1\\0\end{array}\right]=2\left[\begin{array}{c}1\\0\end{array}\right]\)}++ (2,0); + \draw[thick,blue,->] (0,0) -- ++(3,2) node[above] {\(B\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]=4\left[\begin{array}{c}\frac{3}{4}\\\frac{1}{2}\end{array}\right]\)}; + \draw[thick,red,->] (0,0) -- (1,0); + \draw[thick,red,->] (0,0) -- (0.75,0.5); + \draw[red,dashed] (1,0) -- (1.75,0.5) -- (0.75,0.5); + \draw[blue,dashed] (2,0) -- (5,2) -- (3,2); + \end{tikzpicture} + + + The linear transformation B scaling areas by a constant factor, which we call the determinant +
+ + + + + + +

+The transformation of the unit square by the +standard matrix [\vec{e}_1\hspace{0.5em} \vec{e}_2]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]=I is illustrated below. +If \det([\vec{e}_1\hspace{0.5em} \vec{e}_2])=\det(I) is the +area of resulting parallelogram, what is the value of \det([\vec{e}_1\hspace{0.5em} \vec{e}_2])=\det(I)? +

+
+ + +\begin{tikzpicture} +\fill[red!50!white] (0,0) rectangle (1,1); +\draw[thin,gray,<->] (-1,0)-- (3,0); +\draw[thin,gray,<->] (0,-1)-- (0,3); +\draw[thick,blue,->] (0,0) -- node[below] {\(\vec{e}_1=\left[\begin{array}{c}1 \\ 0 \end{array}\right]\)} (1,0); +\draw[thick,blue,->] (0,0) -- node[left] {\(\vec{e}_2=\left[\begin{array}{c} 0 \\ 1 \end{array}\right]\)} (0,1); +\draw[dashed,blue] (1,0) -- (1,1); +\draw[dashed,blue] (0,1) -- (1,1); +\end{tikzpicture} + + + The transformation of the unit square by the identity matrix. +
+

The value for \det([\vec{e}_1\hspace{0.5em} \vec{e}_2])=\det(I) is:

+
    +

  1. 0 +

    2. +

  2. 1 +

    3. +

  3. 2 +

    4. +

  4. 4 +

    5. +
+ + + + + +

+The transformation of the unit square by the +standard matrix [\vec{v}\hspace{0.5em} \vec{v}] is illustrated below: both +T(\vec{e}_1)=T(\vec{e}_2)=\vec{v}. +If \det([\vec{v}\hspace{0.5em} \vec{v}]) is +the area of the generated parallelogram, what is the value of \det([\vec{v}\hspace{0.5em} \vec{v}])? +

+
+ + +\begin{tikzpicture} +%\fill[red!50!white] (0,0) rectangle (1,1); +\draw[thin,gray,<->] (-1,0)-- (4,0); +\draw[thin,gray,<->] (0,-1)-- (0,4); +\draw[thick,blue,->] (0,0) -- node[below] {\(\vec{v}\)} (3,2); +\end{tikzpicture} + + + Transformation of the unit square by a matrix with identical columns. +
+

The value of \det([\vec{v}\hspace{0.5em} \vec{v}]) is:

+
    +

  1. 0 +

    2. +

  2. 1 +

    3. +

  3. 2 +

    4. +

  4. 4 +

    5. +
+ + + + + + +

+The transformations of the unit square by the +standard matrices [\vec{v}\hspace{0.5em} \vec{w}] and +[c\vec{v}\hspace{0.5em} \vec{w}] are illustrated below. +Describe the value of \det([c\vec{v}\hspace{0.5em} \vec{w}]). +

+
+ + +\begin{tikzpicture} +\draw[thin,gray,<->] (-1,0)-- (6,0); +\draw[thin,gray,<->] (0,-1)-- (0,6); +\draw[thick,blue,->] (0,0) -- node[below] {\(\vec{v}\)} (3,1.5); +\draw[thick,purple,->] (0,0) -- node[left] {\(\vec{w}\)} (1,3); +\draw[lightgray,very thick,dashed] (1,3) -- (2,1); +\draw[thick,blue,->] (3,1.5) -- (4,4.5); +\draw[thick,blue,->] (1,3) -- (4,4.5); +\begin{scope}[shift={(8,0)}] +\draw[thin,gray,<->] (-1,0)-- (6,0); +\draw[thin,gray,<->] (0,-1)-- (0,6); +\draw[thick,red,->] (0,0) -- node[below right] {\(c\vec{v}\)} (4,2); +\draw[thick,red,->] (1,3) -- (5,5); +\draw[thick,blue,->] (0,0) -- node[below] {\(\vec{v}\)} (3,1.5); +\draw[thick,purple,->] (0,0) -- node[left] {\(\vec{w}\)} (1,3); +\draw[lightgray,very thick,dashed] (1,3) -- (2,1); +\draw[thick,blue,->] (3,1.5) -- (4,4.5); +\draw[thick,blue,->] (1,3) -- (4,4.5); +\draw[thick,red,->] (4,2) -- (5,5); +\end{scope} +\end{tikzpicture} + + + The parallelograms generated by \vec{v} and \vec{w}/c\vec{w} +
+

Describe the value of \det([c\vec{v}\hspace{0.5em} \vec{w}]):

+
    +

  1. \det([\vec{v}\hspace{0.5em} \vec{w}]) +

    2. +

  2. c\det([\vec{v}\hspace{0.5em} \vec{w}]) +

    3. +

  3. c^2\det([\vec{v}\hspace{0.5em} \vec{w}]) +

    4. +

  4. Cannot be determined from this information. +

    5. +
+ + + + +

Consider the vectors \vec{u}, \vec{v}, \vec{u}+\vec{v}, + and \vec{w} displayed below. Each pair of vectors generates a parallelogram, + and the area of each parallelogram can be described in terms of determinants. +

+
+ + +\begin{tikzpicture} +\draw[thin,gray,<->] (-1,0)-- (6,0); +\draw[thin,gray,<->] (0,-1)-- (0,6); +\draw[thick,blue,->] (0,0) -- node[below] {\(\vec{u}\)} (2,1); +\draw[thick,purple,->] (0,0) -- node[left] {\(\vec{w}\)} (1,3); +\draw[thick,blue,->] (2,1) -- node [below] {\(\vec{v}\)}(3,2); +\draw[thick,red,->] (0,0) -- (3,2)node[above,right] {\(\vec{u}+\vec{v}\)}; +\end{tikzpicture} + + + The vectors \vec{u}, \vec{v}, \vec{u}+\vec{v} and \vec{w} +
+ + + + +

For example, \det([\vec{u}\hspace{0.5em} \vec{w}]) represents the + shaded area shown below.

+
+ + +\begin{tikzpicture} +\path[fill=red,opacity=0.2] (0,0) -- (1,3) -- (3,4) -- (2,1); +\draw[thin,gray,<->] (-1,0)-- (6,0); +\draw[thin,gray,<->] (0,-1)-- (0,6); +\draw[thick,blue,->] (0,0) -- node[below] {\(\vec{u}\)} (2,1); +\draw[thick,purple,->] (0,0) -- node[left] {\(\vec{w}\)} (1,3); +\draw[thick,blue,->] (2,1) -- node [below] {\(\vec{v}\)}(3,2); +\draw[thick,purple,->] (2,1) -- (3,4); +\draw[thick,purple,->] (3,2) -- (4,5); +\draw[thick,blue,->] (1,3) -- (3,4); +\draw[thick,blue,->] (3,4) -- (4,5); +\draw[thick,red,->] (0,0) -- (3,2)node[above,right] {\(\vec{u}+\vec{v}\)}; +\draw[thick,red,->] (1,3) -- (4,5); +\end{tikzpicture} + + + Parallelogram generated by \vec{u} and \vec{w} +
+ + + +

Similarly, \det([\vec{v}\hspace{0.5em} \vec{w}]) represents the shaded area shown below.

+
+ + +\begin{tikzpicture} +\path[fill=red,opacity=0.2] (2,1) -- (3,2) -- (4,5) -- (3,4); +\draw[thin,gray,<->] (-1,0)-- (6,0); +\draw[thin,gray,<->] (0,-1)-- (0,6); +\draw[thick,blue,->] (0,0) -- node[below] {\(\vec{u}\)} (2,1); +\draw[thick,purple,->] (0,0) -- node[left] {\(\vec{w}\)} (1,3); +\draw[thick,blue,->] (2,1) -- node [below] {\(\vec{v}\)}(3,2); +\draw[thick,purple,->] (2,1) -- (3,4); +\draw[thick,purple,->] (3,2) -- (4,5); +\draw[thick,blue,->] (1,3) -- (3,4); +\draw[thick,blue,->] (3,4) -- (4,5); +\draw[thick,red,->] (0,0) -- (3,2)node[above,right] {\(\vec{u}+\vec{v}\)}; +\draw[thick,red,->] (1,3) -- (4,5); +\end{tikzpicture} + + + Parallelogram generated by \vec{v} and \vec{w} +
+ + + + +

+The parallelograms generated by the +standard matrices [\vec{u}\hspace{0.5em} \vec{w}], +[\vec{v}\hspace{0.5em} \vec{w}] and +[\vec{u}+\vec{v}\hspace{0.5em} \vec{w}] are illustrated below. +

+
+ + +\begin{tikzpicture} +\draw[thin,gray,<->] (-1,0)-- (6,0); +\draw[thin,gray,<->] (0,-1)-- (0,6); +\draw[thick,blue,->] (0,0) -- node[below] {\(\vec{u}\)} (2,1); +\draw[thick,purple,->] (0,0) -- node[left] {\(\vec{w}\)} (1,3); +\draw[thick,blue,->] (2,1) -- node [below] {\(\vec{v}\)}(3,2); +\draw[thick,purple,->] (2,1) -- (3,4); +\draw[thick,purple,->] (3,2) -- (4,5); +\draw[thick,blue,->] (1,3) -- (3,4); +\draw[thick,blue,->] (3,4) -- (4,5); +\draw[thick,red,->] (0,0) -- (3,2)node[above,right] {\(\vec{u}+\vec{v}\)}; +\draw[thick,red,->] (1,3) -- (4,5); +\end{tikzpicture} + + + Parallelogram generated by \vec{u}+\vec{v} and \vec{w} +
+

+Describe the value of \det([\vec{u}+\vec{v}\hspace{0.5em} \vec{w}]). +

+
    +

  1. + \det([\vec{u}\hspace{0.5em} \vec{w}])=\det([\vec{v}\hspace{0.5em} \vec{w}]) +

    2. +

  2. + \det([\vec{u}\hspace{0.5em} \vec{w}])+\det([\vec{v}\hspace{0.5em} \vec{w}]) +

    3. +

  3. + \det([\vec{u}\hspace{0.5em} \vec{w}])\det([\vec{v}\hspace{0.5em} \vec{w}]) +

    4. +

  4. Cannot be determined from this information. +

    5. +
+ + + + + + +

+The determinant is the unique function +\det:M_{n,n}\to\IR satisfying these properties: +

    +
  1. \det(I)=1
    2. +
  2. \det(A)=0 whenever two columns of the matrix are identical.
    3. +
  3. + \det[\cdots\hspace{0.5em}c\vec{v}\hspace{0.5em}\cdots]= + c\det[\cdots\hspace{0.5em}\vec{v}\hspace{0.5em}\cdots], assuming no other columns change. +
    4. +
  4. + \det[\cdots\hspace{0.5em}\vec{v}+\vec{w}\hspace{0.5em}\cdots]= + \det[\cdots\hspace{0.5em}\vec{v}\hspace{0.5em}\cdots]+ + \det[\cdots\hspace{0.5em}\vec{w}\hspace{0.5em}\cdots], assuming no other columns change. +
    5. +
+

+

+Note that these last two properties together can be phrased as The determinant is linear in each column. +

+ + + + + + +

+The determinant must also satisfy other properties. +Consider \det([\vec v \hspace{1em}\vec w+c \vec{v}]) and +\det([\vec v\hspace{1em}\vec w]). +

+
+ + +\begin{tikzpicture} +\draw[thin,gray,<->] (-1,0)-- (8,0); +\draw[thin,gray,<->] (0,-1)-- (0,6); +\draw[very thick,blue,->] (0,0) -- node[below right] {\(\vec{v}\)} (3,1); +\draw[very thick,blue,->] (0,0) -- node[left] {\(\vec{w}\)} (1,2); +\draw[dashed,blue,->] (1,2) -- (4,3); +\draw[dashed,blue,->] (3,1) -- (4,3); +\draw[thick,red,->] (0,0) -- (5.5,3.5) node[above] {\(\vec{w}+c\vec{v}\)}; +\draw[dashed,red,->] (3,1) -- (8.5,4.5); +\draw[dashed,blue,->] (5.5,3.5) -- (8.5,4.5); +\draw[thin,dashed,gray] (3,1) -- (2.5,2.5); +\draw[thin,dashed,gray] (4,3) -- (5.5,3.5); +\end{tikzpicture} + + + Parallelogram built by \vec{w}+c\vec{v} and \vec{w} +
+

+The base of both parallelograms is \vec{v}, while the height has not changed, +so the determinant does not change either. This can also be proven using the +other properties of the determinant: +

+ + + \det([\vec{v}+c\vec{w}\hspace{1em}\vec{w}]) +&= + \det([\vec{v}\hspace{1em}\vec{w}])+ + \det([c\vec{w}\hspace{1em}\vec{w}]) + + +&= + \det([\vec{v}\hspace{1em}\vec{w}])+ + c\det([\vec{w}\hspace{1em}\vec{w}]) + + +&= + \det([\vec{v}\hspace{1em}\vec{w}])+ + c\cdot 0 + + +&= + \det([\vec{v}\hspace{1em}\vec{w}]) + + + + + + + +

+Swapping columns may be thought of as a reflection, which is represented by a negative +determinant. For example, the following matrices transform the unit square into +the same parallelogram, but the second matrix reflects its orientation. +

+ + A=\left[\begin{array}{cc}2&3\\0&4\end{array}\right]\hspace{1em}\det A=8\hspace{3em} + B=\left[\begin{array}{cc}3&2\\4&0\end{array}\right]\hspace{1em}\det B=-8 + +
+ + +\begin{tikzpicture}[scale=0.7] +\fill[red!50!white] (0,0) rectangle (1,1); +\draw[thin,gray,<->] (-4,0)-- (4,0); +\draw[thin,gray,<->] (0,-4)-- (0,4); +\draw[thick,blue,->] (0,0) -- node[below] {\(A \vec{e}_1= \left[\begin{array}{c}2 \\ 0 \end{array}\right]\)}++ (2,0); +\draw[thick,blue,->] (0,0) -- ++(3,4) node[above] {\(A \vec{e}_2 = \left[\begin{array}{c} 3 \\ 4 \end{array}\right]\)}; +\draw[thick,red,->] (0,0) -- ++(1,0); +\draw[thick,red,->] (0,0) -- ++(0,1); +\draw[blue,dashed] (2,0) -- (5,4) -- (3,4); +\draw[red,dashed] (1,0) -- (1,1) -- (0,1); +\end{tikzpicture} +\begin{tikzpicture}[scale=0.7] +\fill[red!50!white] (0,0) rectangle (1,1); +\draw[thin,gray,<->] (-4,0)-- (4,0); +\draw[thin,gray,<->] (0,-4)-- (0,4); +\draw[thick,blue,->] (0,0) -- node[below] {\(B \vec{e}_2= \left[\begin{array}{c}2 \\ 0 \end{array}\right]\)}++ (2,0); +\draw[thick,blue,->] (0,0) -- ++(3,4) node[above] {\(B \vec{e}_1 = \left[\begin{array}{c} 3 \\ 4 \end{array}\right]\)}; +\draw[thick,red,->] (0,0) -- ++(1,0); +\draw[thick,red,->] (0,0) -- ++(0,1); +\draw[blue,dashed] (2,0) -- (5,4) -- (3,4); +\draw[red,dashed] (1,0) -- (1,1) -- (0,1); +\end{tikzpicture} + + + Reflection of a parallelogram as a result of swapping columns. +
+ + + + + + +

+The fact that swapping columns multiplies determinants by a negative +may be verified by adding and subtracting columns. +

+ + + \det([\vec{v}\hspace{1em}\vec{w}]) +&= + \det([\vec{v}+\vec{w}\hspace{1em}\vec{w}]) + + +&= + \det([\vec{v}+\vec{w}\hspace{1em}\vec{w}-(\vec{v}+\vec{w})]) + + +&= + \det([\vec{v}+\vec{w}\hspace{1em}-\vec{v}]) + + +&= + \det([\vec{v}+\vec{w}-\vec{v}\hspace{1em}-\vec{v}]) + + +&= + \det([\vec{w}\hspace{1em}-\vec{v}]) + + +&= + -\det([\vec{w}\hspace{1em}\vec{v}]) + + + + + + + + + +

+ To summarize, we've shown that the column versions of the three row-reducing operations + a matrix may be used to simplify a determinant in the following way: +

+
    +

  1. Multiplying a column by a scalar multiplies the + determinant by that scalar: + c\det([\cdots\hspace{0.5em}\vec{v}\hspace{0.5em} \cdots])= + \det([\cdots\hspace{0.5em}c\vec{v}\hspace{0.5em} \cdots]) +

    2. +

  2. Swapping two columns changes the sign of the determinant: + \det([\cdots\hspace{0.5em}\vec{v}\hspace{0.5em} + \cdots\hspace{1em}\vec{w}\hspace{0.5em} \cdots])= + -\det([\cdots\hspace{0.5em}\vec{w}\hspace{0.5em} + \cdots\hspace{1em}\vec{v}\hspace{0.5em} \cdots]) +

    3. +

  3. Adding a multiple of a column to another column does not + change the determinant: + \det([\cdots\hspace{0.5em}\vec{v}\hspace{0.5em} + \cdots\hspace{1em}\vec{w}\hspace{0.5em} \cdots])= + \det([\cdots\hspace{0.5em}\vec{v}+c\vec{w}\hspace{0.5em} + \cdots\hspace{1em}\vec{w}\hspace{0.5em} \cdots]) +

    4. +
+ + + + + +

+ The transformation given by the standard matrix A scales areas by + 4, and the transformation given by the standard matrix B scales + areas by 3. By what factor does the transformation given by the standard matrix + AB scale areas? +

+
+ + +\begin{tikzpicture}[x=0.2in,y=0.2in] +\begin{scope} +\fill[red] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0); +\end{scope} +\draw[->,thick] (2,1) to[bend left=30] node[above] {\(B\)} (3,1); +\begin{scope}[shift={(3,-1)}] +\fill[purple] (0,0) -- (2,1) -- (3,3) -- (1,2) -- (0,0); +\end{scope} +\draw[->,thick] (6.5,1) to[bend left=30] node[above] {\(A\)} (7.5,1); +\begin{scope}[shift={(12,-3)}] +\fill[blue] (0,0) -- (-4,5) -- (-1,7) -- (3,2) -- (0,0); +\end{scope} +\end{tikzpicture} + + + Area changing under the composition of two linear maps +
+
    +

  1. 1 +

    2. +

  2. 7 +

    3. +

  3. 12 +

    4. +

  4. Cannot be determined +

    5. +
+ + + + + +

+Since the transformation given by the standard matrix AB is obtained +by applying the transformations given by A and B, it follows that +\det(AB)=\det(A)\det(B)=\det(B)\det(A)=\det(BA). +

+ + + + + +

+Recall that row operations may be produced by matrix multiplication. +

+
    +

  • Multiply the first row of A by c: + \left[\begin{array}{cccc} + c & 0 & 0 & 0\\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 + \end{array}\right]A + +

    • +

  • Swap the first and second row of A: + \left[\begin{array}{cccc} + 0 & 1 & 0 & 0\\ + 1 & 0 & 0 & 0\\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 + \end{array}\right]A + +

    • +

  • Add c times the third row to the first row of A: + \left[\begin{array}{cccc} + 1 & 0 & c & 0 \\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 + \end{array}\right]A + +

    • +
+ + + + + + +

+The determinants of row operation matrices may be computed +by manipulating columns to reduce each matrix to the identity: +

+
    +

  • Scaling a row: \det + \left[\begin{array}{cccc} + c & 0 & 0 & 0\\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 + \end{array}\right] + = + c\det + \left[\begin{array}{cccc} + 1 & 0 & 0 & 0\\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 + \end{array}\right] + = + c + +

    • +

  • Swapping rows: \det + \left[\begin{array}{cccc} + 0 & 1 & 0 & 0 \\ + 1 & 0 & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 + \end{array}\right] + = + -1\det + \left[\begin{array}{cccc} + 1 & 0 & 0 & 0\\ + 0 & 1 & 0 & 0\\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 + \end{array}\right] + = + -1 + +

    • +

  • Adding a row multiple to another row: \det + \left[\begin{array}{cccc} + 1 & 0 & c & 0\\ + 0 & 1 & 0 & 0\\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 + \end{array}\right] + = + \det + \left[\begin{array}{cccc} + 1 & 0 & c-1c & 0\\ + 0 & 1 & 0-0c & 0\\ + 0 & 0 & 1-0c & 0 \\ + 0 & 0 & 0-0c & 1 + \end{array}\right] + = + \det(I)=1 + +

    • +
+ + + + + +

+Consider the row operation R_1+4R_3\to R_1 applied as follows to show +A\sim B: +

+ +A=\left[\begin{array}{cccc}1&2&3 & 4\\5&6 & 7 & 8\\9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16\end{array}\right] + \sim +\left[\begin{array}{cccc}1+4(9)&2+4(10)&3+4(11) & 4+4(12) \\5&6 & 7 & 8\\9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16\end{array}\right]=B + + +

Find a matrix R such that B=RA, by applying the same row operation to +I=\left[\begin{array}{cccc}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right]. +

 +

Find \det R by comparing with the previous slide. +

 +

If C \in M_{4,4} is a matrix with \det(C)= -3, find +\det(RC)=\det(R)\det(C). +

 + + + + +

+Consider the row operation R_1\leftrightarrow R_3 applied as follows to show +A\sim B: +

+ +A=\left[\begin{array}{cccc}1&2&3&4\\5&6&7&8\\9&10&11&12 \\ 13 & 14 & 15 & 16\end{array}\right] + \sim +\left[\begin{array}{cccc}9&10&11&12\\5&6&7&8\\1&2&3&4 \\ 13 & 14 & 15 & 16\end{array}\right]=B + + +

Find a matrix R such that B=RA, by applying the same row operation to I. +

 +

If C \in M_{4,4} is a matrix with \det(C)= 5, find \det(RC). +

 + + + + +

+Consider the row operation 3R_2\to R_2 applied as follows to show +A\sim B: +

+ +A=\left[\begin{array}{cccc}1&2&3&4\\5&6&7&8\\9&10&11&12 \\ 13 & 14 & 15 & 16\end{array}\right] + \sim +\left[\begin{array}{cccc}1&2&3&4\\3(5)&3(6)&3(7)&3(8)\\9&10&11&12 \\ 13 & 14 & 15 & 16\end{array}\right]=B + + + +

Find a matrix R such that B=RA.

+ +

If C \in M_{4,4} is a matrix with \det(C)= -7, find \det(RC). +

+ + + + + +

Let A be any 4 \times 4 matrix with determinant 2.

+ + + +

Let B be the matrix obtained from A by applying the row operation R_1 - 5 R_3 \to R_1. + What is \mathrm{det}\,B?

+

    +
  1. -4
    2. +
  2. -2
    3. +
  3. 2
    4. +
  4. 10
    5. +

+ + + + +

Let M be the matrix obtained from A by applying the row operation R_3 \leftrightarrow R_1. + What is \mathrm{det}\,M?

+

    +
  1. -4
    2. +
  2. -2
    3. +
  3. 2
    4. +
  4. 10
    5. +

+ + + + +

Let P be the matrix obtained from A by applying the row operation 2 R_4 \to R_4. + What is \mathrm{det}\,P?

+

    +
  1. -4
    2. +
  2. -2
    3. +
  3. 2
    4. +
  4. 10
    5. +

+ + + + + + +

+ Recall that the column versions of the three row-reducing operations + a matrix may be used to simplify a determinant: +

+
    +

  1. Multiplying columns by scalars: + \det([\cdots\hspace{0.5em}c\vec{v}\hspace{0.5em} \cdots])= + c\det([\cdots\hspace{0.5em}\vec{v}\hspace{0.5em} \cdots]) +

    2. +

  2. Swapping two columns: + \det([\cdots\hspace{0.5em}\vec{v}\hspace{0.5em} + \cdots\hspace{1em}\vec{w}\hspace{0.5em} \cdots])= + -\det([\cdots\hspace{0.5em}\vec{w}\hspace{0.5em} + \cdots\hspace{1em}\vec{v}\hspace{0.5em} \cdots]) +

    3. +

  3. Adding a multiple of a column to another column: + \det([\cdots\hspace{0.5em}\vec{v}\hspace{0.5em} + \cdots\hspace{1em}\vec{w}\hspace{0.5em} \cdots])= + \det([\cdots\hspace{0.5em}\vec{v}+c\vec{w}\hspace{0.5em} + \cdots\hspace{1em}\vec{w}\hspace{0.5em} \cdots]) +

    4. +
+ + + + + +

+The determinants of row operation matrices may be computed +by manipulating columns to reduce each matrix to the identity: +

+
    +

  • Scaling a row: + \left[\begin{array}{cccc} + 1 & 0 & 0 &0 \\ + 0 & c & 0 &0\\ + 0 & 0 & 1 &0 \\ + 0 & 0 & 0 & 0 + \end{array}\right] + +

    • +

  • Swapping rows: + \left[\begin{array}{cccc} + 0 & 1 & 0 &0 \\ + 1 & 0 & 0 & 0\\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 0 + \end{array}\right] + +

    • +

  • Adding a row multiple to another row: + \left[\begin{array}{cccc} + 1 & 0 & 0 & 0\\ + 0 & 1 & c & 0\\ + 0 & 0 & 1 & 0 \\ + 0 & 0 & 0 & 1 + \end{array}\right] + +

    • +
+ + + + + +

+Thus we can also use both row operations to simplify determinants: +

+
    +

  • Multiplying rows by scalars: + \det\left[\begin{array}{c}\vdots\\cR\\\vdots\end{array}\right]= + c\det\left[\begin{array}{c}\vdots\\R\\\vdots\end{array}\right] + +

    • +

  • Swapping two rows: + \det\left[\begin{array}{c}\vdots\\R\\\vdots\\S\\\vdots\end{array}\right]= + -\det\left[\begin{array}{c}\vdots\\S\\\vdots\\R\\\vdots\end{array}\right] + +

    • +

  • Adding multiples of rows/columns to other rows: + \det\left[\begin{array}{c}\vdots\\R\\\vdots\\S\\\vdots\end{array}\right]= + \det\left[\begin{array}{c}\vdots\\R+cS\\\vdots\\S\\\vdots\end{array}\right] + +

    • +
+ + + + + +

+ Complete the following derivation for a formula calculating + 2\times 2 determinants: + + + \det\left[\begin{array}{cc} a & b \\ c & d \end{array}\right] + &= + \unknown \det \left[\begin{array}{cc} 1 & b/a \\ c & d \end{array}\right] + + + &= + \unknown \det \left[\begin{array}{cc} 1 & b/a \\ c-c & d-bc/a \end{array}\right] + + + &= + \unknown \det \left[\begin{array}{cc} 1 & b/a \\ 0 & d-bc/a \end{array}\right] + + + &= + \unknown \det \left[\begin{array}{cc} 1 & b/a \\ 0 & 1 \end{array}\right] + + + &= + \unknown \det \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] + + + &= + \unknown \det I + + + &= + \unknown + + +

+ + + +

+ So we may compute the determinant of \left[\begin{array}{cc} 2 & 4 \\ 2 & 3 \end{array}\right] + by using determinant properties to manipulate its rows/columns to reduce the matrix to I: + + + \det\left[\begin{array}{cc} 2 & 4 \\ 2 & 3 \end{array}\right] + &= + 2 \det \left[\begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array}\right] + + + &= + %2 \det \left[\begin{array}{cc} 1 & 2 \\ 2-2(1) & 3-2(2)\end{array}\right]= + 2 \det \left[\begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array}\right] + + + &= + %2(-1) \det \left[\begin{array}{cc} 1 & -2 \\ 0 & +1 \end{array}\right]= + -2 \det \left[\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right] + + + &= + %-2 \det \left[\begin{array}{cc} 1+2(0) & -2+2(1) \\ 0 & 1\end{array}\right] = + -2 \det \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] + + + &= + %-2\det I = + %-2(1) = + -2 + + +

+

+Or we may use a formula: + + \det\left[\begin{array}{cc} 2 & 4 \\ 2 & 3 \end{array}\right] + = + (2)(3)-(4)(2)=-2 + +

+ + + + + + Videos +
+ + Video: Row operations, matrix multiplication, and determinants +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + +
  • Prove or disprove. The determinant is a linear operator on the vector space of n \times n matrices.
    • +
  • Find a matrix that will double the area of a region in \mathbb{R}^2.
    • +
  • Find a matrix that will triple the area of a region in \mathbb{R}^2.
    • +
  • Find a matrix that will halve the area of a region in \mathbb{R}^2.
    • +
+ + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/05-GT/02.ptx b/linear-algebra/source/05-GT/02.ptx new file mode 100644 index 00000000..2cad8fd6 --- /dev/null +++ b/linear-algebra/source/05-GT/02.ptx @@ -0,0 +1,442 @@ + +
+ Computing Determinants (GT2) + + + + +Class Activities + + +

+We've seen that row reducing all the way into RREF gives us a method of computing determinants. +

+

+However, we learned in that this can be tedious for large matrices. Thus, we will try +to figure out how to turn the determinant of a larger matrix +into the determinant of a smaller matrix. +

+ + + + + +

+ The following image illustrates the transformation of the unit cube + by the matrix + \left[\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 0 & 1\end{array}\right]. +

+
+ + + \begin{tikzpicture} + \fill[purple!50!white] (0,0,0) -- (1,0,1) -- (4,0,2) -- (3,0,1) -- (0,0,0); + \draw[thin,gray,->] (0,0,0) -- (3,0,0); + \draw[thin,gray,->] (0,0,0) -- (0,2,0); + \draw[thin,gray,->] (0,0,0) -- (0,0,2); + %(y,z,x) + \draw[blue] (1,0,1) -- (4,0,2) -- (3,0,1); + \draw[blue] (1,1,0) -- (2,1,1) -- (5,1,2) -- (4,1,1) -- (1,1,0); + \draw[blue] (1,0,1) -- +(1,1,0); + \draw[blue] (4,0,2) -- +(1,1,0); + \draw[blue] (3,0,1) -- +(1,1,0); + + \draw[purple,thick,->] (0,0,0) -- (1,1,0) + node[above left]{\tiny\(\left[\begin{array}{c} 0 \\ 1 \\ 1\end{array}\right]\)}; + \draw[purple,thick,->] (0,0,0) -- (1,0,1) + node[below]{\tiny\(\left[\begin{array}{c} 1 \\ 1 \\ 0\end{array}\right]\)}; + \draw[purple,thick,->] (0,0,0) -- (3,0,1) + node[above left]{\tiny\(\left[\begin{array}{c} 1 \\ 3 \\ 0\end{array}\right]\)}; + \draw[purple,dashed,very thick] (0,0,0) -- node[left] {\tiny\(h=1\)} (0,1,0); + \end{tikzpicture} + + + Transformation of the unit cube by the linear transformation. +
+

+ Recall that for this solid V=Bh, where h is the height of the solid + and B is the area of its parallelogram base. + So what must its volume be? +

+
    +

  1. \det \left[\begin{array}{cc} 1 & 1 \\ 1 & 3 \end{array}\right] +

    2. +

  2. \det \left[\begin{array}{cc} 1 & 0 \\ 3 & 1 \end{array}\right] +

    3. +

  3. \det \left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right] +

    4. +

  4. \det \left[\begin{array}{cc} 1 & 3 \\ 0 & 0 \end{array}\right] +

    5. +
+ + + + + +

+If row i contains all zeros except for a 1 on the +main (upper-left to lower-right) diagonal, +then both column and row i +may be removed without changing the value of the determinant. + + \det \left[\begin{array}{cccc} + 3 & {\color{red} 2} & -1 & 3 \\ + {\color{red} 0} & {\color{red} 1} + & {\color{red} 0} & {\color{red} 0} \\ + -1 & {\color{red} 4} & 1 & 0 \\ + 5 & {\color{red} 0} & 11 & 1 + \end{array}\right] = + \det \left[\begin{array}{ccc} + 3 & -1 & 3 \\ + -1 & 1 & 0 \\ + 5 & 11 & 1 + \end{array}\right] + +

+

+Since row and column operations affect the determinants in the same +way, the same technique works for a column of all zeros except for +a 1 on the main diagonal. + + \det \left[\begin{array}{cccc} + 3 & {\color{red} 0} & -1 & 5 \\ + {\color{red} 2} & {\color{red} 1} & {\color{red} 4} & + {\color{red} 0} \\ + -1 & {\color{red} 0} & 1 & 11 \\ + 3 & {\color{red} 0} & 0 & 1 + \end{array}\right] = + \det \left[\begin{array}{ccc} + 3 & -1 & 5 \\ + -1 & 1 & 11 \\ + 3 & 0 & 1 + \end{array}\right] + +

+

+Put another way, if you have either a column or row from the identity matrix, +you can cancel both the column and row containing the 1. +

+ + + + +

+If the 1 is not on the main diagonal, you'll need to use row or column +swaps in order to cancel. + + \det \left[\begin{array}{cccc} + 3 & {\color{red} 0} & -1 & 5 \\ + -1 & {\color{red} 0} & 1 & 11 \\ + {\color{red} 2} & {\color{red} 1} & {\color{red} 4} & + {\color{red} 0} \\ + 3 & {\color{red} 0} & 0 & 1 + \end{array}\right] = + -\det \left[\begin{array}{cccc} + 3 & {\color{red} 0} & -1 & 5 \\ + {\color{red} 2} & {\color{red} 1} & {\color{red} 4} & + {\color{red} 0} \\ + -1 & {\color{red} 0} & 1 & 11 \\ + 3 & {\color{red} 0} & 0 & 1 + \end{array}\right] = + -\det \left[\begin{array}{ccc} + 3 & -1 & 5 \\ + -1 & 1 & 11 \\ + 3 & 0 & 1 + \end{array}\right] + +

+ + + + + +

+ Remove an appropriate row and column of + \det \left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 5 & 12 \\ 3 & 2 & -1 \end{array}\right] + to simplify the determinant to a 2\times 2 determinant. +

+ + + + +

+ Simplify + \det \left[\begin{array}{ccc} 0 & 3 & -2 \\ 2 & 5 & 12 \\ 0 & 2 & -1 \end{array}\right] + to a multiple of a 2\times 2 determinant by first doing the following: +

+ +

Factor out a 2 from a column. +

+

Swap rows or columns to put a 1 on the main diagonal. +

+ + + + +

+ Simplify + \det \left[\begin{array}{ccc} 4 & -2 & 2 \\ 3 & 1 & 4 \\ 1 & -1 & 3\end{array}\right] + to a multiple of a 2\times 2 determinant by first doing the following: +

+ +

Use row/column operations to create two zeroes in the same row or column. +

+

Factor/swap as needed to get a row/column of all zeroes except + a 1 on the main diagonal. +

+ + + + +

+Using row/column operations, you can introduce zeros +and reduce dimension to whittle down the determinant of a large +matrix to a determinant of a smaller matrix. +

+ + + \det\left[\begin{array}{cccc} + 4 & 3 & 0 & 1 \\ + 2 & -2 & 4 & 0 \\ + -1 & 4 & 1 & 5 \\ + 2 & 8 & 0 & 3 + \end{array}\right] + &= + \det\left[\begin{array}{cccc} + 4 & 3 & {\color{red} 0} & 1 \\ + 6 & -18 & {\color{red} 0} & -20 \\ + {\color{red} {-1}} & {\color{red} 4} & + {\color{red} 1} & {\color{red} 5} \\ + 2 & 8 & {\color{red} 0} & 3 + \end{array}\right] + = + \det\left[\begin{array}{ccc} + 4 & 3 & 1 \\ + 6 & -18 & -20 \\ + 2 & 8 & 3 + \end{array}\right] + + + &=\dots= + -2\det\left[\begin{array}{ccc} + {\color{red} 1} & {\color{red} 3} & {\color{red} 4} \\ + {\color{red} 0} & 21 & 43 \\ + {\color{red} 0} & -1 & -10 + \end{array}\right] + = + -2\det\left[\begin{array}{cc} 21 & 43 \\ -1 & -10 \end{array}\right] + + + &= \dots= + -2\det\left[\begin{array}{cc} + -167 & {\color{red}{21}} \\ + {\color{red} 0} & {\color{red} 1} + \end{array}\right] + = -2\det[-167] + + + &=-2(-167)\det(I)= + 334 + + + + + + + +

+Rewrite \det \left[\begin{array}{cccc} 2 & 1 & -2 & 1 \\ 3 & 0 & 1 & 4 +\\ -2 & 2 & 3 & 0 \\ -2 & 0 & -3 & -3 \end{array}\right] +as a multiple of a determinant of a 3\times3 matrix. +

+ + + + + +

+ Compute + + \det\left[\begin{array}{cccc} + 2 & 3 & 5 & 0 \\ + 0 & 3 & 2 & 0 \\ + 1 & 2 & 0 & 3 \\ + -1 & -1 & 2 & 2 + \end{array}\right] + by using any combination of row/column operations. +

+ + + + + +

+Another option is to take advantage of the fact that the determinant is linear +in each row or column. This approach is called +Laplace expansion or cofactor expansion. +

+

+For example, since + + \color{blue}{ + \left[\begin{array}{ccc} 1 & 2 & 4 \end{array}\right] += + 1\left[\begin{array}{ccc} 1 & 0 & 0 \end{array}\right] ++ + 2\left[\begin{array}{ccc} 0 & 1 & 0 \end{array}\right] ++ + 4\left[\begin{array}{ccc} 0 & 0 & 1 \end{array}\right]} +, +

+ + +\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 1} & {\color{blue} 2} & {\color{blue} 4} \end{array}\right] &= +{\color{blue} 1}\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 1} & {\color{blue} 0} & {\color{blue} 0} \end{array}\right] + +{\color{blue} 2}\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 0} & {\color{blue} 1} & {\color{blue} 0} \end{array}\right] + +{\color{blue} 4}\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ {\color{blue} 0} & {\color{blue} 0} & {\color{blue} 1} \end{array}\right] + + +&= -1\det \left[\begin{array}{ccc} 5 & 3 & 2 \\ 5 & 3 & -1 \\ 0 & 0 & 1 \end{array}\right] +-2\det \left[\begin{array}{ccc} 2 & 5 & 3 \\ -1 & 5 & 3 \\ 0 & 0 & 1 \end{array}\right] + +4\det \left[\begin{array}{ccc} 2 & 3 & 5 \\ -1 & 3 & 5 \\ 0 & 0 & 1 \end{array}\right] + + +&= -\det \left[\begin{array}{cc} 5 & 3 \\ 5 & 3 \end{array}\right] +-2 \det \left[\begin{array}{cc} 2 & 5 \\ -1 & 5 \end{array}\right] ++4 \det \left[\begin{array}{cc} 2 & 3 \\ -1 & 3 \end{array}\right] + + + + + + + +

+Recall the formula for a 2\times 2 determinant found in : + + \det \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] + = + ad-bc +. +

+

+There are formulas and algorithms for the determinants of larger matrices, +but they can be pretty tedious to use. For example, writing out a +formula for a 4\times 4 determinant would require 24 different terms! +

+ + \det\left[\begin{array}{cccc} + a_{11} & a_{12} & a_{13} & a_{14} \\ + a_{21} & a_{22} & a_{23} & a_{24} \\ + a_{31} & a_{32} & a_{33} & a_{34} \\ + a_{41} & a_{42} & a_{43} & a_{44} + \end{array}\right] + = + a_{11}(a_{22}(a_{33}a_{44}-a_{43}a_{34})-a_{23}(a_{32}a_{44}-a_{42}a_{34})+\dots)+\dots + + + + + + +

+Based on the previous activities, which technique is easier for computing determinants? +

+
    +

  1. Memorizing formulas. +

    2. +

  2. Using row/column operations. +

    3. +

  3. Laplace expansion. +

    4. +

  4. Some other technique. +

    5. +
+ + + + + +

+ Use your preferred technique to compute + + \det\left[\begin{array}{cccc} + 4 & -3 & 0 & 0 \\ + 1 & -3 & 2 & -1 \\ + 3 & 2 & 0 & 3 \\ + 0 & -3 & 2 & -2 + \end{array}\right] + . +

+ + + +You can check your answers using technology. + +det([4,-3,0,0; 1,-3,2,-1; 3,2,0,3; 0,-3,2,-2]) + + + + + + + Videos +
+ + Video: Simplifying a determinant using row operations +
+
+ + Video: Computing a determinant +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + + Prove that the equation of a line in the plane, through points (x_1,y_1), (x_2,y_2), when x_1 \neq x_2 is given by the equation +\mbox{det}\left(\begin{array}{ccc}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{array}\right) = 0. + + + + + + Prove that the determinant of any diagonal matrix, upper triangular matrix, or lower triangular matrix, is the product of it's diagonal entries. + + + + + + Show that, if an n \times n matrix M has a non-zero determinant, then any \vec{v} \in \mathbb{R}^n can be represented as a linear combination of the columns of M. + + + + + + What is the smallest number of zeros necessary to place in a 4 \times 4 matrix, and the placement of those zeros, such that the matrix has a zero determinant? + + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/05-GT/03.ptx b/linear-algebra/source/05-GT/03.ptx new file mode 100644 index 00000000..7c36bdd3 --- /dev/null +++ b/linear-algebra/source/05-GT/03.ptx @@ -0,0 +1,311 @@ + +
+ Eigenvalues and Characteristic Polynomials (GT3) + + + + +Class Activities + + +

+ An invertible matrix M and its inverse M^{-1} are given below: +

+ + M=\left[\begin{array}{cc}1&2\\3&4\end{array}\right] + \hspace{2em} + M^{-1}=\left[\begin{array}{cc}-2&1\\3/2&-1/2\end{array}\right] + +

+ Which of the following is equal to \det(M)\det(M^{-1})? +

+
    +

  1. -1 +

    2. +

  2. 0 +

    3. +

  3. 1 +

    4. +

  4. 4 +

    5. +
+ + + + + +

For every invertible matrix M, + + \det(M)\det(M^{-1})= \det(I)=1 + + so \det(M^{-1})=\frac{1}{\det(M)}. +

+

Furthermore, + a square matrix M is invertible if and only if \det(M)\not=0.

+ + + + + +

+Consider the linear transformation A : \IR^2 \rightarrow \IR^2 +given by the matrix A = \left[\begin{array}{cc} 2 & 2 \\ 0 & 3 \end{array}\right]. +

+
+ + +\begin{tikzpicture} +\fill[red!50] (0,0) rectangle (1,1); +\draw[thin,gray,<->] (-1,0)-- (4,0); +\draw[thin,gray,<->] (0,-1)-- (0,4); +\draw[thick,blue,->] (0,0) -- node[below right] {\(A \vec{e}_1\)}++ (2,0); +\draw[thick,red,->] (0,0) -- node[below] {\(\vec{e}_1\)}++ (1,0); +\draw[thick,blue,->] (0,0) -- node[above left] {\(A \vec{e}_2\)}++(2,3); +\draw[thick,red,->] (0,0) -- node[left] {\(\vec{e}_2\)}++ (0,1); +\draw[blue,dashed] (2,0) -- (4,3) -- (2,3); +\draw[purple!50!blue,thick,->] (0,0) -- (6,3); +\draw[purple!50!red,thick,->] (0,0) -- (2,1); +\end{tikzpicture} + + + Transformation of the unit square by the linear transformation A +
+

+It is easy to see geometrically that + + A\left[\begin{array}{c}1 \\ 0 \end{array}\right] = + \left[\begin{array}{cc} 2 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{c}1 \\ 0 \end{array}\right]= + \left[\begin{array}{c}2 \\ 0 \end{array}\right]= + 2 \left[\begin{array}{c}1 \\ 0 \end{array}\right] +. +

+

+It is less obvious (but easily checked once you find it) that + + A\left[\begin{array}{c} 2 \\ 1 \end{array}\right] = + \left[\begin{array}{cc} 2 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{c}2 \\ 1 \end{array}\right]= + \left[\begin{array}{c} 6 \\ 3 \end{array}\right] = + 3\left[\begin{array}{c} 2 \\ 1 \end{array}\right] +. +

+ + + + + +

+Let A \in M_{n,n}. +An eigenvectoreigenvector for A +is a vector \vec{x} \in \IR^n such that A\vec{x} is parallel to \vec{x}. +

+
+ + +\begin{tikzpicture} +\fill[gray!50] (0,0) rectangle (1,1); +\draw[thin,gray,<->] (-1,0)-- (4,0); +\draw[thin,gray,<->] (0,-1)-- (0,4); +\draw[thick,blue,->] (0,0) -- node[below right] {\(A \vec{e}_1=2\vec e_1\)}++ (2,0); +\draw[thick,red,->] (0,0) -- node[below] {\(\vec{e}_1\)}++ (1,0); +\draw[thick,gray,->] (0,0) -- node[above left] {\(A \vec{e}_2\)}++(2,3); +\draw[thick,gray,->] (0,0) -- node[left] {\(\vec{e}_2\)}++ (0,1); +\draw[gray,dashed] (2,0) -- (4,3) -- (2,3); +\draw[purple!50!blue,thick,->] (0,0) -- (6,3) + node [below right] {\( + A\left[\begin{array}{c}2\\1\end{array}\right] + = + 3\left[\begin{array}{c}2\\1\end{array}\right] + \)}; +\draw[purple!50!red,thick,->] (0,0) -- (2,1) + node [above] {\(\left[\begin{array}{c}2\\1\end{array}\right]\)}; +\end{tikzpicture} + + + The map A stretches out the eigenvector \left[\begin{array}{c}2 \\ 1 \end{array}\right] by a factor of 3 (the corresponding eigenvalue). +
+

+In other words, A\vec{x}=\lambda \vec{x} for some scalar \lambda. +If \vec x\not=\vec 0, then we say \vec x is a nontrivial eigenvectoreigenvectornontrivial +and we call this \lambda an eigenvalueeigenvalue of A. +

+ + + + + +

+Finding the eigenvalues \lambda that satisfy + + A\vec x=\lambda\vec x=\lambda(I\vec x)=(\lambda I)\vec x + +for some nontrivial eigenvector \vec x is equivalent to finding +nonzero solutions for the matrix equation + + (A-\lambda I)\vec x =\vec 0 +. +

+ + +

+If \lambda is an eigenvalue, and T is the transformation +with standard matrix A-\lambda I, which of these must +contain a non-zero vector? +

+
    +

  1. The kernel of T

    2. +

  2. The image of T

    3. +

  3. The domain of T

    4. +

  4. The codomain of T

    5. +
+ + +

+Therefore, what can we conclude? +

+
    +

  1. A is invertible

    2. +

  2. A is not invertible

    3. +

  3. A-\lambda I is invertible

    4. +

  4. A-\lambda I is not invertible

    5. +
+ + +

+And what else? +

+
    +

  1. \det A=0

    2. +

  2. \det A=1

    3. +

  3. \det(A-\lambda I)=0

    4. +

  4. \det(A-\lambda I)=1

    5. +
+ + + + + +

+ The eigenvalues \lambda for a matrix A are exactly the values + that make A-\lambda I non-invertible. +

+

+ Thus the eigenvalues \lambda for a matrix A + are the solutions to + the equation \det(A-\lambda I)=0. +

+ + + + + +

+The expression \det(A-\lambda I) is called the +characteristic polynomial of A. +

+

+For example, when +A=\left[\begin{array}{cc}1 & 2 \\ 5 & 4\end{array}\right], we have + + A-\lambda I= + \left[\begin{array}{cc}1 & 2 \\ 5 & 4\end{array}\right]- + \left[\begin{array}{cc}\lambda & 0 \\ 0 & \lambda\end{array}\right]= + \left[\begin{array}{cc}1-\lambda & 2 \\ 5 & 4-\lambda\end{array}\right] +. +

+

+Thus the characteristic polynomial of A is + + \det\left[\begin{array}{cc}1-\lambda & 2 \\ 5 & 4-\lambda\end{array}\right] += + (1-\lambda)(4-\lambda)-(2)(5) += + \lambda^2-5\lambda-6 + +and its eigenvalues are the solutions -1,6 to \lambda^2-5\lambda-6=0. +

+ + + + + +

+Let A = \left[\begin{array}{cc} 5 & 2 \\ -3 & -2 \end{array}\right]. +

+ + +

+Compute \det (A-\lambda I) to determine the characteristic polynomial of A. +

+ + +

+Set this characteristic polynomial equal to zero and factor to determine the eigenvalues of A. +

+ + + + + +

+ Find all the eigenvalues for the matrix + A=\left[\begin{array}{cc} 3 & -3 \\ 2 & -4 \end{array}\right]. +

+ + + + + +

+ Find all the eigenvalues for the matrix + A=\left[\begin{array}{cc} 1 & -4 \\ 0 & 5 \end{array}\right]. +

+ + + + + +

+ Find all the eigenvalues for the matrix + A=\left[\begin{array}{ccc} 3 & -3 & 1 \\ 0 & -4 & 2 \\ 0 & 0 & 7 \end{array}\right]. +

+ + + + + + + + Videos +
+ + Video: Finding eigenvalues +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + + What are the maximum and minimum number of eigenvalues associated with an n \times n matrix? Write small examples to convince yourself you are correct, and then prove this in generality. + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/05-GT/04.ptx b/linear-algebra/source/05-GT/04.ptx new file mode 100644 index 00000000..1bf8b42f --- /dev/null +++ b/linear-algebra/source/05-GT/04.ptx @@ -0,0 +1,131 @@ + +
+ Eigenvectors and Eigenspaces (GT4) + + + + +Class Activities + + +

+It's possible to show that -2 is an eigenvalue for +\left[\begin{array}{ccc}-1&4&-2\\2&-7&9\\3&0&4\end{array}\right]. +

+

+Compute the kernel of the transformation with standard matrix + + A-(-2)I + = + \left[\begin{array}{ccc} \unknown & 4&-2 \\ 2 & \unknown & 9\\3&0&\unknown \end{array}\right] + +to find all the eigenvectors \vec x such that A\vec x=-2\vec x. +

+ + + + + + + + +

+ Since the kernel of a linear map is a subspace + of \IR^n, and the kernel obtained from A-\lambda I + contains all the eigenvectors associated with \lambda, + we call this kernel the eigenspaceeigenspace of A associated with \lambda. +

+ + + + + +

+Find a basis for the eigenspace for the matrix + + \left[\begin{array}{ccc} + 0 & 0 & 3 \\ 1 & 0 & -1 \\ 0 & 1 & 3 + \end{array}\right] + +associated with the eigenvalue 3. +

+ + + + + + + + +

+Find a basis for the eigenspace for the matrix + + \left[\begin{array}{cccc} + 5 & -2 & 0 & 4 \\ 6 & -2 & 1 & 5 \\ -2 & 1 & 2 & -3 \\ 4 & 5 & -3 & 6 + \end{array}\right] + +associated with the eigenvalue 1. +

+ + + + + + + + + + +

+Find a basis for the eigenspace for the matrix + + \left[\begin{array}{cccc} + 4 & 3 & 0 & 0 \\ 3 & 3 & 0 & 0 \\ 0 & 0 & 2 & 5 \\ 0 & 0 & 0 & 2 + \end{array}\right] + +associated with the eigenvalue 2. +

+ + + + + + + + + + + Videos +
+ + Video: Finding eigenvectors +
+ + + + + +

Exercises available at .

+ + + Mathematical Writing Explorations + + + Given a matrix A, let \{\vec{v_1},\vec{v_2},\ldots,\vec{v_n}\} be the eigenvectors with associated distinct eigenvalues \{\lambda_1,\lambda_2,\ldots, \lambda_n\}. Prove the set of eigenvectors is linearly independent. + + + + + Sample Problem and Solution +

+ Sample problem . +

+ + +
diff --git a/linear-algebra/source/05-GT/main.ptx b/linear-algebra/source/05-GT/main.ptx new file mode 100644 index 00000000..4cf21668 --- /dev/null +++ b/linear-algebra/source/05-GT/main.ptx @@ -0,0 +1,11 @@ + + + Geometric Properties of Linear Maps (GT) + + + + + + + + diff --git a/linear-algebra/source/05-GT/outcomes/01.ptx b/linear-algebra/source/05-GT/outcomes/01.ptx new file mode 100644 index 00000000..afb226f9 --- /dev/null +++ b/linear-algebra/source/05-GT/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Describe how a row operation affects the determinant of a matrix. +

\ No newline at end of file diff --git a/linear-algebra/source/05-GT/outcomes/02.ptx b/linear-algebra/source/05-GT/outcomes/02.ptx new file mode 100644 index 00000000..ba53de1e --- /dev/null +++ b/linear-algebra/source/05-GT/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Compute the determinant of a 4\times 4 matrix. +

\ No newline at end of file diff --git a/linear-algebra/source/05-GT/outcomes/03.ptx b/linear-algebra/source/05-GT/outcomes/03.ptx new file mode 100644 index 00000000..e0a8b9ca --- /dev/null +++ b/linear-algebra/source/05-GT/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Find the eigenvalues of a 2\times 2 matrix. +

\ No newline at end of file diff --git a/linear-algebra/source/05-GT/outcomes/04.ptx b/linear-algebra/source/05-GT/outcomes/04.ptx new file mode 100644 index 00000000..d511f3eb --- /dev/null +++ b/linear-algebra/source/05-GT/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+Find a basis for the eigenspace of a 4\times 4 matrix associated with a given eigenvalue. +

\ No newline at end of file diff --git a/linear-algebra/source/05-GT/outcomes/main.ptx b/linear-algebra/source/05-GT/outcomes/main.ptx new file mode 100644 index 00000000..26af6362 --- /dev/null +++ b/linear-algebra/source/05-GT/outcomes/main.ptx @@ -0,0 +1,26 @@ + +> + + +

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. + +
+ \ No newline at end of file diff --git a/linear-algebra/source/05-GT/outcomes/question.ptx b/linear-algebra/source/05-GT/outcomes/question.ptx new file mode 100644 index 00000000..7f612f33 --- /dev/null +++ b/linear-algebra/source/05-GT/outcomes/question.ptx @@ -0,0 +1,4 @@ + +

+How do we understand linear maps geometrically? +

diff --git a/linear-algebra/source/05-GT/readiness.ptx b/linear-algebra/source/05-GT/readiness.ptx new file mode 100644 index 00000000..98be940b --- /dev/null +++ b/linear-algebra/source/05-GT/readiness.ptx @@ -0,0 +1,54 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Calculate the area of a parallelogram.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    2. +
  2. +

    Recall and use the definition of a linear transformation.

    +
      +
    • +

      Review:

      +
      • +
    +
    3. +
  3. +

    Find the matrix corresponding to a linear transformation of Euclidean spaces.

    +
      +
    • +

      Review:

      +
      • +
    +
    4. +
  4. +

    Find all roots of quadratic polynomials (including complex ones).

    +
      +
    • +

      Review: Khan Academy, + YouTube (1), + YouTube (2)

      +
      • +
    +
    5. +
  5. +

    Interpret the statement A is an invertible matrix in many equivalent ways in different contexts.

    +
      +
    • +

      Review:

      +
      • +
    +
    6. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/05-GT/samples/01.ptx b/linear-algebra/source/05-GT/samples/01.ptx new file mode 100644 index 00000000..c0b388ff --- /dev/null +++ b/linear-algebra/source/05-GT/samples/01.ptx @@ -0,0 +1,45 @@ + + +GT1 + +

+Let A be a 4 \times 4 matrix with determinant -7. +

+
    +
  1. +

    +Let B be the matrix obtained from A by applying the row operation R_3+3R_4 \to R_3. What is \det(B)? +

    +
    2. +
  2. +

    +Let C be the matrix obtained from A by applying the row operation -3R_2 \to R_2. What is \det(C)? +

    +
    3. +
  3. +

    +Let D be the matrix obtained from A by applying the row operation R_3 \leftrightarrow R_4. What is \det(D)? +

    +
    4. +
+ + +
    +
  1. +

    +Adding a multiple of one row to another row does not change the determinant, so \det(B)=\det(A)=-7. +

    +
    2. +
  2. +

    +Scaling a row scales the determinant by the same factor, so so \det(B)=-3\det(A)=-3(-7)=21. +

    +
    3. +
  3. +

    +Swaping rows changes the sign of the determinant, so \det(B)=-\det(A)=7. +

    +
    4. +
+ + diff --git a/linear-algebra/source/05-GT/samples/02.ptx b/linear-algebra/source/05-GT/samples/02.ptx new file mode 100644 index 00000000..fae2a3c2 --- /dev/null +++ b/linear-algebra/source/05-GT/samples/02.ptx @@ -0,0 +1,83 @@ + + + +GT2 + +

+Show how to compute the determinant of the matrix + + A + = + \left[\begin{array}{cccc} + 1 & 3 & 0 & -1 \\ + 1 & 1 & 2 & 4 \\ + 1 & 1 & 1 & 3 \\ + -3 & 1 & 2 & -5 + \end{array}\right] + +

+ + +

+Here is one possible solution, first applying a single row operation, + and then performing Laplace/cofactor expansions to reduce the determinant + to a linear combination of 2\times 2 determinants: +

+ + +\det + \left[\begin{array}{cccc} + 1 & 3 & 0 & -1 \\ + 1 & 1 & 2 & 4 \\ + 1 & 1 & 1 & 3 \\ + -3 & 1 & 2 & -5 + \end{array}\right] + &= \det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 3 \\ -3 & 1 & 2 & -5 \end{array}\right] = + (-1) \det \left[\begin{array}{ccc} 1 & 3 & -1 \\ 1 & 1 & 3 \\ -3 & 1 & -5 \end{array}\right] + (1) \det \left[\begin{array}{ccc} 1 & 3 & 0 \\ 1 & 1 & 1 \\ -3 & 1 & 2 \end{array}\right] + + + &= (-1) \left( (1) \det \left[\begin{array}{cc} 1 & 3 \\ 1 & -5 \end{array}\right] - (1) \det \left[\begin{array}{cc} 3 & -1 \\ 1 & -5 \end{array}\right] + (-3) \det \left[\begin{array}{cc} 3 & -1 \\ 1 & 3 \end{array}\right] \right) + + + + &\phantom{==} (1) \left( (1) \det \left[\begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array}\right] - (3) \det \left[\begin{array}{cc} 1 & 1 \\ -3 & 2 \end{array}\right] \right) + + + % &= (-1)\left( (1)(-8)-(1)(-14)+(-3)(10) \right) + (1) \left( (1)(1)-(3)(5) \right) + + + &= (-1) \left( -8+14-30 \right) + (1) \left(1-15 \right) + + + &=10 + + +

+Here is another possible solution, using row and column operations to first reduce +the determinant to a 3\times 3 matrix and then applying a formula: +

+ + +\det + \left[\begin{array}{cccc} + 1 & 3 & 0 & -1 \\ + 1 & 1 & 2 & 4 \\ + 1 & 1 & 1 & 3 \\ + -3 & 1 & 2 & -5 + \end{array}\right] +&= +\det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 3 \\ -3 & 1 & 2 & -5 \end{array}\right] = +\det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 2 \\ -3 & 1 & 2 & -7 \end{array}\right] + + +&=-\det \left[\begin{array}{cccc} 1 & 3 & 0 & -1 \\ 1 & 1 & 1 & 2 \\ 0 & 0 & 1 & 0 \\ -3 & 1 & 2 & -7 \end{array}\right] = +-\det \left[\begin{array}{ccc} 1 & 3 & -1 \\ 1 & 1 & 2 \\ -3 & 1 & -7 \end{array}\right] + + + &=-((-7-18-1)-(3+2-21)) + + + &=10 + + + + \ No newline at end of file diff --git a/linear-algebra/source/05-GT/samples/03.ptx b/linear-algebra/source/05-GT/samples/03.ptx new file mode 100644 index 00000000..686b8172 --- /dev/null +++ b/linear-algebra/source/05-GT/samples/03.ptx @@ -0,0 +1,20 @@ + + + + + +GT3 + +

+Explain how to find the eigenvalues of the matrix \left[\begin{array}{cc} -2 & -2 \\ 10 & 7 \end{array}\right] . +

+ + +

+Compute the characteristic polynomial: +\det(A-\lambda I) = \det \left[\begin{array}{cc} -2 - \lambda & -2 \\ 10 & 7-\lambda \end{array}\right] += (-2-\lambda)(7-\lambda)+20 = \lambda ^2 -5\lambda +6 = (\lambda -2)(\lambda -3) +The eigenvalues are the roots of the characteristic polynomial, namely 2 and 3. +

+ + \ No newline at end of file diff --git a/linear-algebra/source/05-GT/samples/04.ptx b/linear-algebra/source/05-GT/samples/04.ptx new file mode 100644 index 00000000..286a98ba --- /dev/null +++ b/linear-algebra/source/05-GT/samples/04.ptx @@ -0,0 +1,19 @@ + + + + GT4 + +

+Explain how to find a basis for the eigenspace associated to the eigenvalue 3 in the matrix \left[\begin{array}{ccc} -7 & -8 & 2 \\ 8 & 9 & -1 \\ \frac{13}{2} & 5 & 2 \end{array}\right]. +

+ + +

+The eigenspace associated to 3 is the kernel of A-3I, so we compute +\RREF(A-3I) = \RREF \left[\begin{array}{ccc} -7-3 & -8 & 2 \\ 8 & 9-3 & -1 \\ \frac{13}{2} & 5 & 2-3 \end{array}\right] = +\RREF \left[\begin{array}{ccc} -10 & -8 & 2 \\ 8 & 6 & -1 \\ \frac{13}{2} & 5 & -1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 0 \end{array}\right]. +Thus we see the kernel is \setBuilder{\left[\begin{array}{c} -a \\ \frac{3}{2} a \\ a \end{array}\right]}{a \in \IR} +which has a basis of \left\{ \left[\begin{array}{c} -1 \\ \frac{3}{2} \\ 1 \end{array}\right] \right\}. +

+ + \ No newline at end of file diff --git a/linear-algebra/source/applications/cryptography.ptx b/linear-algebra/source/applications/cryptography.ptx new file mode 100644 index 00000000..bef2073d --- /dev/null +++ b/linear-algebra/source/applications/cryptography.ptx @@ -0,0 +1,12 @@ + +
+ Cybersecurity: Cryptography + + + +

+TODO port from deprecated LaTeX version of text +

+ + +
diff --git a/linear-algebra/source/applications/geology.ptx b/linear-algebra/source/applications/geology.ptx new file mode 100644 index 00000000..cd6d7846 --- /dev/null +++ b/linear-algebra/source/applications/geology.ptx @@ -0,0 +1,131 @@ + +
+ Geology: Phases and Components + + Activities + + +

+In geology, a phase is any physically separable material in the system, such as various minerals or liquids. +

+

+A component is a chemical compound necessary to make up the phases; these are usually oxides such as Calcium Oxide ({\rm CaO}) or Silicon Dioxide ({\rm SiO_2}). +

+

+In a typical application, a geologist knows how to build each phase from the components, and is interested in determining reactions among the different phases. +

+ + + + + +

+Consider the 3 components +\vec{c}_1={\rm CaO} \hspace{1em} \vec{c}_2={\rm MgO} \hspace{1em} \text{and } \vec{c}_3={\rm SiO_2} +and the 5 phases: + + +\vec{p}_1 &= {\rm Ca_3MgSi_2O_8} & \vec{p}_2 &= {\rm CaMgSiO_4} & \vec{p}_3 &= {\rm CaSiO_3} + + +\vec{p}_4 &= {\rm CaMgSi_2O_6} & \vec{p}_5 &= {\rm Ca_2MgSi_2O_7} + + +

+

+Geologists already know (or can easily deduce) that + + +\vec{p}_1 &= 3\vec{c}_1 + \vec{c}_2 + 2 \vec{c}_3 & \vec{p}_2 &= \vec{c}_1 +\vec{c}_2 + \vec{c}_3 & +\vec{p}_3 &= \vec{c}_1 + 0\vec{c}_2 + \vec{c}_3 + + +\vec{p}_4 &= \vec{c}_1 +\vec{c}_2 + 2\vec{c}_3 & \vec{p}_5 &= 2\vec{c}_1 + \vec{c}_2 + 2 \vec{c}_3 + + +since, for example: + +\vec c_1+\vec c_3 = \mathrm{CaO} + \mathrm{SiO_2} = \mathrm{CaSiO_3} = \vec p_3 + +

+ + + + + +

+To study this vector space, each of the three components \vec c_1,\vec c_2,\vec c_3 may be considered +as the three components of a Euclidean vector. + \vec{p}_1 = \left[\begin{array}{c} 3 \\ 1 \\ 2 \end{array}\right], +\vec{p}_2 = \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right], +\vec{p}_3 = \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right], +\vec{p}_4 = \left[\begin{array}{c} 1 \\ 1 \\ 2 \end{array}\right], +\vec{p}_5 = \left[\begin{array}{c} 2 \\ 1 \\ 2 \end{array}\right]. +

+

+Determine if the set of phases is linearly dependent or linearly independent. +

+ + + + + +

+Geologists are interested in knowing all the possible chemical reactions among the 5 phases: + \vec{p}_1 = \mathrm{Ca_3MgSi_2O_8} = \left[\begin{array}{c} 3 \\ 1 \\ 2 \end{array}\right] \hspace{1em} +\vec{p}_2 = \mathrm{CaMgSiO_4} = \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \hspace{1em} +\vec{p}_3 = \mathrm{CaSiO_3} = \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] +\vec{p}_4 = \mathrm{CaMgSi_2O_6} = \left[\begin{array}{c} 1 \\ 1 \\ 2 \end{array}\right] \hspace{1em} +\vec{p}_5 = \mathrm{Ca_2MgSi_2O_7} = \left[\begin{array}{c} 2 \\ 1 \\ 2 \end{array}\right]. + +That is, they want to find numbers x_1,x_2,x_3,x_4,x_5 such that +x_1\vec{p}_1+x_2\vec{p}_2+x_3\vec{p}_3+x_4\vec{p}_4+x_5\vec{p}_5 = 0. +

+ + +

+Set up a system of equations equivalent to this vector equation. +

+ +

+Find a basis for its solution space. +

+ +

+Interpret each basis vector as a vector equation and a chemical equation. +

+ + + + +

+We found two basis vectors +\left[\begin{array}{c} 1 \\ -2 \\ -2 \\ 1 \\ 0 \end{array}\right] and +\left[\begin{array}{c} 0 \\ -1 \\ -1 \\ 0 \\ 1 \end{array}\right], +corresponding to the vector and chemical equations + + +2\vec{p}_2 + 2 \vec{p}_3 &= \vec{p}_1 + \vec{p}_4 & 2{\rm CaMgSiO_4}+2{\rm CaSiO_3}&={\rm Ca_3MgSi_2O_8}+{\rm CaMgSi_2O_6} + + +\vec{p}_2 +\vec{p}_3 &= \vec{p}_5 & {\rm CaMgSiO_4} + {\rm CaSiO_3} &= {\rm Ca_2MgSi_2O_7} + + +

+

+Combine the basis vectors to produce a chemical equation among the five phases that does not involve \vec{p}_2 = {\rm CaMgSiO_4}. +

+ + + + + + + + + Slideshow +

Slideshow of activities available at .

+ + + +
diff --git a/linear-algebra/source/applications/main.ptx b/linear-algebra/source/applications/main.ptx new file mode 100644 index 00000000..88f8ec18 --- /dev/null +++ b/linear-algebra/source/applications/main.ptx @@ -0,0 +1,8 @@ + + + Applications + + + + + diff --git a/linear-algebra/source/applications/pagerank.ptx b/linear-algebra/source/applications/pagerank.ptx new file mode 100644 index 00000000..42315ceb --- /dev/null +++ b/linear-algebra/source/applications/pagerank.ptx @@ -0,0 +1,490 @@ + +
+ Computer Science: PageRank + + Activities + + + +The $978,000,000,000 Problem + + +

+

+

+In the picture below, each circle represents a webpage, and each arrow +represents a link from one page to another. +

+
+ + +\begin{tikzpicture} +\begin{scope}[every node/.style={circle,thick,draw}] + \node (1) at (0,4) {1}; + \node (2) at (0,0) {2}; + \node (3) at (4,0) {3}; + \node (4) at (4,4) {4}; + \node (5) at (6,4) {5}; + \node (6) at (10,4) {6} ; + \node (7) at (6,0) {7} ; +\end{scope} +\begin{scope}[>={Stealth[red]}, every edge/.style={draw=red,very thick}] + \draw [->] (1) edge[bend right=30] (2); + \draw [->] (1) edge (4); + \path [->] (2) edge[bend right=30] (1); + \draw [->] (2) edge (3); + \draw [->] (3) edge (4); + \draw [->] (3) edge (7); + \draw [->] (4) edge (2); + \draw [->] (5) edge[bend right=30] (6); + \draw [->] (5) edge (7); + \draw [->] (6) edge[bend right=30] (5); + \draw [->] (6) edge (7); + \draw [->] (7) edge (4); + \draw [->] (7) edge[bend left=30] (2); +\end{scope} +\end{tikzpicture} + + + A seven-webpage network +
+

+Based on how these pages link to each other, +write a list of the 7 webpages in order from +most important to least important. +

+ + + + + +The $978,000,000,000 Idea + + +

+Links are endorsements. That is: +

+
    +
  1. +

    +A webpage is important if it is linked to (endorsed) by important pages. +

    +
    2. +
  2. +

    +A webpage distributes its importance equally among all the pages it links to +(endorses). +

    +
    3. +
+ + + + + +

+Consider this small network with only three pages. Let x_1, x_2, x_3 be the importance of the three pages respectively. +

+
+ + +\begin{tikzpicture} +\begin{scope}[every node/.style={circle,thick,draw}] + \node (1) at (0,4) {1}; + \node (2) at (0,0) {2}; + \node (3) at (4,0) {3}; +\end{scope} +\begin{scope}[>={Stealth[red]}, every edge/.style={draw=red,very thick}] + \draw [->] (1) edge[bend right=30] (2); + \draw [->] (1) edge (3); + \path [->] (2) edge[bend right=30] (1); + \draw [->] (3) edge (2); +\end{scope} +\end{tikzpicture} + + + A three-webpage network +
+
    +
  1. x_1 splits its endorsement in half between x_2 and x_3
    2. +
  2. x_2 sends all of its endorsement to x_1
    3. +
  3. x_3 sends all of its endorsement to x_2.
    4. +
+ +

+This corresponds to the page rank system: +

+ + && x_2 && &=& x_1 +\frac{1}{2} x_1&& &+&x_3 &=& x_2 +\frac{1}{2} x_1&& && &=& x_3 + + + + + + + + +
+ + +\begin{tikzpicture} +\begin{scope}[every node/.style={circle,thick,draw}] + \node (1) at (0,4) {1}; + \node (2) at (0,0) {2}; + \node (3) at (4,0) {3}; +\end{scope} +\begin{scope}[>={Stealth[red]}, every edge/.style={draw=red,very thick}] + \draw [->] (1) edge[bend right=30] (2); + \draw [->] (1) edge (3); + \path [->] (2) edge[bend right=30] (1); + \draw [->] (3) edge (2); +\end{scope} +\end{tikzpicture} + + + A three-webpage network +
+ +

+ + && x_2 && &=& x_1 +\frac{1}{2} x_1&& &+&x_3 &=& x_2 +\frac{1}{2} x_1&& && &=& x_3 + +

+

+ + \left[\begin{array}{ccc}0&1&0\\\frac{1}{2}&0 & 1\\\frac{1}{2}&0&0\end{array}\right] + \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] + = + \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] + +

+ +

+By writing this linear system in terms of matrix multiplication, we obtain the +page rank matrix + + A + = + \left[\begin{array}{ccc} + 0 & 1 & 0 \\ + \frac{1}{2} & 0 & 1 \\ + \frac{1}{2} & 0 & 0 + \end{array}\right] + and page rank vector + \vec{x}=\left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right]. +

+Thus, computing the importance of pages on a network is equivalent to solving +the matrix equation A\vec{x}=1\vec{x}. +

+ + + + + + +

+Thus, our $978,000,000,000 problem is what kind of problem? + + \left[\begin{array}{ccc}0&1&0\\\frac{1}{2}&0&\frac{1}{2}\\\frac{1}{2}&0&0\end{array}\right] + \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] + = + 1\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right] + +

+
    +
  1. An antiderivative problem
    2. +
  2. A bijection problem
    3. +
  3. A cofactoring problem
    4. +
  4. A determinant problem
    5. +
  5. An eigenvector problem
    6. +
+ + + + +

+Find a page rank vector \vec x satisfying A\vec x=1\vec x +for the following network's page rank matrix A. +

+

+That is, find the eigenspace associated with \lambda=1 for the matrix +A, and choose a vector from that eigenspace. +

+ +
+ + +\begin{tikzpicture} +\begin{scope}[every node/.style={circle,thick,draw}] + \node (1) at (0,4) {1}; + \node (2) at (0,0) {2}; + \node (3) at (4,0) {3}; +\end{scope} +\begin{scope}[>={Stealth[red]}, every edge/.style={draw=red,very thick}] + \draw [->] (1) edge[bend right=30] (2); + \draw [->] (1) edge (3); + \path [->] (2) edge[bend right=30] (1); + \draw [->] (3) edge (2); +\end{scope} +\end{tikzpicture} + + + A three-webpage network +
+

+ + A + = + \left[\begin{array}{ccc} + 0 & 1 & 0 \\ + \frac{1}{2} & 0 & 1 \\ + \frac{1}{2} & 0 & 0 + \end{array}\right] + +

+ + + + + + + + +

+Row-reducing + + A-I + = + \left[\begin{array}{ccc} + -1 & 1 & 0 \\ + \frac{1}{2} & -1 & 1 \\ + \frac{1}{2} & 0 & -1 + \end{array}\right] + \sim + \left[\begin{array}{ccc} + 1 & 0 & -2 \\ + 0 & 1 & -2 \\ + 0 & 0 & 0 + \end{array}\right] + +yields the basic eigenvector \left[\begin{array}{c} 2 \\ 2 \\1 \end{array}\right]. +

+Therefore, we may conclude that pages 1 and 2 are equally important, +and both pages are twice as important as page 3. +

+ + + + + +

+ Compute the 7 \times 7 page rank matrix for the following network. +

+
+ + +\begin{tikzpicture} +\begin{scope}[every node/.style={circle,thick,draw}] + \node (1) at (0,4) {1}; + \node (2) at (0,0) {2}; + \node (3) at (4,0) {3}; + \node (4) at (4,4) {4}; + \node (5) at (6,4) {5}; + \node (6) at (10,4) {6} ; + \node (7) at (6,0) {7} ; +\end{scope} +\begin{scope}[>={Stealth[red]}, every edge/.style={draw=red,very thick}] + \draw [->] (1) edge[bend right=30] (2); + \draw [->] (1) edge (4); + \path [->] (2) edge[bend right=30] (1); + \draw [->] (2) edge (3); + \draw [->] (3) edge (4); + \draw [->] (3) edge (7); + \draw [->] (4) edge (2); + \draw [->] (5) edge[bend right=30] (6); + \draw [->] (5) edge (7); + \draw [->] (6) edge[bend right=30] (5); + \draw [->] (6) edge (7); + \draw [->] (7) edge (4); + \draw [->] (7) edge[bend left=30] (2); +\end{scope} +\end{tikzpicture} + + + A seven-webpage network +
+

+ For example, since website 1 distributes its endorsement equally between 2 and 4, the first column is + \left[\begin{array}{c} 0 \\ \frac{1}{2} \\ 0 \\ \frac{1}{2} \\ 0 \\ 0 \\ 0 \end{array}\right]. +

+ + + + + +

+Find a page rank vector for the given page rank matrix. +

+ +

+ +A=\left[\begin{array}{ccccccc} +0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 \\ +\frac{1}{2} & 0 & 0 & 1 & 0 & 0 & \frac{1}{2} \\ +0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 \\ +\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} \\ +0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 \\ +0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 \\ +0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & \frac{1}{2} & 0 +\end{array}\right] + +

+
+ + +\begin{tikzpicture} +\begin{scope}[every node/.style={circle,thick,draw}] + \node (1) at (0,4) {1}; + \node (2) at (0,0) {2}; + \node (3) at (4,0) {3}; + \node (4) at (4,4) {4}; + \node (5) at (6,4) {5}; + \node (6) at (10,4) {6} ; + \node (7) at (6,0) {7} ; +\end{scope} +\begin{scope}[>={Stealth[red]}, every edge/.style={draw=red,very thick}] + \draw [->] (1) edge[bend right=30] (2); + \draw [->] (1) edge (4); + \path [->] (2) edge[bend right=30] (1); + \draw [->] (2) edge (3); + \draw [->] (3) edge (4); + \draw [->] (3) edge (7); + \draw [->] (4) edge (2); + \draw [->] (5) edge[bend right=30] (6); + \draw [->] (5) edge (7); + \draw [->] (6) edge[bend right=30] (5); + \draw [->] (6) edge (7); + \draw [->] (7) edge (4); + \draw [->] (7) edge[bend left=30] (2); +\end{scope} +\end{tikzpicture} + + + A seven-webpage network +
+ +

+Which webpage is most important? +

+ + + + + +

+Since a page rank vector for the network is given by \vec x, +it's reasonable to consider page 2 as the most important page. + + \vec{x} + = + \left[\begin{array}{c} 2 \\ 4 \\2 \\ 2.5 \\ 0 \\ 0 \\ 1\end{array}\right] + +

+

+Based upon this page rank vector, +here is a complete ranking of all seven pages from most important to least important: + + 2, 4, 1, 3, 7, 5, 6 + +

+
+ + +\begin{tikzpicture} +\begin{scope}[every node/.style={circle,thick,draw}] + \node (1) at (0,4) {1}; + \node (2) at (0,0) {2}; + \node (3) at (4,0) {3}; + \node (4) at (4,4) {4}; + \node (5) at (6,4) {5}; + \node (6) at (10,4) {6} ; + \node (7) at (6,0) {7} ; +\end{scope} +\begin{scope}[>={Stealth[red]}, every edge/.style={draw=red,very thick}] + \draw [->] (1) edge[bend right=30] (2); + \draw [->] (1) edge (4); + \path [->] (2) edge[bend right=30] (1); + \draw [->] (2) edge (3); + \draw [->] (3) edge (4); + \draw [->] (3) edge (7); + \draw [->] (4) edge (2); + \draw [->] (5) edge[bend right=30] (6); + \draw [->] (5) edge (7); + \draw [->] (6) edge[bend right=30] (5); + \draw [->] (6) edge (7); + \draw [->] (7) edge (4); + \draw [->] (7) edge[bend left=30] (2); +\end{scope} +\end{tikzpicture} + + + A seven-webpage network +
+ + + + + + + +

+Given the following diagram, +use a page rank vector to rank the pages 1 through 7 in order +from most important to least important. +

+
+ + +\begin{tikzpicture} +\begin{scope}[every node/.style={circle,thick,draw}] + \node (1) at (0,4) {1}; + \node (2) at (3,4) {2}; + \node (3) at (6,4) {3}; + \node (4) at (10,4) {4}; + \node (5) at (0,0) {5}; + \node (6) at (3,0) {6}; + \node (7) at (6,0) {7} ; +\end{scope} + +\begin{scope}[>={Stealth[red]}, every edge/.style={draw=red,very thick}] + \draw [->] (1) edge (2); + \draw [->] (1) edge (5); + \draw [->] (2) edge (5); + \draw [->] (2) edge[bend left=30] (7); + \draw [->] (3) edge[bend right=30] (1); + \draw [->] (3) edge (2); + \draw [->] (4) edge (3); + \draw [->] (5) edge[bend right=30] (6); + \draw [->] (6) edge[bend right=30] (5); + \draw [->] (6) edge (2); + \draw [->] (7) edge (4); + \draw [->] (7) edge[bend left=30] (2); +\end{scope} +\end{tikzpicture} + + + Another seven-webpage network +
+ + + + + + Slideshow +

Slideshow of activities available at .

+ +
diff --git a/linear-algebra/source/applications/truss.ptx b/linear-algebra/source/applications/truss.ptx new file mode 100644 index 00000000..c1580ce1 --- /dev/null +++ b/linear-algebra/source/applications/truss.ptx @@ -0,0 +1,479 @@ + +
+ Civil Engineering: Trusses and Struts + + Activities + + + +

+In engineering, a truss is a structure designed from several beams +of material called struts, assembled to behave as a single object. +

+
+ + A simple truss +
+
+ + +\drawtruss{} + + + A simple truss +
+ + + + + +

+Consider the representation of a simple truss pictured below. +All of the seven struts are of equal length, affixed to two anchor points +applying a normal force to nodes C and E, and +with a 10000 N load applied to the node given by D. +

+
+ + +\drawtruss{} + + + A simple truss +
+

+Which of the following must hold for the truss to be stable? +

+
    +

  1. +All of the struts will experience compression. +

    2. +

  2. +All of the struts will experience tension. +

    3. +

  3. +Some of the struts will be compressed, but others will be tensioned. +

    4. +
+ + + + + +

+Since the forces must balance at each node for the truss to be stable, +some of the struts will be compressed, while others will be tensioned. +

+
+ + +\drawtruss{\trussCompletion} + + + Completed truss +
+

+By finding vector equations that must hold at each node, we may +determine many of the forces at play. +

+ + + + + +

+For example, at the bottom left node there are 3 forces acting. +

+
+ + +\drawtruss{\trussCForces} + + + Truss with forces +
+

+Let \vec F_{CA} be the force on C given by the compression/tension +of the strut CA, let \vec F_{CD} be defined similarly, and let +\vec N_C be the normal force of the anchor point on C. +

+

+For the truss to be stable, we must have: +

+ +\vec F_{CA}+\vec F_{CD}+\vec N_C=\vec 0 + + + + + + +

+Using the conventions of the previous remark, and where \vec L +represents the load vector on node D, find four more vector equations +that must be satisfied for each of the other four nodes of the truss. +

+
+ + +\drawtruss{} + + + A simple truss +
+ +A: \unknown + + +B: \unknown + + +C: \vec F_{CA}+\vec F_{CD}+\vec N_C=\vec 0 + +D:\unknown +E:\unknown + + + + + +

+The five vector equations may be written as follows. +

+ +A: \vec F_{AC}+\vec F_{AD}+\vec F_{AB}=\vec 0 + + +B: \vec F_{BA}+\vec F_{BD}+\vec F_{BE}=\vec 0 + + +C: \vec F_{CA}+\vec F_{CD}+\vec N_C=\vec 0 + + +D: \vec F_{DC}+\vec F_{DA}+\vec F_{DB} +\vec F_{DE}+\vec L=\vec 0 + + +E: \vec F_{EB}+\vec F_{ED}+\vec N_E=\vec 0 + + + + + + + + +

+Each vector has a vertical and horizontal component, +so it may be treated as a vector in \IR^2. +Note that \vec F_{CA} must have the same magnitude (but opposite +direction) as \vec F_{AC}. +

+ + \vec{F}_{CA} + = + x\begin{bmatrix} \cos(60^\circ) \\ \sin(60^\circ) \end{bmatrix} + = + x\begin{bmatrix} 1/2 \\ \sqrt{3}/2\end{bmatrix} + + + + \vec{F}_{AC} + = + x\begin{bmatrix} \cos(-120^\circ) \\ \sin(-120^\circ) \end{bmatrix} + = + x\begin{bmatrix} -1/2 \\ -\sqrt{3}/2\end{bmatrix} + + + + + + +

+To write a linear system that models the truss under consideration +with constant load 10000 newtons, +how many scalar variables will be required? +

+
    +
  • +

    +7: 5 from the nodes, 2 from the anchors +

    +
    • +
  • +

    +9: 7 from the struts, 2 from the anchors +

    +
    • +
  • +

    +11: 7 from the struts, 4 from the anchors +

    +
    • +
  • +

    +12: 7 from the struts, 4 from the anchors, 1 from the load +

    +
    • +
  • +

    +13: 5 from the nodes, 7 from the struts, 1 from the load +

    +
    • +
+
+ + +\drawtruss{} + + + A simple truss +
+ + + + + + +

+Since the angles for each strut are known, +one variable may be used to represent each. +

+
+ + +\drawtruss{\trussStrutVariables} + + + Variables for the truss +
+

+ For example: +

+ +\vec F_{AB}=-\vec F_{BA}=x_1\begin{bmatrix}\cos(0)\\\sin(0)\end{bmatrix} +=x_1\begin{bmatrix}1\\0\end{bmatrix} + + +\vec F_{BE}=-\vec F_{EB}=x_5\begin{bmatrix}\cos(-60^\circ)\\\sin(-60^\circ)\end{bmatrix} +=x_5\begin{bmatrix}1/2\\-\sqrt{3}/2\end{bmatrix} + + + + + + +

+Since the angle of the normal forces for each anchor point are unknown, +two variables may be used to represent each. +

+
+ + +\drawtruss{\trussNormalForces} + + + Truss with normal forces +
+ +\vec N_C=\begin{bmatrix}y_1\\y_2\end{bmatrix} +\hspace{3em} +\vec N_D=\begin{bmatrix}z_1\\z_2\end{bmatrix} + +

+The load vector is constant. +

+ +\vec L = \begin{bmatrix}0\\-10000\end{bmatrix} + + + + + + +

+Each of the five vector +equations found previously represent two linear equations: +one for the horizontal component and one for the vertical. +

+
+ + +\drawtruss{\trussStrutVariables} + + + Variables for the truss +
+ +C: \vec F_{CA}+\vec F_{CD}+\vec N_C=\vec 0 + + +\Leftrightarrow +x_2\begin{bmatrix}\cos(60^\circ)\\\sin(60^\circ)\end{bmatrix}+ +x_6\begin{bmatrix}\cos(0^\circ)\\\sin(0^\circ)\end{bmatrix}+ +\begin{bmatrix}y_1\\y_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix} + +Using the approximation \sqrt{3}/2\approx 0.866, we have + +\Leftrightarrow +x_2\begin{bmatrix}0.5\\0.866\end{bmatrix}+ +x_6\begin{bmatrix}1\\0\end{bmatrix}+ +y_1\begin{bmatrix}1\\0\end{bmatrix}+ +y_2\begin{bmatrix}0\\1\end{bmatrix}= +\begin{bmatrix}0\\0\end{bmatrix} + + + + + + +

+Expand the vector equation given below using sine and cosine of appropriate angles, +then compute each component (approximating \sqrt{3}/2\approx 0.866). +

+
+ + +\drawtruss{\trussStrutVariables} + + + Variables for the truss +
+ +D:\vec F_{DA}+\vec F_{DB}+\vec F_{DC}+\vec F_{DE}=-\vec L + + +\Leftrightarrow +x_3\begin{bmatrix}\cos(\unknown)\\\sin(\unknown)\end{bmatrix}+ +x_4\begin{bmatrix}\cos(\unknown)\\\sin(\unknown)\end{bmatrix}+ +x_6\begin{bmatrix}\cos(\unknown)\\\sin(\unknown)\end{bmatrix}+ +x_7\begin{bmatrix}\cos(\unknown)\\\sin(\unknown)\end{bmatrix}= +\begin{bmatrix}\unknown\\\unknown\end{bmatrix} + + +\Leftrightarrow +x_3\begin{bmatrix}\unknown\\\unknown\end{bmatrix}+ +x_4\begin{bmatrix}\unknown\\\unknown\end{bmatrix}+ +x_6\begin{bmatrix}\unknown\\\unknown\end{bmatrix}+ +x_7\begin{bmatrix}\unknown\\\unknown\end{bmatrix}= +\begin{bmatrix}\unknown\\\unknown\end{bmatrix} + + + + + + +

+The full augmented matrix given by the ten equations in this linear system +is given below, where the elevent columns correspond to x_1,\dots,x_7,y_1,y_2,z_1,z_2, +and the ten rows correspond to the horizontal and vertical components of the +forces acting at A,\dots,E. +

+ +\left[\begin{array}{ccccccccccc|c} +1&-0.5&0.5&0&0&0&0&0&0&0&0&0\\ +0&-0.866&-0.866&0&0&0&0&0&0&0&0&0\\ +-1&0&0&-0.5&0.5&0&0&0&0&0&0&0\\ +0&0&0&-0.866&-0.866&0&0&0&0&0&0&0\\ +0&0.5&0&0&0&1&0&1&0&0&0&0\\ +0&0.866&0&0&0&0&0&0&1&0&0&0\\ +0&0&-0.5&0.5&0&-1&1&0&0&0&0&0\\ +0&0&0.866&0.866&0&0&0&0&0&0&0&10000\\ +0&0&0&0&-0.5&0&-1&0&0&1&0&0\\ +0&0&0&0&0.866&0&0&0&0&0&1&0\\ +\end{array}\right] + + + + + + +

+This matrix row-reduces to the following. +

+\sim +\left[\begin{array}{ccccccccccc|c} +1&0&0&0&0&0&0&0&0&0&0&-5773.7\\ +0&1&0&0&0&0&0&0&0&0&0&-5773.7\\ +0&0&1&0&0&0&0&0&0&0&0&5773.7\\ +0&0&0&1&0&0&0&0&0&0&0&5773.7\\ +0&0&0&0&1&0&0&0&0&0&0&-5773.7\\ +0&0&0&0&0&1&0&0&0&-1&0&2886.8\\ +0&0&0&0&0&0&1&0&0&-1&0&2886.8\\ +0&0&0&0&0&0&0&1&0&1&0&0\\ +0&0&0&0&0&0&0&0&1&0&0&5000\\ +0&0&0&0&0&0&0&0&0&0&1&5000\\ +\end{array}\right] + + + + + + +

+Thus we know the truss must satisfy the following conditions. +

+ + +x_1=x_2=x_5&=-5882.4 + + +x_3=x_4&=5882.4 + + +x_6=x_7&=2886.8+z_1 + + +y_1&=-z_1 + + +y_2=z_2&=5000 + + +

+In particular, the negative x_1,x_2,x_5 represent tension (forces pointing into the nodes), +and the postive x_3,x_4 represent compression (forces pointing out of the nodes). +The vertical normal forces y_2+z_2 counteract the 10000 load. +

+
+ + +\drawtruss{\trussCompletion} + + + Completed truss +
+ + + + + + Slideshow +

Slideshow of activities available at .

+ + + +
diff --git a/linear-algebra/source/future-ON/01.ptx b/linear-algebra/source/future-ON/01.ptx new file mode 100644 index 00000000..fd2a19fc --- /dev/null +++ b/linear-algebra/source/future-ON/01.ptx @@ -0,0 +1,142 @@ + +
+ Dot Products (ON1) + + + + + + + + +

+ Consider the binary operation ''\circledast'' defined on vectors \vec v, \vec w in \IR^n by + \vec v \circledast \vec w = v_1 w_1 + v_2 w_2 + v_3 w_3 + \cdots + v_n w_n. + +

+ + +

+Let \vec v =\left[\begin{array}{c} 1 \\ -1 \\ 0 \\ 2 \\ 3 \end{array}\right] and \vec w=\left[\begin{array}{c} 5 \\ 12 \\ -1 \\ 1 \\ 2 \end{array}\right]. What is \vec v \circledast \vec w? +

+
    +
  1. 25
    2. +
  2. \left[\begin{array}{c} 5 \\ -12 \\ 0 \\ 2 \\ 6 \end{array}\right]
    3. +
  3. 1
    4. +
  4. \left[\begin{array}{c} 6 \\ 11 \\ -1 \\ 3 \\ 5 \end{array}\right]
    5. +
+ + +

Let \vec v=\left[\begin{array}{c} 48 \\ 55 \end{array}\right].

+
  1. Graph \vec v and use the Pythagorean Theorem to determine the length of \vec v.
    2. +
  2. What is \vec v \circledast \vec v?
+ + +

Let \vec v=\left[\begin{array}{c} v_1 \\ v_2 \end{array}\right].

+
  1. Graph \vec v and use the Pythagorean Theorem to determine the length of \vec v.
    2. +
  2. What is \vec v \circledast \vec v?
    3. +
  3. For a vector \vec v in \mathbb{R}^2, how is the length of \vec v related to\vec v \circledast \vec v?
    4. +
+ + +The Dot Product +

The dot product is a binary operation on vectors that helps us measure the length of vectors and the angle formed by a pair of vectors.

+ + + +

+Given two n-dimensional vectors \vec v and \vec w, the dot productdot product \vec v \cdot \vec w is defined by + + \vec v \cdot \vec w = \left[\begin{array}{c} v_1 \\ v_2 \\ v_3 \\ \vdots \\ v_n \end{array}\right] \cdot \left[\begin{array}{c} w_1 \\ w_2 \\ w_3 \\ \vdots \\ w_n \end{array}\right] + = v_1 w_1 + v_2 w_2 + v_3 w_3 + \cdots + v_n w_n + +The dot product combines two vectors and creates a scalar that gives us geometric information about the input vectors. If both vectors are the same, then \vec{v} \cdot \vec{v} gives us the square of the length of \vec{v}. The length of a vector \vec v in \IR^n, denoted \lvert \vec v \rvert, is defined as + \lvert \vec v \rvert = \sqrt{\vec v \cdot \vec v}=\sqrt{v_1^2 + v_2^2 + v_3 ^2 + \cdots + v_n^2} + Vectors of length 1 are called unit vectors. +

+ + + + + +

Consider each of the following properties of the dot product. Label each property as valid if the property holds for Euclidean vectors \vec u, \vec v and \vec w from \IR^n, and scalars a,b \in \IR, and invalid if it does not. +

+
    +

  1. \left(\vec u \cdot \vec v\right) \cdot \vec w=\vec u \cdot \left(\vec v \cdot \vec w\right).

    2. +

  2. \left(a \vec v\right) \cdot \vec w=a \left(\vec v \cdot \vec w\right).

    3. +

  3. \vec w\cdot \vec v=\vec v \cdot \vec w.

    4. +

  4. \left(a+b\right) \left(\vec v \cdot \vec w\right)= \left(a \vec v\right) \cdot \left(b \vec w\right).

    5. +

  5. \left(a \vec u+b \vec v\right) \cdot \vec w= \left(a \vec u\right) \cdot \vec w + \left(b \vec v\right) \cdot \vec w.

    6. +
+ + + + +

Like arithmetic of real numbers, the dot product on vectors satisfies some familiar properties. Let \vec u, \vec v and \vec w be vectors from \IR^n, and let a,b \in \IR be scalars. +

    +

  1. \vec u \cdot \vec v = \vec v \cdot \vec u.

    2. +

  2. \left( a\vec u\right) \cdot \vec v = a\left(\vec u \cdot \vec v\right).

    3. +

  3. \left(a\vec u + b \vec v\right)\cdot \vec w =a \vec u \cdot \vec w + b \vec v \cdot \vec w.

    4. +
+

+ + + + + +

Given the linear transformation S:\IR^2 \to \IR^2 whose standard matrix is + \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] + and vector \vec v = \left[\begin{array}{c} 3 \\ 4 \end{array}\right], +

+

Graph \vec v and S( \vec v ).

 +

For an unspecified vector \vec w = \left[\begin{array}{c} w_1 \\ w_2 \end{array}\right], describe the relationship between \vec w and S( \vec w ).

 + + + +

Consider \vec v = \left[\begin{array}{c} 3 \\ 4 \end{array}\right].

+ +

What vector \vec w = \left[\begin{array}{c} ? \\ ? \end{array}\right] is the result of rotating \vec v by 90^{\circ} counter-clockwise?

 +

Find the dot product \vec v \cdot \vec w.

 +

For an arbitrary vector \vec x = \left[\begin{array}{c} x_1 \\ x_2 \end{array}\right], what vector \vec y = \left[\begin{array}{c} ? \\ ? \end{array}\right] is the result of rotating \vec x by 90^{\circ} counter-clockwise?

+

Find the dot product \vec x \cdot \vec y.

 +

Suppose two vectors are perpendicular. What can you say about their dot product?

 + + + + +

Two vectors \vec u and \vec v in \IR^n are orthogonalorthogonal provided \vec u \cdot \vec v = 0. +

+ +

A bunch of stuff goes here. So it turns out you can find the angle between vectors using the Law of Cosines as a starting place.

+ + + + + + +

Suppose that \vec u =\left[\begin{array}{c} 4 \\ -1 \\ 0 \end{array}\right] and \vec v = \left[\begin{array}{c} 2 \\ 3 \\ 1 \end{array}\right].

+ +

Find the length of \vec u and the length of \vec v.

 +

Describe all vectors \vec w that are orthogonal to \vec u.

 +

Find the angle between \vec u and \vec v.

 + + + + Videos + + + + +

Exercises available at checkit.clontz.org.

+ + +
diff --git a/linear-algebra/source/future-ON/02.ptx b/linear-algebra/source/future-ON/02.ptx new file mode 100644 index 00000000..e76e3f53 --- /dev/null +++ b/linear-algebra/source/future-ON/02.ptx @@ -0,0 +1,54 @@ + +
+ Projections (ON2) + + + + + + +

+ A set S of vectors is called an orthogonal set provided \vec u \cdot \vec v = 0 for any pair of vectors \vec u, \vec v \in S with \vec u \ne \vec v. +

+ + + + + +

+ If we rotate the standard basis of \IR^2 by 45^{\circ}, we produce the set B=\left\{\vec b_1, \vec b_2\right\}=\left\{ \left[\begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array}\right], \left[\begin{array}{c} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array}\right]\right\} +

+ +

Determine the length of \vec b_1 and \vec b_2.

 +

Determine the dot product \vec b_1 \cdot \vec b_2.

 +

Is B is an orthogonal set? Explain why or why not.

 +

Is B is a basis of \IR^2? Explain why or why not.

 + + + + +

+ A basis B of a vector space is called an orthogonal basis provided every pair of vectors in B is orthogonal. If, in addition, each vector in B is a unit vector, then B is called an orthonormal basis. +

+ + + + + Videos + + + + +

Exercises available at checkit.clontz.org.

+ + +
diff --git a/linear-algebra/source/future-ON/03.ptx b/linear-algebra/source/future-ON/03.ptx new file mode 100644 index 00000000..3a66b5cc --- /dev/null +++ b/linear-algebra/source/future-ON/03.ptx @@ -0,0 +1,33 @@ + +
+ Orthonormal Bases (ON3) + + + + + +

+ TODO +

+ + + + + Videos + + + + +

Exercises available at checkit.clontz.org.

+ + +
diff --git a/linear-algebra/source/future-ON/05.ptx b/linear-algebra/source/future-ON/05.ptx new file mode 100644 index 00000000..9b07effc --- /dev/null +++ b/linear-algebra/source/future-ON/05.ptx @@ -0,0 +1,92 @@ + +
+ Diagonalization (GT5 - Coming Soon!) + + + + + + +

+ Scaffolding for the general power below +

+ + + + + + +

+Prove that if D = \left[\begin{array}{ccccc}\lambda_1 & \cdots & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & \cdots & 0\\ +\vdots & & \ddots & & \vdots\\ +\vdots & & & & \vdots\\ +0 & \cdots & 0 & \cdots & \lambda_n\end{array}\left] + then +D^k = \left[\begin{array}{ccccc}\lambda_1^k & \cdots & 0 & \cdots & 0\\ 0 & \lambda_2^k & \cdots & \cdots & 0\\ +\vdots & & \ddots & & \vdots\\ +\vdots & & & & \vdots\\ +0 & \cdots & 0 & \cdots & \lambda_n^k\end{array}\left] +

 + + + +

+A matrix A is diagonalizablediagonalizable if it is similar to a diagonal matrix, that is, if there exist a diagonal matrix D and an invertible matrix C such that +A = CDC^{-1} +

+ + + +

+A is diagonalizable iff A has n linearly independent eigenvectors. +

 + +\begin{itemize} + \item Assume $A$ has $n$ linearly independent eigenvectors $\vec{v_1},\ldots, \vec{v_n}$ with associated eigenvalues $\lambda_1,\ldots, \lambda_n$. Construct the matrix $C$ using the eigenvectors as columns. \pause + \item Let $D = C^{-1}AC$, so that $A = CDC^{-1}$.\pause + \item Let $\vec{e_i}$ be the $i$-th standard basis vector for $\R^n$. What would occur if we tried to multiply $$D\vec{e_i} = C^{-1}AC\vec{e_i}?$$ + \pause $$D\vec{e_i} = C^{-1}AC\vec{e_i} = C^{-1}A\vec{v_i} = C^{-1}\lambda_i\vec{v_i} = \lambda_i C^{-1}\vec{v_i} = \lambda_i \vec{e_i}$$\\ + What can we conclude from this? +\end{itemize} +\end{frame} + +\begin{frame}{Examples} +Diagonalize the matrix $$A = \m{\frac{1}{2}&\frac{3}{2}\\\frac{3}{2}&\frac{1}{2}}$$ +\pause +\begin{itemize} + \item Find the eigenvalues of $A$. \pause + \item Find the associated eigenvectors of $A$. \pause + \item Verify the eigenvectors are linearly independent \pause + \item Then $A = CDC^{-1}$. Is this representation unique? +\end{itemize} +\end{frame} + +\begin{frame}{Examples} +Diagonalize $$A = \m{2/3&-4/3 \\ -2/3&4/3 }$$ + +Diagonalize $$A = \m{1&1\\0&1}$$ + +\end{frame} + + + + + Videos + + + + + +

Exercises available at checkit.clontz.org.

+ + +
diff --git a/linear-algebra/source/future-ON/main.ptx b/linear-algebra/source/future-ON/main.ptx new file mode 100644 index 00000000..1b285e22 --- /dev/null +++ b/linear-algebra/source/future-ON/main.ptx @@ -0,0 +1,9 @@ + + + Orthogonal and Orthonormal Vectors (ON - Coming Soon!) + + + + + + diff --git a/linear-algebra/source/future-ON/not03.ptx b/linear-algebra/source/future-ON/not03.ptx new file mode 100644 index 00000000..83973080 --- /dev/null +++ b/linear-algebra/source/future-ON/not03.ptx @@ -0,0 +1,33 @@ + +
+ Orthogonal Vectors (ON3) + + + + + +

+ TODO +

+ + + + + Videos + + + + +

Exercises available at checkit.clontz.org.

+ + +
diff --git a/linear-algebra/source/future-ON/outcomes/01.ptx b/linear-algebra/source/future-ON/outcomes/01.ptx new file mode 100644 index 00000000..f4298b6e --- /dev/null +++ b/linear-algebra/source/future-ON/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Use the dot product to determine norms, distances, and angles. +

diff --git a/linear-algebra/source/future-ON/outcomes/02.ptx b/linear-algebra/source/future-ON/outcomes/02.ptx new file mode 100644 index 00000000..51e48f35 --- /dev/null +++ b/linear-algebra/source/future-ON/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Compute projections of Euclidean vectors. +

\ No newline at end of file diff --git a/linear-algebra/source/future-ON/outcomes/03.ptx b/linear-algebra/source/future-ON/outcomes/03.ptx new file mode 100644 index 00000000..d7caa577 --- /dev/null +++ b/linear-algebra/source/future-ON/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Use the Gram-Schmidt algorithm to compute an orthonormal basis. +

\ No newline at end of file diff --git a/linear-algebra/source/future-ON/outcomes/main.ptx b/linear-algebra/source/future-ON/outcomes/main.ptx new file mode 100644 index 00000000..d1f6ad07 --- /dev/null +++ b/linear-algebra/source/future-ON/outcomes/main.ptx @@ -0,0 +1,20 @@ + +> + + +

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
+ \ No newline at end of file diff --git a/linear-algebra/source/future-ON/outcomes/not03.ptx b/linear-algebra/source/future-ON/outcomes/not03.ptx new file mode 100644 index 00000000..0efd5e1c --- /dev/null +++ b/linear-algebra/source/future-ON/outcomes/not03.ptx @@ -0,0 +1,4 @@ + +

+Determine the orthogonality of a set of Euclidean vectors. +

\ No newline at end of file diff --git a/linear-algebra/source/future-ON/outcomes/question.ptx b/linear-algebra/source/future-ON/outcomes/question.ptx new file mode 100644 index 00000000..cfd5f370 --- /dev/null +++ b/linear-algebra/source/future-ON/outcomes/question.ptx @@ -0,0 +1,4 @@ + +

+ TODO big question? +

diff --git a/linear-algebra/source/future-ON/readiness.ptx b/linear-algebra/source/future-ON/readiness.ptx new file mode 100644 index 00000000..d2ca3cfa --- /dev/null +++ b/linear-algebra/source/future-ON/readiness.ptx @@ -0,0 +1,28 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Add numbers up to ten.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    2. +
  2. +

    Subtract numbers up to ten.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    3. +
+ + \ No newline at end of file diff --git a/linear-algebra/source/main.ptx b/linear-algebra/source/main.ptx new file mode 100644 index 00000000..b00812a1 --- /dev/null +++ b/linear-algebra/source/main.ptx @@ -0,0 +1,21 @@ + + + + + Linear Algebra for Team-Based Inquiry Learning + 2024 Edition PREVIEW + 2024 Edition PREVIEW (Instructor Version) + + + + + + + + + + + + + + diff --git a/linear-algebra/source/meta/backmatter.ptx b/linear-algebra/source/meta/backmatter.ptx new file mode 100644 index 00000000..0a772573 --- /dev/null +++ b/linear-algebra/source/meta/backmatter.ptx @@ -0,0 +1,16 @@ + + + Back Matter + + + + Appendix + + + + + Index + + + + diff --git a/linear-algebra/source/meta/copyright.ptx b/linear-algebra/source/meta/copyright.ptx new file mode 100644 index 00000000..0ec159a3 --- /dev/null +++ b/linear-algebra/source/meta/copyright.ptx @@ -0,0 +1,13 @@ + + + + 20172023 + Steven Clontz and Drew Lewis + +This work is freely available for noncommerical, educational purposes. +For specific licensing information, including the terms for licensing of derivative works, please visit + + GitHub.com/TeamBasedInquiryLearning +. + + diff --git a/linear-algebra/source/meta/definitions.ptx b/linear-algebra/source/meta/definitions.ptx new file mode 100644 index 00000000..3211f6cd --- /dev/null +++ b/linear-algebra/source/meta/definitions.ptx @@ -0,0 +1,5 @@ + +
+ Definitions + +
\ No newline at end of file diff --git a/linear-algebra/source/meta/docinfo.ptx b/linear-algebra/source/meta/docinfo.ptx new file mode 100644 index 00000000..fd3f3dd9 --- /dev/null +++ b/linear-algebra/source/meta/docinfo.ptx @@ -0,0 +1,69 @@ + + + tbilla + TBIL-LA + Activities and exercises for easily implementing Team-Based Inquiry Learning in a single-variable calculus classroom. + +\newcommand{\markedPivot}[1]{\boxed{#1}} +\newcommand{\IR}{\mathbb{R}} +\newcommand{\IC}{\mathbb{C}} +\renewcommand{\P}{\mathcal{P}} +\renewcommand{\Im}{\operatorname{Im}} +\newcommand{\RREF}{\operatorname{RREF}} +\newcommand{\vspan}{\operatorname{span}} +\newcommand{\setList}[1]{\left\{#1\right\}} +\newcommand{\setBuilder}[2]{\left\{#1\,\middle|\,#2\right\}} +\newcommand{\unknown}{\,{\color{gray}?}\,} +\newcommand{\drawtruss}[2][1]{% +\begin{tikzpicture}[scale=#1, every node/.style={scale=#1}] +\draw (0,0) node[left,magenta]{C} -- + (1,1.71) node[left,magenta]{A} -- + (2,0) node[above,magenta]{D} -- cycle; +\draw (2,0) -- + (3,1.71) node[right,magenta]{B} -- + (1,1.71) -- cycle; +\draw (3,1.71) -- (4,0) node[right,magenta]{E} -- (2,0) -- cycle; +\draw[blue] (0,0) -- (0.25,-0.425) -- (-0.25,-0.425) -- cycle; +\draw[blue] (4,0) -- (4.25,-0.425) -- (3.75,-0.425) -- cycle; +\draw[thick,red,->] (2,0) -- (2,-0.75); +#2 +\end{tikzpicture} +} +\newcommand{\trussNormalForces}{% +\draw [thick, blue,->] (0,0) -- (0.5,0.5); +\draw [thick, blue,->] (4,0) -- (3.5,0.5); +} +\newcommand{\trussCompletion}{% +\trussNormalForces +\draw [thick, magenta,<->] (0.4,0.684) -- (0.6,1.026); +\draw [thick, magenta,<->] (3.4,1.026) -- (3.6,0.684); +\draw [thick, magenta,<->] (1.8,1.71) -- (2.2,1.71); +\draw [thick, magenta,->] (1.6,0.684) -- (1.5,0.855); +\draw [thick, magenta,<-] (1.5,0.855) -- (1.4,1.026); +\draw [thick, magenta,->] (2.4,0.684) -- (2.5,0.855); +\draw [thick, magenta,<-] (2.5,0.855) -- (2.6,1.026); +} +\newcommand{\trussCForces}{% +\draw [thick, blue,->] (0,0) -- (0.5,0.5); +\draw [thick, magenta,->] (0,0) -- (0.4,0.684); +\draw [thick, magenta,->] (0,0) -- (0.5,0); +} +\newcommand{\trussStrutVariables}{% +\node[above] at (2,1.71) {\(x_1\)}; +\node[left] at (0.5,0.866) {\(x_2\)}; +\node[left] at (1.5,0.866) {\(x_3\)}; +\node[right] at (2.5,0.866) {\(x_4\)}; +\node[right] at (3.5,0.866) {\(x_5\)}; +\node[below] at (1,0) {\(x_6\)}; +\node[below] at (3,0) {\(x_7\)}; +} + + +\usepackage{tikz} +\usepackage{tikz-cd} + + Learning Outcomes + + + + \ No newline at end of file diff --git a/linear-algebra/source/meta/frontmatter.ptx b/linear-algebra/source/meta/frontmatter.ptx new file mode 100644 index 00000000..78c8a0b2 --- /dev/null +++ b/linear-algebra/source/meta/frontmatter.ptx @@ -0,0 +1,123 @@ + + + + + Steven Clontz + Department of Mathematics and Statistics + University of South Alabama + sclontz@southalabama.edu + + + Drew Lewis + drew.lewis@gmail.com + + + Contributing Authors + + Jessalyn Bolkema + Department of Mathematics + California State University, Dominguez Hills + jbolkema@csudh.edu + + + Jeff Ford + Department of Mathematics, Computer Science, and Statistics + Gustavus Adolphus College + jford@gustavus.edu + + + Jordan Kostiuk + Department of Mathematics + Brown University + jordan_kostiuk@brown.edu + + + Sharona Krinsky + Department of Mathematics + California State University, Los Angeles + skrinsk@calstatela.edu + + + Jennifer Nordstrom + Department of Mathematics + Linfield University + jnordstrom@linfield.edu + + + Kate Owens + Department of Mathematics + College of Charleston + owensks@cofc.edu + + + + + + + + Linear Algebra for Team-Based Inquiry Learning + + + + + +

+This work includes materials used under license from the following works: +

+
    +
  • +

    Understanding Linear Algebra

    +
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    • +

      +
      • +
    • +

      CC BY 4.0

      +
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    +
    • +
  • +

    +
      +
    • +

      Copyright 2023 Reeve Hunter

      +
      • +
    • +

      CC BY 4.0

      +
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+ + + + TBIL Resource Library +

+This work is made available as part of the +TBIL Resource Library, +a product of NSF DUE Award #2011807. +

+ + + For Instructors +

If you are adopting this text in your class, please fill out this short form so we can track usage, let you know about updates, etc.

+ + + + Video Resources +

+Videos are available at the end of each section. A complete playlist of videos +aligned with this text is + + available on YouTube +. +

+ + + + diff --git a/linear-algebra/source/meta/sample-exercises.ptx b/linear-algebra/source/meta/sample-exercises.ptx new file mode 100644 index 00000000..7d97e37c --- /dev/null +++ b/linear-algebra/source/meta/sample-exercises.ptx @@ -0,0 +1,40 @@ + +
+ Sample Exercises with Solutions +

+Here we model one exercise and solution for each learning objective. +Your solutions should not look identical to those shown below, but +these solutions can give you an idea of the level of detail required +for a complete solution. +

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
diff --git a/precalculus/assets/Images/Mapping-notation-example.jpg b/precalculus/assets/Images/Mapping-notation-example.jpg new file mode 100644 index 00000000..96946ebf Binary files /dev/null and b/precalculus/assets/Images/Mapping-notation-example.jpg differ diff --git a/precalculus/source/01-EQ/01.ptx b/precalculus/source/01-EQ/01.ptx new file mode 100644 index 00000000..5841f1a2 --- /dev/null +++ b/precalculus/source/01-EQ/01.ptx @@ -0,0 +1,503 @@ + + +
+ Linear Equations and Inequalities (EQ1) + + + + + Activities + + + Recall that when solving a linear equation, you use addition, subtraction, multiplication and division to isolate the variable. + + + +

Solve the linear equations. +

+ + + +

3x-8=5x+2 +

    +

  1. x=2

    2. +

  2. x=5

    3. +

  3. x=-5

    4. +

  4. x=-2

    5. +

+ + + + +

5(3x-4)=2x-(x+3) +

    +

  1. x=\frac{17}{14}

    2. +

  2. x=\frac{14}{17}

    3. +

  3. x=\frac{23}{14}

    4. +

  4. x=\frac{14}{23}

    5. +

+ + + + + +

Solve the linear equation. + \frac{2}{3}x-8=\frac{5x+1}{6} +

+ + + +

Which equation is equivalent to \frac{2}{3}x-8=\frac{5x+1}{6} but does not contain any fractions? +

    +

  1. 12x-48=15x+3

    2. +

  2. 3x-24=10x+2

    3. +

  3. 4x-8=5x+1

    4. +

  4. 4x-48=5x+1

    5. +

+ + + + +

Use the simplified equation from part (a) to solve \frac{2}{3}x-8=\frac{5x+1}{6}. +

    +

  1. x=-17

    2. +

  2. x=-\frac{26}{7}

    3. +

  3. x=-9

    4. +

  4. x=-49

    5. +

+ + + + + +

It is not always the case that a linear equation has exactly one solution. Consider the following linear equations which appear similar, but their solutions are very different. +

+ + + +

Which of these equations has one unique solution? +

    +

  1. 4(x-2)=4x+6

    2. +

  2. 4(x-1)=4x+4

    3. +

  3. 4(x-1)=x+4

    4. +

+ + + + +

Which of these equations has no solutions? +

    +

  1. 4(x-2)=4x+6

    2. +

  2. 4(x-1)=4x+4

    3. +

  3. 4(x-1)=x+4

    4. +

+ + + + +

Which of these equations has many solutions? +

    +

  1. 4(x-2)=4x+6

    2. +

  2. 4(x-1)=4x+4

    3. +

  3. 4(x-1)=x+4

    4. +

+ + + + +

+ What happens to the x variable when a linear equation has no solution or many solutions? +

+ + + + + A linear equation with one unique solution is a conditional equation. A linear equation that is true for all values of the variable is an identity equation. A linear equation with no solutions is an inconsistent equation. + + + +

+ An inequality is a relationship between two values that are not equal. +

+ + + +

What is the solution to the linear equation 3x-1=5?

+ + + + +

Which of these values is a solution of the inequality 3x-1 \ge 5? +

    +

  1. x=0

    2. +

  2. x=2

    3. +

  3. x=4

    4. +

  4. x=10

    5. +

+ + + + +

Express the solution of the inequality 3x-1 \ge 5 in interval notation. +

    +

  1. (-\infty, 2]

    2. +

  2. (-\infty, 2)

    3. +

  3. (2,\infty)

    4. +

  4. [2,\infty)

    5. +

+ + + + +

+ Draw the solution to the inequality on a number line. +

    +

  1. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/-1,0.2/0,0.4/1,0.6/2,0.8/3, 1/4} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{-]}, ultra thick, blue] (0.59,0) -- (0.6,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (-.1,-.16ex) -- (0.6, -.16ex) -- (0.6, .16ex) -- (-.1,.16ex) -- cycle; +\end{tikzpicture} + + +

    2. +

  2. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/-1,0.2/0,0.4/1,0.6/2,0.8/3, 1/4} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{-)}, ultra thick, blue] (0.59,0) -- (0.6,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (-.1,-.16ex) -- (0.6, -.16ex) -- (0.6, .16ex) -- (-.1,.16ex) -- cycle; +\end{tikzpicture} + + +

    3. +

  3. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/-1,0.2/0,0.4/1,0.6/2,0.8/3, 1/4} +\draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{(-}, ultra thick, blue] (0.59,0) -- (0.6,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0.6,-.16ex) -- (1.1, -.16ex) -- (1.1, .16ex) -- (0.6,.16ex) -- cycle; +\end{tikzpicture} + + +

    4. +

  4. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/-1,0.2/0,0.4/1,0.6/2,0.8/3, 1/4} +\draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{[-}, ultra thick, blue] (0.59,0) -- (0.6,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0.6,-.16ex) -- (1.1, -.16ex) -- (1.1, .16ex) -- (0.6,.16ex) -- cycle; +\end{tikzpicture} + + +

    5. +

+ + + + + + +

+ Let's consider what happens to the inequality when the variable has a negative coefficient. +

+ + + +

Which of these values is a solution of the inequality -x\lt 8? +

    +

  1. x=-10

    2. +

  2. x=-8

    3. +

  3. x=4

    4. +

  4. x=10

    5. +

+ + + + +

Solve the linear inequality -x\lt 8. How does your solution compare to the values chosen in part (a)? +

+ + + + +

Expression the solution of the inequality -x\lt 8 in interval notation. +

    +

  1. (-\infty,-8]

    2. +

  2. (-\infty,-8)

    3. +

  3. (-8, \infty)

    4. +

  4. [-8,\infty)

    5. +

+ + + + +

+ Draw the solution to the inequality on a number line. +

    +

  1. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/5,0.2/6,0.4/7,0.6/8,0.8/9, 1/10} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{-]}, ultra thick, blue] (0.59,0) -- (0.6,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (-.1,-.16ex) -- (0.6, -.16ex) -- (0.6, .16ex) -- (-.1,.16ex) -- cycle; +\end{tikzpicture} + + +

    2. +

  2. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/5,0.2/6,0.4/7,0.6/8,0.8/9, 1/10} +\draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{-)}, ultra thick, blue] (0.59,0) -- (0.6,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (-.1,-.16ex) -- (0.6, -.16ex) -- (0.6, .16ex) -- (-.1,.16ex) -- cycle; +\end{tikzpicture} + + +

    3. +

  3. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/5,0.2/6,0.4/7,0.6/8,0.8/9, 1/10} +\draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{(-}, ultra thick, blue] (0.59,0) -- (0.6,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0.6,-.16ex) -- (1.1, -.16ex) -- (1.1, .16ex) -- (0.6,.16ex) -- cycle; +\end{tikzpicture} + + +

    4. +

  4. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/5,0.2/6,0.4/7,0.6/8,0.8/9, 1/10} +\draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{[-}, ultra thick, blue] (0.59,0) -- (0.6,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0.6,-.16ex) -- (1.1, -.16ex) -- (1.1, .16ex) -- (0.6,.16ex) -- cycle; +\end{tikzpicture} + + +

    5. +

+ + + + + + + You can treat solving linear inequalities, just like solving an equation. The one exception is when you multiply or divide by a negative value, reverse the inequality symbol. + + + +

Solve the following inequalities. Express your solution in interval notation and graphically on a number line. +

+ + + +

+ -3x-1 \le 5 +

+ + + + +

+ 3(x+4) \gt 2x-1 +

+ + + + +

+ -\frac{1}{2}x \ge -\frac{2}{4}+\frac{5}{4}x +

+ + + + + + +

+ A compound inequality includes multiple inequalities in one statement. +

+ + + + + +

Consider the statement 3 \le x \lt 8. This really means that 3 \le x and x \lt 8. +

+ + + +

+ Which of the following inequalities are equivalent to the compound inequality 3 \le 2x-3 \lt 8 ? +

    +

  1. 3 \le 2x-3

    2. +

  2. 3 \ge 2x-3

    3. +

  3. 2x-3 \lt 8

    4. +

  4. 2x-3 \gt 8

    5. +

+ + + + +

+ Solve the inequality 3 \le 2x-3. +

    +

  1. x \le 0

    2. +

  2. x \ge 0

    3. +

  3. x \le 3

    4. +

  4. x \ge 3

    5. +

+ + + + +

+ Solve the inequality 2x-3 \lt 8. +

    +

  1. x \gt \frac{11}{2}

    2. +

  2. x \lt \frac{11}{2}

    3. +

  3. x \gt \frac{5}{2}

    4. +

  4. x \lt \frac{5}{2}

    5. +

+ + + + +

Which compound inequality describes how the two solutions overlap? +

    +

  1. 0 \le x \lt \frac{11}{2}

    2. +

  2. 0 \le x \lt \frac{5}{2} +

    3. +

  3. \frac{5}{2} \lt x \le 3

    4. +

  4. 3 \le x \lt \frac{11}{2}

    5. +

+ + + + +

+ Draw the solution to the inequality on a number line. +

    +

  1. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.3,0); +\foreach \x/\xtext in {0/0,0.2/1,0.4/2,0.6/3,0.8/4, 1/5, 1.2/6} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{[-}, ultra thick, blue] (-0.01,0) -- (0,0); +\draw[{-)}, ultra thick, blue] (1.09,0) -- (1.1,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0,-.16ex) -- (1.1, -.16ex) -- (1.1, .16ex) -- (0,.16ex) -- cycle; +\end{tikzpicture} + + +

    2. +

  2. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.3,0); +\foreach \x/\xtext in {0/0,0.2/1,0.4/2,0.6/3,0.8/4, 1/5, 1.2/6} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; + \draw[{[-}, ultra thick, blue] (-0.01,0) -- (0,0); + \draw[{-)}, ultra thick, blue] (.49,0) -- (.5,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0,-.16ex) -- (0.5, -.16ex) -- (0.5, .16ex) -- (0,.16ex) -- cycle; +\end{tikzpicture} + + +

    3. +

  3. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.3,0); +\foreach \x/\xtext in {0/0,0.2/1,0.4/2,0.6/3,0.8/4, 1/5, 1.2/6} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{(-}, ultra thick, blue] (0.29,0) -- (0.3,0); +\draw[{-]}, ultra thick, blue] (0.39,0) -- (0.4,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0.3,-.16ex) -- (0.4, -.16ex) -- (0.4, .16ex) -- (0.3,.16ex) -- cycle; +\end{tikzpicture} + + +

    4. +

  4. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.3,0); +\foreach \x/\xtext in {0/0,0.2/1,0.4/2,0.6/3,0.8/4, 1/5, 1.2/6} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; + \draw[{[-}, ultra thick, blue] (0.59,0) -- (0.6,0); + \draw[{-)}, ultra thick, blue] (1.09,0) -- (1.1,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0.6,-.16ex) -- (1.1, -.16ex) -- (1.1, .16ex) -- (0.6,.16ex) -- cycle; +\end{tikzpicture} + + +

    5. +

+ + + + + + + Solving a compound linear inequality, uses the same methods as a single linear inequality ensuring that you perform the same operations on all three parts. Alternatively, you can break the compound inquality up into two and solve separately. + + + +

+ Solve the following inequalities. Express your solution in interval notation and graphically on a number line. +

+ + + +

+ 8 \lt -3x-1 \le 11 +

+ + + + +

+ -6 \le \frac{x-12}{4} \lt -2 +

+ + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/01-EQ/02.ptx b/precalculus/source/01-EQ/02.ptx new file mode 100644 index 00000000..72d3b8ef --- /dev/null +++ b/precalculus/source/01-EQ/02.ptx @@ -0,0 +1,509 @@ + + +
+ Applications of Linear Equations (EQ2) + + + + + Activities + + + +

+ Linear equations can be used to solve many types of real-world applications. We'll investigate some of those in this section. +

+ + + + +

+ Distance, rate, and time problems are a standard example of an application of a linear equation. For these, it's important to remember that + d=rt + where d is distance, r is the rate (or speed), and t is time. +

+

+ Often we will have more than one moving object, so it is helpful to denote which object's distance, rate, or time we are referring to. One way we can do this is by using a subscript. For example, if we are describing an eastbound train (as we will in the first example), it may be helpful to denote its distance, rate, and time as d_E, r_E, and t_E respectively. Notice that the subscript E is a label reminding us that we are referring to the eastbound train. +

+ + + + + + + +

Two trains leave a station at the same time. One is heading east at a speed of 75 mph, while the other is heading west at a speed of 85 mph. After how long will the trains be 400 miles apart?

+ + + + + + +

+ How fast is each train traveling? +

    +
  1. r_E=85 mph, r_W=75 mph
    2. +
  2. r_E=75 mph, r_W=85 mph
    3. +
  3. r_E=400 mph, r_W=400 mph
    4. +
  4. r_E=75 mph, r_W=400 mph
    5. +
  5. r_E=400 mph, r_W=85 mph
    6. + +
+

+ + +

+ B. +

+ + + + + +

+ Which of the statements describes how the times of the eastbound and westbound train are related? +

    +
  1. The eastbound train is slower than the westbound train, so 75+t_E=85+t_W.
    2. +
  2. The eastbound train left an hour before the westbound train, so if we let t_E=t, then t_W=t-1.
    3. +
  3. Both trains have been traveling the same amount of time, so t_E=t_W. Since they are the same, we can just call them both t.
    4. +
  4. We don't know how the times relate to each other, so we must denote them separately as t_E and t_W.
    5. +
  5. Since the trains are traveling at different speeds, we need the proportion \frac{r_E}{r_W}=\frac{t_E}{t_W}.
    6. +
+

+ + +

+ C. +

+ + + + + +

+ Fill in the following table using the information you've just determined about the trains' rates and times since they left the station. Some values are there to help you get started. + + + + + + rate + time + distance from station + + + eastbound train + + + 75t + + + westbound train + + t + + + +

+ + +

+ + + + rate + time + distance from station + + + eastbound train + 75 + t + 75t + + + westbound train + 85 + t + 85t + + +

+ + + + + +

+ At the moment in question, the trains are 400 miles apart. How does that total distance relate to the distance each train has traveled? +

    +
  1. The 400 miles is irrelevant. They've been traveling the same amount of time so they must be the same distance away from the station. That tells us d_E = d_W .
    2. +
  2. The 400 miles is the difference between the distance each train traveled, so d_E - d_W = 400.
    3. +
  3. The 400 miles represents the sum of the distances that each train has traveled, so d_E + d_W = 400.
    4. +
  4. The 400 miles is the product of the distance each train traveled, so (d_E)( d_W) = 400.
    5. +
+

+ + +

+ C. +

+ + + + + +

+ Now plug in the expressions from your table for d_E and d_w. What equation do you get? +

    +
  1. 75t=85t
    2. +
  2. 75t-85t=400
    3. +
  3. 75t+85t=400
    4. +
  4. (75t)(85t)=400
    5. + +
+

+ + +

+ C. +

+ + + + + +

+ Notice that we now have a linear equation in one variable, t. Solve for t, and put that answer in context of the problem. +

    +
  1. The trains are 400 miles apart after 2 hours.
    2. +
  2. The trains are 400 miles apart after 2.5 hours.
    3. +
  3. The trains are 400 miles apart after 3 hours.
    4. +
  4. The trains are 400 miles apart after 3.5 hours.
    5. +
  5. The trains are 400 miles apart after 4 hours.
    6. +
+

+ + +

+ B. +

+ + + + + + + In we examined the motion of two objects moving at the same time in opposite directions. In we will examine a different perspective, but still apply d=rt to solve. + + + + +

Jalen needs groceries, so decides to ride his bike to the store. It takes him half an hour to get there. After finishing his shopping, he sees his friend Alex who offers him a ride home. He takes the same route home as he did to the store, but this time it only takes one-fifth of an hour. If his average speed was 18 mph faster on the way home, how far away does Jalen live from the grocery store?

+ +

+ We'll use the subscript b to refer to variables relating to Jalen's trip to the store while riding his bike and the subscript c to refer to variables relating to Jalen's trip home while riding in his friend's car. +

+ + + + + +

How long does his bike trip from home to the store and his car trip from the store back home take? + +

    +
  1. t_b=18 hours, t_c=18 hours
    2. +
  2. t_b=\frac{1}{5} of an hour, t_c=\frac{1}{2} of an hour
    3. +
  3. t_b=\frac{1}{2} of an hour, t_c=\frac{1}{5} of an hour
    4. +
  4. t_b=2 hours, t_c=5 hours
    5. +
  5. t_b=5 hours, t_c=2 hours
    6. +
+

+ + +

+ C. +

+ + + + + +

+ Which of the statements describes how the speed (rate) of the bike trip and the car trip are related? +

    + +
  1. Both the trip to the store and the trip home covered the same distance, so r_b=r_c. Since they are the same, we can just call them both r.
    2. +
  2. We don't know how the two rates relate to each other, so cannot write an equation comparing them and must leave them as separate variables r_b and r_c.
    3. +
  3. Jalen's rate on the trip home in the car was 18 mph faster than his trip to the store on his bike, so if we let r_b=r, then r_c=r-18.
    4. +
  4. Jalen's rate on the trip home in the car was 18 mph faster than his trip to the store on his bike, so if we let r_b=r, then r_c=r+18.
    5. +
+

+ + +

+ D. +

+ + + + + +

+ Fill in the following table using the information you've just determined about the Jalen's rates and times on each leg of his grocery store trip. Then fill in the distance column based on how distance relates to rate and time in each case. + + + + + + rate + time + distance covered + + + bike trip (to the store) + + + car trip (going back home) + + + +

+ + +

+ + + + rate + time + distance covered + + + bike trip (to the store) + r + \frac{1}{2} + \frac{1}{2}r + + + car trip (going back home) + r+18 + \frac{1}{5} + \frac{1}{5}(r+18) + + +

+ + + + + +

+ Our goal is to figure out how far away Jalen lives from the store. To help us get there, write an equation relating d_b and d_c. +

    +
  1. The distance he traveled by bike is the same as the distance he traveled by car, so d_b = d_c
    2. +
  2. The distance he traveled by bike took longer than the distance he traveled by car, so d_b + \frac{1}{2} = d_c + \frac{1}{5}
    3. +
  3. The distance, d, between his house and the grocery store is sum of the distance he traveled on his bike and the distance he traveled in the car, so d_b + d_c = d.
    4. +
  4. The distance, d, between his house and the grocery store is sum of the difference he traveled on his bike and the distance he traveled in the car, so d_b - d_c = d.
    5. + +
+

+ + +

+ A. +

+ + + + + +

+ Now plug in the expressions from your table for d_b and d_c into the equation you just found. Notice that it is a linear equation in one variable, r. Solve for r. +

+ + +

+ r=12 +

+ + + + + +

+ Our goal was to determine the distance between Jalen's house and the grocery store. Solving for r did not tell us that distance, but it did get us one step closer. Use that value to help you determine the distance between his house and the store, and write your answer using the context of the problem. (Hint: can you find an expression involving r that we made that represents that distance? ) +

    +
  1. The grocery store is 6 miles away from Jalen's house.
    2. +
  2. The grocery store is 8 miles away from Jalen's house.
    3. +
  3. The grocery store is 10 miles away from Jalen's house.
    4. +
  4. The grocery store is 12 miles away from Jalen's house.
    5. +
  5. The grocery store is 14 miles away from Jalen's house.
    6. + +
+

+ + +

+ A. +

+ + + + + + + Another type of application of linear equations is called a mixture problem. In these we will mix together two things, like two types of candy in a candy store or two solutions of different concentrations of alcohol. + + + + +

Ammie's favorite snack to share with friends is candy salad, which is a mixture of different types of candy. Today she chooses to mix Nerds Gummy Clusters, which cost $8.38 per pound, and Starburst Jelly Beans, which cost $7.16 per pound. If she makes seven pounds of candy salad and spends a total of $55.61, how many pounds of each candy did she buy?

+ + + + + +

There are two "totals" in this situation: the total weight (in pounds) of candy Ammie bought and the total amount of money (in dollars) Ammie spent. Let's begin with the total weight. If we let N represent the pounds of Nerds Gummy Clusters and S represent the pounds of Starburst Jelly Beans, which of the following equations can represent the total weight? + +

    +
  1. N-S=7
    2. +
  2. NS=7
    3. +
  3. N+S=7
    4. +
  4. \frac{N}{S}=7
    5. +
+

+ + +

+ C. +

+ + + + + +

Which expressions represent the amount she spent on each candy? Again, we will let N represent the pounds of Nerds Gummy Clusters and S represent the pounds of Starburst Jelly Beans. + +

    +
  1. N spent on Nerds Gummy Clusters; S spent on Starburst Jelly Beans
    2. +
  2. 8.38N spent on Nerds Gummy Clusters; 7.16S spent on Starburst Jelly Beans
    3. +
  3. 8.38 + N spent on Nerds Gummy Clusters; 7.16 + S spent on Starburst Jelly Beans
    4. +
  4. 8.38 - N spent on Nerds Gummy Clusters; 7.16 - S spent on Starburst Jelly Beans
    5. +
+

+ + +

+ B. +

+ + + + + +

Now we focus on the total cost. Which of the following equations can represent the total amount she spent? + +

    +
  1. N+S=55.61
    2. +
  2. 8.38N+7.16S=55.61
    3. +
  3. 8.38+ N + 7.16 + S=55.61
    4. +
  4. 8.38- N + 7.16 - S=55.61
    5. +
+

+ + +

+ B. +

+ + + + + +

We are almost ready to solve, but we have two variables in our weight equation and our cost equation. We will get the cost equation to one variable by using the weight equation as a substitution. Which of the following is a way to express one variable in terms of the other? (Hint: More than one answer may be correct here!) + +

    +
  1. If N is the total weight of the Nerds Gummy Clusters, then 7-N could represent the weight of the Starburst Jelly Beans.
    2. +
  2. If N is the total weight of the Nerds Gummy Clusters, then 7+N could represent the weight of the Starburst Jelly Beans.
    3. +
  3. If S is the total weight of the Starburst Jelly Beans, then 7-S could represent the weight of the Nerds Gummy Clusters.
    4. + +
  4. If S is the total weight of the Starburst Jelly Beans, then 7+S could represent the weight of the Nerds Gummy Clusters.
    5. + +
+

+ + +

+ A. or C. +

+ + + + + +

Plug your expressions in to the total cost equation. (Hint: More than one of these may be correct!) + +

    +
  1. 8.38N+7.16(7-N)=55.61
    2. +
  2. 8.38S+7.16(7-S)=55.61
    3. +
  3. 8.38(7-N)+7.16N=55.61
    4. +
  4. 8.38(7-S)+7.16S=55.61
    5. + +
+

+ + +

+ A. or D., depending on the substitution chosen +

+ + + + + +

Now solve for N and S, and put your answer in the context of the problem. + +

    +
  1. Ammie bought 2.5 lbs of Nerds Gummy Clusters and 4.5 lbs of Starburst Jelly Beans.
    2. +
  2. Ammie bought 3.5 lbs of Nerds Gummy Clusters and 3.5 lbs of Starburst Jelly Beans.
    3. +
  3. Ammie bought 4.5 lbs of Nerds Gummy Clusters and 2.5 lbs of Starburst Jelly Beans.
    4. +
  4. Ammie bought 5.5 lbs of Nerds Gummy Clusters and 1.5 lbs of Starburst Jelly Beans.
    5. +
+

+ + +

+ C. +

+ + + + + + + + +

+ A chemist needs to mix two solutions to create a mixture consisting of 30% alcohol. She uses 20 liters of the first solution, which has a concentration of 21% alcohol. How many liters of the second solution (that is 45% alcohol) should she add to the first solution to create the mixture that is 30% alcohol? +

+ + + +

+ 12 liters +

+ + + + + + + Videos +

Coming eventually.....

+ +
diff --git a/precalculus/source/01-EQ/03.ptx b/precalculus/source/01-EQ/03.ptx new file mode 100644 index 00000000..067cd933 --- /dev/null +++ b/precalculus/source/01-EQ/03.ptx @@ -0,0 +1,323 @@ + + +
+ Distance and Midpoint (EQ3) + + + + + Activities + + + +

The points A and B are shown in the graph below. Use the graph to answer the following questions: +

+
+ + A = point((2,6),size=40)+text('A',(2.05,6),horizontal_alignment="left", gridlines=[[2,2.5,3.0,3.5,4],[2,2.5,3.0,3.5,4,4.5,5.0,5.5,6]]) + B = point((4,2),size=40)+text('B',(4.05,2),horizontal_alignment="left") + L=line([(2,6),(4,2)],color="purple") + A+B+L + +
+ + + +

+ Draw a right triangle so that the hypotenuse is the line segment between points A and B . Label the third point of the triangle C. +

+ + + Point C should be at the point (2,2). + + + + +

+ Find the lengths of line segments AC and BC . +

+ + + AC is 4 units long. + BC is 2 units long. + Make sure students pay attention to the scale. + + + + +

+ Now that you know the lengths of AC and BC , how can you find the length of AB ? Find the length of AB. +

+ + + Students should see that they can use the Pythagorean Theorem to find the length of side AB (which is \sqrt{20} or approximately 4.5). + + + + + +

Using the Pythagorean Theorem (a^2+b^2=c^2) can be helpful in finding the distance of a line segment (as long as you create a right triangle!). +

+ + + + +

Suppose you are given two points (x_{1},y_{1}) and (x_{2},y_{2}). Let's investigate how to find the length of the line segment that connects these two points! +

+ + + +

+ Draw a sketch of a right triangle so that the hypotenuse is the line segment between the two points. +

+ + + Students may need help in their drawing. Make sure the hypotenuse is the line segment that connects the two points. + + + + +

+ Find the lengths of the legs of the right triangle. +

+ + + The lengths of the legs of the triangle should be y_{2}-y_{1} and x_{2}-x_{1} (or y_{1}-y_{2} and x_{1}-x_{2} depending on how students created their drawing). + + + + +

+ Find the length of the line segment that connects the two original points. +

+ + + Students should see the connection to the previous activity and apply the Pythagorean Theorem. They should get either \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} or \sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2} to represent the length of the side that connects the two points. + + + + + + +

+ The distance, d, between two points, (x_{1},y_{1}) and (x_{2}, y_{2}), can be found by using the distance formula: d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} Notice that the distance formula is an application of the Pythagorean Theorem! +

+ + + + + +

+ Apply to calculate the distance between the given points. +

+ + + +

What is the distance between (4,6) and (9,15)? +

    +
  1. 10.2
    2. +
  2. 10.3
    3. +
  3. \sqrt{106}
    4. +
  4. \sqrt{56}

+ + + B and C + + + + +

What is the distance between (-2,5) and (-7,-1)? +

    +
  1. \sqrt{11}
    2. +
  2. 7.8
    3. +
  3. 3.3
    4. +
  4. \sqrt{61}

+ + + B and D + + + + +

+ Suppose the line segment AB has one endpoint, A, at the origin. For which coordinate of B would make the line segment AB the longest? +

    +
  1. (3,7)
    2. +
  2. (2,-8)
    3. +
  3. (-6,4)
    4. +
  4. (-5,-5)

+ + + B + + + + + +

+ Notice in , you can give a distance in either exact form (leaving it with a square root) or as an approximation (as a decimal). Make sure you can give either form as sometimes one form is more useful than another! +

+ + +

A midpoint refers to the point that is located in the middle of a line segment. In other words, the midpoint is the point that is halfway between the two endpoints of a given line segment. +

+ + + + +

Two line segments are shown in the graph below. Use the graph to answer the following questions: +

+
+ + A = point((8,2),size=40)+text('A',(8.1,2),horizontal_alignment="left", gridlines=[[2,3,4,5,6,7,8],[2,2.5,3,3.5,4,4.5,5,5.5,6]]) + B = point((8,6),size=40)+text('B',(8.1,6),horizontal_alignment="left") + C = point ((2,2),size=40)+text('C',(2,2.1),horizontal_alignment="center") + L=line([(8,2),(2,2)],color="black") + M=line([(8,2),(8,6)],color="purple") + A+B+C+L+M + +
+ + + +

What is the midpoint of the line segment AB? +

    +
  1. (16,4)
    2. +
  2. (8,4)
    3. +
  3. (8,8)
    4. +
  4. (10,2)

+ + + B + + + + +

What is the midpoint of the line segment AC? +

    +
  1. (6,0)
    2. +
  2. (4,4)
    3. +
  3. (6,4)
    4. +
  4. (5,2)

+ + + D + + + + +

Suppose we connect the two endpoints of the two line segments together, to create the new line segment, BC. Can you make an educated guess to where the midpoint of BC is? +

    +
  1. (10,8)
    2. +
  2. (6,4)
    3. +
  3. (5,4)
    4. +
  4. (5,2)

+ + + C + + + + +

+ How can you test your conjecture? Is there a mathematical way to find the midpoint of any line segment? +

+ + + + + + +

+ The midpoint of a line segment with endpoints (x_{1},y_{1}) and (x_{2}, y_{2}), can be found by taking the average of the x and y values. Mathematically, the midpoint formula states that the midpoint of a line segment can be found by: \left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right) +

+ + + + + +

+ Apply to calculate the midpoint of the following line segments. +

+ + + +

What is the midpoint of the line segment with endpoints (-4,5) and (-2,-3)? +

    +
  1. (3,1)
    2. +
  2. (-3,1)
    3. +
  3. (1,1)
    4. +
  4. (1,4)

+ + + B + + + + +

What is the midpoint of the line segment with endpoints (2,6) and (-6,-8)? +

    +
  1. (-3,-1)
    2. +
  2. (-2,0)
    3. +
  3. (-2,-1)
    4. +
  4. (4,7)

+ + + C + + + + +

Suppose C is the midpoint of AB and is located at (9,8). The coordinates of A are (10,10). What are the coordinates of B? +

    +
  1. (9.5,9)
    2. +
  2. (11,12)
    3. +
  3. (18,16)
    4. +
  4. (8,6)

+ + + D + + + + + + +

+ On a map, your friend Sarah's house is located at (-2, 5) and your other friend Austin's house is at (6,-2). +

+ + + +

+ How long is the direct path from Sarah's house to Austin's house? +

+ + + \sqrt{113} or 10.6 + + + + +

+ Suppose your other friend, Micah, lives in the middle between Sarah and Austin. What is the location of Micah's house on the map? +

+ + + (2,1.5) + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
diff --git a/precalculus/source/01-EQ/04.ptx b/precalculus/source/01-EQ/04.ptx new file mode 100644 index 00000000..57e81110 --- /dev/null +++ b/precalculus/source/01-EQ/04.ptx @@ -0,0 +1,384 @@ + + +
+ Absolute Value Equations and Inequalities (EQ4) + + + + + + Activities + + An absolute value, written \lvert x \rvert, is the non-negative value of x. If x is a positive number, then \lvert x \rvert=x. If x is a negative number, then \lvert x \rvert=-x. + + + +

Let's consider how to solve an equation when an absolute value is involved. +

+ + + +

Which values are solutions to the absolute value equation \lvert x \rvert = 2? +

    +

  1. x=2

    2. +

  2. x=0

    3. +

  3. x=-1

    4. +

  4. x=-2

    5. +

+ + + + +

Which values are solutions to the absolute value equation \lvert x-7 \rvert = 2? +

    +

  1. x=9

    2. +

  2. x=7

    3. +

  3. x=5

    4. +

  4. x=-9

    5. +

+ + + + +

Which values are solutions to the absolute value equation 3\lvert x-7 \rvert +5= 11? It may be helpful to rewrite the equation to isolate the absolute value. +

    +

  1. x=7

    2. +

  2. x=-9

    3. +

  3. x=5

    4. +

  4. x=9

    5. +

+ + + + + + +

Absolute value represents the distance a value is from 0 on the number line. So, \lvert x-7 \rvert = 2 means that the expression x-7 is 2 units away from 0. +

+ + + +

What values on the number line could x-7 equal? +

    +

  1. x=-7

    2. +

  2. x=-2

    3. +

  3. x=0

    4. +

  4. x=2

    5. +

  5. x=7

    6. +

+ + + + +

This gives us two separate equations to solve. What are those two equations? +

    +

  1. x-7=-7

    2. +

  2. x-7=-2

    3. +

  3. x-7=0

    4. +

  4. x-7=2

    5. +

  5. x-7=7

    6. +

+ + + + +

Solve each equation for x. +

+ + + + + + When solving an absolute value equation, begin by isolating the absolute value expression. Then rewrite the equation into two linear equations and solve. + + If c \gt 0, + + \lvert ax+b \rvert = c + + becomes the following two equations + ax+b =c \quad \text{and} \quad ax+b=-c + + + +

Solve the following absolute value equations. +

+ + + +

\lvert 3x+4 \rvert = 10 +

    +

  1. \{-2, 2\}

    2. +

  2. \left\{-\frac{14}{3}, 2\right\}

    3. +

  3. \{-10, 10\}

    4. +

  4. No solution

    5. +

+ + + + +

3\lvert x-7 \rvert+5 = 11 +

    +

  1. \{-2, 2\}

    2. +

  2. \{-9, 9\}

    3. +

  3. \{5, 9\}

    4. +

  4. No solution

    5. +

+ + + + +

2\lvert x+1 \rvert+8 = 4 +

    +

  1. \{-4, 4\}

    2. +

  2. \{-6, 6\}

    3. +

  3. \{5, 7\}

    4. +

  4. No solution

    5. +

+ + + + + Since the absolute value represents a distance, it is always a positive number. Whenever you encounter an isolated absolute value equation equal to a negative value, there will be no solution. + + + +

Just as with linear equations and inequalities, we can consider absolute value inequalities from equations. +

+ + + +

Which values are solutions to the absolute value inequality \lvert x-7 \rvert \le 2? +

    +

  1. x=9

    2. +

  2. x=7

    3. +

  3. x=5

    4. +

  4. x=-9

    5. +

+ + + + +

Rewrite the absolute value inequality \lvert x-7 \rvert \le 2 as a compound inequality. +

    +

  1. 0 \le x-7 \le 2

    2. +

  2. -2 \le x-7 \le 2

    3. +

  3. -2 \le x-7 \le 0

    4. +

  4. 2 \le x \le 7

    5. +

+ + + + +

Solve the compound inequality that is equivalent to \lvert x-7 \rvert \le 2 found in part (b). Write the solution in interval notation. +

    +

  1. [7,9]

    2. +

  2. [5,9]

    3. +

  3. [5,7]

    4. +

  4. [2,7]

    5. +

+ + + + + +

+ Draw the solution to \lvert x-7 \rvert \le 2 on the number line. +

    +

  1. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/1,0.2/3,0.4/5,0.6/7,0.8/9, 1/11} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{[-}, ultra thick, blue] (0.59,0) -- (0.6,0); +\draw[{-]}, ultra thick, blue] (0.79,0) -- (0.8,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0.6,-.16ex) -- (0.8, -.16ex) -- (0.8, .16ex) -- (0.6,.16ex) -- cycle; +\end{tikzpicture} + + +

    2. +

  2. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/1,0.2/3,0.4/5,0.6/7,0.8/9, 1/11} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{[-}, ultra thick, blue] (0.39,0) -- (0.4,0); +\draw[{-]}, ultra thick, blue] (0.79,0) -- (0.8,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0.4,-.16ex) -- (0.8, -.16ex) -- (0.8, .16ex) -- (0.4,.16ex) -- cycle; +\end{tikzpicture} + + +

    3. +

  3. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/1,0.2/3,0.4/5,0.6/7,0.8/9, 1/11} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{[-}, ultra thick, blue] (0.39,0) -- (0.4,0); +\draw[{-]}, ultra thick, blue] (0.59,0) -- (0.6,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0.4,-.16ex) -- (0.6, -.16ex) -- (0.6, .16ex) -- (0.4,.16ex) -- cycle; +\end{tikzpicture} + + +

    4. +

  4. + + + \begin{tikzpicture}[scale=7] +\draw[->, thick] (-0.1,0) -- (1.1,0); +\foreach \x/\xtext in {0/1,0.2/3,0.4/5,0.6/7,0.8/9, 1/11} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\draw[{[-}, ultra thick, blue] (0.09,0) -- (0.1,0); +\draw[{-]}, ultra thick, blue] (0.59,0) -- (0.6,0); +\fill[opacity = 0.2, blue,rounded corners=1ex] (0.1,-.16ex) -- (0.6, -.16ex) -- (0.6, .16ex) -- (0.1,.16ex) -- cycle; +\end{tikzpicture} + + +

    5. +

+ + + + + + + +

Now let's consider another type of absolute value inequality. +

+ + + +

Which values are solutions to the absolute value inequality \lvert x-7 \rvert \ge 2? +

    +

  1. x=9

    2. +

  2. x=7

    3. +

  3. x=5

    4. +

  4. x=-9

    5. +

+ + + + +

Which two of the following inequalities are equivalent to \lvert x-7 \rvert \ge 2. +

    +

  1. x-7 \le 2

    2. +

  2. x-7 \le -2

    3. +

  3. x-7 \ge 2

    4. +

  4. x-7 \ge -2

    5. +

+ + + + +

Solve the two inequalities found in part (b). Write the solution in interval notation and graph on the number line. +

    +

  1. (-\infty,7] \cup [9,\infty) + + + \begin{tikzpicture}[scale=7] + \draw[->, thick] (-0.1,0) -- (1.1,0); + \foreach \x/\xtext in {0/1,0.2/3,0.4/5,0.6/7,0.8/9, 1/11} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; + \draw[{-]}, ultra thick, blue] (0.59,0) -- (0.6,0); + \draw[{[-}, ultra thick, blue] (0.79,0) -- (0.8,0); + \fill[opacity = 0.2, blue,rounded corners=1ex] (-0.1,-.16ex) -- (0.6, -.16ex) -- (0.6, .16ex) -- (-0.1,.16ex) -- cycle; + \fill[opacity = 0.2, blue,rounded corners=1ex] (.8,-.16ex) -- (1.1, -.16ex) -- (1.1, .16ex) -- (0.8,.16ex) -- cycle; + \end{tikzpicture} + + +

    2. +

  2. (-\infty,5] \cup [9,\infty) + + + \begin{tikzpicture}[scale=7] + \draw[->, thick] (-0.1,0) -- (1.1,0); + \foreach \x/\xtext in {0/1,0.2/3,0.4/5,0.6/7,0.8/9, 1/11} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; + \draw[{-]}, ultra thick, blue] (0.39,0) -- (0.4,0); + \draw[{[-}, ultra thick, blue] (0.79,0) -- (0.8,0); + \fill[opacity = 0.2, blue,rounded corners=1ex] (-0.1,-.16ex) -- (0.4, -.16ex) -- (0.4, .16ex) -- (-0.1,.16ex) -- cycle; + \fill[opacity = 0.2, blue,rounded corners=1ex] (.8,-.16ex) -- (1.1, -.16ex) -- (1.1, .16ex) -- (0.8,.16ex) -- cycle; + \end{tikzpicture} + + +

    3. +

  3. (-\infty,5] \cup [7,\infty) + + + \begin{tikzpicture}[scale=7] + \draw[->, thick] (-0.1,0) -- (1.1,0); + \foreach \x/\xtext in {0/1,0.2/3,0.4/5,0.6/7,0.8/9, 1/11} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; + \draw[{-]}, ultra thick, blue] (0.39,0) -- (0.4,0); + \draw[{[-}, ultra thick, blue] (0.59,0) -- (0.6,0); + \fill[opacity = 0.2, blue,rounded corners=1ex] (-0.1,-.16ex) -- (0.4, -.16ex) -- (0.4, .16ex) -- (-0.1,.16ex) -- cycle; + \fill[opacity = 0.2, blue,rounded corners=1ex] (.6,-.16ex) -- (1.1, -.16ex) -- (1.1, .16ex) -- (0.6,.16ex) -- cycle; + \end{tikzpicture} + + +

    4. +

  4. (-\infty,2] \cup [7,\infty) + + + \begin{tikzpicture}[scale=7] + \draw[->, thick] (-0.1,0) -- (1.1,0); + \foreach \x/\xtext in {0/1,0.2/3,0.4/5,0.6/7,0.8/9, 1/11} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; + \draw[{-]}, ultra thick, blue] (0.09,0) -- (0.1,0); + \draw[{[-}, ultra thick, blue] (0.59,0) -- (0.6,0); + \fill[opacity = 0.2, blue,rounded corners=1ex] (-0.1,-.16ex) -- (0.1, -.16ex) -- (0.1, .16ex) -- (-0.1,.16ex) -- cycle; + \fill[opacity = 0.2, blue,rounded corners=1ex] (.6,-.16ex) -- (1.1, -.16ex) -- (1.1, .16ex) -- (0.6,.16ex) -- cycle; + \end{tikzpicture} + + +

    5. +

+ + + + +

+ When solving an absolute value inequality, rewrite it as compound inequalities. Assume k is positive. + + \lvert x \rvert \lt k \text{ becomes } -k \lt x \lt k. + + \lvert x \rvert \gt k \text{ becomes } x\gt k \text{ and } x\lt-k. +

+ + + +

Solve the following absolute value inequalities. Write your solution in interval notation and graph on a number line. +

+ + + +

\lvert 3x+4 \rvert \lt 10 +

+ + + + +

3\lvert x-7 \rvert+5 \gt 11 +

+ + + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
diff --git a/precalculus/source/01-EQ/05.ptx b/precalculus/source/01-EQ/05.ptx new file mode 100644 index 00000000..a5eacf8d --- /dev/null +++ b/precalculus/source/01-EQ/05.ptx @@ -0,0 +1,639 @@ + + +
+ Quadratic Equations (EQ5) + + + + + Activities + + + +

+ A quadratic equation is of the form: ax^2+bx+c=0 + where a and b are coefficients (and a \neq0), x is the variable, and c is the constant term. +

+ + + + + +

+ Before beginning to solve quadratic equations, we need to be able to identify all various forms of quadratics. +

+ +

Which of the following is a quadratic equation? +

    +
  1. 6-x^2=3x
    2. +
  2. (2x-1)(x+3)=0
    3. +
  3. 4(x-3)+7=0
    4. +
  4. (x-4)^2+1=0
    5. +
  5. 5x^2-3x=17-4x

+ + + + +

+ To solve a quadratic equation, we will need to apply the zero product property, which states that if a*b=0, then either a=0 or b=0. In other words, you can only have a product of 0 if one (or both!) of the factors is 0. +

+ + + + + +

+ In this activity, we will look at how to apply the zero product property when solving quadratic equations. +

+ + + +

Which of the following equations can you apply as your first step in solving? +

    +
  1. 2x^2-3x+1=0
    2. +
  2. (2x+1)(x+1)=0
    3. +
  3. 3x^2-4=6x
    4. +
  4. x(3x+5)=0

+ + + + +

Suppose you are given the quadratic equation, (2x-1)(x-1)=0. Applying , would give you: +

    +
  1. 2x^2-3x+1=0
    2. +
  2. (2x+1)=0 and (x+1)=0
    3. +
  3. (2x-1)=0 and (x-1)=0

+ + + + +

After applying the zero product property, what are the solutions to the quadratic equation (2x-1)(x-1)=0? +

    +
  1. x=-\frac{1}{2} and x=1
    2. +
  2. x=\frac{1}{2} and x=-1
    3. +
  3. x=-\frac{1}{2} and x=-1
    4. +
  4. x=\frac{1}{2} and x=1

+ + + + + +

+ Notice in and , that not all equations are set up "nicely." You will need to do some manipulation to get everything on one side (AND in factored form!) and 0 on the other *before* applying the zero product property. +

+ + + + +

+ Suppose you want to solve the equation 2x^2+5x-12=0, which is NOT in factored form. +

+ + + +

Which of the following is the correct factored form of 2x^2+5x-12=0? +

    +
  1. (2x-3)(x-4)=0
    2. +
  2. (2x+3)(x-4)=0
    3. +
  3. (2x+3)(x+4)=0
    4. +
  4. (2x-3)(x+4)=0

+ + + + +

After applying , which of the following will be a solution to 2x^2+5x-12=0? +

    +
  1. x=-\frac{3}{2} and x=-4
    2. +
  2. x=\frac{3}{2} and x=4
    3. +
  3. x=-\frac{3}{2} and x=4
    4. +
  4. x=\frac{3}{2} and x=-4

+ + + + + + +

+ Solve each of the following quadratic equations: +

+ + + + (2x-5)(x+7)=0 + + + + + 3x(4x-1)=0 + + + + + 3x^2-14x-5=0 + + + + + 6-x^2=3x + + + + + + +

+ Suppose you are given the equation, x^2=9: +

+ + + +

+ How many solutions does this equation have? +

    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 3
+

+ + + + +

+ What are the solutions to this equation? +

    +
  1. x=0
    2. +
  2. x=3
    3. +
  3. x=3,-3
    4. +
  4. x=9,-9
+

+ + + + +

+ How is this quadratic equation different than the equations we've solved thus far? +

+ + + + + + +

+ The square root property states that a quadratic equation of the form x^2=k^2 (where k is a nonzero number) will give solutions x=k and x=-k. +

+ In other words, if we have an equation with a perfect square on one side and a number on the other side, we can take the square root of both sides to solve the equation. +

+

+ + + + + +

+ Suppose you are given the equation, 3x^2-8=4: +

+ + + +

+ What would be the first step in solving 3x^2-8=4? +

    +
  1. Divide by 3 on both sides
    2. +
  2. Subtract 4 on both sides
    3. +
  3. Add 8 on both sides
    4. +
  4. Multiply by 3 on both sides
+

+ + + + +

+ Isolate the x^2 term and apply solve for x. +

+ + + + +

+ What are the solution(s) to 3x^2-8=4? +

    +
  1. x=6,-6
    2. +
  2. x=2,-2
    3. +
  3. x=0
    4. +
  4. x=2
+

+ + + + + + +

+ Solve the following quadratic equations by applying the square root property (). +

+ + + + 5x^2+7=47 + + + + + 2x^2=-144 + + + + + 3x^2+1=25 + + + + + (x+2)^2+3=19 + + + + + 3(x-4)^2=15 + + + + + + Not all quadratic equations can be factored or can be solved by using the square root property. In the next few activities, we will learn two additional methods in solving quadratics. + + + + +

+ Another method for solving a quadratic equation is known as completing the square. With this method, we add or subtract terms to both sides of an equation until we have a perfect square trinomial on one side of the equal sign and a constant on the other side. We then apply the square root property. + +

Note: A perfect square trinomial is a trinomial that can be factored into a binomial squared. For example, x^2+4x+4 is a perfect square trinomial because it can be factored into (x+2)(x+2) or (x+2)^2.

+

+ + + + +

+ Let's work through an example together to solve x^2+6x=4. (Notice that the methods of factoring and the square root property do not work with this equation.) +

+ + + +

+ In order to apply , we first need to have a perfect square trinomial on one side of the equals sign. Which of the following number(s) could we add to the left side of the equation to create a perfect square trinomial? +

    +
  1. 4
    2. +
  2. 9
    3. +
  3. -9
    4. +
  4. 2
+

+ + + + +

+ Add your answer from part a to the right side of the equation as well (i.e. whatever you do to one side of an equation you must do to the other side too!) and then factor the perfect square trinomial on the left side. Which equation best represents the equation now? +

    +
  1. (x+3)^2=-5
    2. +
  2. (x-3)^2=13
    3. +
  3. (x+3)^2=13
    4. +
  4. (x-3)^2=-5
+

+ + + + +

+ Apply the square root property () to both sides of the equation to determine the solution(s). Which of the following is the solution(s) of x^2+6x=4? +

    +
  1. 3+\sqrt{13} and 3-\sqrt{13}
    2. +
  2. -3+\sqrt{13} and -3-\sqrt{13}
    3. +
  3. 3+\sqrt{5} and 3-\sqrt{5}
    4. +
  4. -3+\sqrt{5} and -3-\sqrt{5}
+

+ + + + + +

To complete the square, the leading coefficient, a (i.e., the coefficient of the x^2 term), must equal 1. If it does not, then factor the entire equation by a and then follow similar steps as in .

+ + + + +

+ Let's solve the equation 2x^2+8x-5=0 by completing the square. +

+ + + +

+ Rewrite the equation so that all the terms with the variable x is on one side of the equation and a constant is on the other. +

+ + + + +

+ Notice that the coefficient of the x^2 term is not 1. What could we factor the left side of the equation by so that the coefficient of the x^2 is 1? +

+ + + + +

+ Once you factor the left side, what equation represents the equation you now have? +

    +
  1. 2(x^2-8x)=-5
    2. +
  2. 2(x^2-4x)=-5
    3. +
  3. 2(x^2+4x)=5
    4. +
  4. 2(x^2+8x)=5
+

+ + + + +

+ Just like in , let's now try and create the perfect square trinomial (inside the parentheses) on the left side of the equation. Which of the following number(s) could we add to the left side of the equation to create a perfect square trinomial? +

    +
  1. 4
    2. +
  2. 8
    3. +
  3. -8
    4. +
  4. 2
+

+ + + + +

+ What would we need to add to the right-hand side of the equation to keep the equation balanced? +

    +
  1. 4
    2. +
  2. 8
    3. +
  3. -8
    4. +
  4. 2
+

+ + + + +

+ Which of the following equation represents the quadratic equation you have now? +

    +
  1. 2(x+2)^2=9
    2. +
  2. 2(x-2)^2=9
    3. +
  3. 2(x+2)^2=13
    4. +
  4. 2(x-2)^2=13
+

+ + + + +

+ Apply the square root property and solve the quadratic equation. +

+ + + + + + +

+ Solve the following quadratic equations by completing the square. +

+ + + + x^2-12x=-11 + + + + + x^2+2x-33=0 + + + + + 4x^2+16x=65 + + + + + + The last method for solving quadratic equations is the quadratic formula - a formula that will solve all quadratic equations! +

+ A quadratic equation of the form ax^2+bx+c=0 can be solved by the quadratic formula: + x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} + where a, b, and c are real numbers and a\neq0. +

+ + + + +

+ Use the quadratic formula () to solve x^2+4x=-3. +

+ + + +

+ When written in standard form, what are the values of a, b, and c? +

+ + + + +

+ When applying the quadratic formula, what would you get when you substitute a, b, and c? +

    +
  1. x=\frac{-4\pm\sqrt{4^2-4(1)(3)}}{2(1)}
    2. +
  2. x=\frac{4\pm\sqrt{(-4)^2-4(1)(3)}}{2(1)}
    3. +
  3. x=\frac{-4\pm\sqrt{(-4)^2-4(1)(-3)}}{2(1)}
    4. +
  4. x=\frac{4\pm\sqrt{4^2-4(1)(-3)}}{2(1)}
+

+ + + + +

+ What is the solution(s) to x^2+4x=-3? +

    +
  1. x=-1,3
    2. +
  2. x=1,3
    3. +
  3. x=-1,-3
    4. +
  4. x=1,-3
+

+ + + + + + +

+ Use the quadratic formula () to solve 2x^2-13=7x. +

+ + + +

+ When written in standard form, what are the values of a, b, and c? +

+ + + + +

+ When applying the quadratic formula, what would you get when you substitute a, b, and c? +

    +
  1. x=\frac{-7\pm\sqrt{7^2-4(2)(13)}}{2(1)}
    2. +
  2. x=\frac{7\pm\sqrt{(-7)^2-4(2)(-13)}}{2(2)}
    3. +
  3. x=\frac{-7\pm\sqrt{(-7)^2-4(1)(-13)}}{2(1)}
    4. +
  4. x=\frac{7\pm\sqrt{7^2-4(2)(-13)}}{2(2)}
+

+ + + + +

+ What is the solution(s) to 2x^2-13=7x? +

    +
  1. x=\frac{7+\sqrt{73}}{4} and \frac{7-\sqrt{73}}{4}
    2. +
  2. x=\frac{7+\sqrt{101}}{4} and \frac{-7-\sqrt{101}}{4}
    3. +
  3. x=\frac{-7+\sqrt{55}}{4} and \frac{7-\sqrt{55}}{4}
    4. +
  4. x=\frac{-7+\sqrt{155}}{4} and \frac{-7-\sqrt{155}}{4}
+

+ + + + + + +

+ Use the quadratic formula () to solve x^2=6x-12. +

+ + + +

+ When written in standard form, what are the values of a, b, and c? +

+ + + + +

+ When applying the quadratic formula, what would you get when you substitute a, b, and c? +

    +
  1. x=\frac{-6\pm\sqrt{6^2-4(1)(12)}}{2(1)}
    2. +
  2. x=\frac{6\pm\sqrt{(-6)^2-4(1)(12)}}{2(1)}
    3. +
  3. x=\frac{-6\pm\sqrt{(-6)^2-4(1)(-12)}}{2(1)}
    4. +
  4. x=\frac{6\pm\sqrt{6^2-4(1)(-12)}}{2(1)}
+

+ + + + +

+ Notice that the number under the square root is a negative. Recall that when you have a negative number under a square root, that gives an imaginary number (\sqrt{-1}=i. What is the solution(s) to x^2=6x-12? +

    +
  1. x=3+i\sqrt{3} and 3-i\sqrt{3}
    2. +
  2. x=6+i\sqrt{12} and 6-i\sqrt{12}
    3. +
  3. x=-3+i\sqrt{3} and -3-i\sqrt{3}
    4. +
  4. x=-6+i\sqrt{12} and -6-i\sqrt{12}
+

+ + + + + + +

+ Solve the following quadratic equations by applying the quadratic formula (). +

+ + + + 2x^2-3x=5 + + + + + 4x^2-1=-8x + + + + + 2x^2-7x-13=-10 + + + + + x^2-6x+12=0 + + + + + + +

+ Now that you have seen all the different ways to solve a quadratic equation, you will need to know WHEN to use which method. Are some methods better than others? +

+ + + +

+ Which is the best method to use to solve 5x^2=80? +

    +
  1. Factoring and Zero Product Property
    2. +
  2. Square Root Property
    3. +
  3. Completing the Square
    4. +
  4. Quadratic Formula
+

+ + + + +

+ Which is the best method to use to solve 5x^2+9x=-4? +

    +
  1. Factoring and Zero Product Property
    2. +
  2. Square Root Property
    3. +
  3. Completing the Square
    4. +
  4. Quadratic Formula
+

+ + + + +

+ Which is the best method to use to solve 3x^2+9x=0? +

    +
  1. Factoring and Zero Product Property
    2. +
  2. Square Root Property
    3. +
  3. Completing the Square
    4. +
  4. Quadratic Formula
+

+ + + + +

+ Go back to parts a, b, and c and solve each of the quadratic equations. Would you still use the same method? +

+ + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/01-EQ/06.ptx b/precalculus/source/01-EQ/06.ptx new file mode 100644 index 00000000..2274c88c --- /dev/null +++ b/precalculus/source/01-EQ/06.ptx @@ -0,0 +1,440 @@ + + +
+ Rational Equations (EQ6) + + + + + Activities + + + +

An algebraic expression is called a rational expressionrational expression if it can be written as the ratio of two polynomials, p and q.

+

+ An equation is called a rational equationrational equation if it consists of only rational expressions and constants. +

+ + + + + +

+ Technically, linear and quadratic equations are also rational equations. They are a special case where the denominator of the rational expressions is 1. We will focus in this section on cases where the denominator is not a constant; that is, rational equations where there are variables in the denominator. +

+

+ With variables in the denominator, there will often be values that cause the denominator to be zero. This is a problem because division by zero is undefined. Thus, we need to be sure to exclude any values that would make those denominators equal to zero. +

+ + + + + +

Which value(s) should be excluded as possible solutions to the following rational equations? Select all that apply. +

+ + + + + +

+ \frac{2}{x+5}=\frac{x-3}{x-8}-7 +

    +
  1. -7
    2. +
  2. -5
    3. +
  3. 2
    4. +
  4. 3
    5. +
  5. 8
    6. + +
+

+ + + + + + +

+ \frac{x^2-6x+8}{x^2-4x+3}=0 +

    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 3
    5. +
  5. 4
    6. + +
+

+ + + + + + + +

Consider the rational equation + 5 = -\frac{6}{x-2} + +

+ + + + + +

What value should be excluded as a possible solution? +

    +
  1. 5
    2. +
  2. 6
    3. +
  3. -6
    4. +
  4. 2
    5. +
  5. -2
    6. +
+

+ + + + + +

To solve, we begin by clearing out the fraction involved. What can we multiply each term by that will clear the fraction? +

    +
  1. x-5
    2. +
  2. x-6
    3. +
  3. x+6
    4. +
  4. x-2
    5. +
  5. x+2
    6. +
+

+ + + + + +

Multiply each term by the expression you chose and simplify. Which of the following linear equations does the rational equation simplify to? +

    +
  1. 5(x-5) = -6
    2. +
  2. 5(x-6) = -6
    3. +
  3. 5(x+6) = -6
    4. +
  4. 5(x-2) = -6
    5. +
  5. 5(x+2) = -6
    6. +
+

+ + + + + +

Solve the linear equation. Check your answer using the original rational equation.

+ + + + + + + +

Consider the rational equation + \frac{4}{x+1} = -\frac{2}{x+6} + +

+ + + + + +

What values should be excluded as possible solutions? +

    +
  1. 2 and 4
    2. +
  2. 1 and 6
    3. +
  3. -1 and -6
    4. +
  4. 1 and 4
    5. +
  5. 2 and 6
    6. +
+

+ + + + + +

To solve, we'll once again begin by clearing out the fraction involved. Which of the following should we multiply each term by to clear out all of the fractions? +

    +
  1. x-2 and x-4
    2. +
  2. x-1 and x-6
    3. +
  3. x+1 and x+6
    4. +
  4. x-1 and x-4
    5. +
  5. x-2 and x-6
    6. +
+

+ + + + + +

Multiply each term by the expressions you chose and simplify. Which of the following linear equations does the rational equation simplify to? +

    +
  1. 4(x+1) = -2(x+6)
    2. +
  2. 4(x+6) = -2(x+1)
    3. +
  3. 4(x+1)(x+6) = -2(x+1)(x+6)
    4. +
  4. 4(x+1) = -2(-x-6)
    5. +
  5. 4(x+6) = -2(-x-1)
    6. +
+

+ + + + + +

Solve the linear equation. Check your answer using the original rational equation.

+ + + + + + + +

+ In , you may have noticed that the resulting linear equation looked like the result of cross-multiplying. This is no coincidence! Cross-multiplying is a method of clearing out fractions that works specifically when the equation is in proportional form: \frac{a}{b}=\frac{c}{d}. +

+ + + + + +

Consider the rational equation + \frac{x}{x+2} = -\frac{2}{x+2}-\frac{2}{5} + +

+ + + + + +

What value(s) should be excluded as possible solutions?

+ + + + + +

To solve, we'll once again begin by clearing out the fraction involved. Which of the following should we multiply each term by to clear out all of the fractions? +

    +
  1. x+2, x+2, and 5
    2. +
  2. x+2 and 5
    3. +
  3. x+2
    4. +
  4. 5
    5. +
+

+ + + + + +

Multiply each term by the expressions you chose and simplify. You should end up with a linear equation. +

+ + + + + +

Solve the linear equation. Check your answer using the original rational equation.

+ + + + + + + +

+ demonstrates why it is so important to determine excluded values and check our answers when solving rational equations. Just because a number is a solution to the linear equation we found, it doesn't mean it is automatically a solution to the rational equation we started with. +

+ + + + + +

Consider the rational equation + \frac{2x}{x-1}- \frac{3}{x-3} = \frac{x^2-11x+18}{x^2-4x+3} + +

+ + + + + +

What values should be excluded as possible solutions? Select all that apply. +

    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 3
    5. +
  5. 9
    6. + + +
+

+ + + + + +

To solve, we'll begin by clearing out any fractions involved. Which of the following should we multiply each term by to clear out all of the fractions? +

    +
  1. x-1
    2. +
  2. x-1 and x-3
    3. +
  3. x-1, x-3, and x^2-4x+3
    4. +
  4. x-1 and x^2-4x+3
    5. +
  5. x-3 and x^2-4x+3
    6. +
+

+ + + + + +

Multiply each term by the expressions you chose and simplify. Notice that the result is a quadratic equation. Which of the following quadratic equations does the rational equation simplify to? +

    +
  1. x^2+2x-15=0
    2. +
  2. x^2-11x+18=0
    3. +
  3. x^2-9x-9=0
    4. +
  4. x^2-13x+21=0
    5. + +
+

+ + + + + +

Solve the quadratic equation. Check your answer using the original rational equation. What are the solutions to the rational equation? +

    +
  1. x=3 and x=-5
    2. +
  2. x=-3 and x=5
    3. +
  3. x=3
    4. +
  4. x=-5
    5. +
  5. x=-3
    6. +
  6. x=5
    7. +
+

+ + + + + + + +

Consider the rational equation + \frac{2x}{x-2}- \frac{x^2+21x-15}{x^2+3x-10} = \frac{-6}{x+5} + +

+ + + + + +

What values should be excluded as possible solutions? + + +

+ + + + + +

What expression(s) should we multiply by to clear out all of the fractions? + +

+ + + + + +

Multiply each term by the expressions you chose and simplify. Your result should be a quadratic equation. + +

+ + + + + +

Solve the quadratic equation. Check your answer using the original rational equation. What are the solutions to the rational equation? + +

+ + + + + + + +

Solve the following rational equations. +

+ + + + +

\displaystyle \frac{4}{x}+9=16

+ + + + + +

\displaystyle -5=\frac{2}{x-4}

+ + + + + +

\displaystyle \frac{-3}{x-10}=\frac{x}{x-6}

+ + + + + +

\displaystyle \frac{x+2}{x-3}+\frac{x}{2x-1}=6

+ + + + + + + + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/01-EQ/07.ptx b/precalculus/source/01-EQ/07.ptx new file mode 100644 index 00000000..e334f4df --- /dev/null +++ b/precalculus/source/01-EQ/07.ptx @@ -0,0 +1,261 @@ + + +
+ Quadratic and Rational Inequalities (EQ7) + + + + + Activities + + +

In and we learned how to solve quadratic and rational equations. In this section, we use these skills + to solve quadratic and rational inequalities. +

+ + + + + +

Consider the quadratic inequality x^2-4x-32\gt0.

+ + + +

Use a graphing utility to graph the function f(x)=x^2-4x-32. Which part of the graph represents where x^2-4x-32\gt0?

+ + + + + f(x)=x^2-4*x-32 + p=plot(f,xmin=-10,xmax=10,ymin=-40,ymax=40,color="blue",thickness=3,gridlines=[[-10..10],[-40,-35..40]]) + p+=polygon([[8,0],[10,0],[10,45],[8,45]],color='blue',alpha=0.2,aspect_ratio="automatic") + p+=polygon([[-4,0],[-10,0],[-10,45],[-4,45]],color='blue',alpha=0.2,aspect_ratio="automatic") + p + + + + + + +

Which pieces of information about f(x)=x^2-4x-32 were needed to answer part (a)?

+
    +
  1. The y-intercept
    2. +
  2. The x-intercepts
    3. +
  3. The minimum value
    4. +
+ +

B.

 + + + +

Use algebra to find the x-intercepts of f(x)=x^2-4x-32 and mark them on a number line. + Then, shade the part of the number line where x^2-4x-32>0. +

+ + + \begin{tikzpicture}[scale=5] +\draw[<->, thick] (-1.1,0) -- (1.1,0); +\foreach \x/\xtext in {-1/-10,-0.8/-8,-0.6/-6,-0.4/-4,-0.2/-2,0/0,0.2/2,0.4/4,0.6/6,0.8/8, 1/10} +\draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\end{tikzpicture} + + + + +

+ + + \begin{tikzpicture}[scale=5] +\draw[<->, thick] (-1.1,0) -- (1.1,0); +\foreach \x/\xtext in {-1/-10,-0.8/-8,-0.6/-6,-0.4/-4,-0.2/-2,0/0,0.2/2,0.4/4,0.6/6,0.8/8, 1/10} +\draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; +\fill[opacity = 0.2, blue,rounded corners=1ex] (-1.1,-.16ex) -- (-0.4, -.16ex) -- (-0.4, .16ex) -- (-1.1,.16ex) -- cycle; +\fill[opacity = 0.2, blue,rounded corners=1ex] (1.1,-.16ex) -- (0.8, -.16ex) -- (0.8, .16ex) -- (1.1,.16ex) -- cycle; +\end{tikzpicture} + + +

+ + + + +

Now use interval notation to express where x^2-4x-32>0. +

    +

  1. (-4,8)

    2. +

  2. [ -4,8]

    3. +

  3. (-\infty,-4)\cup(8,\infty)

    4. +

  4. (-\infty,-4]\cup[8,\infty)

    5. +

+ +

C.

 + + + + +

A sign chart is a number line representing the x-axis that shows where a + function is positive or negative. Instead of shading, which can be ambiguous, it is often decorated with + a '+' or a '-' to indicate which regions are positive or negative. For example, a sign chart for + f(x)=x^2-4x-32 is below. +

+
+ + + \begin{tikzpicture}[scale=5] + \draw[<->, thick] (-1.1,0) -- (1.1,0); + \foreach \x/\xtext in {-1/-10,-0.8/-8,-0.6/-6,-0.4/-4,-0.2/-2,0/0,0.2/2,0.4/4,0.6/6,0.8/8, 1/10} + \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; + \draw[thick,red,dashed] (-0.4,-0.16) -- (-0.4,0.16); + \draw[thick,red,dashed] (0.8,-0.16) -- (0.8,0.16); + \node[above] at (0.2,0.02) {\Large \bf $-$}; + \node[above] at (1,0.02) {\Large \bf $+$}; + \node[above] at (-0.7,0.02) {\Large \bf $+$}; + \end{tikzpicture} + + + A sign chart for the function f(x)=x^2-4x-32. +
+ + + + +

Solve the quadratic inequality algebraically2x^2-28 \lt 10x. + Write your solution using interval notation.

+
    +

  1. \left(-\infty,-2\right)\cup\left(7,\infty\right)

    2. +

  2. \left(-\infty,-7\right)\cup\left(2,\infty\right)

    3. +

  3. \left(-7,2\right)

    4. +

  4. \left(-2,7\right)

    5. +
+ +

D.

 + + + + +

Solve the inequality -2x^2-10x-10 \ge 6x+20. + Write your solution using interval notation.

+
    +

  1. [-5,-3]

    2. +

  2. (-\infty,-5]\cup[-3,\infty)

    3. +

  3. [3,5]

    4. +

  4. (-\infty,3]\cup[5,\infty)

    5. +
+ +

A.

 + + + + +

Consider the rational inequality \frac{4x+3}{x+2} \gt x. +

+ + + +

For which of the following functions will a graph help us solve the rational inequality above?

+
    +
  1. f(x)=\frac{4x+3}{x+2}
    2. +
  2. g(x)=\frac{4x+3}{x+2}-x
    3. +
  3. h(x)=x-\frac{4x+3}{x+2}
    4. +
+ +

B or C is most helpful.

 + + + +

Use a graphing utility to graph the function g(x)=\frac{4x+3}{x+2}-x. + Which part of the graph represents where \frac{4x+3}{x+2}-x\gt0?

+ + + + +

Simplify \frac{4x+3}{x+2} - x into a single rational expression. +

    +

  1. \frac{4x+3}{x+2}

    2. +

  2. \frac{3x+3}{x+2}

    3. +

  3. \frac{x^2+6x+3}{x+2}

    4. +

  4. \frac{-x^2+2x+3}{x+2}

    5. +

+ +

D.

 + + + +

How do these values you found in part (b) relate to the numerator and the denominator + of the combined rational function in part (c)? +

+ +

Factor the numerator.

 +

The places where the numerator or denominator vanishes are the potential places where the sign can change.

 + + + +

For what values is the original inequality a true statement?

+
    +
  1. x \lt -2 and -1 \lt x \lt 3
    2. +
  2. -2 \lt x \lt -1 and x \gt 3
    3. +
  3. -2\lt x \lt -1
    4. +
  4. 1 \lt x \lt 3
    5. +
+ +

A.

 + + + +

How can we express the answers to part (e) for the rational inequality using interval notation?

+
    +
  1. \left(-\infty,-2\right) \cup (-1,3)
    2. +
  2. \left(-2,-1\right) \cup (3,\infty)
    3. +
  3. \left(-2,-1\right)
    4. +
  4. \left(1,3\right)
    5. +
+ +

A.

 + + + + +

+ The values on the x-axis where a function is equal to zero or undefined are called + partition values. +

+ + + + +

Solve the rational inequality \frac{x+8}{x-2} \le \frac{x+10}{x+5}. +

+ + + +

Write the solution using interval notation.

+
    +
  1. (-\infty, -12)\cup[-5,2]
    2. +
  2. (-\infty, -12]\cup(-5,2)
    3. +
  3. (-12,-5]\cup[2,\infty)
    4. +
  4. [-12,-5)\cup(2,\infty)
    5. +
+ +

B.

 + + + +

+ Compare the interval notation from to the interval + notation for this activity. When do we include the partition values in the answer with a bracket? +

+ +

Values that make the function undefined cannot be solutions, while values that make + the simplified rational function zero are solutions.

 + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/01-EQ/main.ptx b/precalculus/source/01-EQ/main.ptx new file mode 100644 index 00000000..dfb512cd --- /dev/null +++ b/precalculus/source/01-EQ/main.ptx @@ -0,0 +1,19 @@ + + + + Equations, Inequalities, and Applications (EQ) + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/precalculus/source/01-EQ/outcomes/01.ptx b/precalculus/source/01-EQ/outcomes/01.ptx new file mode 100644 index 00000000..c7b1a52c --- /dev/null +++ b/precalculus/source/01-EQ/outcomes/01.ptx @@ -0,0 +1,5 @@ + +

+ Solve a linear equation in one variable. + Solve a linear inequality in one variable and express the solution graphically and using interval notation. +

\ No newline at end of file diff --git a/precalculus/source/01-EQ/outcomes/02.ptx b/precalculus/source/01-EQ/outcomes/02.ptx new file mode 100644 index 00000000..a55d47f8 --- /dev/null +++ b/precalculus/source/01-EQ/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Solve application problems involving linear equations. +

\ No newline at end of file diff --git a/precalculus/source/01-EQ/outcomes/03.ptx b/precalculus/source/01-EQ/outcomes/03.ptx new file mode 100644 index 00000000..75b7c463 --- /dev/null +++ b/precalculus/source/01-EQ/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Given two points, determine the distance between them and the midpoint of the line segment connecting them. +

diff --git a/precalculus/source/01-EQ/outcomes/04.ptx b/precalculus/source/01-EQ/outcomes/04.ptx new file mode 100644 index 00000000..c6b238dd --- /dev/null +++ b/precalculus/source/01-EQ/outcomes/04.ptx @@ -0,0 +1,5 @@ + +

+Solve a linear equation involving an absolute value. +Solve a linear inequality involving absolute values and express the answers graphically and using interval notation. +

\ No newline at end of file diff --git a/precalculus/source/01-EQ/outcomes/05.ptx b/precalculus/source/01-EQ/outcomes/05.ptx new file mode 100644 index 00000000..02e4ab0f --- /dev/null +++ b/precalculus/source/01-EQ/outcomes/05.ptx @@ -0,0 +1,5 @@ + +

+Solve quadratic equations using factoring, the square root property, completing the square, +and the quadratic formula and express these answers in exact form. +

\ No newline at end of file diff --git a/precalculus/source/01-EQ/outcomes/06.ptx b/precalculus/source/01-EQ/outcomes/06.ptx new file mode 100644 index 00000000..ebc86ae4 --- /dev/null +++ b/precalculus/source/01-EQ/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+Solve a rational equation. +

\ No newline at end of file diff --git a/precalculus/source/01-EQ/outcomes/07.ptx b/precalculus/source/01-EQ/outcomes/07.ptx new file mode 100644 index 00000000..33017941 --- /dev/null +++ b/precalculus/source/01-EQ/outcomes/07.ptx @@ -0,0 +1,5 @@ + +

+Solve quadratic inequalities and express the solution graphically and with interval notation. +Solve rational inequalities and express the solution graphically and using interval notation. +

\ No newline at end of file diff --git a/precalculus/source/01-EQ/outcomes/main.ptx b/precalculus/source/01-EQ/outcomes/main.ptx new file mode 100644 index 00000000..32316009 --- /dev/null +++ b/precalculus/source/01-EQ/outcomes/main.ptx @@ -0,0 +1,34 @@ + +> + +

+ BIG IDEA for the chapter goes here, in outcomes/main.ptx +

+

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
  7. + +
    8. +
+ \ No newline at end of file diff --git a/precalculus/source/01-EQ/readiness.ptx b/precalculus/source/01-EQ/readiness.ptx new file mode 100644 index 00000000..b34d873f --- /dev/null +++ b/precalculus/source/01-EQ/readiness.ptx @@ -0,0 +1,98 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Plot points on the coordinate plane.

    +
      +

    • Review and Practice: Khan Academy

      +
      • +
    +
    2. +
  2. +

    Determine whether a given value is a solution to an equation.

    +
      +

    • Review: YouTube (Megan MathTeacher Snow)

      +
      • +

    • Review and Practice: Khan Academy

      +
      • +
    +
    3. + +
  3. +

    Solve two step linear equations.

    +
      +

    • Review: YouTube (Mathispower4u)

      • +

    • Review and Practice: Khan Academy

      • +
    +
    4. + +
  4. +

    Use interval notation.

    +
      +

    • Review: YouTube (The Organic Chemistry Tutor)

      • +

    • Practice: IXL

      • +
    +
    5. + +
  5. +

    Find the absolute value of real numbers.

    +
      +

    • Review and Practice: Khan Academy

      • +
    +
    6. + +
  6. +

    Use the Pythagorean Theorem to find side lengths of right triangles.

    +
      +

    • Review and Practice: Khan Academy

      • +
    +
    7. + +
  7. +

    Factor quadratic equations.

    +
      +

    • Review and Practice: Khan Academy

      • +
    +
    8. + +
  8. +

    Solve one-step problems involving distance, rate, and time.

    +
      +

    • Review: YouTube (Krista King Math)

      • + +

    • Review: YouTube (Mathispower4u)

      • + +
    +
    9. + +
  9. +

    Use the percent equation.

    +
      +

    • Review and Practice: Khan Academy

      • +
    +
    10. +
  10. +

    Simplify square roots of positive numbers.

    +
      +

    • Review and Practice: Khan Academy

      • +
    +
    11. +
  11. +

    Simplify square roots of negative numbers using i.

    +
      +

    • Review and Practice: Khan Academy

      • +
    +
    12. + + + + +
+ + \ No newline at end of file diff --git a/precalculus/source/02-FN/01.ptx b/precalculus/source/02-FN/01.ptx new file mode 100644 index 00000000..9ad39fd6 --- /dev/null +++ b/precalculus/source/02-FN/01.ptx @@ -0,0 +1,766 @@ + + +
+ Introduction to Functions (FN1) + + + + + Activities + + + +

A relation functionrelation is a relationship between sets of values. Relations in mathematics are usually represented as ordered pairs: (input, output) or (x, y). When observing relations, we often refer to the x-values as the domain and the y-values as the range.

+ + + + + +

+ Mapping Notation functionmapping notation (also known as an arrow diagram) is a way to show relationships visually between sets. For example, suppose you are given the following ordered pairs: (3, -8), (4,6), and (2,-1). Each of the x-values "map onto" a y-value and can be visualized in the following way:

+
+ + Every x-value from the ordered pair list is listed in the input set and every y-value is listed in the output set. An arrow is drawn from every x-value to its corresponding y-value. +
+

Notice that an arrow is used to indicate which x-value is mapped onto its corresponding y-value. +

+ + + + + +

+ Use mapping notation to create a visual representation of the following relation. + (-1,5), (2,6), (4,-2) +

+ + + +

+ What is the domain? +

    +
  1. {5, 6, -2}
    2. +
  2. {-1, 2, 4}
    3. +
  3. {-2, -1, 2, 4, 5, 6}

+ + +

+ B: {-1, 2, 4} +

+ + + + +

+ What is the range? +

    +
  1. {5, 6, -2}
    2. +
  2. {-1, 2, 4}
    3. +
  3. {-2, -1, 2, 4, 5, 6}

+ + +

+ A: {5, 6, -2} +

+ + + + + + +

+ Use mapping notation to create a visual representation of the following relation. + (6,4), (3,4), (6,5) +

+ + + +

+ What is the domain? +

    +
  1. {3, 6}
    2. +
  2. {6, 3, 6}
    3. +
  3. {3, 4, 5, 6}
    4. +
  4. {4, 5}

+ + +

+ A: {3, 6} +

+ + + + +

+ What is the range? +

    +
  1. {3, 6}
    2. +
  2. {6, 3, 6}
    3. +
  3. {3, 4, 5, 6}
    4. +
  4. {4, 5}

+ + +

+ D: {4, 5} +

+ + + + + + +

+ Use mapping notation to create a visual representation of the following relation. + (1,2), (-5,2), (-7,2) +

+ + + +

+ What is the domain? +

    +
  1. {2, 2, 2}
    2. +
  2. {-7, -5, 1, 2}
    3. +
  3. {-7, -5, 1}
    4. +
  4. {2}

+ + +

+ C: {-7, -5, 1} +

+ + + + +

+ What is the range? +

    +
  1. {2, 2, 2}
    2. +
  2. {-7, -5, 1, 2}
    3. +
  3. {-7, -5, 1}
    4. +
  4. {2}

+ + +

+ D: {2} +

+ + + + + + Notice that in , , and , each set represents a very different relationship. Many concepts in mathematics will depend on particular relationships, so it is important to be able to visualize relationships and compare them. + + + + +

+ A function is a relation where every input (or x-value) is mapped onto exactly one output (or y-value).

+

+ Note that all functions are relations but not all relations are functions! +

+ + + + + +

+ Relations can be expressed in multiple ways (ordered pairs, tables, and verbal descriptions). +

+ + + +

+ Let's revisit some of the sets of ordered pairs we've previously explored in , , and . Which of the following sets of ordered pairs represent a function? +

    +
  1. (-1,5), (2,6), (4,-2)
    2. +
  2. (6,4), (3,4), (6,5)
    3. +
  3. (1,2), (-5,2), (-7,2)
    4. +
  4. (-1,2), (-1,9), (1,9)

+ + +

+ A and C +

+ + + + + +

Note that relations can be expressed in a table. A table of values is shown below. Is this an example of a function? Why or why not? +

+ + + x + y + + + -5 + -2 + + + -4 + -5 + + + -2 + 8 + + + 8 + -4 + + + 8 + 1 + + + + +

+ This relation is not a function. The x-value, 8, has two outputs. +

+ + + + +

+ Relations can also be expressed in words. Suppose you are looking at the amount of time you spend studying versus the grade you earn in your Algebra class. Is this an example of a function? Why or why not? +

+ + +

+ Yes, this is an example of a function. +

+ + + + + + Notice that when trying to determine if a relation is a function, we often have to rely on looking at the domain and range values. Thus, it is important to be able to idenfity the domain and range of any relation! + + + + +

+ For each of the given functions, determine the domain and range. +

+ + + +

+ (-4,3), (-1,8), (7,4), (1,9) +

+ + +

+ Domain: {-4, -1, 1, 7}; + Range: {3, 4, 8, 9} +

+ + + + +

+ +

+ + +

+ Domain: {2, 3, 4}; + Range: {-8, -1, 6} +

+ + + + +

+ + + x + y + + + -2 + 5 + + + 0 + 4 + + + 3 + 6 + + + 8 + 1 + + +

+ + +

+ Domain: {-2, 0, 3, 8}; + Range: {1, 4, 5, 6} +

+ + + + +

+ The amount of time you spend studying versus the grade you earn in your Algebra class. +

+ + +

+ Domain: {time studying}; + Range: {grade in Algebra class} +

+ + + + + + + Determine whether each of the following relations is a function. + + + +

+ + + A = point((-1,5),size=40,xmin=-10,xmax=10,ymin=-10,ymax=10) + B = point((2,6),size=40) + C = point((4,-2),size=40) + A+B+C + + +

+ + +

+ This relation is a function. +

+ + + + +

+ + + A = point((6,4),size=40,xmin=-10,xmax=10,ymin=-10,ymax=10) + B = point((3,4),size=40) + C = point((6,5),size=40) + A+B+C + + +

+ + +

+ This relation IS NOT a function. +

+ + + + +

+ + + A = point((1,2),size=40,xmin=-10,xmax=10,ymin=-10,ymax=10) + B = point((-5,2),size=40) + C = point((-7,2),size=40) + A+B+C + + +

+ + +

+ This relation is a function. +

+ + + + + + You probably noticed (in ) that when the graph has points that "line up" or are on top of each other, they have the same x-values. When this occurs, this shows that the same x-value has two different outputs (y-values) and that the relation is not a function. + + + + +

+ The vertical line test is a method used to determine whether a relation on a graph is a function.

+

Start by drawing a vertical line anywhere on the graph and observe the number of times the relation on the graph intersects with the vertical line. If every possible vertical line intersects the graph at only one point, then the relation is a function. If, however, the graph of the relation intersects a vertical line more than once (anywhere on the graph), then the relation is not a function. +

+ + + + + +

+ Use the vertical line test () to determine whether each graph of a relation represents a function. +

+ + + + + + f(x)=sqrt(9-x^2) + g(x)=-(sqrt(9-x^2)) + p=plot(f, (x, -3, 3), ymin=-4, ymax=4, color='blue', thickness=3)+ plot(g, (x, -3, 3), ymin=-4, ymax=4, color='blue', thickness=3) + p + + + + +

+ This graph does not pass the vertical line test and is therefore NOT a function. +

+ + + + + + + f(x) = -4*x^3+2*x^2+3 + p=plot(f, (x, -5, 5), ymin=-15, ymax=15, color='blue', thickness=3) + p + + + + +

+ This graph passes the vertical line test and is therefore a function. +

+ + + + + + + f(x)=5*x-1 + p=plot(f, (x, -5, 5), ymin=-10, ymax=10, color='blue', thickness=3) + p + + + + +

+ This graph passes the vertical line test and is therefore a function. +

+ + + + + + + f(x)=sqrt(x) + g(x)=-(sqrt(x)) + p=plot(f, (x, 0, 5), ymin=-10, ymax=10, color='blue', thickness=3)+ plot(g, (x, 0, 5), ymin=-10, ymax=10, color='blue', thickness=3) + p + + + + +

+ This graph does not pass the vertical line test and is therefore NOT a function. +

+ + + + + + +

+ Let's explore how to determine whether an equation represents a function. +

+ + + +

+ Suppose you are given the equation x=y^2. +

    +
  • + If x=4, what kind of y-values would you get for x=y^2? +
    • +
  • +

    + Based on this information, do you think x=y^2 is a function? +

    +
    • +
+

+ + + If x=4, y could be -2 or 2. Given that there are two outputs for a given x, then x=y^2 is not a function. + + + + + +

+ Suppose you are given the equation y=3x^2+2. +

    +
  • + If x=4, what kind of y-values would you get for y=3x^2+2? +
    • +
  • +

    + Based on this information, do you think y=3x^2+2 is a function? +

    +
    • +
+

+ + + If x=4, y=50. For every input (x-value), there is exactly one output (y-value) so y=3x^2+2 is a function. + + + + + +

+ Suppose you are given the equation x^2+y^2=25. +

    +
  • + If x=4, what kind of y-values would you get for x^2+y^2=25? +
    • +
  • +

    + Based on this information, do you think x^2+y^2=25 is a function? +

    +
    • +
+

+ + + If x=4, y could be -3 or 3. Given that there are two outputs for a given x, then x^2+y^2=25 is not a function. + + + + + +

+ Suppose you are given the equation y=-4x-3. +

    +
  • + If x=4, what kind of y-values would you get for y=-4x-3? +
    • +
  • +

    + Based on this information, do you think y=-4x-3 is a function? +

    +
    • +
+

+ + + If x=4, y=-19. For every input (x-value), there is exactly one output (y-value) so y=-4x-3 is a function. + + + + +

+ How can you look at an equation to determine whether or not it is a function? +

+ + + Students may have different answers. Ideally, you want students to see that they can check whether a given relation is a function in more than one way. They can test values (as this activity did by leading them to plug in values) or by using the vertical line test on a graph. + + + + + + Notice that shows that equations with a y^2 term generally do not define functions. This is because to solve for a squared variable, you must consider both positive and negative inputs. For example, both 2^2=4 and (-2)^2=4. + + + + +

+ It's important to be able to determine the domain of any equation, especially when thinking about functions. Answer the following questions given the equation y=\sqrt{x}. +

+ + + +

+ What values of x would give an error (if any)? +

    +
  1. -2
    2. +
  2. 0
    3. +
  3. 4
    4. +
  4. -5

+ + +

+ A and D +

+ + + + +

+ Based on this information, for what values of x would the equation exist? +

    +
  1. -2
    2. +
  2. 0
    3. +
  3. 4
    4. +
  4. -5
+

+ + +

+ B and C +

+ + + + +

+ How can we represent the domain of this equation in interval notation? +

    +
  1. (-\infty,0)
    2. +
  2. [0,\infty)
    3. +
  3. (0, 0)
    4. +
  4. (-\infty,\infty)
+

+ + +

+ B: [0,\infty) +

+ + + + + + +

+ Answer the following questions given the equation y=-5x+1. +

+ + + +

+ What values of x would give an error (if any)? +

    +
  1. -2
    2. +
  2. 0
    3. +
  3. 4
    4. +
  4. -5

+ + +

+ None of the values will give an error. +

+ + + + +

+ Based on this information, for what values of x would the equation exist? +

    +
  1. -2
    2. +
  2. 0
    3. +
  3. 4
    4. +
  4. -5
+

+ + +

+ A, B, C, and D +

+ + + + +

+ How can we represent the domain of this equation in interval notation? +

    +
  1. (-\infty,0)
    2. +
  2. (0,\infty)
    3. +
  3. (-5, 1)
    4. +
  4. (-\infty,\infty)
+

+ + +

+ D: (-\infty,\infty) +

+ + + + + + +

+ Answer the following questions given the equation y=\frac{3}{x-5}. +

+ + + +

+ What values of x would give an error (if any)? +

    +
  1. -3
    2. +
  2. 0
    3. +
  3. -4
    4. +
  4. 5

+ + +

+ D: 5 +

+ + + + +

+ Based on this information, for what values of x would the equation exist? +

    +
  1. -3
    2. +
  2. 0
    3. +
  3. -4
    4. +
  4. 5
+

+ + +

+ A, B, and C +

+ + + + +

+ How can we represent the domain of this equation in interval notation? +

    +
  1. (-\infty,5)
    2. +
  2. (5,\infty)
    3. +
  3. (-5, 5)
    4. +
  4. (-\infty,5)U(5,\infty)
+

+ + +

+ D: (-\infty,5)U(5,\infty) +

+ + + + + + When determining the domain of an equation, it is often easier to first find values of x that make the function undefined. Once you have those values, then you know that x can be any value but those. + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/02-FN/02.ptx b/precalculus/source/02-FN/02.ptx new file mode 100644 index 00000000..57f8fb5c --- /dev/null +++ b/precalculus/source/02-FN/02.ptx @@ -0,0 +1,524 @@ + + +
+ Function Notation (FN2) + + + + + Activities + + +

+ As we saw in the last section, we can represent functions in many ways, like using a set of ordered pairs, a graph, a description, or an equation. When describing a function with an equation, we will often use function notation. +

+

+ If y is written as a function of x, like in the equation y=x+5, we can replace the y with f(x) and get the function notation f(x)=x+5. +

+ +

+ The x is the input variable, and f(x) is the y-value or output that corresponds to x. +

+ +

+ Generally, we use the letter f for functions. Other letters are okay as well; g(x) and h(x) are common. If we are using multiple functions at one time, we often denote them with different letters so we can refer to one without any confusion as to which function we mean. +

+ + + + +

Rewrite the following equations using function notation. In each case, assume y is a function of the variable x. +

+ + + + + +

y=2x+14

+ + +

+ f(x)=2x+14 +

+ + + + + +

y+x=3x^2-5

+ + +

+ f(x)=3x^2-x-5 +

+ + + + + +

\displaystyle \frac{2}{x}-x^4 = y-5

+ + +

+ f(x)=\displaystyle \frac{2}{x}-x^4 +5 +

+ + + + + + + + + + + + +

Let f(x)=3x^2-4x+1. Find the value of f(x) for the given values of x. +

+ + + + + x + f(x) + + + -5 + + + -\frac{1}{2} + + + 0 + + + 2 + + + 10 + + +
+ + + +

+ + + + x + f(x) + + + -5 + 96 + + + -\frac{1}{2} + \frac{15}{4} + + + 01 + + + 25 + + + 10261 + + +
+

+ + + + + If we are asked to find the value of f(x) for a certain x-value, say x=5, we use the notation f(5) to indicate that. + + + + +

Let f(x), g(x), and h(x) be defined as shown. + + f(x)&=3x^2-4x+1 + g(x)&=\sqrt{13-x^2} + h(x)&=\frac{x^2-6x+8}{x^2-4x+3} + + Find the following, if they exist. +

+ + + + +

+ f(-4), f(0) and f(2) +

+ + +

+ f(-4)=65, f(0)=1 and f(2)=5 +

+ + + + + +

+ g(0), g(2), and g(8) +

+ + +

+ g(0)=\sqrt{13}, g(2)=3, and g(8) is not defined +

+ + + + + +

+ h(3), h(4), and h(10) +

+ + +

+ h(3) is not defined, h(4)=0, and h(10)=\frac{16}{21} +

+ + + + + + + + +

+ Sometimes functions are made up of multiple functions put together. We call these piecewise functions. Each piece is defined for only a certain interval, and these intervals do not overlap. When evaluating a piecewise function at a given x-value, we first need to find the interval that includes the x-value, and then plug in to the corresponding function piece. +

+ + + + +

Let f(x) be a piecewise function as shown below. + + f(x)=\begin{cases} + x^2+3, & x < 5 \\ + 9-2x, & x \geq 5 + \end{cases} + +

+ + + +

+ On which interval from the piecewise function does the value x=1 belong? +

    +
  1. x < 5
    2. +
  2. x \leq 5
    3. +
  3. x > 5
    4. +
  4. x \geq 5
    5. +
+

+ + +

+ A. x < 5 +

+ + + + + +

+ Find f(1). +

    +
  1. 3
    2. +
  2. 4
    3. +
  3. 5
    4. +
  4. 6
    5. +
  5. 7
    6. +
+

+ + +

+ B. x< 4 +

+ + + + + +

+ On which interval from the piecewise function does the value x=5 belong? +

    +
  1. x < 5
    2. +
  2. x \leq 5
    3. +
  3. x > 5
    4. +
  4. x \geq 5
    5. +
+

+ + +

+ D. x \geq 5 +

+ + + + + +

+ Find f(5). +

    +
  1. -10
    2. +
  2. -5
    3. +
  3. -1
    4. +
  4. 17
    5. +
  5. 28
    6. +
+

+ + +

+ C. -1 +

+ + + + + + + +

+ We've been practicing evaluating functions at specific numeric values. It's also possible to evaluate a function given an expression involving variables. +

+ + + + +

Let g(x)=x^2-3x. +

+ + + +

+ Find g(a). +

    +
  1. (ax)^2-3ax
    2. +
  2. a^2-3a
    3. +
  3. a(x^2-3x)
    4. +
  4. ax^2-3ax
    5. +
  5. a-3
    6. +
+

+ + +

+ B. a^2-3a +

+ + + + + +

+ Find g(x+h). +

    +
  1. x^2-3x+h
    2. +
  2. (x+h)^2-3x
    3. +
  3. (x+h)^2-3(x+h)
    4. +
  4. x^2-3(x+h)
    5. + +
+

+ + +

+ C. (x+h)^2-3(x+h) +

+ + + + + +

+ We should also be able to look at a graph of a function and evaluate it for different values of x. The next activity explores that. +

+ + + + +

Let f(x) be the function graphed below. + + + f(x) = (.5*(x-2)*(x-4)*(x+1)+1) + p=plot(f, (x, -3, 5.5), ymin=-6, ymax=12, color='blue', thickness=3, gridlines=[[-5,-4,..,6],[-5,-4,..,12]]) + p + + +

+ + + +

+ Find f(1). +

    + +
  1. -4
    2. +
  2. -2
    3. +
  3. 0
    4. +
  4. 2
    5. +
  5. 4
    6. + +
+

+ + +

+ E. 4 +

+ + + + + + + +

+ Find f(3). +

    + +
  1. -1
    2. +
  2. 0
    3. +
  3. 1
    4. +
  4. 2
    5. +
  5. 3
    6. +
  6. 4
    7. + +
+

+ + +

+ A. -1 +

+ + + + + +

+ For which x-value(s) does f(x)=1? +

    + +
  1. -1
    2. +
  2. 0
    3. +
  3. 1
    4. +
  4. 2
    5. +
  5. 3
    6. +
  6. 4
    7. + +
+

+ + +

+ A. -1, D. 2, and F. 4 +

+ + + + + +

+ For which x-value(s) does f(x)=4. Estimate as needed! + +

+ + +

+ approximately -0.5, 1, and approximately 4.5 +

+ + + + + + + + +

+ In these activities, we are flipping the question around. This time we know what the function equals at some x-value, and we want to recover that x-value (or values!). +

+ + + +

Let h(x)=5x+7. + Find the x-value(s) such that h(x)=-13.

+ + +

+ -4 +

+ + + + + +

Let f(x)=x^2-3x-9. + Find the x-value(s) such that f(x)=9.

+ + +

+ -3 and 6 +

+ + + + + + + +

+ Ellie has \$13 in her piggy bank, and she gets an additional \$1.50 each week for her allowance. Assuming she does not spend any money, the total amount of allowance, A(w), she has after w weeks can be modeled by the function A(w)=13+1.50w. +

+ + + +

How much money will be in her piggy bank after 5 weeks?

+ + +

+ $20.50 +

+ + + + + +

After how many weeks will she have $40 in her piggy bank?

+ + +

+ 18 weeks +

+ + + + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/02-FN/03.ptx b/precalculus/source/02-FN/03.ptx new file mode 100644 index 00000000..7ab2d99e --- /dev/null +++ b/precalculus/source/02-FN/03.ptx @@ -0,0 +1,867 @@ + + +
+ Characteristics of a Function's Graph (FN3) + + + + + Activities + + + In this section, we will be looking at different kinds of graphs and will identify various characteristics. These ideas can span all kinds of functions, so you will see these come up multiple times! + + + + +

+ One of the easiest things to identify from a graph are the intercepts, which are points at which the graph crosses the axes. + An x-intercept is a point at which the graph crosses the x-axis and a y-intercept is a point at which the graph crosses the y-axis. Because intercepts are points, they are typically written as an ordered pair: (x,y). +

+ + + + + +

+ Use the following graphs to answer the questions. +

+ + + + f(x) = -1/2*x-2 + p=plot(f, (x, -5, 5), ymin=-10, ymax=10, color='blue', thickness=3) + p+=text('f(x)', (4,-2), fontsize=18, color='black') + p + + + + + f(x) = (x+1)*(x-3) + p=plot(f, (x, -5, 5), ymin=-10, ymax=10, color='blue', thickness=3) + p+=text('g(x)', (4, 1), fontsize=18, color='black') + p + + + + + + +

What are the x-intercept(s) of f(x)? +

    +
  1. (0, -4)
    2. +
  2. (-2, 0)
    3. +
  3. (-4, 0)
    4. +
  4. (0, -2)

+ + +

+ C +

+ + + + +

What are the x-intercept(s) of g(x)? +

    +
  1. (0, -3)
    2. +
  2. (-1, 0)
    3. +
  3. (3, 0)
    4. +
  4. (-3, 0)

+ + +

+ Choices B and C +

+ + + + +

What are the y-intercept(s) of f(x)? +

    +
  1. (0, -4)
    2. +
  2. (-2, 0)
    3. +
  3. (-4, 0)
    4. +
  4. (0, -2)

+ + +

+ D +

+ + + + +

What are the y-intercept(s) of g(x)? +

    +
  1. (0, -3)
    2. +
  2. (-1, 0)
    3. +
  3. (3, 0)
    4. +
  4. (-3, 0)

+ + +

+ A +

+ + + + +

+ Sketch a graph of a function with the following intercepts: +

    +
  • + x-intercepts: (-2,0) and (6,0) +
    • +
  • + y-intercept: (0,4) +
    • +
+

+ + +

+ Answers will vary. +

+ + + + +

+ Sketch a graph of a function with the following intercepts: +

    +
  • + x-intercept: (-1,0) +
    • +
  • + y-intercept: (0,6) and (0,-2) +
    • +
+

+ + +

+ You could draw a variety of graphs, but they would not be functions. +

+ + + + + + Notice in , that a function can have multiple x-intercepts, but only one y-intercept. Having more than one y-intercept would create a graph that is not a function! + + + + +

+ The domain refers to the set of possible input values and the range refers to the set of possible output values. If given a graph, however, it would be impossible to list out all the values for the domain and range so we use interval notation to represent the set of values. +

+ + + + + + + +

+ Use the following graph to answer the questions below. +

+
+ + + f(x) = 9/8*(x-4)+4 + p=plot(f, (x, -4, 4), ymin=-7, ymax=7, xmin=-5,xmax=5,color='blue', thickness=3,gridlines=[[-8..8],[-8..8]]) + p+=point((4,4),pointsize=75,color='blue') + p+=point((-4,-5),pointsize=75,color='blue') + p + + +
+ + + +

+ Draw on the x-axis all the values in the domain. +

+ + +

+ Students should shade all values of x from -4 to 4. The intent here is to help students visualize that the domain consists of more than x-values that are integers. +

+ + + + +

+ What interval represents the domain you drew in part a? +

    +
  1. [4, -4]
    2. +
  2. [-4, 4]
    3. +
  3. (-4, 4)
    4. +
  4. (4, -4)

+ + +

+ B +

+ + + + +

+ Draw on the y-axis all the values in the range. +

+ + +

+ Students should shade all values of y from -5 to 4. The intent here is to help students visualize that the range consists of more than y-values that are integers. +

+ + + + +

+ What interval represents the range you drew in part c? +

    +
  1. (-5, 4)
    2. +
  2. [-4, 4]
    3. +
  3. [-5, 4]
    4. +
  4. (4, -5)

+ + +

+ C +

+ + + + + + +

+ Use the following graph to answer the questions below. +

+
+ + + p=arrow((0,4), (9,-2),xmin=-2,xmax=9,ymin=-5,ymax=6, gridlines=[[-8..8],[-8..8]]) + p+=point((0,4),pointsize=75,color='blue') + p + + +
+ + + +

+ What is the domain of this graph? +

    +
  1. [4, \infty)
    2. +
  2. (-\infty, 0]
    3. +
  3. (-\infty, 4]
    4. +
  4. [0, \infty)

+ + +

+ D +

+ + + + +

+ What is the range of this graph? +

    +
  1. [4, \infty)
    2. +
  2. (-\infty, 0]
    3. +
  3. (-\infty, 4]
    4. +
  4. [0, \infty)

+ + +

+ C +

+ + + + + + When writing your intervals for domain and range, notice that you will need to write them from the smallest values to the highest values. For example, we wouldn't write [4,-\infty) as an interval because -\infty is smaller than 4. +

+ For domain, read the graph from left to right. For range, read the graph from bottom to top. +

+ + + + +

+ Use the following graph to answer the questions below. +

+
+ + + p=arrow((3,-4), (-4,8),xmin=-6,xmax=5,ymin=-6,ymax=8, gridlines=[[-8..8],[-8..8]]) + p+=point((3,-4), pointsize=75, markeredgecolor='blue',color='white',zorder=2) + p + + +
+ + + +

+ What is the domain of this graph? +

    +
  1. (-\infty, 3)
    2. +
  2. (\infty, -4]
    3. +
  3. (-4, \infty)
    4. +
  4. (-\infty, 3]

+ + +

+ A +

+ + + + +

+ What is the range of this graph? +

    +
  1. (-\infty, 3)
    2. +
  2. (\infty, -4]
    3. +
  3. (-4, \infty)
    4. +
  4. (-\infty, 3]

+ + +

+ C +

+ + + + + + +

+ Use the following graph to answer the questions below. +

+
+ + + p=line([(5,-1), (4,-5)],thickness=3,xmin=-6,xmax=8,ymin=-6,ymax=8, gridlines=[[-8..8],[-8..8]]) + p+=line([(4,-5),(-1,6)],thickness=3) + p+=line([(-1,6),(-3,1)],thickness=3) + + p+=point((-3,1),pointsize=75,color='blue') + p+=point((5,-1),pointsize=75,markeredgecolor='blue',color='white',zorder=2) + + p + + +
+ + + +

+ What is the domain of this graph? +

    +
  1. (-3, 5)
    2. +
  2. (-5, 7)
    3. +
  3. [-5, 7]
    4. +
  4. [-3,5)

+ + +

+ D +

+ + + + +

+ What is the range of this graph? +

    +
  1. (-3, 5)
    2. +
  2. (-5, 6)
    3. +
  3. [-5, 6]
    4. +
  4. [-3,5)

+ + +

+ C +

+ + + + + + Notice that finding the domain and range can be tricky! Be sure to pay attention to the x- and y-values of the entire graph - not just the endpoints! + + + + +

+ In this activity, we will look at where the function is increasing and decreasing. Use the following graph to answer the questions below. +

+ + + f(x) = (-x^2-2*x+3) + p=plot(f, (x, -4, 2), ymin=-1, ymax=5, color='blue', thickness=3, gridlines=[[-8..8],[-8..8]]) + + + + + +

+ Where do you think the graph is increasing? +

+ + +

+ The function is increasing from (-\infty, -1). Instructors can emphasize this by having students think about where the function is "going up." It may be helpful to also note that the function is no longer "going up" once you get to the "top". This could help students think about how to write their answer in interval notation (with parentheses and not brackets). +

+ + + + +

+ Which interval best represents where the function is increasing? +

    +
  1. (-\infty, -1]
    2. +
  2. (-\infty, -1)
    3. +
  3. (-1,\infty)
    4. +
  4. [-1,\infty)

+ + +

+ B +

+ + + + +

+ Where do you think the graph is decreasing? +

+ + +

+ The function is decreasing from (-1, \infty). Instructors can emphasize this by having students think about where the function is "going down." It may be helpful to also note that the function is no longer "going down" once you get to the "bottom". In this case, the "bottom" does not exist...This might be a good opportunity to discuss with students how to address this when writing the range in interval notation. +

+ + + + +

+ Which interval best represents where the function is decreasing? +

    +
  1. (-\infty, -1]
    2. +
  2. (-\infty, -1)
    3. +
  3. (-1,\infty)
    4. +
  4. [-1,\infty)

+ + +

+ C +

+ + + + +

+ Based on what you see on the graph, do you think this graph has any maxima or minima? +

+ + + + + + +

+ As you noticed in , functions can increase or decrease (or even remain constant!) for a period of time. The interval of increase is when the y-values of the function increase as the x-values increase. The interval of decrease is when the y-values of the function decrease as the x-values increase. The function is constant when the y-values remain constant as x-values increase (also known as the constant interval). +

+

+ The easiest way to identify these intervals is to read the graph from left to right and look at what is happening to the y-values. +

+ + + + + +

+ The maximum value, or global maximum, of a graph is the point where the y-coordinate has the largest value. The minimum value, or global minimum is the point on the graph where the y-coordinate has the smallest value. +

+

+ Graphs can also have relative/local maximums and relative/local minimums. A relative maximum point is a point where the function value (i.e, y-value) is larger than all others in some neighborhood around the point. Similarly, a relative minimum point is a point where the function value (i.e, y-value) is smaller than all others in some neighborhood around the point. +

+ + + + + +

+ Use the following graph to answer the questions below. +

+
+ + + p=plot(spline([(-4,4),(-2,7),(2,-4),(5,10)]),(x,-4,5),thickness=3, gridlines=[[-8..11],[-8..11]]) + p+=point((-4,4),pointsize=50,color='blue',xmin=-7,xmax=8,ymin=-5,ymax=10) + p+=point((5,10),pointsize=50,color='blue') + p + + +
+ + + +

+ At what value of x is there a global maximum? +

    +
  1. x=-4
    2. +
  2. x=-2
    3. +
  3. x=2
    4. +
  4. x=5

+ + +

+ D +

+ + + + +

+ What is the global maximum? +

    +
  1. 10
    2. +
  2. 7
    3. +
  3. 4
    4. +
  4. -4

+ + +

+ A +

+ + + + +

+ At what value of x is there a global minimum? +

    +
  1. x=-4
    2. +
  2. x=-2
    3. +
  3. x=2
    4. +
  4. x=5

+ + +

+ C +

+ + + + +

+ What is the global minimum? +

    +
  1. 10
    2. +
  2. 7
    3. +
  3. 4
    4. +
  4. -4

+ + +

+ D +

+ + + + +

+ At what value of x is there a local or relative maximum? +

    +
  1. x=-4
    2. +
  2. x=-2
    3. +
  3. x=2
    4. +
  4. x=5

+ + +

+ B +

+ + + + +

+ What is the local or relative maximum? +

    +
  1. 10
    2. +
  2. 7
    3. +
  3. 4
    4. +
  4. -4

+ + +

+ B +

+ + + + +

+ At what value of x is there a local or relative minimum? +

    +
  1. x=-4
    2. +
  2. x=-2
    3. +
  3. x=2
    4. +
  4. x=5

+ + +

+ C +

+ + + + +

+ What is the local or relative minimum? +

    +
  1. 10
    2. +
  2. 7
    3. +
  3. 4
    4. +
  4. -4

+ + +

+ D +

+ + + + + + Notice that in , there are two ways we talk about max and min. We might want to know the location of where the max or min are (i.e., determining at which x-value the max or min occurs at) or we might want to know what the max or min are (i.e., the y-value). +

Also, note that in , a local minimum is also a global minimum. +

+ + + + +

+ Sometimes, it is not always clear what the maxima or minima are or if they exist. Consider the following graph of f(x): +

+ + + f(x) = abs(1*x) + p=plot(f, (x, -5, 5), ymin=-5, ymax=5, color='blue', thickness=3) + p+=point((0,0), pointsize=75, markeredgecolor='blue',color='white',zorder=3) + p+=point((0,1),pointsize=75,color='blue') + p + + + + + +

+ What is the value of f(0)? +

    +
  1. 1
    2. +
  2. 0
    3. +
  3. There is no relative minimum

+ + +

+ A +

+ + + + + +

+ What is the relative minimum of f(x)? +

    +
  1. 1
    2. +
  2. 0
    3. +
  3. There is no relative minimum

+ + +

+ C +

+ + + + +

+ What is the global minimum of f(x)? +

    +
  1. 1
    2. +
  2. 0
    3. +
  3. There is no global minimum

+ + +

+ C +

+ + + + + + +

+ Use the following graph to answer the questions below. +

+
+ + + p=line([(-8,4),(-2,4)],thickness=3, gridlines=[[-8..11],[-8..11]]) + p+=plot(spline([(-2,4.1),(-0.5,8),(2,-3),(4.9,9.4)]),(x,-2,5),thickness=3) + p+=arrow((4.9,9.4),(5,10),thickness=3) + p+=point((-8,4),pointsize=50,color='blue',xmin=-9,xmax=6,ymin=-5,ymax=10) + p + + +
+ + + +

+ What is the domain? +

+ + +

+ [-8,\infty) +

+ + + + +

+ What is the range? +

+ + +

+ [-3.5, \infty) +

+ + + + +

+ What is the x-intercept(s)? +

+ + +

+ (1.25,0) and (3.5,0) +

+ + + + +

+ What is the y-intercept? +

+ + +

+ (0, 6.5) +

+ + + + +

+ Where is the function increasing? +

+ + +

+ (-2,-0.5)U(2.25,\infty) +

+ + + + +

+ Where is the function decreasing? +

+ + +

+ (-0.5,2,25) +

+ + + + +

+ Where is the constant interval? +

+ + +

+ (-8,-2) +

+ + + + +

+ At what x-values do the local maxima occur? +

+ + +

+ At x=-0,5 +

+ + + + +

+ At what x-values do the local minima occur? +

+ + +

+ At x=2.25 +

+ + + + +

+ What are the global max and min? +

+ + +

+ Global minimum is -3.5. There is no global maximum. +

+ + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/02-FN/04.ptx b/precalculus/source/02-FN/04.ptx new file mode 100644 index 00000000..e9717829 --- /dev/null +++ b/precalculus/source/02-FN/04.ptx @@ -0,0 +1,828 @@ + + +
+ Transformation of Functions (FN4) + + + + + Activities + +

Informally, a transformation of a given function is an algebraic process by which we change the function to a related function that has the same fundamental shape, but may be shifted, reflected, and/or stretched in a systematic way.

+ + + + +

Consider the following two graphs. + + + + p=plot(spline([(-2,-2),(0,2),(1,1),(2,3)]),(x,-2,2),thickness=3, gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-2,-2),pointsize=50,color='blue',xmin=-4,xmax=4,ymin=-5,ymax=5) + p+=point((2,3),pointsize=50,color='blue') + p+=text('$f(x)$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(spline([(-2,-1),(0,3),(1,2),(2,4)]),(x,-2,2),thickness=3, gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-2,-1),pointsize=50,color='blue',xmin=-4,xmax=4,ymin=-5,ymax=5) + p+=point((2,4),pointsize=50,color='blue') + p+=text('$f(x)+1$', (1, 4), fontsize=18, color='black') + p + + + +

+ + + +

How is the graph of f(x)+1 related to that of f(x)? +

    +

  1. Shifted up 1 unit

    2. +

  2. Shifted left 1 unit

    3. +

  3. Shifted down 1 unit

    4. +

  4. Shifted right 1 unit

    5. +

+ + +

+ Choice A: shifted up 1 unit +

+ + + + + + +

Consider the following two graphs. + + + + p=plot(spline([(-2,-2),(0,2),(1,1),(2,3)]),(x,-2,2),thickness=3, gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-2,-2),pointsize=50,color='blue',xmin=-4,xmax=4,ymin=-5,ymax=5) + p+=point((2,3),pointsize=50,color='blue') + p+=text('$f(x)$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(spline([(-2,-4),(0,0),(1,-1),(2,1)]),(x,-2,2),thickness=3, gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-2,-4),pointsize=50,color='blue',xmin=-4,xmax=4,ymin=-5,ymax=5) + p+=point((2,1),pointsize=50,color='blue') + p+=text('$f(x)-2$', (1, 4), fontsize=18, color='black') + p + + + +

+ + + +

How is the graph of f(x)-2 related to that of f(x)? +

    +

  1. Shifted up 2 units

    2. +

  2. Shifted left 2 units

    3. +

  3. Shifted down 2 units

    4. +

  4. Shifted right 2 units

    5. +

+ + +

+ Choice C: shifted down 2 units +

+ + + + + +

+ Notice that in and the y-values of the transformed graph are changed while the x-values remain the same. +

+ + + + +

+ Given a function f(x) and a constant c, the transformed function g(x)=f(x)+c is a vertical translation of the graph of f(x). That is, all the outputs change by c units. If c is positive, the graph will shift up. If c is negative, the graph will shift down. +

+ + + + + +

Consider the following two graphs. + + + + p=plot(spline([(-2,-2),(0,2),(1,1),(2,3)]),(x,-2,2),thickness=3, gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-2,-2),pointsize=50,color='blue',xmin=-4,xmax=4,ymin=-5,ymax=5) + p+=point((2,3),pointsize=50,color='blue') + p+=text('$f(x)$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(spline([(-3,-2),(-1,2),(0,1),(1,3)]),(x,-3,1),thickness=3, gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-3,-2),pointsize=50,color='blue',xmin=-4,xmax=4,ymin=-5,ymax=5) + p+=point((1,3),pointsize=50,color='blue') + p+=text('$f(x+1)$', (1, 4), fontsize=18, color='black') + p + + + +

+ + + +

How is the graph of f(x+1) related to that of f(x)? +

    +

  1. Shifted up by 1 unit

    2. +

  2. Shifted left 1 unit

    3. +

  3. Shifted down 1 unit

    4. +

  4. Shifted right 1 unit

    5. +

+ + +

+ Choice B: shifted left 1 unit +

+ + + + + + +

Consider the following two graphs. + + + + p=plot(spline([(-2,-2),(0,2),(1,1),(2,3)]),(x,-2,2),thickness=3, gridlines=[[-5, -4,-3,-2,-1,1,2,3,4,5],[-5, -4,-3,-2,-1,1,2,3,4, 5]]) + p+=point((-2,-2),pointsize=50,color='blue',xmin=-5,xmax=5,ymin=-5,ymax=5) + p+=point((2,3),pointsize=50,color='blue') + p+=text('$f(x)$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(spline([(1,-2),(3,2),(4,1),(5,3)]),(x,1,5),thickness=3, gridlines=[[-5, -4,-3,-2,-1,1,2,3,4,5],[-5,-4,-3,-2,-1,1,2,3,4,5]]) + p+=point((1,-2),pointsize=50,color='blue',xmin=-5,xmax=5,ymin=-5,ymax=5) + p+=point((5,3),pointsize=50,color='blue') + p+=text('$f(x-3)$', (1, 4), fontsize=18, color='black') + p + + + +

+ + + +

How is the graph of f(x-3) related to that of f(x)? +

    +

  1. Shifted up by 3 units

    2. +

  2. Shifted left 3 units

    3. +

  3. Shifted down 3 units

    4. +

  4. Shifted right 3 units

    5. +

+ + +

+ Choice D: shifted right 3 units +

+ + + + + +

+ Notice that in and the x-values of the transformed graph are changed while the y-values remain the same. +

+ + + + +

+ Given a function f(x) and a constant c, the transformed function g(x)=f(x+c) is a horizontal translation of the graph of f(x). If c is positive, the graph will shift left. If c is negative, the graph will shift right. +

+ + + + + +

+ Describe how the graph of the function is a transformation of the graph of the original function f. +

+ + + +

f(x-4)+1 +

    +

  1. Shifted down 4 units

    2. +

  2. Shifted left 4 units

    3. +

  3. Shifted down 1 unit

    4. +

  4. Shifted right 4 units

    5. +

  5. Shifted up 1 unit

    6. +

+ + +

+ Choices D and E: shifted right 4 units and up 1 unit +

+ + + + +

f(x+3)-2 +

    +

  1. Shifted down 2 units

    2. +

  2. Shifted left 3 units

    3. +

  3. Shifted up 3 unit

    4. +

  4. Shifted right 3 units

    5. +

  5. Shifted up 2 unit

    6. +

+ + +

+ Choices A and B: shifted left 3 units and down 2 units +

+ + + + + + +

Consider the following two graphs. + + + + p=plot(spline([(-2,-2),(0,2),(1,1),(2,3)]),(x,-2,2),thickness=3, gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-2,-2),pointsize=50,color='blue',xmin=-4,xmax=4,ymin=-5,ymax=5) + p+=point((2,3),pointsize=50,color='blue') + p+=text('$f(x)$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(spline([(-2,2),(0,-2),(1,-1),(2,-3)]),(x,-2,2),thickness=3, gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-2,2),pointsize=50,color='blue',xmin=-4,xmax=4,ymin=-5,ymax=5) + p+=point((2,-3),pointsize=50,color='blue') + p+=text('$-f(x)$', (1, 4), fontsize=18, color='black') + p + + + +

+ + + +

How is the graph of -f(x) related to that of f(x)? +

    +

  1. Shifted down 2 units

    2. +

  2. Reflected over the x-axis

    3. +

  3. Reflected over the y-axis

    4. +

  4. Shifted right 2 units

    5. +

+ + +

+ Choice B: reflected over the x-axis +

+ + + + + + + +

Consider the following two graphs. + + + + p=plot(spline([(-2,-2),(0,2),(1,1),(2,3)]),(x,-2,2),thickness=3, gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-2,-2),pointsize=50,color='blue',xmin=-4,xmax=4,ymin=-5,ymax=5) + p+=point((2,3),pointsize=50,color='blue') + p+=text('$f(x)$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(spline([(2,-2),(0,2),(-1,1),(-2,3)]),(x,-2,2),thickness=3, gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((2,-2),pointsize=50,color='blue',xmin=-4,xmax=4,ymin=-5,ymax=5) + p+=point((-2,3),pointsize=50,color='blue') + p+=text('$f(-x)$', (1, 4), fontsize=18, color='black') + p + + + +

+ + + +

How is the graph of f(-x) related to that of f(x)? +

    +

  1. Shifted down 2 units

    2. +

  2. Reflected over the x-axis

    3. +

  3. Reflected over the y-axis

    4. +

  4. Shifted left 2 units

    5. +

+ + +

+ Choice C: reflected over the y-axis +

+ + + + + +

+ Notice that in the y-values of the transformed graph are changed while the x-values remain the same. While in , the x-values of the transformed graph are changed while the y-values remain the same. +

+ + + + +

+ Given a function f(x), the transformed function g(x)=-f(x) is a vertical reflection of the graph of f(x). That is, all the outputs, are multiplied by -1. The new graph is a reflection of the old graph about the x-axis. +

+ + + + + +

+ Given a function f(x), the transformed function y=g(x)=f(-x) is a horizontal reflection of the graph of f(x). That is, all the inputs, are multiplied by -1. The new graph is a reflection of the old graph about the y-axis. +

+ + + + + +

Consider the following graph of the function f(x). + + + + p=plot(sqrt(-1*x**2+9),(x,-3,3),thickness=3) + p+=point((-3,0),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-4,ymax=4) + p+=point((3,0),pointsize=50,color='blue') + p+=text('$f(x)$', (1, 4), fontsize=18, color='black') + p + + + +

+ + + +

How is the graph of -f(x+2)+3 related to that of f(x)? +

    +

  1. Shifted up 2 units

    2. +

  2. Shifted up 3 units

    3. +

  3. Reflected over the x-axis

    4. +

  4. Reflected over the y-axis

    5. +

  5. Shifted left 3 units

    6. +

  6. Shifted left 2 units

    7. +

+ + +

+ Choices B, C and F: shifted left 2 units, reflected over the x-axis, and shifted up 3 units +

+ + + + +

Which of the following represents the graph of the transformed function g(x)=-f(x+2)+3? +

    +

  1. + + p=plot(-1*sqrt(-1*(x+2)**2+9)+3,(x,-5,1),thickness=3, gridlines=[[-6,-5,-4,-3,-2,-1,1,2,3,4,5,6],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-5,3),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-4,ymax=4) + p+=point((1,3),pointsize=50,color='blue') + p + +

    2. +

  2. + + p=plot(sqrt(-1*(x+2)**2+9)-3,(x,-5,1),thickness=3, gridlines=[[-6,-5,-4,-3,-2,-1,1,2,3,4,5,6],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-5,-3),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-4,ymax=4) + p+=point((1,-3),pointsize=50,color='blue') + p + +

    3. +

  3. + + p=plot(-1*sqrt(-1*(x-2)**2+9)+3,(x,-1,5),thickness=3,gridlines=[[-6,-5,-4,-3,-2,-1,1,2,3,4,5,6],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-1,3),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-4,ymax=4) + p+=point((5,3),pointsize=50,color='blue') + p + +

    4. +

  4. + + p=plot(-1*sqrt(-1*(x-2)**2+9),(x,-1,5),thickness=3, gridlines=[[-6,-5,-4,-3,-2,-1,1,2,3,4,5,6],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-1,0),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-4,ymax=4) + p+=point((5,0),pointsize=50,color='blue') + p + +

    5. +

+ + +

+ Choice A +

+ + + + + +

+ Notice that in the resulting graph is different if you perform the reflection first and then the vertical shift, versus the other order. When combining transformations, it is very important to consider the order of the transformations. Be sure to follow the order of operations. +

+ + + +

Consider the following two graphs. + + + + p=plot(abs(x+2)-4,(x,-5,-1),thickness=3, gridlines=[[-5,-4,-3,-2,-1,1,2,3,4,5],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-5,-1),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-5,ymax=5) + p+=point((-1,-3),pointsize=50,color='blue') + p+=text('$f(x)$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(abs(-x+2)-1,(x,1,5),thickness=3,gridlines=[[-5,-4,-3,-2,-1,1,2,3,4,5],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((1,0),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-5,ymax=5) + p+=point((5,2),pointsize=50,color='blue') + p+=text('$g(x)$', (1, 4), fontsize=18, color='black') + p + + + +

+ + + +

How is the graph of g(x) related to that of f(x)? +

    +

  1. Shifted up 3 units

    2. +

  2. Shifted up 1 unit

    3. +

  3. Reflected over the x-axis

    4. +

  4. Reflected over the y-axis

    5. +

  5. Shifted left 1 unit

    6. +

  6. Shifted right 4 units

    7. +

+ + +

+ Choices A and D: reflected over the y-axis, and shifted up 3 units +

+ + + + +

+ List the order the transformations must be applied. +

+ + + + +

Write an equation for the graphed function g(x) using transformations of the graph f(x). +

    +

  1. g(x)=-f(x)+3

    2. +

  2. g(x)=f(-x)+3

    3. +

  3. g(x)=f(-x+3)

    4. +

  4. g(x)=-f(x+3)

    5. +

+ + +

+ Choice B: +

+ + + + + + +

Consider the following two graphs. + + + + p=plot(2,(x,-1,1),thickness=3, gridlines=[[-6, -5,-4,-3,-2,-1,1,2,3,4,5,6],[-4,-3,-2,-1,1,2,3,4]]) + p+=plot(-x+3,(x,1,4), thickness =3) + p+=plot(x+3,(x,-4,-1), thickness =3) + p+=point((-4,-1),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-5,ymax=5) + p+=point((4,-1),pointsize=50,color='blue') + p+=text('$f(x)$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(4,(x,-1,1),thickness=3, gridlines=[[-6,-5,-4,-3,-2,-1,1,2,3,4,5,6],[-4,-3,-2,-1,1,2,3,4]]) + p+=plot(2*(-x+3),(x,1,4), thickness =3) + p+=plot(2*(x+3),(x,-4,-1), thickness =3) + p+=point((-4,-2),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-5,ymax=5) + p+=point((4,-2),pointsize=50,color='blue') + p+=text('$2f(x)$', (3, 4), fontsize=18, color='black') + p + + + +

+ + + +

Consider the y-value of the two graphs at x=1. How do they compare? +

    +

  1. The y-value of 2f(x) is twice that of f(x).

    2. +

  2. The y-value of 2f(x) is half that of f(x).

    3. +

  3. The y-value of 2f(x) and f(x) are the same.

    4. +

  4. The y-value of 2f(x) is negative that of f(x).

    5. +

+ + + + +

How is the graph of 2f(x) related to that of f(x)? +

    +

  1. Vertically stretched by a factor of 2

    2. +

  2. Vertically compressed by a factor of 2

    3. +

  3. Horizontally stretched by a factor of 2

    4. +

  4. Horizontally compressed by a factor of 2

    5. +

+ + +

+ Choice A: vertically stretched by a factor of 2 +

+ + + + + +

Consider the following two graphs. + + + + p=plot(2,(x,-1,1),thickness=3, gridlines=[[-6,-5,-4,-3,-2,-1,1,2,3,4,5,6],[-4,-3,-2,-1,1,2,3,4]]) + p+=plot(-x+3,(x,1,4), thickness =3) + p+=plot(x+3,(x,-4,-1), thickness =3) + p+=point((-4,-1),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-5,ymax=5) + p+=point((4,-1),pointsize=50,color='blue') + p+=text('$f(x)$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(2,(x,-.5,.5),thickness=3, gridlines=[[-6, -5,-4,-3,-2,-1,1,2,3,4,5,6],[-4,-3,-2,-1,1,2,3,4]]) + p+=plot((-2*x+3),(x,.5,2), thickness =3) + p+=plot((2*x+3),(x,-2,-.5), thickness =3) + p+=point((-2,-1),pointsize=50,color='blue',xmin=-6,xmax=6,ymin=-5,ymax=5) + p+=point((2,-1),pointsize=50,color='blue') + p+=text('$f(2x)$', (3, 4), fontsize=18, color='black') + p + + + +

+ + + +

How is the graph of f(2x) related to that of f(x)? +

    +

  1. Vertically stretched by a factor of 2

    2. +

  2. Vertically compressed by a factor of 2

    3. +

  3. Horizontally stretched by a factor of 2

    4. +

  4. Horizontally compressed by a factor of 2

    5. +

+ + +

+ Choice D: horizontally compressed by a factor of 2 +

+ + + + +

+ Notice that in the y-values are doubled while the x-values remain the same. While, in the x-values are cut in half while the y-values remain the same. +

+ + + +

+ Given a function f(x), the transformed function g(x)=af(x) is a vertical stretch or vertical compression of the graph of f(x). That is, all the outputs, are multiplied by a. If a \gt 1 the new graph is a vertical stretch of the old graph away from the x-axis. If 0 \lt a \lt 1 the new graph is a vertical compression of the old graph towards the x-axis. Points on the x-axis are unchanged. +

+ + + + + +

+ Given a function f(x), the transformed function g(x)=f(ax) is a horizontal stretch or horizontal compression of the graph of f(x). That is, all the inputs, are divided by a. If a \gt 1 the new graph is a horizontal compression of the old graph toward the y-axis. If 0 \lt a \lt 1 the new graph is a horizontal stretch of the old graph away from the y-axis. Points on the y-axis are unchanged. +

+ + + + +

We often use a set of basic functions with which to begin transformations. We call these parent functions. + + + + p=plot(x,thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5) + p+=text('$f(x)=x$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(x**2,thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5) + p+=text('$f(x)=x^2$', (1, 4), fontsize=18, color='black') + p + + + + + + + + p=plot(abs(x),thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5) + p+=text('$f(x)=|x|$', (1, 4), fontsize=18, color='black') + p + + + + + p=plot(sqrt(x),thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5) + p+=text(r"$f(x)=\sqrt{x}$", (1, 4), fontsize=18, color='black') + p + + + + + + + + p=plot(x**3,thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5) + p+=text('$f(x)=x^3$', (-2, 4), fontsize=18, color='black') + p + + + + + p=plot(1/x, thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5, detect_poles=True) + p+=text(r"$f(x)=\frac{1}{x}$", (-2, 4), fontsize=18, color='black') + p + + + +

+ + + + +

Consider the function g(x)=3\sqrt{-x}+2 +

+ + + +

Identify the parent function f(x). +

    +

  1. f(x)=x^{2}

    2. +

  2. f(x)=\lvert x \rvert

    3. +

  3. f(x)=\sqrt{x}

    4. +

  4. f(x)=x

    5. +

+ + +

+ Choice C: f(x)=\sqrt{x} +

+ + + + +

+ Graph the parent function f(x). +

+ + + + +

How is the graph of g(x) related to that of the parent functionf(x)? +

    +

  1. Reflected over the x-axis

    2. +

  2. Reflected over the y-axis

    3. +

  3. Shifted down 2 units

    4. +

  4. Shifted up 2 units

    5. +

  5. Vertically stretched by a factor of 3

    6. +

  6. Horizontally compressed by a factor of 3

    7. +

+ + +

+ Choices B, D, E: reflected over the y-axis, vertically stretched by a factor of 3, shifted up 2 units +

+ + + + +

+ Graph the transformed function g(x). +

+ + + + + + +

Consider the following graph of the function g(x). + + + + p=plot(-1*(x+2)**2-3,thickness=3, xmin=-6,xmax=6,ymin=-10,ymax=10, gridlines=[[-6,-6,-4,-3,-2,-1,1,2,3,4,5,6],[-10,-8,-6,-4,-2,2,4,6,8,10]]) + p+=text('$g(x)$', (1, 4), fontsize=18, color='black') + p + + + +

+ + + +

Identify the parent function. +

    +

  1. f(x)=x^{2}

    2. +

  2. f(x)=\lvert x \rvert

    3. +

  3. f(x)=\sqrt{x}

    4. +

  4. f(x)=x

    5. +

+ + +

+ Choice A: f(x)=x^{2} +

+ + + + +

How is the graph of g(x) related to that of the parent functionf(x)? +

    +

  1. Reflected over the x-axis

    2. +

  2. Reflected over the y-axis

    3. +

  3. Shifted down 3 units

    4. +

  4. Shifted up 3 units

    5. +

  5. Shifted left 2 units

    6. +

  6. Shifted right 2 units

    7. +

+ + +

+ Choices A, B, D: shifted left 2 units, reflected over the x-axis, shifted down 3 units +

+ + + + +

Write an equation to represent the transformed function g(x). +

    +

  1. g(x)=-(x-2)^{2}-3

    2. +

  2. g(x)=-(x+2)^{2}+3

    3. +

  3. g(x)=(-x+2)^{2}-3

    4. +

  4. g(x)=-(x+2)^{2}-3

    5. +

+ + +

+ Choice D: g(x)=-(x+2)^2+3 +

+ + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/02-FN/05.ptx b/precalculus/source/02-FN/05.ptx new file mode 100644 index 00000000..2644b6b6 --- /dev/null +++ b/precalculus/source/02-FN/05.ptx @@ -0,0 +1,662 @@ + + +
+ Combining and Composing Functions (FN5) + + + + + Activities + + + +

Let f(x)=x^2-3x and g(x)= x^3-4x^2+7.

+ + + + +

+ Which of the following seems likely to be the most simplified form of f(x)+g(x)? + +

    +
  1. x^2-3x+x^3-4x^2+7
    2. +
  2. x^5-7x^3+7
    3. +
  3. -x^3+5x^2-3x-7
    4. +
  4. x^3-3x^2-3x+7
    5. +
+

+ + +

+ D. +

+

A isn't simplified, B added exponents and coefficients down the line, C is subtraction +

+ + + + + +

+ Which of the following seems likely to be the most simplified form of f(x)-g(x)? + +

    +
  1. x^3-3x^2-3x+7
    2. +
  2. -x^3+5x^2-3x-7
    3. +
  3. -x^3-3x^2-3x+7
    4. +
  4. x^2-3x-x^3+4x^2-7
    5. +
+

+ + +

+ B. +

+

A is added, C didn't distribute the negative to all terms of g(x), D isn't simplified +

+ + + + + + + + + +

Let f(x)=\sqrt{x+1} and g(x)= 5x.

+ + + + +

+ Which of the following seems likely to be the most simplified form of f(x)\cdot g(x)? + +

    +
  1. \sqrt{5x+1}
    2. +
  2. 5\sqrt{x+1}
    3. +
  3. \sqrt{5x^2+5x}
    4. +
  4. 5x\sqrt{x+1}
    5. +
+

+ + +

+ D. +

+

+ A is f(g(x)), B is g(f(x)), C distributed under the radical +

+ + + + + +

+ Which of the following seems likely to be the most simplified form of \frac{f(x)}{g(x)}? + +

    +
  1. \frac{5x}{\sqrt{x+1}} +
    2. +
  2. \frac{\sqrt{x+1}}{5x}
    3. +
  3. \sqrt{\frac{x}{5x}+\frac{1}{5x}}
    4. +
  4. \sqrt{\frac{5x}{x}+\frac{5x}{1}}
    5. +
+

+ + +

+ B. +

+

+ A is g/f, C divides each term under the radical by g, D flips the fractions in C +

+ + + + + + +

+ In and , we have found the sum, difference, product, and quotient of two functions. We can use the following notation for these newly created functions: + + (f+g)(x) \amp = f(x)+g(x) + (f-g)(x) \amp = f(x)-g(x) + (f\cdot g)(x) \amp = f(x)\cdot g(x) + \left(\frac{f}{g}\right)(x) \amp = \frac{f(x)}{g(x)} + + + With \left(\frac{f}{g}\right)(x) , we note that the quotient is only defined when g(x)\neq 0. + +

+ + + + +

Let \displaystyle f(x)=\frac{1}{3x-5}.

+ + + + +

+ Find f(4). + + +

    +
  1. \frac{4}{3x-5}
    2. +
  2. \frac{1}{4(3x-5)}
    3. +
  3. \frac{1}{7}
    4. +
  4. 7
    5. +
+

+ + +

+ See for a reminder of what this notation means! +

+ + +

+ C. +

+ + + + + + + +

+ If you were asked to find f(x^3-2), how do you think you would proceed? + +

    +
  1. Multiply the original function \frac{1}{3x-5} by x^3-2.
    2. +
  2. Plug the expression x^3-2 in for all the x-values in \frac{1}{3x-5}.
    3. +
  3. Plug the original function \frac{1}{3x-5} in for all the x-values in x^3-2.
    4. +
  4. Multiply 3x-5 by + x^3-2.
    5. + +
+

+ + +

+ B. +

+

+ A is multiplying, C is g(f) +

+ + + + + +

+ Find f(x^3-2). + +

    +
  1. \frac{1}{3x-5} \cdot (x^3-2)
    2. +
  2. \frac{1}{3(x^3-2)-5}
    3. +
  3. \left(\frac{1}{3x-5} \right)^3-2
    4. +
  4. (3x-5)(x^3-2)
    5. + +
+

+ + +

+ B. +

+

+ These mirror the previous answer choices. +

+ + + + + +

+ What if we gave the expression x^3-2 a name? Let's define g(x)=x^3-2. What's another way we could denote f(x^3-2)? +

    +
  1. f(x) \cdot g(x)
    2. +
  2. g(f(x))
    3. +
  3. f(g(x))
    4. +
  4. \frac{f(x)}{g(x)}
    5. + +
+

+ + +

+ C. +

+ + + + + + + + + +

+ Given the functions f(x) and g(x), we define the composition of f and g to be the new function h(x) given by + + h(x) = f(g(x)) + . + We also sometimes use the notation f \circ g or (f \circ g)(x) to refer to f(g(x)). +

+ + + + +

+ When discussing the composite function f(g(x)), also written as (f\circ g)(x), we often call g(x) the "inner function" and f(x) the "outer function". It is important to note that the inner function is actually the first function that gets applied to a given input, and then the outer function is applied to the output of the inner function. + + +

+ + + + +

Let \displaystyle f(x)=\frac{1}{3x-5} and g(x)=x^3-2.

+ + + + +

+ Find f(g(x)). + +

    +
  1. \frac{x^3-2}{3x-5}
    2. +
  2. \frac{1}{(3x-5)(x^3-2)}
    3. +
  3. \frac{1}{3(x^3-2)-5}
    4. +
  4. \left( \frac{1}{3x-5} \right)^{3} -2
    5. + +
+

+ + +

+ C. +

+ + + + + +

+ Find g(f(x)). + +

    +
  1. \frac{x^3-2}{3x-5}
    2. +
  2. \frac{1}{(3x-5)(x^3-2)}
    3. +
  3. \frac{1}{3(x^3-2)-5}
    4. +
  4. \left( \frac{1}{3x-5} \right)^{3} -2
    5. + +
+

+ + +

+ D. +

+ + + + + + +

+ We can also evaluate the composition of two functions at a particular value just as we did with one function. For example, we may be asked to find something like f(g(2)) or (g\circ f)(-3). +

+ + + + +

Let \displaystyle f(x)=2x^3 and g(x)=\sqrt{6-x}.

+ + + + +

+ Find f(g(2)). + +

    +
  1. 14
    2. +
  2. 16
    3. +
  3. 18
    4. +
  4. 20
    5. +
  5. undefined
    6. + +
+

+ + +

+ B. +

+ + + + + +

+ Find (g \circ f)(-3). + +

    +
  1. 50
    2. +
  2. 54
    3. +
  3. \sqrt{60}
    4. +
  4. \sqrt{-48}
    5. +
  5. undefined
    6. + +
+

+ + +

+ C. +

+

+ B is f \circ g, D has a mistake in subtraction +

+ + + + + +

+ Find (f \circ g)(10). + +

    +
  1. 2(\sqrt{-4})^3
    2. +
  2. 16
    3. +
  3. \sqrt{-1994}
    4. +
  4. -16
    5. +
  5. undefined
    6. + +
+

+ + +

+ E. +

+

+ A ignores domain, B assumes \sqrt{-4} is 2, C is g \circ f, but ignoring domain, D assumes \sqrt{-4} is -2 +

+ + + + + + + + +

+ As we saw in , in order for a composite function to make sense, we need to ensure that the range of the inner function lies within the domain of the outer function so that the resulting composite function is defined at every possible input. + +

+ + + +

+ In addition to the possibility that functions are given by formulas, + functions can be given by tables or graphs. + We can think about composite functions in these settings as well, + and the following activities prompt us to consider functions given in this way. +

+ + + + + + + +

+Let functions p and q be given by the graphs below.

+ + + + p=line([(-4,5), (-1,-1)],color="blue",thickness=3,aspect_ratio=1,xmin=-3.5,xmax=3.5,ymin=-3.5,ymax=3.5,gridlines=[[-4,-3,..,4],[-4,-3,..,4]]) + p+=line([(-1,-1),(1,0)],color="blue",thickness=3) + p+=line([(1,0),(4,6)],color="blue",thickness=3) + p+=line([(-3,-4),(-1,2)],color="#080",thickness=3) + p+=line([(-1,2),(1,2)],color="#080",thickness=3) + p+=line([(1,2),(4,-1)],color="#080",thickness=3) + p+=text('$p(x)$', (-2, 3), fontsize=18, color='blue') + p+=text('$q(x)$', (3, 1), fontsize=18, color='#080') + p + + + + +

+ Find each of the following. If something is not defined, explain why. +

+ + + + + + +

+ (p \circ q)(0) +

+ + +

+ 1 +

+ + + + + +

+ q(p(0)) +

+ + +

+ 2 +

+ + + + + +

+ p(p(1)) +

+ + +

+ -1/2 +

+ + + + + +

+ (q \circ p)(-3) +

+ + +

+ 0 +

+ + + + + +

+ Find two values of x such that q(p(x)) = 2. +

+ + +

+ x-values on the interval [-2,1.5] +

+ + + + + + + + +

+ + Let functions f and g be given by the tables below. + + + + + + + x + f(x) + + + 0 + 6 + + + 1 + 4 + + + 2 + 3 + + + 3 + 4 + + + 4 + 7 + + +
+ + + + x + g(x) + + + 0 + 1 + + + 1 + 3 + + + 2 + 0 + + + 3 + 5 + + + 4 + 2 + + +
+ + + Find each of the following. If something is not defined, explain why. +

+ + + + + + +

+ (f \circ g)(2) +

+ + +

+ 6 +

+ + + + + +

+ (g \circ f)(3) +

+ + +

+ 2 +

+ + + + + +

+ g(f(4)) +

+ + +

+ not defined because f(4)=7 and 7 isn't in the domain of g(x) +

+ + + + + +

+ For what value(s) of x is f(g(x)) = 4? +

+ + +

+ 0, 1 +

+ + + + + +

+ What are the domain and range of (f\circ g)(x)? +

+ + +

+ Domain: \{0,1,2,3,4 \} and + Range: \{3,4,6 \} +

+ + + + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
diff --git a/precalculus/source/02-FN/06.ptx b/precalculus/source/02-FN/06.ptx new file mode 100644 index 00000000..2c28020a --- /dev/null +++ b/precalculus/source/02-FN/06.ptx @@ -0,0 +1,398 @@ + + +
+ Finding the Inverse Function (FN6) + + + + + Activities + + A function is a process that converts a collection of inputs to a corresponding collection of outputs. One question we can ask is: for a particular function, can we reverse the process and think of the original function's outputs as the inputs? + + + +

Temperature can be measured using many different units such as Fahrenheit, Celsius, and Kelvin. Fahrenheit is what is usually reported on the news each night in the United States, while Celsius is commonly used for scientific work. We will begin by converting between these two units. + + To convert from degrees Fahrenheit to Celsius use the following formula. + C=\frac{5}{9} (F-32) +

+ + + + + +

Room temperature is around 68 degrees Fahrenheit. Use the above equation to convert this temperature to Celsius. +

    +
  1. 5.8
    2. +
  2. 20
    3. +
  3. 155.4
    4. +
  4. 293
    5. + + +
+

+ + + + +

Solve the equation C=\frac{5}{9} (F-32) for F in terms of C. +

    +
  1. F=\frac{5}{9} C + 32
    2. +
  2. F=\frac{5}{9} C - 32
    3. +
  3. F=\frac{9}{5}( C + 32)
    4. +
  4. F=\frac{9}{5} C + 32
    5. +
+

+ + + + + +

Alternatively, 20 degrees Celsius is a fairly comfortable temperature. Use your solution for F in terms of C to convert this temperature to Fahrenheit. +

    +
  1. 43.1
    2. +
  2. -20.9
    3. +
  3. 93.6
    4. +
  4. 68
    5. +
+

+ + + + + + Notice that when you converted 68 degrees Fahrenheit, you got a value of 20 degrees Celsius. Alternatively, when you converted 20 degrees Celsius, you got 68 degrees Fahrenheit. This indicates that the equation you were given for C and the equation you found for F are inverses. + + + +

+ Let f be a function. If there exists a function g such that + + f(g(x))=x \quad \text{and} \quad g(f(x))=x + + for all x, then we say f has an inverse function, or that g is the inverse of f. + + When a given function f has an inverse function, we usually denote it as f^{-1}, which is read as "f inverse". +

+ + + + An inverse is a function that "undoes" another function. For any input in the domain, the function g will reverse the process of f. + + + +

It is important to note that in we say "if there exists a function," but we don't guarantee that this is always the case. How can we determine whether a function has a corresponding inverse or not? + + Consider the following two functions f and g represented by the tables. + + + + x + f(x) + + + 0 + 6 + + + 1 + 4 + + + 2 + 3 + + + 3 + 4 + + + 4 + 6 + + +
+ + + + + x + g(x) + + + 0 + 3 + + + 1 + 1 + + + 2 + 4 + + + 3 + 2 + + + 4 + 0 + + +
+

+ + + + + + +

Use the definition of g(x) in to find an x such that g(x)=4. +

    +
  1. x=0
    2. +
  2. x=1
    3. +
  3. x=2
    4. +
  4. x=3
    5. +
  5. x=4
    6. +
+

+ + + + +

+ Is it possible to reverse the input and output rows of the function g(x) and have the new table result in a function? +

+ + + + +

Use the definition of f(x) in to find an x such that f(x)=4. +

    +
  1. x=0
    2. +
  2. x=1
    3. +
  3. x=2
    4. +
  4. x=3
    5. +
  5. x=4
    6. +
+

+ + + + +

+ Is it possible to reverse the input and output rows of the function f(x) and have the new table result in a function? +

+ + + + + + Some functions, like f(x) in , have a given output value that corresponds to two or more input values: f(0)=6 and f(4)=6 . If we attempt to reverse the process of this function, we have a situation where the new input 6 would correspond to two potential outputs. + + + + +

+ A one-to-one function is a function in which each output value corresponds to exactly one input. +

+ + + + + A function must be one-to-one in order to have an inverse. + + + + +

Consider the function f(x)=\dfrac{x-5}{3}. +

+ + + + + +

When you evaluate this expression for a given input value of x, what operations do you perform and in what order? +

    +
  1. divide by 3, subtract 5
    2. +
  2. subtract 5, divide by 3
    3. +
  3. add 5, multiply by 3
    4. +
  4. multiply by 3, add 5
    5. +
+

+ + + + +

When you "undo" this expression to solve for a given ouput value of y, what operations do you perform and in what order? +

    +
  1. divide by 3, subtract 5
    2. +
  2. subtract 5, divide by 3
    3. +
  3. add 5, multiply by 3
    4. +
  4. multiply by 3, add 5
    5. +
+

+ + + + +

This set of operations reverses the process for the original function, so can be considered the inverse function. Write an equation to express the inverse function f^{-1}. +

    +
  1. f^{-1}(x)=\frac{x}{3}-5
    2. +
  2. f^{-1}(x)=\frac{x-5}{3}
    3. +
  3. f^{-1}(x)=5(x+3)
    4. +
  4. f^{-1}(x)=3x+5
    5. +
+

+ + + + +

+ Check your answer to the previous question by finding f(f^{-1}(x)) and f^{-1}(f(x)). +

+ + + + + + +

+ To find the inverse of a one-to-one function, perform the reverse operations in the opposite order. + +

+ + + + + + +

Let's look at an alternate method for finding an inverse by solving the function for x and then interchanging the x and y. + + h(x)=\dfrac{x}{x+1} + +

+ + + + +

+ Interchange the variables x and y. +

    +
  1. y=\frac{x}{x+1}
    2. +
  2. x=\frac{y}{x+1}
    3. +
  3. x=\frac{y}{y+1}
    4. +
  4. x=\frac{x}{y+1}
    5. +
+

+ + + +

+ Choice C: x=\frac{y}{y+1} +

+ + + +

+ Eliminate the denominator. +

    +
  1. y(x+1)=x
    2. +
  2. x(x+1)=y
    3. +
  3. x(y+1)=y
    4. +
  4. x(y+1)=x
    5. +
+

+ + + +

+ Choice C: x(y+1)=y +

+ + + +

+ Distribute and gather the y terms together. +

    +
  1. yx+y=x
    2. +
  2. x^{2}+x=y
    3. +
  3. xy-y=-x
    4. +
  4. xy=0
    5. +
+

+ + + +

+ Choice C: x=\frac{y}{y+1} +

+ + + + +

+ Write the inverse function, by factoring and solving for y. +

    +
  1. h^{-1}(x)= \frac{x}{x-1}
    2. +
  2. h^{-1}(x)=\frac{x}{1-x}
    3. +
  3. h^{-1}(x)= \frac{-x}{1-x}
    4. +
  4. h^{-1}(x)= \frac{x+1}{x}
    5. +
+

+ + + +

+ Choice B: h^{-1}(x)=\frac{x}{1-x} +

+ + + + + +

Find the inverse of each function, using either method. Check your answer using function composition. +

+ + + + +

g(x)=\dfrac{4x-1}{7} +

    +
  1. g^{-1}(x)=\frac{7x+1}{4}
    2. +
  2. g^{-1}(x)= \frac{7x}{4}+1
    3. +
  3. g^{-1}(x)= \frac{4x+1}{7}
    4. +
  4. g^{-1}(x)=\frac{7}{4x-1}
    5. +
+

+ + + + + +

f(x)=3-\sqrt{x+5} +

    +
  1. f^{-1}(x)=3+\sqrt{x-5}
    2. +
  2. f^{-1}(x)=(x-3)^{2}+5
    3. +
  3. f^{-1}(x)= \frac{1}{3-\sqrt{x+5}}
    4. +
  4. f^{-1}(x)=(3-x)^{2}-5
    5. +
+

+ + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/02-FN/main.ptx b/precalculus/source/02-FN/main.ptx new file mode 100644 index 00000000..b5e4d5b8 --- /dev/null +++ b/precalculus/source/02-FN/main.ptx @@ -0,0 +1,18 @@ + + + + Functions (FN) + + + + + + + + + + + + + + \ No newline at end of file diff --git a/precalculus/source/02-FN/outcomes/01.ptx b/precalculus/source/02-FN/outcomes/01.ptx new file mode 100644 index 00000000..562bef42 --- /dev/null +++ b/precalculus/source/02-FN/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+ Determine if a relation, equation, or graph defines a function using the definition as well as the vertical line test. +

\ No newline at end of file diff --git a/precalculus/source/02-FN/outcomes/02.ptx b/precalculus/source/02-FN/outcomes/02.ptx new file mode 100644 index 00000000..4acfa23e --- /dev/null +++ b/precalculus/source/02-FN/outcomes/02.ptx @@ -0,0 +1,5 @@ + +

+ Use and interpret function notation to evaluate a function for a given input value and + find a corresponding input value given an output value. +

\ No newline at end of file diff --git a/precalculus/source/02-FN/outcomes/03.ptx b/precalculus/source/02-FN/outcomes/03.ptx new file mode 100644 index 00000000..15b8c5b6 --- /dev/null +++ b/precalculus/source/02-FN/outcomes/03.ptx @@ -0,0 +1,5 @@ + +

+ Use the graph of a function to find the domain and range in interval notation, the x- and y-intercepts, + the maxima and minima, and where it is increasing and decreasing using interval notation. +

\ No newline at end of file diff --git a/precalculus/source/02-FN/outcomes/04.ptx b/precalculus/source/02-FN/outcomes/04.ptx new file mode 100644 index 00000000..3ba4d1c6 --- /dev/null +++ b/precalculus/source/02-FN/outcomes/04.ptx @@ -0,0 +1,5 @@ + +

+ Apply transformations including horizontal and vertical shifts, stretches, and reflections to a function. + Express the result of these transformations graphically and algebraically. +

\ No newline at end of file diff --git a/precalculus/source/02-FN/outcomes/05.ptx b/precalculus/source/02-FN/outcomes/05.ptx new file mode 100644 index 00000000..3b1e7e88 --- /dev/null +++ b/precalculus/source/02-FN/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+ Find the sum, difference, product, quotient, and composition of two or more functions and evaluate them. +

\ No newline at end of file diff --git a/precalculus/source/02-FN/outcomes/06.ptx b/precalculus/source/02-FN/outcomes/06.ptx new file mode 100644 index 00000000..7cf2a706 --- /dev/null +++ b/precalculus/source/02-FN/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+ Find the inverse of a one-to-one function. +

\ No newline at end of file diff --git a/precalculus/source/02-FN/outcomes/main.ptx b/precalculus/source/02-FN/outcomes/main.ptx new file mode 100644 index 00000000..45c3190f --- /dev/null +++ b/precalculus/source/02-FN/outcomes/main.ptx @@ -0,0 +1,31 @@ + +> + +

+ BIG IDEA for the chapter goes here, in outcomes/main.ptx +

+

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
+ \ No newline at end of file diff --git a/precalculus/source/02-FN/readiness.ptx b/precalculus/source/02-FN/readiness.ptx new file mode 100644 index 00000000..365ec9a5 --- /dev/null +++ b/precalculus/source/02-FN/readiness.ptx @@ -0,0 +1,103 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Use interval notation for solutions.

    +
      +
    • +

      Review:

      +
        +

      • Introduction to Interval Notation

        • +

      • Inequalities and Interval Notation

        • +
      +
      • +
    • +

      Practice:

      +
        +
      • +

        Linear Inequalities

        + + +
        • +
      +
      • +
    +
    2. + +
  2. + +

    Use substitution.

    +
      +
    • +

      Review:Substitution

      +
      • +
    • +

      Practice: Substitution

      +
      • +
    +
    3. +
  3. +

    Apply order of operations.

    +
      +
    • +

      Review:

      +

      • Introduction to order of operations

        • +

      • Order of operations

        • +
      +
      • + +
    • +

      Practice:

      +

      • Order of operations

        • +

      • Order of operations 2

        • + +
      + +
      • +
    +
    4. +
  4. +

    Create a table from a function.

    +
      +
    • +

      Review:

      +
        +

      • Create table

        • +
      +
      • +
    • +

      Practice:

      +

      • Graph points

        • +
      +
      • +
    +
    5. +
  5. +

    Combine like terms.

    +
      +
    • +

      Review:

      +

      • Combine like terms

        • +

      • Combine like terms 2

        • +
      +
      • +
    • +

      Practice:

      +

      • Combine like terms

        • +

      • Combine like terms 2

        • +
      +
      • +
    +
    6. + + +
+ + + diff --git a/precalculus/source/03-LF/01.ptx b/precalculus/source/03-LF/01.ptx new file mode 100644 index 00000000..18fd24ea --- /dev/null +++ b/precalculus/source/03-LF/01.ptx @@ -0,0 +1,766 @@ + + +
+ Slope and Average Rate of Change (LF1) + + + + + Activities + + + This section will explore ideas around average rate of change and slope. To help us get started, let's take a look at a context in which these ideas can be helpful. + + + + +

Robert came home one day after school to a very hot house! When he got home, the temperature on the thermostat indicated that it was 85 degrees! Robert decided that was too hot for him, so he turned on the air conditioner. The table of values below indicate the temperature of his house after turning on the air conditioner. + + + + Time (minutes) + Temperature (degrees Fahrenheit) + + + 0 + 85 + + + 1 + 84.3 + + + 2 + 83.6 + + + 3 + 82.9 + + + 4 + 82.2 + + + 5 + 81.5 + + + 6 + 80.8 + + +
+

+ + + +

How much did the temperature change from 0 to 2 minutes? +

    +
  1. The temperature decreased by 0.7 degrees
    2. +
  2. The temperature decreased by 1.4 degrees
    3. +
  3. The temperature increased by 0.7 degrees
    4. +
  4. The temperature increased by 1.4 degrees
    5. >

+ + +

+ B: The temperature decreased by 1.4 degrees +

+ + + + +

How much did the temperature change from 4 to 6 minutes? +

    +
  1. The temperature decreased by 0.7 degrees
    2. +
  2. The temperature decreased by 1.4 degrees
    3. +
  3. The temperature increased by 0.7 degrees
    4. +
  4. The temperature increased by 1.4 degrees
    5. >

+ + +

+ B: The temperature decreased by 1.4 degrees +

+ + + + +

+ If Robert wanted to know how much the temperature was decreasing each minute, how could he figure that out? +

+ + +

+ Choosing any two points that have a time difference of 1 minute should show that the temperature decreases by 0.7 degrees every minute. +

+ + + + +

+ How would you describe the overall behavior of the temperature of Robert's house? +

+ + +

+ Overall, the temperature of Robert's house is decreasing (the air conditioner appears to be working!). Notice that in parts a and b, the temperature decreased 1.4 degrees in 2 minutes. +

+ + + + + + Notice in that the temperature appears to be decreasing at a constant rate (i.e., the temperature decreased 1.4 degrees for every 2-minute interval). Upon further investigation, you might have also noticed that the temperature decreased by 0.7 degrees every minute. + + + + +

+ Refer back to the data Robert collected of the temperature of his house after turning on the air conditioner (). +

+ + + +

+ If this pattern continues, what will the temperature be after 8 minutes? +

    +
  1. 80.1
    2. +
  2. 78.7
    3. +
  3. 80.8
    4. +
  4. 79.4
+

+ + +

+ D: 79.4 +

+ + + + +

+ If this pattern continues, how long will it take for Robert's house to reach 78 degrees? +

    +
  1. 12 minutes
    2. +
  2. 9 minutes
    3. +
  3. 10 minutes
    4. +
  4. 11 minutes
+

+ + +

+ C: 10 minutes +

+ + + + + An average rate of change helps us to see and understand how a function is generally behaving. For example, in and , we began to see how the temperature of Robert’s house was decreasing every minute the air conditioner was on. In other words, when looking at average rate of change, we are comparing how one quantity is changing with respect to something else changing. + + + + +

+ An average rate of change of a function calculates the amount of change in one item divided by the corresponding amount of change in another. +

+

+ To calculate the average rate of change for any function f(x), we pick two points, a and b, and evaluate the function at those two points. We then find the difference between the y-values and x-values to calcuate the average rate of change: + +

+ + \frac{f(b)-f(a)}{b-a}. +

+ + + + + +

+ Use the table below to answer the questions. + + + + x + f(x) + + + -5 + 28 + + + -4 + 19 + + + -3 + 12 + + + -2 + 7 + + + -1 + 4 + + +
+

+ + + +

+ Applying , what is the average rate of change when x=-5 to x=-2? +

    +
  1. \frac{1}{7}
    2. +
  2. -3
    3. +
  3. -7
    4. +
  4. 7
+

+ + +

+ C: -7 +

+ + + + +

+ What is the average rate of change on the interval [-4,-1]? +

    +
  1. -5
    2. +
  2. -3
    3. +
  3. 5
    4. +
  4. 3
+

+ + +

+ A: -5 +

+ + + + +

+ Does this function have a constant average rate of change? +

+ + +

+ No. This would be a great opportunity for students to make observations of the average rate of change across different intervals. +

+ + + + + + +

+ Use the graph to calculate the average rate of change on the given intervals. +

+ + + f(x) = sqrt(4+x) + p=plot(f, (x, -5, 15), ymin=-15, ymax=15, color='blue', thickness=3, gridlines=[[-10..15],[-10..10]])+ point((-4,0),size=40,xmin=-10,xmax=15,ymin=-10,ymax=10)+ point((0,2),size=40,xmin=-10,xmax=15,ymin=-10,ymax=10)+point((12,4),size=40,xmin=-10,xmax=15,ymin=-10,ymax=10) + p + + + + + +

+ Applying , what is the average rate of change on the interval [-4,0]? +

    +
  1. -\frac{1}{2}
    2. +
  2. \frac{1}{2}
    3. +
  3. -2
    4. +
  4. 2
+

+ + +

+ B: \frac{1}{2} +

+ + + + +

+ What is the average rate of change on the interval [0,12]? +

    +
  1. \frac{1}{6}
    2. +
  2. -6
    3. +
  3. -\frac{1}{6}
    4. +
  4. 6
+

+ + +

+ A: \frac{1}{6} +

+ + + + + + + +

+ Just like with tables and graphs, you should be able to find the average rate of change when given a function. For this activity, use the function f(x)=-3x^2-1 to answer the following questions. +

+ + + +

+ Applying , what is the average rate of change on the interval [-2,3]? +

    +
  1. \frac{41}{5}
    2. +
  2. -3
    3. +
  3. -\frac{1}{3}
    4. +
  4. \frac{5}{41}
+

+ + +

+ B: -3 +

+ + + + + +

+ What is the average rate of change on the interval [0,4]? +

    +
  1. \frac{2}{25}
    2. +
  2. -50
    3. +
  3. -\frac{1}{12}
    4. +
  4. -12
+

+ + +

+ D: -12 +

+ + + + + + +

+ Use the given graph of the function, f(x)=3x-4, to investigate the average rate of change of a linear function. +

+ + + f(x) = 3*x-4 + p=plot(f, (x, -10, 10), ymin=-10, ymax=10, color='blue', thickness=3, gridlines=[[-10..10],[-10..10]]) + p + + + + + +

+ What is the average rate of change on the interval [-2,0]? +

    +
  1. \frac{1}{3}
    2. +
  2. 3
    3. +
  3. -\frac{1}{7}
    4. +
  4. -7
+

+ + +

+ B: 3 +

+ + + + +

+ What is the average rate of change on the interval [-1,5]? Notice that you cannot see the point at x=5. How could you use the equation of the line to determine the y-value when x=5? +

    +
  1. 3
    2. +
  2. \frac{1}{3}
    3. +
  3. \frac{2}{3}
    4. +
  4. -3
+

+ + +

+ A: 3 +

+ + + + +

+ Based on your observations in parts a and b, what do you think will be the average rate of change on the interval [5,25]? +

+ + + + + + Notice in , the average rate of change was the same regardless of which interval you were given. But in , the average rate of change was not the same across different intervals. + + + + +

+ The slope of a line has a constant that represents the direction and steepness of the line. For a linear function, the slope never changes - meaning it has a constant average rate of change. +

+ + + + + +

+ The steepness of a line depends on the vertical and horizontal distances between two points on the line. Use the graph below to compare the steepness, or slope, of the two lines. +

+ + + f(x) = (1/2)*x-1 + g(x) = 4*x-3 + p=plot(f, (x, -10, 10), ymin=-10, ymax=10, color='blue', thickness=3, gridlines=[[-10..10],[-10..10]]) + point((0,-1),size=40,xmin=-10,xmax=10,ymin=-10,ymax=10)+ point((4,1),size=40,xmin=-10,xmax=10,ymin=-10,ymax=10) + plot(g, (x, -10, 10), ymin=-10, ymax=10, color='red', thickness=3) + point((0,-3),size=40,xmin=-10,xmax=10,ymin=-10,ymax=10, color='red')+ point((2,5),size=40,xmin=-10,xmax=10,ymin=-10,ymax=10, color='red') + p + + + + + +

+ What is the vertical distance between the two points on the red line? +

    +
  1. 2
    2. +
  2. 4
    3. +
  3. 8
    4. +
  4. \frac{1}{2}
+

+ + +

+ C +

+ + + + +

+ What is the horizontal distance between the two points on the red line? +

    +
  1. 2
    2. +
  2. 4
    3. +
  3. 8
    4. +
  4. \frac{1}{2}
+

+ + +

+ A +

+ + + + +

+ Using information from parts a and b, what value could we use to describe the steepness of the red line? +

    +
  1. 2
    2. +
  2. 4
    3. +
  3. 8
    4. +
  4. \frac{1}{2}
+

+ + +

+ B. This could be a good place to discuss where 4 comes from (change of y over the change in x). +

+ + + + +

+ What is the vertical distance between the two points on the blue line? +

    +
  1. 2
    2. +
  2. 4
    3. +
  3. 8
    4. +
  4. \frac{1}{2}
+

+ + +

+ A +

+ + + + +

+ What is the horizontal distance between the two points on the blue line? +

    +
  1. 2
    2. +
  2. 4
    3. +
  3. 8
    4. +
  4. \frac{1}{2}
+

+ + +

+ B +

+ + + + +

+ Using information from parts d and e, what value could we use to describe the steepness of the blue line? +

    +
  1. 2
    2. +
  2. 4
    3. +
  3. 8
    4. +
  4. \frac{1}{2}
+

+ + +

+ D. This could be a good place to discuss where \frac{1}{2} comes from (change of y over the change in x). +

+ + + + +

+ Which line is the steepest? +

+ + +

+ The red line. The slope of the red line is 4 and the slope of the blue line is \frac{1}{2}. Because 4 is larger than \frac{1}{2}, the red line is steeper. +

+ + + + + + The steepness, or slope, of a line can be found by the change in y (the vertical distance between two points on the line) divided by the change in x (the horizontal distance between two points on the line). Slope can be calculated as "rise over run." +

+ Slope is a way to describe the steepness of a line. The red line in has a larger value for it's slope than the blue line. Thus, the red line is steeper than the blue line. +

+ + + + +

+ Now that we know how to find the slope (or steepness) of a line, let's look at other properties of slope. Use the graph below to answer the following questions. +

+ + + f(x) = (1/2)*x-1 + g(x) = -(1/2)*x-1 + p=plot(f, (x, -10, 10), ymin=-10, ymax=10, color='blue', thickness=3, gridlines=[[-10..10],[-10..10]]) + plot(g, (x, -10, 10), ymin=-10, ymax=10, color='green', thickness=3, gridlines=[[-10..10],[-10..10]]) + p + + + + + +

+ What is the slope of the blue line? +

    +
  1. \frac{1}{2}
    2. +
  2. 2
    3. +
  3. -\frac{1}{2}
    4. +
  4. -2
+

+ + +

+ A +

+ + + + +

+ What is the slope of the green line? +

    +
  1. \frac{1}{2}
    2. +
  2. 2
    3. +
  3. -\frac{1}{2}
    4. +
  4. -2
+

+ + +

+ C +

+ + + + +

+ How are the slopes of the lines similar? +

+ + +

+ Students will probably notice that both slopes have the same steepness (\frac{1}{2}). +

+ + + + +

+ How are the slopes of the lines different? +

+ + +

+ Although both slopes have the same steepness (\frac{1}{2}), one line has a positive slope (blue line) and the other line has a negative slope (green line). +

+ + + + + + Notice in that the slope does not just indicate how steep a line is, but also it's direction. A negative slope indicates that the line is decreasing (from left to right) and a positive slope indicates that the line is increasing (from left to right). + + + + +

+ Suppose (-3,7) and (7,2) are two points on a line. +

+ + + +

+ Plot these points on a graph and find the slope by using "rise over run." +

    +
  1. \frac{1}{2}
    2. +
  2. 2
    3. +
  3. -\frac{1}{2}
    4. +
  4. -2
+

+ + +

+ C +

+ + + + +

+ Now calculate the slope by using the change in y over the change in x. +

    +
  1. \frac{1}{2}
    2. +
  2. 2
    3. +
  3. -\frac{1}{2}
    4. +
  4. -2
+

+ + +

+ C +

+ + + + +

+ What do you notice about the slopes you got in parts a and b? +

+ + +

+ Students should notice that they get the same answer. Both methods can be used to find the slope. +

+ + + + + + We can calculate slope (m) by finding the change in y and dividing by the change in x. Mathematically, this means that when given (x_{1},y_{1}) and (x_{2}, y_{2}), m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}. + + + + +

+ Calculate the slope of each representation of a line using the slope formula. +

+ + + +

+ + + x + f(x) + + + -2 + -7 + + + -1 + -4 + + + 0 + -1 + + + 1 + 2 + + + 2 + 5 + + +

+ + + + + + + f(x) = 0*x+5 + p=plot(f, (x, -10, 10), ymin=-10, ymax=10, color='blue', thickness=3, gridlines=[[-10..10],[-10..10]]) + p + + + + + + +

+ (-4,7) and (-4,1) +

+ + + + + + In , there were slopes that were 0 and undefined. When a line is vertical, the slope is undefined. This means that there is only a vertical distance between two points and there is no horizontal distance. When a line is horizontal, the slope is 0. This means that the line never rises vertically, giving a vertical distance of zero. + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/03-LF/02.ptx b/precalculus/source/03-LF/02.ptx new file mode 100644 index 00000000..f95786ba --- /dev/null +++ b/precalculus/source/03-LF/02.ptx @@ -0,0 +1,660 @@ + + +
+ Equations of Lines (LF2) + + + + + Activities + + + +

+ Consider the graph of two lines. + + + p=plot(2*x+3,(x,-5,5),thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5, color='blue', gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=plot(2*x-2,(x,-5,5),thickness=3, color='red') + p+=text('line A', (-1, 4), fontsize=18, color='black') + p+=text('line B', (1, -2), fontsize=18, color='black') + p + + +

+ + + + +

Find the slope of line A. +

    +

  1. 1

    2. +

  2. 2

    3. +

  3. \frac{1}{2}

    4. +

  4. -2

    5. +

+ + + +

+ Choice B: slope is 2 +

+ + + + + +

Find the slope of line B. +

    +

  1. 1

    2. +

  2. 2

    3. +

  3. \frac{1}{2}

    4. +

  4. -2

    5. +

+ + + +

+ Choice B: slope is 2 +

+ + + + + + +

Find the y-intercept of line A. +

    +

  1. -2

    2. +

  2. -1.5

    3. +

  3. 1

    4. +

  4. 3

    5. +

+ + + +

+ Choice D: y-intercept is 3 +

+ + + + + +

Find the y-intercept of line B. +

    +

  1. -2

    2. +

  2. -1.5

    3. +

  3. 1

    4. +

  4. 3

    5. +

+ + + +

+ Choice A: y-intercept is -2 +

+ + + + +

+ What is the same about the two lines? +

+ + + + +

+ What is different about the two lines? +

+ + + + + + +

+ Notice that in the lines have the same slope but different y-intercepts. It is not enough to just know one piece of information to determine a line, you need both a slope and a point. +

+ + + + +

+ Linear functions can be written in slope-intercept form + f(x)=mx+b + where b is the y-intercept (or starting value) and m is the slope (or constant rate of change). +

+ + + + + +

+ Write the equation of each line in slope-intercept form. +

+ + + +

+ + + p=plot(-1/3*x+1,(x,-5,5),thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5, color='blue', gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p + + +

+

+

    +

  1. y= -3x+1

    2. +

  2. y= -x+3

    3. +

  3. y= -\frac{1}{3}x+1

    4. +

  4. y= -\frac{1}{3}x+3

    5. +

+ + + +

+ Choice C: y=-\frac{1}{3}x+1 (A has wrong slope, B has wrong everything, D has wrong y-intercept) +

+ + + + + +

+ The slope is 4 and the y-intercept is (0,-3). +

+

+

    +

  1. f(x)= 4x-3

    2. +

  2. f(x)= 3x-4

    3. +

  3. f(x)= -4x+3

    4. +

  4. f(x)= 4x+3

    5. +

+ + + +

+ Choice A: y=4x-3 (B has wrong slope, C has wrong everything, D has wrong y-intercept) +

+ + + + + +

+ Two points on the line are (0,1) and (2,4). +

+

+

    +

  1. y= 2x+1

    2. +

  2. y= -\frac{3}{2}x+4

    3. +

  3. y= \frac{3}{2}x+1

    4. +

  4. y= \frac{3}{2}x+4

    5. +

+ + + +

+ Choice C: y=\frac{3}{2}x+1 +

+ + + + + +

+ + + + x + f(x) + + + -2 + -8 + + + 0 + -2 + + + 1 + 1 + + + 4 + 10 + + + + +

+

+

    +

  1. f(x)= -3x-2

    2. +

  2. f(x)= -\frac{1}{3}x-2

    3. +

  3. f(x)= 3x+1

    4. +

  4. f(x)= 3x-2

    5. +

+ + + +

+ Choice D: f(x)=3x-2 +

+ + + + + + +

+ Let's try to write the equation of a line given two points that don't include the y-intercept. +

+ + + +

+ Plot the points (2,1) and (-3,4). +

+ + + + +

+ Find the slope of the line joining the points. +

+

+

    +

  1. -\frac{5}{3}

    2. +

  2. -\frac{3}{5}

    3. +

  3. \frac{3}{5}

    4. +

  4. -3

    5. +

+ + +

+ Choice B: \frac{4-1}{-3-2} +

+ + + + +

+ When you draw a line connecting the two points, it's often hard to draw an accurate enough graph to determine the y-intercept of the line exactly. Let's use the slope-intercept form and one of the given points to solve for the y-intercept. Try using the slope and one of the points on the line to solve the equation y=mx+b for b. +

+

+

    +

  1. 2

    2. +

  2. \frac{11}{5}

    3. +

  3. \frac{5}{2}

    4. +

  4. 3

    5. +

+ + +

+ Choice B: +

+ + + + +

+ Write the equation of the line in slope-intercept form. +

+ + + + + It would be nice if there was another form of the equation of a line that works for any points and does not require the y-intercept. + + + +

+ Linear functions can be written in point-slope form + y-y_0=m(x-x_0) + where (x_0, y_0) is any point on the line and m is the slope. +

+ + + + + +

+ Write an equation of each line in point-slope form. +

+ + + +

+ + + p=plot(1/3*x+2/3,(x,-5,5),thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5, color='blue', gridlines=[[-4,-3,-2,-1,1,2,3,4],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((1,1),pointsize=50,color='blue') + p+=point((-2,0),pointsize=50,color='blue') + p + + +

+

+

    +

  1. y= \frac{1}{3}x+\frac{2}{3}

    2. +

  2. y-1= 3(x-1)

    3. +

  3. y-1= \frac{1}{3}(x-1)

    4. +

  4. y+2= \frac{1}{3}(x+2)

    5. +

  5. y= \frac{1}{3}(x+2)

    6. +

+ + + +

+ Choice C or E +

+ + + + + +

+ The slope is 4 and (-1,-7) is a point on the line. +

+

+

    +

  1. y+7= 4(x+1)

    2. +

  2. y-7= 4(x-1)

    3. +

  3. y+1= 4(x+7)

    4. +

  4. y-4= 7(x-1)

    5. +

+ + + +

+ Choice A +

+ + + + + +

+ Two points on the line are (1,0) and (2,-4). +

+

+

    +

  1. y= -4x+1

    2. +

  2. y-0=-2(x-1)

    3. +

  3. y+4=-4(x-2)

    4. +

  4. y+4=-3(x-2)

    5. +

+ + + +

+ Choice C: +

+ + + + + +

+ + + + x + f(x) + + + -2 + -8 + + + 1 + 1 + + + 4 + 10 + + + + +

+

+

    +

  1. y+8= 3(x-2)

    2. +

  2. y-1= -\frac{1}{3}(x-1)

    3. +

  3. y+8= -\frac{1}{3}(x+2)

    4. +

  4. y-10= 3(x-4)

    5. +

+ + + +

+ Choice D +

+ + + + + + + +

+ Consider again the two points from , (2,1) and (-3,4). +

+ + + + + +

+ Use point-slope form to find an equation of the line. +

+

+

    +

  1. y=-\frac{3}{5}x+\frac{11}{5}

    2. +

  2. y-1=-\frac{3}{5}(x-2)

    3. +

  3. y-4 =-\frac{3}{5}(x+3)

    4. +

  4. y-2=-\frac{3}{5}(x-1)

    5. +

+ + +

+ Choice B or C +

+ + + + +

+ Solve the point-slope form of the equation for y to rewrite the equation in slope-intercept form. Identify the slope and intercept of the line. +

+ + + + + Notice that it was possible to use either point to find an equation of the line in point-slope form. But, when rewritten in slope-intercept form the equation is unique. + + + + +

+ For each of the following lines, determine which form (point-slope or slope-intercept) would be "easier" and why. Then, write the equation of each line. +

+ + + +

+ + + p=plot(3/4*x+2,(x,-5,5),thickness=3, xmin=-5,xmax=5,ymin=-6,ymax=6, color='blue', gridlines=[[-6,-5,-4,-3,-2,-1,1,2,3,4,5,6],[-4,-3,-2,-1,1,2,3,4]]) + p+=point((-4,-1),pointsize=50,color='blue') + p+=point((4,5),pointsize=50,color='blue') + p + + +

+ + + +

+ Slope-intercept: y=\frac{3}{4}x+2 +

+ + + + + +

+ The slope is -\frac{1}{2} and (1,-3) is a point on the line. +

+ + + +

+ Point-slope: y+3=-\frac{1}{2}(x-1) +

+ + + + + +

+ Two points on the line are (0,3) and (2,0). +

+ + + +

+ Slope-intercept: y=-\frac{3}{2}x+3 +

+ + + + + + It is always possible to use both forms to write the equation of a line and they are both valid. Although, sometimes the given information lends itself to make one form easier. + + + + +

+ Write the equation of each line. +

+ + + +

+ The slope is 0 and (-1,-7) is a point on the line. +

+

+

    +

  1. y=-7

    2. +

  2. y=7x

    3. +

  3. y=-x

    4. +

  4. x=-1

    5. +

+ + + +

+ Choice A +

+ + + + +

+ Two points on the line are (3,0) and (3,5). +

+

+

    +

  1. y=3x+3

    2. +

  2. y=3x+5

    3. +

  3. x=3

    4. +

  4. y=3

    5. +

+ + + +

+ Choice C +

+ + + + + +

+ + + p=plot(-2,(x,-5,5),thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5, gridlines=[[-6,-5,-4,-3,-2,-1,1,2,3,4,5,6],[-4,-3,-2,-1,1,2,3,4]],color='blue') + p + + +

+

+

    +

  1. x=-2

    2. +

  2. y-2=x

    3. +

  3. y=-2x-2

    4. +

  4. y=-2

    5. +

+ + + +

+ Choice D +

+ + + + + + + +

+ A horizontal line has a slope of zero and has the form y=k where k is a constant. A vertical line has an undefined slope and has the form x=h where h is a constant. +

+ + + + +

+ The equation of a line can also be written in standard form. Standard form looks like Ax+By=C. +

+ + + + It is possible to rearrange a line written in standard form to slope-intercept form, by solving for y. + + + +

+ Given a line in standard form + 5x+4y=2. + Find the slope and y-intercept of the line. +

+ + +

+ Slope: -\frac{5}{4} y-intercept: \frac{2}{5} +

+ + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/03-LF/03.ptx b/precalculus/source/03-LF/03.ptx new file mode 100644 index 00000000..9dece5ec --- /dev/null +++ b/precalculus/source/03-LF/03.ptx @@ -0,0 +1,750 @@ + + +
+ Graphs of Linear Equations (LF3) + + + + + Activities + + + +

+

+ + + + +

+ Draw a line that goes through the point (1,4). +

+ + + p=point((1,4),pointsize=50,color='blue',ymin=-7, ymax=7, xmin=-5,xmax=5) + + + + + Many answers possible. + + + + + +

+ Was this the only possible line that goes through the point (1,4)? + +

    +

  1. Yes. The line is unique.

    2. +

  2. No. There is exactly one more line possible.

    3. +

  3. No. There are a lot of lines that go through (1,4).

    4. +

  4. No. There are an infinite number of lines that go through (1,4).

    5. +
+

+ + +

D

+

(Though C is also true. D is just more descriptive.)

+ + + + + +

+ Now draw a line that goes through the points (1,4) and (-3,-2). +

+ + + p=point([(1,4),(-3,-2)],pointsize=50,color='blue',ymin=-7, ymax=7, xmin=-5,xmax=5) + + + + + Just the one possible line. + + + + + +

+ Was this the only possible line that goes through the points (1,4) and (-3,-2)? + +

    +

  1. Yes. The line is unique.

    2. +

  2. No. There is exactly one more line possible.

    3. +

  3. No. There are a lot of lines that go through (1,4) and (-3,-2).

    4. +

  4. No. There are an infinite number of lines that go through (1,4) and (-3,-2).

    5. +
+

+ + +

A

+

(Discussion could include that we would have to make the line curve to connect the points in more than one way. But, a line has to have the same slope everywhere.)

+ + + + + +

+ If you are given two points, then you can always graph the line containing them by plotting them and connecting them with a line. +

+ + + + +

+

+ + + + +

+ Graph the line containing the points (-7,1) and (6,-2). +

+ + +

+ + + f(x) = (-3/13*(x+7)+1) + p=point({(-7,1),(6,-2)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+ + + + + +

+ Graph the line containing the points (-3,0) and (0,8). +

+ + +

+ + + f(x) = ((8/3)*x+8) + p=point({(-3,0),(0,8)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+ + + + + +

+ Graph the line given by the table below. + + + + x + y + + + -3 + -12 + + + -2 + -9 + + + -1 + -6 + + + 0 + -3 + + + 1 + 0 + + + 2 + 3 + + + + + +

+ + +

+ + + f(x) = 3*x-3 + p=point({(-3,-12),(-2,-9),(-1,-6),(0,-3),(1,0),(2,3)},pointsize=50,color='blue') + p+=plot(f, (x, -12, 12), ymin=-12, ymax=12, color='blue', thickness=3, gridlines=[[-12,-11,..,12],[-12,-11,..,12]]) + p + + +

+ + + + + +

+ Let's say you are given a table that listed six points that are on the same line. How many of those points are necessary to use to graph the line? + +

    +

  1. One point is enough.

    2. +

  2. Two points are enough.

    3. +

  3. Three points are enough.

    4. +

  4. You need to plot all six points.

    5. +

  5. You can use however many you want.

    6. + +
+

+ + +

B

+

Discussion could include pointing out that only two are necessary, but we can include more if we want. Also, including more is a good way to catch an error in plotting. One will stick out!

+ + + + + + + + + +

+ In , we were given at least two points in each question. However, sometimes we are not directly given two points to graph a line. Instead we are given some combination of characteristics about the line that will help us find two points. These characteristics could include a point, the intercepts, the slope, or an equation. +

+ + + + +

A line has a slope of -\frac{1}{3} and its y-intercept is 4. +

+ + + + + +

+ We were given the y-intercept. What point does that correspond to? + +

    +
  1. (4,0)
    2. +
  2. (0,4)
    3. +
  3. \left(4,-\frac{1}{3}\right)
    4. +
  4. \left(-\frac{1}{3},4\right)
    5. +
+

+ + +

B

+ + + + + +

+ After we plot the y-intercept, how can we use the slope to find another point? + +

    + +
  1. Start at the y-intercept, then move up one space and to the left three spaces to find another point.
    2. +
  2. Start at the y-intercept, then move up one space and to the right three spaces to find another point.
    3. +
  3. Start at the y-intercept, then move down one space and to the left three spaces to find another point.
    4. +
  4. Start at the y-intercept, then move down one space and to the right three spaces to find another point.
    5. + +
+

+ + +

A and D

+ + + + +

+ Graph the line that has a slope of -\frac{1}{3} and its y-intercept is 4. +

+ + +

+ + + f(x) = ((-1/3)*x+4) + p=point({(0,4)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+ + + + + + + + +

A line is given by the equation y=-2x+5. +

+ + + + + +

+ What form is the equation given in? + +

    +
  1. Standard form
    2. +
  2. Point-slope form
    3. +
  3. Slope-intercept form
    4. +
  4. The form it is in doesn't have a name.
    5. +
+

+ + +

C

+ + + + + +

+ The form gives us one point right away: the y-intercept. Which of the following is the y-intercept? +

    + +
  1. (-2,0)
    2. +
  2. (0,-2)
    3. +
  3. (5,0)
    4. +
  4. (0,5)
    5. + +
+

+ + +

D

+ + + + + After we plot the y-intercept, we can use the slope to find another point. Find another point and graph the resulting line. + + +

+ + + f(x) = ((-2)*x+5) + p=point({(0,5)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+ + + + + + + + + +

A line contains the point (-3,-2) and has slope \frac{1}{5} . Which of the following is the graph of that line? +

+
    +
  1. + + f(x) = (-.2*(x+3)-2) + p=point({(-3,-2),(2,-3)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +
    2. +
  2. + + f(x) = (.2*(x+3)-2) + p=point({(-3,-2),(2,-1)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +
    3. +
  3. + + f(x) = (5*(x+3)-2) + p=point({(-3,-2),(-2,3)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +
    4. +
  4. + + f(x) = (-5*(x+3)-2) + p=point({(-3,-2),(-2,-7)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +
    5. +
+ + +

B

+ + + + + + + + + + + +

A line is given by the equation y-6=-4(x+2). +

+ + + + + +

+ What form is the equation given in? + +

    +
  1. Standard form
    2. +
  2. Point-slope form
    3. +
  3. Slope-intercept form
    4. +
  4. The form it is in doesn't have a name.
    5. +
+

+ + +

B

+ + + + + +

+ The form gives us one point right away. Which of the following is a point on the line? +

    + +
  1. (-2,-6)
    2. +
  2. (-2,6)
    3. +
  3. (2,-6)
    4. +
  4. (2,6)
    5. + +
+

+ + +

B

+ + + + + + After we plot this point, we can use the slope to find another point. Find another point and graph the resulting line. + + +

+ + + f(x) = ((-4)*(x+2)+6) + p=point({(-2,6)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+ + + + + + +

Recall from that the equation of a horizontal line has the form y=k where k is a constant and a vertical line has the form x=h where h is a constant. +

+ + + + +

+ Which type of line has a slope of zero? +

    +

  1. Horizontal

    2. +

  2. Vertical

    3. + +
+

+ + +

+ A +

+ + + + +

+ Which type of line has an undefined slope? +

    +

  1. Horizontal

    2. +

  2. Vertical

    3. + +
+

+ + +

+ B +

+ + + + + +

+ Graph the vertical line that goes through the point (4,-2). +

+ + +

+ + f(x) = (7/2*(x)+7) + p=point({(4,-2)},pointsize=50,color='blue') + p+=line([(4,-8),(4,8)], xmin=-8, xmax=8, ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + +

+ What is the equation of the vertical line through the point (4,-2)? +

    +
  1. x=4
    2. +
  2. y=4
    3. +
  3. x=-2
    4. +
  4. y=-2
    5. +
+

+ + +

+ A +

+ + + + +

+ Graph the horizontal line that goes through the point (4,-2). +

+ + +

+ + f(x) = -2 + p=point({(4,-2)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + +

+ What is the equation of the horizontal line through the point (4,-2)? +

    +
  1. x=4
    2. +
  2. y=4
    3. +
  3. x=-2
    4. +
  4. y=-2
    5. +
+

+ + +

+ D +

+ + + + + + +

Graph each line described below. +

+ + + +

+ The line containing the points (-3,4) and (5,-2). +

+ + +

+ + f(x) = (-3/4*(x+3)+4) + p=point({(-3,4),(5,-2)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + +

+ The line whose x-intercept is -2 and whose y-intercept is 7. +

+ + +

+ + f(x) = (7/2*(x)+7) + p=point({(-2,0),(0,7)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + +

+ The line whose slope is \frac{2}{5} that goes through the point (4,6). +

+ + +

+ + f(x) = (2/5*(x-4)+6) + p=point({(4,6)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + +

+ The line whose slope is -\frac{1}{3} and whose y-intercept is -4. +

+ + +

+ + f(x) = (-1/3*(x)-4) + p=point({(0,-4)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + +

+ The vertical line through the point (-2,-7). +

+ + +

+ + p=point({(-2,-7)},pointsize=50,color='blue') + p+=line([(-2,-8),(-2,8)], xmin=-8,xmax=8,ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + +

+ The horizontal line through the point (-6,3). +

+ + +

+ + f(x) = 3 + p=point({(-6,3)},pointsize=50,color='blue') + p+=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + +

+ The line with equation y=-\frac{5}{3}x-6. +

+ + +

+ + f(x) = (-5/3*x-6) + p=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + +

+ The line with equation y-5=\frac{7}{2}(x-2). +

+ + +

+ + f(x) = (7/2*(x-2)+5) + p=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + +

+ The line with equation 3x-6y=8. +

+ + +

+ + f(x) = (1/2*x-(4/3)) + p=plot(f, (x, -8, 8), ymin=-8, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + +

+ + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/03-LF/04.ptx b/precalculus/source/03-LF/04.ptx new file mode 100644 index 00000000..6aa53758 --- /dev/null +++ b/precalculus/source/03-LF/04.ptx @@ -0,0 +1,549 @@ + + +
+ Parallel and Perpendicular Lines (LF4) + + + + + Activities + + + +

+ Let's revisit to investigate special types of lines. +

+ + + p=plot(2*x+3,(x,-5,5),thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5, color='blue', gridlines=[[-4..4],[-4..4]]) + p+=plot(2*x-2,(x,-5,5),thickness=3, color='red') + p+=text('Line A', (-1, 3), fontsize=18, color='black') + p+=text('Line B', (1, -2), fontsize=18, color='black') + p + + + + + +

What is the slope of line A? +

    +

  1. 1

    2. +

  2. 2

    3. +

  3. \frac{1}{2}

    4. +

  4. -2

    5. +

+ + +

+ B +

+ + + + +

What is the slope of line B? +

    +

  1. 1

    2. +

  2. 2

    3. +

  3. \frac{1}{2}

    4. +

  4. -2

    5. +

+ + +

+ B +

+ + + + +

What is the y-intercept of line A? +

    +

  1. -2

    2. +

  2. -1.5

    3. +

  3. 1

    4. +

  4. 3

    5. +

+ + +

+ D +

+ + + + +

What is the y-intercept of line B? +

    +

  1. -2

    2. +

  2. -1.5

    3. +

  3. 1

    4. +

  4. 3

    5. +

+ + +

+ A +

+ + + + +

+ What is the same about the two lines? +

+ + +

+ Both lines have the same slope (m=2). +

+ + + + +

+ What is different about the two lines? +

+ + +

+ The lines have different y-intercepts. +

+ + + + + + Notice that in the two lines never touch. + + + + +

+ Parallel lines are lines that always have the same distance apart (equidistant) and will never meet. Parallel lines have the same slope, but different y-intercepts. +

+ + + + + +

+ Suppose you have the function, f(x)=-\frac{1}{2}x-1 +

+ + + +

+ What is the slope of f(x)? +

    +
  1. -1
    2. +
  2. 2
    3. +
  3. 1
    4. +
  4. -\frac{1}{2}
+

+ + +

+ D +

+ + + + +

+ Applying , what would the slope of a line parallel to f(x) be? +

    +
  1. -1
    2. +
  2. 2
    3. +
  3. 1
    4. +
  4. -\frac{1}{2}
+

+ + +

+ D +

+ + + + +

+ Find the equation of a line parallel to f(x) that passes through the point (-4,2). +

+ + +

+ y-2=-\frac{1}{2}(x+4) or

y=-\frac{1}{2}x

+

+ + + + + + +

+ Consider the graph of the two lines below. +

+ + + p=plot(2*x+3,(x,-5,5),thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5, color='blue', gridlines=[[-4..4],[-4..4]],aspect_ratio=1) + p+=plot(-1/2*x+1,(x,-5,5),thickness=3, color='red') + p+=text('Line A', (-2.5, 3), fontsize=18, color='red') + p+=text('Line B', (-1.8, -3), fontsize=18, color='blue') + p + + + + + +

What is the slope of line A? +

    +

  1. 3

    2. +

  2. 2

    3. +

  3. -\frac{1}{2}

    4. +

  4. -2

    5. +

+ + +

+ C +

+ + + + +

What is the slope of line B? +

    +

  1. 3

    2. +

  2. 2

    3. +

  3. -\frac{1}{2}

    4. +

  4. -2

    5. +

+ + +

+ B +

+ + + + +

What is the y-intercept of line A? +

    +

  1. -2

    2. +

  2. -\frac{1}{2}

    3. +

  3. 2

    4. +

  4. 3

    5. +

+ + +

+ D +

+ + + + +

What is the y-intercept of line B? +

    +

  1. -2

    2. +

  2. -\frac{1}{2}

    3. +

  3. 1

    4. +

  4. 3

    5. +

+ + +

+ C +

+ + + + +

+ If you were to think of slope as "rise over run," how would you write the slope of each line? +

+ + +

+ Line A could be written as -\frac{1}{2} and Line B could be written as \frac{2}{1}. +

+ + + + +

+ How would you compare the slopes of the two lines? +

+ + +

+ Students might notice that when writing the slopes of Line A and Line B, the slopes are negative reciprocals of each other. +

+ + + + + + Notice in , that even though the two lines have different slopes, the slopes are somewhat similar. For example, if you take the slope of Line A \left(-\frac{1}{2}\right) and flip and negate it, you will get the slope of Line B \left(\frac{2}{1}\right). + + + + +

+ Perpendicular lines are two lines that meet or intersect each other at a right angle. The slopes of two perpendicular lines are negative reciprocals of each other (given that the slope exists!). +

+ + + + + +

+ Suppose you have the function, f(x)=3x+5 +

+ + + +

+ What is the slope of f(x)? +

    +
  1. -\frac{1}{3}
    2. +
  2. 3
    3. +
  3. 5
    4. +
  4. -\frac{1}{5}
+

+ + +

+ B +

+ + + + +

+ Applying , what would the slope of a line perpendicular to f(x) be? +

    +
  1. -\frac{1}{3}
    2. +
  2. 3
    3. +
  3. 5
    4. +
  4. -\frac{1}{5}
+

+ + +

+ A +

+ + + + +

+ Find the equation of a line perpendicular to f(x) that passes through the point (3,6). +

+ + +

+ y-6=-\frac{1}{3}(x-3) or

y=-\frac{1}{3}x+7

+

+ + + + + + + +

+ For each pair of lines, determine if they are parallel, perpendicular, or neither. +

+ + + +

+ f(x)=-3x+4 + g(x)=5-3x +

+ + +

+ Parallel. The slope of f(x) is -3 and the slope of g(x) is -3. +

+ + + + +

+ f(x)=2x-5 + g(x)=6x-5 +

+ + +

+ Neither. The slope of f(x) is 2 and the slope of g(x) is 6. These lines do, however, have the same y-intercept. +

+ + + + +

+ f(x)=6x-5 + g(x)=\frac{1}{6}x+8 +

+ + +

+ Neither. The slope of f(x) is 6 and the slope of g(x) is \frac{1}{6}. Although they are reciprocals of one another, they are not negative reciprocals. +

+ + + + +

+ f(x)=\frac{4}{5}x+3 + g(x)=-\frac{5}{4}x-1 +

+ + +

+ Perpendicular. The slope of f(x) is \frac{4}{5} and the slope of g(x) is -\frac{5}{4} (and are negative reciprocals of one another). +

+ + + + + + +

+ Consider the linear equation, f(x)=-\frac{2}{3}x-4 and the point A: (-6,4). +

+ + + +

+ Find the equation of a line that is parallel to f(x) and passes through the point A. +

+ + +

+ y-4=-\frac{2}{3}(x+6) or

+ y=-\frac{2}{3}x +

+

+ + + + +

+ Find the equation of a line that is perpendicular to f(x) and passes through the point A. +

+ + +

+ y-4=\frac{3}{2}(x+6) or

+ y=\frac{3}{2}x+13 +

+

+ + + + + + +

+ Consider the line, y=2, as shown in the graph below. +

+ + + p=plot(2,(x,-5,5),thickness=3, xmin=-5,xmax=5,ymin=-5,ymax=5, color='blue', gridlines=[[-4..4],[-4..4]]) + p + + + + + +

+ What is the slope of the line y=2? +

    +
  1. undefined
    2. +
  2. 0
    3. +
  3. 1
    4. +
  4. -\frac{1}{2}
+

+ + +

+ B +

+ + + + +

+ What is the slope of a line that is parallel to y=2? +

    +
  1. undefined
    2. +
  2. 0
    3. +
  3. 1
    4. +
  4. -\frac{1}{2}
+

+ + +

+ B +

+ + + + +

+ Find the equation of a line that is parallel to y=2 and passes through the point (-1,-4). +

+ + +

+ y=-4. Students might need the graph to help them visualize why the equation is in the form y= number. +

+ + + + +

+ What is the slope of a line that is perpendicular to y=2? +

    +
  1. undefined
    2. +
  2. 0
    3. +
  3. 1
    4. +
  4. -\frac{1}{2}
+

+ + +

+ A. You might need to help students see why the slope is undefined by showing that -\frac{1}{0} is not defined. +

+ + + + +

+ Find the equation of a line that is perpendicular to y=2 and passes through the point (-1,2). +

+ + +

+ x=-1. Students might need the graph to help them visualize why the equation is in the form x= number. +

+ + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
diff --git a/precalculus/source/03-LF/05.ptx b/precalculus/source/03-LF/05.ptx new file mode 100644 index 00000000..65475f33 --- /dev/null +++ b/precalculus/source/03-LF/05.ptx @@ -0,0 +1,482 @@ + + +
+ Linear Models and Meanings (LF5) + + + + + Activities + + +

+ We begin by revisiting from . +

+ + + + +

+ Ellie has \$13 in her piggy bank, and she gets an additional \$1.50 each week for her allowance. Assuming she does not spend any money, the total amount of allowance, A(w), she has after w weeks can be modeled by the function A(w)=13+1.50w. +

+ + + +

How much money will be in her piggy bank after 5 weeks?

+ + +

+ \$20.50 +

+ + + + + +

After how many weeks will she have \$40 in her piggy bank?

+ + +

+ 18 weeks +

+ + + + + + + + The function in the previous activity is an example of a linear model. A linear model is a linear function that describes, or models, a real-life application. In this section we will build and use linear models. + + + + +

+ Jack bought a package containing 40 cookies. Each day he takes two in his lunch to work. + +

+ + + +

How many cookies are left in the package after 3 days? +

    +
  1. 46
    2. +
  2. 42
    3. +
  3. 38
    4. +
  4. 36
    5. +
  5. 34
    6. +
+

+ + +

+ E +

+ + + + +

+ Fill out the following table to represent the number of cookies left in the package after the given number of days. +

+ + + number of days + number of cookies left + + + 0 + + + 2 + + + 5 + + + 10 + + + 20 + + + + +

+ + + number of days + number of cookies left + + + 0 + 40 + + + 2 + 38 + + + 5 + 30 + + + 10 + 20 + + + 20 + 0 + + +

+ + + + +

+ What is the y-intercept? Explain what it represents in the context of the problem. +

+ + +

+ 40; It represents the initial amount of cookies in the package. +

+ + + + +

+ What is the rate of change? Explain what it represents in the context of the problem. +

+ + +

+ -2; It represents how the number of cookies in the package changes per day. +

+ + + + +

+ Write a linear function to model the situation. Let d represent the number of days elapsed and C(d) represent the number of cookies in the package. (Hint: Use the previous two questions to help!) +

+ + +

+ C(d)=40-2d +

+ + + + +

+ Find C(6). Explain what that means in the context of the problem. +

+ + +

+ 28; After 6 days, there will be 28 cookies left in the package. +

+ + + + +

+ How many days will it take to empty the package? What does this correspond to on the graph? +

+ + +

+ 20; This corresponds to the x-intercept. +

+ + + + + + +

+ Daisy's Doughnut Shop sells delicious doughnuts. Each month, they incur a fixed cost of \$2000 for rent, insurance, and other expenses. Then, for each doughnut they produce, it costs them an additional \$0.25. + +

+ + + +

In January, Daisy's Doughnut Shop produced 1000 doughnuts. What was their total monthy cost to run the shop? +

    +
  1. \$2000.25
    2. +
  2. \$2002.50
    3. +
  3. \$2025.00
    4. +
  4. \$2250.00
    5. +
  5. \$4500.00
    6. +
+

+ + +

+ D +

+ + + + +

+ Fill out the following table to represent the cost for producing various amounts of doughnuts. +

+ + + number of doughnuts + cost + + + 0 + + + 500 + + + 1000 + + + 1500 + + + 2000 + + + + + + + number of doughnuts + cost + + + 0 + \$2000 + + + 500 + \$2125 + + + 1000 + \$2250 + + + 1500 + \$2375 + + + 2000 + \$2500 + + + + + + +

+ What is the y-intercept? Explain what it represents in the context of the problem. +

+ + +

+ \$2000; It is the fixed cost that must be paid regardless. +

+ + + + +

+ What is the rate of change? Explain what it represents in the context of the problem. +

+ + +

+ 0.25; This how much the total cost changes for each additional doughnut produced. +

+ + + + +

+ Write a linear function to model the situation. Let x represent the number of doughnuts produced and C(x) represent the total cost. (Hint: Use the previous two questions to help!) +

+ + +

+ C(x)=2000+0.25x +

+ + + + +

+ Find C(1300). Explain what that means in the context of the problem. +

+ + +

+ \$2325; The cost to produce 1300 doughnuts is \$2325. +

+ + + + +

+ Find the x-intercept. Explain what it means in the context of the problem. +

+ + +

+ -8000; This doesn't make sense in the context in the problem. Daisy can't produce a negative number of doughnuts. +

+ + + + + + + + + +

+ A taxi costs \$5.00 up front, and then charges \$0.73 per mile traveled. +

+ + + + +

+ Write a linear function to model this situation. +

+ + +

+ C(x)=5+0.73x +

+ + + + + +

+ How much will it cost for a 13 mile taxi ride? +

+ + +

+ \$14.49 +

+ + + + +

+ If the taxi ride cost \$11.06, how many miles did it travel? +

+ + +

+ 8.3 miles +

+ + + + + + + +

+ When reporting the weather, temperature is given in degrees Fahrenheit (F) or degrees Celsius (C). The two scales are related linearly, which means we can find a linear model to describe their relationship. This model lets us convert between the two scales. +

+ + + + +

+ Water freezes at 0C and 32F. Water boils at 100C and 212F. Use this information to write two ordered pairs. +

+ + +

+ Choose Celsius to be your input value, and Fahrenheit to be the output value. +

+ + +

+ (0,32) and (100,212) +

+ + + + + +

+ Use the two points to write a linear model for this situation. Use C and F as your variables. +

+ + +

+ F=\frac{9}{5}C+32 + +

+ + + + +

+ If the temperature outside is 25C, what is the temperature in Fahrenheit? +

+ + +

+ 77F +

+ + + + +

+ If the temperature outside is 50F, what is the temperature in Celsius? +

+ + +

+ 10C +

+ + + + +

+ What temperature value is the same in Fahrenheit as it is in Celsius? +

+ + +

+ -40C=-40F +

+ + + + + + + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
diff --git a/precalculus/source/03-LF/06.ptx b/precalculus/source/03-LF/06.ptx new file mode 100644 index 00000000..d68a2a06 --- /dev/null +++ b/precalculus/source/03-LF/06.ptx @@ -0,0 +1,570 @@ + + +
+ Systems of Linear Equations (LF6) + + + + + Activities + + + +

+ Often times when solving a real-world application, more than one equation is necessary to describe the information. We'll investigate some of those in this section. +

+ + + + +

+ Admission into a carnival for 4 children and 2 adults is \$128.50. For 6 children and 4 adults, the admission is \$208. Assuming a different price for children and adults, what is the price of the child's admission and the price of the adult admission? +

+ + + +

+ Let c represent the cost of a child's admission and a the cost of an adult admission. Write an equation to represent the total cost for 4 children and 2 adults. +

+

+

    +

  1. 4a+2c=128.50

    2. +

  2. a+c=128.50 +

    3. +

  3. 4c+2a=128.50

    4. +

  4. a+c=336.50

    5. +

+ + +

+ C: 4c+2a=128.50 +

+ + + + +

+ Now write an equation to represent the total cost for 6 children and 4 adults. +

+

+

    +

  1. 6c+4a=208

    2. +

  2. a+c=208 +

    3. +

  3. 4a+6c=208

    4. +

  4. a+c=336.50

    5. +

+ + +

+ A: 6c+4a=208 +

+ + + + +

+ Using the above equations, check by substitution which admission prices would satisfy both equations? +

+

+

    +

  1. c=\$20 and a=\$24.25

    2. +

  2. c=\$24.50 and a=\$15.25

    3. +

  3. c=\$20 and a=\$22

    4. +

  4. c=\$14.50 and a=\$30.25

    5. +

+ + +

+ B: c=\$24.50 and a=\$15.25 +

+ + + + + +

+ A system of linear equations consists of two or more linear equations made up of two or more variables. + + A solution to a system of equations is a value for each of the variables that satisfies all the equations at the same time. +

+ + + + +

+ Consider the following system of equations. + \begin{cases} + y=2x+4\\ + 3x+2y=1 + \end{cases} +

+ + + + +

+ Which of the ordered pairs is a solution to the system? + +

+

+

    +

  1. (3,10)

    2. +

  2. (0,4)

    3. +

  3. (1,-1)

    4. +

  4. (-1,2)

    5. +

+ + +

+ D: (-1,2) +

+ + + + + + While we can test points to determine if they are solutions, it is not feasible to test every possible point to find a solution. We need a method to solve a system. + + + +

+ Consider the following system of equations. + \begin{cases} + 3x-y=2\\ + x+4y=5 + \end{cases} +

+ + + +

+ Rewrite the first equation in terms of y. +

+ + + + +

+ Rewrite the second equation in terms of y. +

+ + + + +

+ Graph the two equations on the same set of axes. Where do the lines intersect? +

+ + + + +

+ Check that the point of intersection of the two lines is a solution to the system of equations. +

+ + + + + Sometimes it is difficult to determine the exact intersection point of two lines using a graph. Let's explore another possible method for solving a system of equations. + + + +

+ Consider the following system of equations. + \begin{cases} + 3x+y=4\\ + x+3y=10 + \end{cases} +

+ + + +

+ Graph the two equations on the same set of axes. Is it possible to determine exactly where the lines intersect? +

+ + + + +

+ Solve the first equation for y and substitute into the second equation. What is the resulting equation? +

+

+

    +

  1. x+4-3x=10

    2. +

  2. x+3(4-3x)=10

    3. +

  3. 4-3x+3y=10

    4. +

  4. 3x+(4-3x)=4

    5. +

+ + +

+ B: x+3(4-3x)=10 +

+ + + + +

+ Solve the resulting equation from part (b) for x. +

+

+

    +

  1. x=-3

    2. +

  2. x=\frac{1}{4}

    3. +

  3. x=\frac{7}{3}

    4. +

  4. x=0

    5. +

+ + + +

+ B: x=\frac{1}{4} +

+ + + + +

+ Substitute the value of x into the first equation to find the value of y. +

+

+

    +

  1. y=-5

    2. +

  2. y=\frac{13}{4}

    3. +

  3. y=-3

    4. +

  4. y=\frac{4}{3}

    5. +

+ + +

+ B: y=\frac{13}{4} +

+ + + +

+ B: y=\frac{13}{4} +

+ + + +

+ Write the solution to the system of equations + (the found values of x and y) as an ordered pair. +

+ + + + + + This method of solving a system of equations is referred to as the Substitution Method. +
    +
  1. +

    + Solve one of the equations for one variable. +

    +
    2. +
  2. +

    + Substitute the expression into the other equation to solve for the remaining variable. +

    +
    3. +
  3. +

    + Substitute that value into either equation to find the value of the first variable. +

    +
    4. +
+ + + + + +

+ Solve the following system of equations using the substitution method. + \begin{cases} + x+2y=-1\\ + -x+y=3 + \end{cases} +

+ + +

+ \left( -\dfrac{7}{3}, \dfrac{2}{3} \right) +

+ + + + While the substitution method will always work, sometimes the resulting equations will be difficult to solve. Let's explore a third method for solving a system of two linear equations with two variables. + + + +

+ Consider the following system of equations. + \begin{cases} + 5x+7y=12\\ + 3x-7y=37 + \end{cases} +

+ + + +

+ Add the two equations together. What is the resulting equation? +

+

+

    +

  1. 2x=-15

    2. +

  2. 14y=49

    3. +

  3. 8x+14y=49

    4. +

  4. 8x=49

    5. +

+ + +

+ D: 8x=49 +

+ + + + +

+ Use the resulting equation after addition, to solve for the variable. +

+

+

    +

  1. x=-\frac{15}{2}

    2. +

  2. y=\frac{49}{14}

    3. +

  3. x=\frac{49}{22}

    4. +

  4. x=\frac{49}{8}

    5. +

+ + +

+ D: x=\frac{49}{8} +

+ + + + + +

+ Use the value to find the solution to the system of equations. +

+ + +

+ \left( \dfrac{49}{8}, -\dfrac{149}{56}\right) +

+ + + + + + This method of solving a system of equations is referred to as the Elimination Method. +
    +
  1. +

    + Combine the two equations using addition or subtraction to eliminate one of the variables. +

    +
    2. +
  2. +

    + Solve the resulting equation. +

    +
    3. +
  3. +

    + Substitute that value into either equation to find the value of the other variable. +

    +
    4. +
+ + + +

+ Solve the following system of equations using the elimination method. + \begin{cases} + 7x-4y=3\\ + 3y-7x=8 + \end{cases} +

+ + +

+ \left(-\frac{41}{7} ,-11 \right) +

+ + + + + + +

+ Consider the following system of equations. + \begin{cases} + 5x-9y=6\\ + -10x+4y=2 + \end{cases} + Notice that if you simply add the two equations together, it will not eliminate a variable. Substitution will also be difficult since it involves fractions. +

+ + + +

+ What value can you multiply the first equation by so that when you add the result to the second equation one variable cancels? +

+

+

    +

  1. -1 and the x will cancel

    2. +

  2. 2 and the x will cancel

    3. +

  3. 3 and the y will cancel

    4. +

  4. -2 and the y will cancel

    5. +

+ + +

+ B: Multiply the second equation by 2 and the x will cancel +

+ + + + +

+ Perform the multiplication and add the two equations. What is the resulting equation? +

+

+

    +

  1. -5y=8

    2. +

  2. -14y=14

    3. +

  3. -14y=8

    4. +

  4. -5x=4

    5. +

+ + +

+ B: -14y=14 +

+ + + + +

+ What is the solution to the system of equations? +

+

+

    +

  1. (-1,-1.6)

    2. +

  2. (-1,-0.6)

    3. +

  3. (-0.6,-1)

    4. +

  4. (-1.6,-1)

    5. +

+ + +

+ C: (3,1) +

+ + + + + +

+ For each system of equations, determine which method (graphical, substitution, or elimination) might be best for solving. +

+ + + +

+ \begin{cases} + 5x+9y=16\\ + x+2y=4 + \end{cases} +

+

+

    +

  1. Graphical

    2. +

  2. Substitution

    3. +

  3. Elimination

    4. +

+ + +

+ B: Substitution +

+ + + + +

+ \begin{cases} + y=4x-6\\ + y=-5x+21 + \end{cases} +

+

+

    +

  1. Graphical

    2. +

  2. Substitution

    3. +

  3. Elimination

    4. +

+ + +

+ A: Graphical +

+ + + + +

+ \begin{cases} + x+y=10\\ + x-y=12 + \end{cases} +

+

+

    +

  1. Graphical

    2. +

  2. Substitution

    3. +

  3. Elimination

    4. +

+ + +

+ C: Elimination +

+ + + + + +

Solve each of the systems of equations from using the method you chose. +

+ + +

+ (a) (-4,4) + + (b) (3,6) + + (c) (11,-1) +

+ + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/03-LF/07.ptx b/precalculus/source/03-LF/07.ptx new file mode 100644 index 00000000..4e9e6292 --- /dev/null +++ b/precalculus/source/03-LF/07.ptx @@ -0,0 +1,246 @@ + + +
+ Applications of Systems of Linear Equations (LF7) + + + + + Activities + + + Now that we have explored multiple methods for solving systems of linear equations, let's put those in to practice using some real-world application problems. + + + +

+ Let's begin by revisiting the carnival admission problem from .

+

+ Admission into a carnival for 4 children and 2 adults is $128.50. For 6 children and 4 adults, the admission is $208. Assuming a different price for children and adults, what is the price of the child's admission and the price of the adult admission? +

+ + + + +

+ First, set up a system of equations representing the given information. Use x to represent the child admission price and y for the adult admission price. +

+

+

    +

  1. \begin{cases} + x+y=128.50\\ + x+y=208 + \end{cases}

    2. +

  2. \begin{cases} + 2x+4y=128.50 \\ + 4x+6y=208 + \end{cases}

    3. +

  3. \begin{cases} + 4x+2y=128.50\\ + 6x+4y=208 + \end{cases}

    4. +

  4. \begin{cases} + 6x+4y=128.50 \\ + 4x+2y=208 + \end{cases}

    5. +

+ + +

+ C: \begin{cases} + 4x+2y=128.50\\ + 6x+4y=208 + \end{cases} +

+ + + + +

+ Solve the system of equations. +

+

+

    +

  1. (27, 10.25)

    2. +

  2. (15.25,24.5)

    3. +

  3. (24.5, 15.25)

    4. +

  4. (10,37)

    5. +

+ + +

+ C: (24.5, 15.25) +

+ + + + + +

+ Write your solution in terms of the price of admission for children and adults. +

+ + +

+ The child admission price is \$24.50 and the adult admission price is \$15.25. +

+ + + + + + +

Let's revisit another application we've encountered before in , .

+

Ammie's favorite snack to share with friends is candy salad, which is a mixture of different types of candy. Today she chooses to mix Nerds Gummy Clusters, which cost $8.38 per pound, and Starburst Jelly Beans, which cost \$7.16 per pound. If she makes seven pounds of candy salad and spends a total of \$55.61, how many pounds of each candy did she buy?

+ + + +

+ Set up a system of equations to represent the mixture problem. Let N represent the pounds of Nerds Gummy Clusters and S represent the pounds of Starburst Jelly Beans in the mixture. +

+

+

    +

  1. \begin{cases} + N+S=7 \\ + 7.16N+8.38S=55.61 + \end{cases}

    2. +

  2. \begin{cases} + N+S=7 \\ + 8.38N+7.16S=55.61 + \end{cases}

    3. +

  3. \begin{cases} + N+S=55.61 \\ + 8.38N+7.16S=389.27 + \end{cases}

    4. +

  4. \begin{cases} + N+S=7 \\ + 7N+7S=55.61 + \end{cases}

    5. +

+ + +

+ B: \begin{cases} + N+S=7 \\ + 8.38N+7.16S=55.61 + \end{cases} +

+ + + + +

Now solve the system of equations and put your answer in the context of the problem. +

+

+

    +
  1. Ammie bought 2.5 lbs of Nerds Gummy Clusters and 4.5 lbs of Starburst Jelly Beans.
    2. +
  2. Ammie bought 3.5 lbs of Nerds Gummy Clusters and 3.5 lbs of Starburst Jelly Beans.
    3. +
  3. Ammie bought 4.5 lbs of Nerds Gummy Clusters and 2.5 lbs of Starburst Jelly Beans.
    4. +
  4. Ammie bought 5.5 lbs of Nerds Gummy Clusters and 1.5 lbs of Starburst Jelly Beans.
    5. + +
+

+ + +

+ C: Ammie bought 4.5 lbs of Nerds Gummy Clusters and 2.5 lbs of Starburst Jelly Beans. +

+ + + + + +

+ A couple has a total household income of \$104{,}000. The wife earns \$16{,}000 less than twice what the husband earns. How much does the wife earn? +

+ + + +

+ Set up a system of equations to represent the situation. Let w represent the wife's income and h represent the husband's income. +

+

+

    +

  1. \begin{cases} + w+h=104000 \\ + h=2w-16000 + \end{cases}

    2. +

  2. \begin{cases} + w+h=104000 \\ + 2h=w-16000 + \end{cases}

    3. +

  3. \begin{cases} + w+2h=104000 \\ + w=h-16000 + \end{cases}

    4. +

  4. \begin{cases} + w+h=104000 \\ + w=2h-16000 + \end{cases}

    5. +

+ + +

+ D: \begin{cases} + w+h=104000 \\ + w=2h-16000 + \end{cases} +

+ + + + +

+ Solve the system of equations. How much does the wife earn? +

+

+

    +
  1. The wife earns \$29{,}300.
    2. +
  2. The wife earns \$40{,}000.
    3. +
  3. The wife earns \$64{,}000.
    4. +
  4. The wife earns \$74{,}600.
    5. + +
+

+ + +

+ C: \$64{,}000 +

+ + + + + +

+ Kenneth currently sells suits for Company A at a salary of \$22{,}000 plus a \$10 commission for each suit sold. Company B offers him a position with a salary of \$28{,}000 plus a \$4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal? +

+ +

+ Set-up and solve a system of equations to represent the situation. +

+ + +

+ \begin{cases} + y=10x+22000 \\ + y=4x+28000 + \end{cases} + + 1000 suits +

+ + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/03-LF/main.ptx b/precalculus/source/03-LF/main.ptx new file mode 100644 index 00000000..3cdb2cb1 --- /dev/null +++ b/precalculus/source/03-LF/main.ptx @@ -0,0 +1,19 @@ + + + + Linear Functions (LF) + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/precalculus/source/03-LF/outcomes/01.ptx b/precalculus/source/03-LF/outcomes/01.ptx new file mode 100644 index 00000000..8d871c35 --- /dev/null +++ b/precalculus/source/03-LF/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+ Determine the average rate of change of a given function over a given interval. Find the slope of a line. +

\ No newline at end of file diff --git a/precalculus/source/03-LF/outcomes/02.ptx b/precalculus/source/03-LF/outcomes/02.ptx new file mode 100644 index 00000000..1c3e4244 --- /dev/null +++ b/precalculus/source/03-LF/outcomes/02.ptx @@ -0,0 +1,5 @@ + +

+ Determine an equation for a line when given two points on the line and when given the slope and one point on the line. + Express these equations in slope-intercept or point-slope form and determine the slope and y-intercept of a line given an equation. +

\ No newline at end of file diff --git a/precalculus/source/03-LF/outcomes/03.ptx b/precalculus/source/03-LF/outcomes/03.ptx new file mode 100644 index 00000000..48ebbb5d --- /dev/null +++ b/precalculus/source/03-LF/outcomes/03.ptx @@ -0,0 +1,5 @@ + +

+ Graph a line given its equation or some combination of characteristics, such as points on the graph, + a table of values, the slope, or the intercepts. +

\ No newline at end of file diff --git a/precalculus/source/03-LF/outcomes/04.ptx b/precalculus/source/03-LF/outcomes/04.ptx new file mode 100644 index 00000000..7ac2a5a9 --- /dev/null +++ b/precalculus/source/03-LF/outcomes/04.ptx @@ -0,0 +1,5 @@ + +

+ Use slope relationships to determine whether two lines are parallel or perpendicular, and find the + equation of lines parallel or perpendicular to a given line through a given point. +

\ No newline at end of file diff --git a/precalculus/source/03-LF/outcomes/05.ptx b/precalculus/source/03-LF/outcomes/05.ptx new file mode 100644 index 00000000..cb2247e6 --- /dev/null +++ b/precalculus/source/03-LF/outcomes/05.ptx @@ -0,0 +1,5 @@ + +

+ Build linear models from verbal descriptions, and use the models to establish conclusions, + including by contextualizing the meaning of slope and intercept parameters. +

\ No newline at end of file diff --git a/precalculus/source/03-LF/outcomes/06.ptx b/precalculus/source/03-LF/outcomes/06.ptx new file mode 100644 index 00000000..798537bf --- /dev/null +++ b/precalculus/source/03-LF/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+ Solve a system of two linear equations in two variables. +

\ No newline at end of file diff --git a/precalculus/source/03-LF/outcomes/07.ptx b/precalculus/source/03-LF/outcomes/07.ptx new file mode 100644 index 00000000..bce9d58c --- /dev/null +++ b/precalculus/source/03-LF/outcomes/07.ptx @@ -0,0 +1,4 @@ + +

+ Solve questions involving applications of systems of equations. +

\ No newline at end of file diff --git a/precalculus/source/03-LF/outcomes/main.ptx b/precalculus/source/03-LF/outcomes/main.ptx new file mode 100644 index 00000000..32316009 --- /dev/null +++ b/precalculus/source/03-LF/outcomes/main.ptx @@ -0,0 +1,34 @@ + +> + +

+ BIG IDEA for the chapter goes here, in outcomes/main.ptx +

+

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
  7. + +
    8. +
+ \ No newline at end of file diff --git a/precalculus/source/03-LF/readiness.ptx b/precalculus/source/03-LF/readiness.ptx new file mode 100644 index 00000000..fb4cf760 --- /dev/null +++ b/precalculus/source/03-LF/readiness.ptx @@ -0,0 +1,66 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Plot points on the coordinate plane.

    +
      +

    • Review and Practice: Khan Academy

      +
      • +
    +
    2. +
  2. +

    Evaluate a function at a given value.

    +
      +

    • Review: Khan Academy

      +
      • +

    • Review: Function Notation:

      +
      • +

    • Review and Practice: Khan Academy

      +
      • +
    +
    3. + +
  3. +

    Find and plot x- and y-intercepts.

    +
      +
    • +

      Review and Practice:Khan Academy

      +
      • +
    +
    4. + +
  4. +

    Define and find reciprocals.

    +
      +
    • +

      Review: Khan Academy

      +
      • +
    +
    5. + +
  5. +

    Solve a linear equation for a variable.

    +
      +
    • +

      Review: MrFullerHelp

      +
      • +
    • +

      Review: TabletClass Math

      +
      • +
    • +

      Review: Khan Academy: Multiple Variables

      +
      • +

    • Review and Practice: Khan Academy: Multi-Step Equations

      +
      • +
    +
    6. + +
+ + \ No newline at end of file diff --git a/precalculus/source/04-PR/01.ptx b/precalculus/source/04-PR/01.ptx new file mode 100644 index 00000000..d10d49f6 --- /dev/null +++ b/precalculus/source/04-PR/01.ptx @@ -0,0 +1,355 @@ + + +
+ Graphing Quadratic Functions (PR1) + + + + + Activities + + Quadratic functions have many different applications in the real world. For example, say we want to identify a point at which the maximum profit or minimum cost occurs. Before we can interpret some of these situations, however, we will first need to understand how to read the graphs of quadratic functions to locate these least and greatest values. + + + +

Use the graph of the quadratic function f(x)=3(x-2)^2-4 to answer the questions below.

+
+ + Quadratic function that opens upward + + f(x) = (3*(x-2)^2-4) + p=plot(f, (x, -1, 5), ymin=-7, ymax=10, color='blue', thickness=3) + + +
+ + + +

Make a table for values of f(x) corresponding to the given x -values. What is happening to the y-values as the x-values increase? Do you notice any other patterns of the y-values of the table?

+ + + + x + f(x) + + + -2 + + + -1 + + + 0 + + + 1 + + + 2 + + + 3 + + + 4 + + + 5 + + +
+ + + + +

At which point (x,y) does f(x) have a minimum value? That is, is there a point on the graph that is lower than all other points? + +

    +

  1. The minimum value appears to occur near (0, 8) .

    2. +

  2. The minimum value appears to occur near (-\frac {1}{5}, 10) .

    3. +

  3. The minimum value appears to occur near (2, -4) .

    4. +

  4. There is no minimum value of this function.

    5. +
+

+ + + + +

At which point (x,y) does f(x) have a maximum value? That is, is there a point on the graph that is higher than all other points? + +

    +

  1. The maximum value appears to occur near (-2, 44) .

    2. +

  2. The maximum value appears to occur near (-\frac {1}{5}, 10) .

    3. +

  3. The maximum value appears to occur near (2, -4) .

    4. +

  4. There is no maximum value of this function.

    5. +
+

+ + + + + + +

The maximum or minimum of a quadratic function is also known as its vertex. The vertex form quadratic functionvertex form of a quadratic function is given by f(x)=a(x-h)^2+k, where (h, k) is the vertex quadratic functionvertex from vertex form of the parabola.

+

Vertex form can be used to identify the axis of symmetry, also known as the line of symmetry (the line that makes the shape of an object symmetrical). For a quadratic function, the axis of symmetry always passes through the vertex (h,k) and so x = h is the axis of symmetry. quadratic functionaxis of symmetry from vertex form +

+ + + + + +

Use the given quadratic function, f(x)=3(x-2)^2-4, to answer the following:

+ + + +

Applying , what is the vertex and axis of symmetry of f(x)?

+
    +

  1. vertex: (2,-4); axis of symmetry: x=2

    2. +

  2. vertex: (-2,4); axis of symmetry: x=-2

    3. +

  3. vertex: (-2,-4); axis of symmetry: x=-2

    4. +

  4. vertex: (2,4); axis of symmetry: x=2

    5. +
+ + + + +

Compare what you got in part a with the values you found in . What do you notice?

+ + + + + + +

The standard form quadratic functionstandard form of a quadratic function is given by f(x)=ax^2+bx+c, where a, b , and c are real coefficients.

+

Just as with the vertex form of a quadratic, we can use the standard form of a quadratic to find the axis of symmetry quadratic functionaxis of symmetry from standard form and the vertex quadratic functionvertex from standard formby using the values of a, b , and c . Given the standard form of a quadratic, the axis of symmetry is x=\frac {-b}{2a} and has a vertex at the point (\frac{-b}{2a},f(\frac{-b}{2a})). +

+ + + + + +

Use the graph of the quadratic function to answer the questions below.

+
+ + Quadratic function that opens downward with vertex (-1,4) + + f(x) = (-x^2-2*x+3) + p=plot(f, (x, -4, 2), ymin=-1, ymax=5, color='blue', thickness=3) + + +
+ + + +

Which of the following quadratic functions could be the graph shown in the figure? +

    +

  1. f(x)=x^2+2x+3

    2. +

  2. f(x)=-(x+1)^2+4

    3. +

  3. f(x)=-x^2-2x+3

    4. +

  4. f(x)=(x+1)^2+4

    5. +
+

+ + + + +

What is the maximum or minimum value?

    +

  1. -1

    2. +

  2. 4

    3. +

  3. -3

    4. +

  4. 1

    5. +
+

+ + + + + + +

Consider the following four graphs of quadratic functions:

+ + + + + f(x) = -2*(x+3)^2+1 + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3) + p+=text('Graph A', (0, 8), fontsize=18, color='black') + p + + + + + f(x) = (x+3)^2+2 + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3) + p+=text('Graph B', (0, 8), fontsize=18, color='black') + p + + + + + + + f(x) = 1*(x-2)^2-3 + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3) + p+=text('Graph C', (0, 8), fontsize=18, color='black') + p + + + + + f(x) = -1*(x-2)^2+4 + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3) + p+=text('Graph D', (0, 8), fontsize=18, color='black') + p + + + + + + + +

Which of the graphs above have a maximum?

    +

  1. Graph A

    2. +

  2. Graph B

    3. +

  3. Graph C

    4. +

  4. Graph D

    5. +
+

+ + + + +

Which of the graphs above have a minimum?

    +

  1. Graph A

    2. +

  2. Graph B

    3. +

  3. Graph C

    4. +

  4. Graph D

    5. +
+

+ + + + +

Which of the graphs above have an axis of symmetry at x=2?

    +

  1. Graph A

    2. +

  2. Graph B

    3. +

  3. Graph C

    4. +

  4. Graph D

    5. +
+

+ + + + +

Which of the graphs above represents the function f(x)=-(x-2)^2+4? +

    +

  1. Graph A

    2. +

  2. Graph B

    3. +

  3. Graph C

    4. +

  4. Graph D

    5. +
+

+ + + + +

Which of the graphs above represents the function f(x)=x^2-4x+1? +

    +

  1. Graph A

    2. +

  2. Graph B

    3. +

  3. Graph C

    4. +

  4. Graph D

    5. +
+

+ + + + + + +

Notice that the maximum or minimum value of the quadratic function is the y-value of the vertex. +

+ + + + + +

A function f(x) has a maximum value at 7 and its axis of symmetry at x=-2.

+ + + + +

Sketch a graph of a function that meets the criteria for f(x).

+ + + + +

Was your graph the only possible answer? Try to sketch another graph that meets this criteria.

+ + + + + + +

Other points, such as x- and y-intercepts, may be helpful in sketching a more accurate graph of a quadratic function. +

+ + + + + +

Consider the following two quadratic functions f(x)=x^2-4x+20 and g(x)=2x^2-8x+24 and answer the following questions:

+ + + + +

Applying , what is the vertex and axis of symmetry of f(x)?

+
    +

  1. vertex: (2,-16); axis of symmetry: x=2

    2. +

  2. vertex: (-2,16); axis of symmetry: x=-2

    3. +

  3. vertex: (-2,-16); axis of symmetry: x=-2

    4. +

  4. vertex: (2,16); axis of symmetry: x=2

    5. +
+ + + + + +

Applying , what is the vertex and axis of symmetry of g(x)?

+
    +

  1. vertex: (2,-16); axis of symmetry: x=2

    2. +

  2. vertex: (-2,16); axis of symmetry: x=-2

    3. +

  3. vertex: (-2,-16); axis of symmetry: x=-2

    4. +

  4. vertex: (2,16); axis of symmetry: x=2

    5. +
+ + + + + +

What do you notice about f(x) and g(x)? +

+ + + + +

Now graph both f(x) and g(x) and draw a sketch of each graph on one coordinate plane. How are they similar/different? +

+ + + + + + +

Exercises available at .

+ + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/04-PR/02.ptx b/precalculus/source/04-PR/02.ptx new file mode 100644 index 00000000..e6c29014 --- /dev/null +++ b/precalculus/source/04-PR/02.ptx @@ -0,0 +1,240 @@ + + +
+ Quadratic Models and Meanings (PR2) + + + + + Activities +

Objects launched into the air follow a path that can be described by a quadratic function. We can also use quadratic functions to model area, profit, and population. Knowing the key components of a quadratic function allow us to find maximum profit, the point where an object hits the ground, or how much of an object to make for a minimum cost.

+ + + +

A water balloon is tossed vertically from a fifth story window. It's height h(t), in feet, at a time t, in seconds, is modeled by the function + + h(t)=-16t^2+40t+50 \amp + +

+ + + +

Complete the following table. Do all the values have meaning in terms of the model? + + + + t + h(t) + + + 0 + + + 1 + + + 2 + + + 3 + + + 4 + + + 5 + + +
+

+ + + + +

Compute the slope of the line joining t=0 and t=1. Then, compute the slope of the line joining t=1 and t=2. What do you notice about the slopes?

+ + + + +

What is the meaning of h(0)=50?

+
    +

  1. The initial height of the water balloon is 50 feet.

    2. +

  2. The water balloon reaches a maximum height of 50 feet.

    3. +

  3. The water balloon hits the ground after 50 seconds.

    4. +

  4. The water balloon travels 50 feet before hitting the ground.

    5. +
+ + + + +

Find the vertex of the quadratic function h(t). +

    +

  1. (0,50)

    2. +

  2. (1,74)

    3. +

  3. (1.25,75)

    4. +

  4. (3.4,0)

    5. +
+

+ + + + +

What is the meaning of the vertex? +

    +

  1. The water balloon reaches a maximum height of 50 feet at the start.

    2. +

  2. After 1 second, the water balloon reaches a maximum height of 74 feet.

    3. +

  3. After 1.25 seconds, the water balloon reaches the maximum height.

    4. +

  4. After 3.4 seconds, the water balloon hits the ground.

    5. +
+

+ + + + + +

The population of a small city is given by the function P(t)=-50t^2+1200t+32000, where t is the number of years after 2015. + +

+ + + +

When will the population of the city reach a maximum? +

    +

  1. 2020

    2. +

  2. 2022

    3. +

  3. 2025

    4. +

  4. 2027

    5. +

+ + + + +

Determine when the population of the city is increasing and when it is decreasing. +

+ + + + +

When will the population of the city reach 36{,}000 people? +

    +

  1. 2019

    2. +

  2. 2025

    3. +

  3. 2027

    4. +

  4. 2035

    5. +

+ + + + + + +

The unit price of an item affects its supply and demand. That is, if the unit price increases, the demand for the item will usually decrease. For example, an online streaming service currently has 84 million subscribers at a monthly charge of $6. Market research has suggested that if the owners raise the price to $8, they would lose 4 million subscribers. Assume that subscriptions are linearly related to the price. + +

+ + + +

Which of the following represents a linear function which relates the price of the streaming service p to the number of subscribers Q? +

    +

  1. Q(p)=-2p

    2. +

  2. Q(p)=-2p+84

    3. +

  3. Q(p)=-2p-4

    4. +

  4. Q(p)=-2p+96

    5. +

+ + + + +

Using the fact that revenue R is price times the number of items sold, R=pQ, which of the following represents the revenue in terms of the price? +

    +

  1. R(p)=-2p^2

    2. +

  2. R(p)=-2p^2+84p

    3. +

  3. R(p)=-2p^2-4p

    4. +

  4. R(p)=-2p^2+96p

    5. +
+

+ + + + +

What price should the streaming service charge for a monthly subscription to maximize their revenue?

    +

  1. \$10

    2. +

  2. \$19.50

    3. +

  3. \$24

    4. +

  4. \$28.25

    5. +

+ + + + +

How many subscribers would the company have at this price? +

    +

  1. 39.5 million

    2. +

  2. 48 million

    3. +

  3. 57 million

    4. +

  4. 76 million

    5. +

+ + + + +

What is the maximum revenue? +

    +

  1. 760 million

    2. +

  2. 1112 million

    3. +

  3. 1152 million

    4. +

  4. 1116 million

    5. +

+ + + + + + + The owner of a ranch decides to enclose a rectangular region with 240 feet of fencing. To help the fencing cover more land, he plans to use one side of his barn as part of the enclosed region. What is the maximum area the rancher can enclose? + + + +

+ Draw a picture to represent the fenced area against the barn. Use l to represent the length of fence parallel to the barn and w to represent the two sides perpendicular to the barn. +

+ + + + + +

Find an equation for the area of the fence in terms of the length l. It may be useful to find an equation for the total amount of fencing in terms of the length l and width w. +

    +

  1. A=lw

    2. +

  2. A=l^{2}

    3. +

  3. A=l(240-l)

    4. +

  4. A=l\left(120-\frac{l}{2}\right)

    5. +

+ + + + +

+ Use the area equation to find the maximum area the rancher can enclose. +

+ + + + + + + + + +

Exercises available at .

+ + + + Videos +

Coming eventually.....

+ +
diff --git a/precalculus/source/04-PR/03.ptx b/precalculus/source/04-PR/03.ptx new file mode 100644 index 00000000..dd81e7fb --- /dev/null +++ b/precalculus/source/04-PR/03.ptx @@ -0,0 +1,385 @@ + + +
+ Polynomial Long Division (PR3) + + + + + + Activities + +

We have seen previously that we can reduce rational functions by factoring, for example + \frac{x^2+5x+4}{x^3+3x^2+2x}=\frac{(x+1)(x+4)}{x(x+2)(x+1)}=\frac{x+4}{x(x+2)}. + In this section, we will explore what can we do to simplify rational functions if we are not able to + reduce by easily factoring? +

+ + + +

Recall that a fraction is called proper if its numerator is smaller than its denominator, and improper + if the numerator is larger than the denominator (so \frac{3}{5} is a proper fraction, but \frac{32}{7} is an + improper fraction). Similarly, we define a proper rational function to be a rational function where + the degree of the numerator is less than the degree of the denominator.

+ + + + + + Label each of the following rational functions as either proper or improper. +
    +
  1. \frac{x^3+x}{x^2+4}
    2. +
  2. \frac{3}{x^2+3x+4}
    3. +
  3. \frac{7+x^3}{x^2+x+1}
    4. +
  4. \frac{x^4+x+1}{x^4+4x^2}
    5. +
+ +

A, C, and D are improper, while B is proper.

 + + + +

When dealing with an improper fraction such as \frac{32}{7}, it is sometimes useful to rewrite this + as an integer plus an improper fraction, e.g. \frac{32}{7}=4+\frac{4}{7}. Similarly, it will sometimes be + useful to rewrite an improper rational function as the sum of a polynomial and a proper rational function, such + as \frac{x^3+x}{x^2+4}=x-\frac{3x}{x^2+4}. +

+ + + + + Consider the improper fraction \frac{357}{11}. + + + +

Use long division to write \frac{357}{11} as an integer plus a proper fraction.

+ + + + + \( + \begin{array}{ r *{19}{c@{}} } + &3 & 2 & \\ + \cline{2-4}\\[-14.5pt] + \multicolumn{1}{r}{11\rlap{\kern4.5pt\smash{\raise.2ex\hbox{)}}}} & 3 &5 &7 \\ + & 3&3 & \\ + \cline{2-3} + & & 2 & 7 \\ + & & 2 & 2 \\ + \cline{3-4} + & & & 5 \\ + \end{array} + \) + + +

So \frac{357}{11}=32+\frac{5}{11}.

+ + + + +

Now we will carefully redo this process in a way that we can generalize to rational functions. + Note that we can rewrite 357 as 357=3\cdot10^2+5\cdot10+7, and 11 as 11=1\cdot10+1. + By comparing the leading terms in these expansions, we see that to knock off the leading term of 357, we need to + multiply 11 by 3\cdot10^1. +

+

Using the fact that 357=11\cdot30+27, rewrite \frac{357}{11} as \frac{357}{11}=30+\frac{?}{11}. +

+ +

\frac{357}{11}=30+\frac{27}{11}.

 + + + +

Note now that if we can rewrite \frac{27}{11} as an integer plus a proper fraction, we will be done, + since \frac{357}{11}=30+\frac{27}{11}. +

+

+ Rewrite \frac{27}{11}=?+\frac{?}{11} as an integer plus a proper fraction. +

+ +

\frac{27}{11}=2+\frac{5}{11}.

 + + + +

Combine your work in the previous two parts to rewrite \frac{357}{11} as an integer plus a proper fraction. + How does this compare to what you obtained in part (a)? +

+ +

\frac{357}{11}=30+\frac{27}{11}=30+2+\frac{5}{11}=32+\frac{5}{11}.

 + + + + + +

+ Now let's consider the rational function \frac{3x^2+5x+7}{x+1}. We want to rewrite this as a polynomial + plus a proper rational function. +

+ + + +

Looking at the leading terms, what do we need to multiply x+1 by so that it would have the same leading term + as 3x^2+5x+7? +

+
    +
  1. 3
    2. +
  2. x
    3. +
  3. 3x
    4. +
  4. 3x+5
    5. +
+ +

C

 + + + +

Rewrite 3x^2+5x+7=3x(x+1)+?, and use this to rewrite \frac{3x^2+5x+7}{x+1}=3x+\frac{?}{x+1}. +

+ +

\frac{3x^2+5x+7}{x+1}=3x+\frac{2x+7}{x+1}

 + + + +

+ Now focusing on \frac{2x+7}{x+1}, what do we need to multiply x+1 by so that it would have the same leading term + as 2x+7? +

+
    +
  1. 2
    2. +
  2. x
    3. +
  3. 2x
    4. +
  4. 2x+7
    5. +
+ +

A

 + + + +

Rewrite \frac{2x+7}{x+1}=2+\frac{?}{x+1}.

+ +

\frac{2x+7}{x+1}=2+\frac{5}{x+1}

 + + + +

Combine this with the previous parts to rewrite \frac{3x^2+5x+7}{x+1}=3x+?+\frac{?}{x+1}.

+ +

\frac{3x^2+5x+7}{x+1}=3x+2+\frac{5}{x+1}

 + + + + + +

Next we will use the notation of long division to rewrite the rational function \frac{3x^2+5x+7}{x+1} as a + polynomial plus a proper rational function.

+ + + + +

First, let's use long division notation to write the quotient.

+ + + \( + \begin{array}{ r *{19}{c@{}} } + & & ? & & & & & & \\ + \cline{2-9}\\[-14.5pt] + \multicolumn{1}{r}{(x+1)\rlap{\kern4.5pt\smash{\raise.2ex\hbox{)}}}} & 3 &x^2 &+ &5 &x &+ & 7 \\ + \end{array} + \) + + +

What do we need to multiply x+1 by so that it would have the same leading term + as 3x^2+5x+7?

+ + + + + + \( + \begin{array}{ r *{19}{c@{}} } + &3 & x & & & & & & \\ + \cline{2-9}\\[-14.5pt] + \multicolumn{1}{r}{(x+1)\rlap{\kern4.5pt\smash{\raise.2ex\hbox{)}}}} & 3 &x^2 &+ &5 &x &+ & 7 \\ + \end{array} + \) + + + + + + +

Now to rewrite 3x^2+5x+7 as 3x(x+1)+?, place the product 3x(x+1) below and subtract.

+ + + \( + \begin{array}{ r *{19}{c@{}} } + & 3 & x & & & & & & \\ + \cline{2-9}\\[-14.5pt] + \multicolumn{1}{r}{(x+1)\rlap{\kern4.5pt\smash{\raise.2ex\hbox{)}}}} & 3 &x^2 &+ &5 &x &+ & 7 \\ + & 3 & x^2 & + & 3 & x & & & \\ + \cline{2-6} + & & & & ? & & + & ? & \\ + \end{array} + \) + + + + + + + \( + \begin{array}{ r *{19}{c@{}} } + &3 & x & & & & & & \\ + \cline{2-9}\\[-14.5pt] + \multicolumn{1}{r}{(x+1)\rlap{\kern4.5pt\smash{\raise.2ex\hbox{)}}}} & 3 &x^2 &+ &5 &x &+ & 7 \\ + & 3 & x^2 & + & 3 & x & & & \\ + \cline{2-6} + & & & & 2 & x & + & 7 & \\ + \end{array} + \) + + + + + + + +

Now focusing on 2x+7, what do we need to multiply x+1 by so that it would have the same leading term as 2x+7?

+ + + \( + \begin{array}{ r *{19}{c@{}} } + &3 & x &+ &? & & & & \\ + \cline{2-9}\\[-14.5pt] + \multicolumn{1}{r}{(x+1)\rlap{\kern4.5pt\smash{\raise.2ex\hbox{)}}}} & 3 &x^2 &+ &5 &x &+ & 7 \\ + & 3 & x^2 & + & 3 & x & & & \\ + \cline{2-6} + & & & & 2 & x & + & 7 & \\ + \end{array} + \) + + + + + + + \( + \begin{array}{ r *{19}{c@{}} } + &3 & x &+ &2 & & & & \\ + \cline{2-9}\\[-14.5pt] + \multicolumn{1}{r}{(x+1)\rlap{\kern4.5pt\smash{\raise.2ex\hbox{)}}}} & 3 &x^2 &+ &5 &x &+ & 7 \\ + & 3 & x^2 & + & 3 & x & & & \\ + \cline{2-6} + & & & & 2 & x & + & 7 & \\ + \end{array} + \) + + + + + + +

Now, subtract 2(x+1) to finish the long division.

+ + + \( + \begin{array}{ r *{19}{c@{}} } + &3 & x &+ &2 & & & & \\ + \cline{2-9}\\[-14.5pt] + \multicolumn{1}{r}{(x+1)\rlap{\kern4.5pt\smash{\raise.2ex\hbox{)}}}} & 3 &x^2 &+ &5 &x &+ & 7 \\ + & 3 & x^2 & + & 3 & x & & & \\ + \cline{2-6} + & & & & 2 & x & + & 7 & \\ + & & & & ? & &+ & ? & \\ + \cline{5-9} + & & & & & & & ? & \\ + \end{array} + \) + + + + + + + \( + \begin{array}{ r *{19}{c@{}} } + &3 & x &+ &2 & & & & \\ + \cline{2-9}\\[-14.5pt] + \multicolumn{1}{r}{(x+1)\rlap{\kern4.5pt\smash{\raise.2ex\hbox{)}}}} & 3 &x^2 &+ &5 &x &+ & 7 \\ + & 3 & x^2 & + & 3 & x & & & \\ + \cline{2-6} + & & & & 2 & x & + & 7 & \\ + & & & & 2 & x &+ & 2 & \\ + \cline{5-9} + & & & & & & & 5 & \\ + \end{array} + \) + + + + + + +

This long division calculation has shown that 3x^2+5x+7 = (x+1)(3x+2)+5. + Use this to rewrite \frac{3x^2+5x+7}{x+1} as a polynomial plus a proper rational function.

+ + +

\frac{3x^2+5x+7}{x+1} = 3x+2+\frac{5}{x+1}

+ + + + +

Note that in and we performed the same computations, but just organized our work a little differently.

+ + + + +

Rewrite \frac{x^2+1}{x-1} as a polynomial plus a proper rational function.

+ +

Note that x^2+1=x^2+0x+1.

 +

x+1+\frac{3}{x-1}.

 + + + + +

Rewrite \frac{x^5+x^3+2x^2-6x+7}{x^2+x-1} as a polynomial plus a proper rational function.

+ + +

x^3-x^2+3x-2+\frac{-x+5}{x^2+x-1}.

+ + + + + +

Rewrite \frac{3x^4-5x^2+2}{x-1} as a polynomial plus a proper rational function.

+ +

3x^3+3x^2-2x-2.

 + + + + + + +

Exercises available at .

+ + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
+ diff --git a/precalculus/source/04-PR/04.ptx b/precalculus/source/04-PR/04.ptx new file mode 100644 index 00000000..fb4bb5fc --- /dev/null +++ b/precalculus/source/04-PR/04.ptx @@ -0,0 +1,401 @@ + + +
+ Zeros of Polynomial Functions (PR4) + + + + + Activities + + + +

Recall that to find the x-intercepts of a function f(x), we need to find the values + of x that make f(x)=0. + We saw in that the zero product property () was helpful when f(x) + is a polynomial that we can factor. +

+ + + +

+ Consider the following polynomials. +

+ + + +

What are the x-intercepts of the function f(x)=(x+5)(x-3)?

+ + +

-5 and 3

+ + + + +

What are the x-intercepts of the function g(x)=(x+5)(x-3)(x-7)?

+ + +

-5, 3, and 7

+ + + + +

What are the x-intercepts of the function h(x)=x^3-x^2-56x?

+ +

h(x)=x(x^2-x-56).

 + +

-7, 0, and 8

+ + + + + +

Real zeros of a polynomial function are the same as the x-intercepts. +

+ + + + + Factor Theorem + +

A number c is a zero of a polynomial function f(x) (that is, f(c)=0) if and only if + x-c is a factor of f(x).

+ + + + + +

Use to find the zeros of the polynomial f(x)=x^4-x^3. +

+ +

0 and 1.

 + + + + +

Consider the polynomial f(x)=2x^3-7x^2-33x+108.

+ + + +

It turns out that x-3 is a factor of f(x). Use this fact to find a polynomial g(x) such that + f(x)=(x-3)g(x). +

+
    +
  1. g(x)=2x^2-x-36
    2. +
  2. g(x)=2x^2+x-36
    3. +
  3. g(x)=2x^2-15x-27
    4. +
  4. g(x)=2x^2+15x-27
    5. +
+ + +

A.

+ + + + +

Since f(x)=(x-3)(2x^2-x-36), how can we rewrite f(x)?.

+
    +
  1. f(x)=(x-3)(x+4)(2x-9)
    2. +
  2. f(x)=(x-3)(x-4)(2x+9)
    3. +
  3. f(x)=(x-3)(2x+3)(x-9)
    4. +
  4. f(x)=(x-3)(2x-3)(x+9)
    5. +
+ + +

A.

+ + + + +

How many zeros does f(x) have? +

    +
  1. 1
    2. +
  2. 2
    3. +
  3. 3
    4. +
  4. 4
    5. +
+

+ +

3

 + + + + + +

The degree polynomial functionmultiplicity of a polynomial function is the highest power of x in the expanded form of the function.

+ + + +

Consider the function f(x)=5x-4.

+ +

What is the degree of f(x)?

+
    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 5
    5. +
+ +

1.

 + + + +

Find the zeros of f(x). How many are there?

+
    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 5
    5. +
+ +

1.

 + + + +

Consider the function g(x)=5x^2+7x-6.

+ + + +

What is the degree of g(x)?

+
    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 5
    5. +
+ +

2.

 + + + +

Find the zeros of g(x). How many are there?

+
    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 5
    5. +
+ +

2.

 + + + + +

Consider the function h(x)=x^5+x^4.

+ + + +

What is the degree of g(x)?

+
    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 5
    5. +
+ +

5.

 + + + +

Find the zeros of h(x). How many are there?

+
    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 5
    5. +
+ +

There are 2 distinct real zeros.

 + + + + + +

The multiplicity polynomial functionmultiplicity of a zero is the number of times the corresponding linear factor appears in the factored form of the polynomial function. +

+ + + + +

In the polynomial h(x)=x^5+x^4=x^4(x+1), 0 was a zero with multiplicity 4.

+ + + +

How many zeros does the polynomial f(x)=x^2+4 have?

+
    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 3
    5. +
+ +

There are no real zeros, but two distinct complex zeros.

 + + + + + Fundamental Theorem of Algebra + +

A polynomial function of degree n > 0 has at least one, possibly complex zero. + Nonzero polynomials of degree n have exactly n zeros, counted with multiplicity.

+ + + + + +

Consider the polynomial f(x)=(3x-2)(x+1)^2(x-6)^3.

+ + + +

Find all of the zeros of f(x) with their corresponding multiplicities. + +

+ +

The zeros are \frac{2}{3} with multiplicty 1, -1 with multiplicity 2, and 6 with multiplicity 3.

 + + + +

Find the degree of f(x). +

    +
  1. 1
    2. +
  2. 2
    3. +
  3. 3
    4. +
  4. 6
    5. +
+ +

D

 + + + +

How do all of the exponents on the factors relate to the degree of f(x)?

+ + +

The degree is the sum of the exponents.

+ + + + +

How do the multiplicities of the zeros relate to the degree of f(x)?

+ + + +

Each exponent is the multiplicity of the corresponding zero.

+ + + + + + + +

Find a polynomial that satisfies the following properties:

+
    +
  • -1 is a zero with multiplicity 2
    • +
  • 4 is a zero with multiplicity 1
    • +
  • 7 is a zero with multiplicity 3
    • +
+ + +

One such polynomial is (x+1)^2(x-4)(x-7)^3.

+ + + + + + + +

Find the (complex) zeros for the function, f(x)=x^2+16.

+ +

The zeros are 4i and -4i.

 + + + + Conjugate Zeros Theorem + +

Let f(x) be a polynomial function with real coefficients. If a+bi, is a complex zero of the function, + then the conjugate a-bi is also a zero of the function. + These two zeroes are called conjugate zeros, or a conjugate pair of zeros. +

+ + + + + +

Consider the following information about a polynomial f(x):

+
    +
  • x=2 is a zero with multiplicity 1
    • +
  • x=-1 is a zero with multiplicity 2
    • +
  • x=i is a zero with multiplicity 1
    • +
+ + + +

What is the smallest possible degree of such a polynomial f(x) with real coefficients?

+
    +
  1. 2
    2. +
  2. 3
    3. +
  3. 4
    4. +
  4. 5
    5. +
  5. 6
    6. +
+ +

5

 + + + +

+ Write an express for such a polynomial f(x) with real coefficients of smallest possible degree. +

+ + +

f(x)=(x-2)(x-1)^2(x^2+1)=x^5-4x^4+6x^3-6x^2+5x-2 is one such polynomial.

+ + + + + + +

Consider the function f(x)=x^4+x^3+2x^2+4x-8.

+ + + +

Use a graphing utility to graph f(x).

+ + + + + f(x)=x^4+x^3+2*x^2+4*x-8 + p=plot(f, xmin=-5,xmax=5,ymin=-10,ymax=10,color="blue",thickness=3,gridlines=[[-5..5],[-10..10]]) + p+=text("$f(x)$",(3,3),color="blue",fontsize=20) + p + + + + + + +

Find all the zeros of f(x) and their corresponding multiplicities.

+ + +

f(x) has zeros at -2, 1, -2i, and 2i, all of multiplicity 1.

+ + + + + + + + + +

Exercises available at .

+ + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/04-PR/05.ptx b/precalculus/source/04-PR/05.ptx new file mode 100644 index 00000000..2edc643c --- /dev/null +++ b/precalculus/source/04-PR/05.ptx @@ -0,0 +1,499 @@ + + +
+ Graphs of Polynomial Functions (PR5) + + + + + Activities + + + +

Just like with quadratic functions, we should be able to determine key characteristics that will help guide us in creating a sketch of any polynomial function. We can start by finding both x and y -intercepts and then explore other characteristics polynomial functions can have. Recall that the zeros of a function are the x-intercepts - i.e., the values of x that cross or touch the x-axis. Just like with quadratic functions, we can find the zeros of a function by setting the function equal to 0 and solving for x.

+ + + + + +

The end behavior polynomial functionend behavior of a polynomial function describes the behavior of the graph at the "ends" of the function. In other words, as we move to the right of the graph (as the x values increase), what happens to the y values? Similarly, as we move to the left of the graph (as the x values decrease), what happens to the y values?

+ + + + + + +Use the graphs of the following polynomial functions to answer the questions below. + + + + + f(x) = x^4-11*x^3+42*x^2-64*x+32 + p=plot(f, (x, -5, 5), ymin=-15, ymax=15, color='blue', thickness=3) + p+=text('Graph A', (0, 8), fontsize=18, color='black') + p + + + + + f(x) = -4*x^3+2*x^2+3 + p=plot(f, (x, -5, 5), ymin=-15, ymax=15, color='blue', thickness=3) + p+=text('Graph B', (0, 8), fontsize=18, color='black') + p + + + + + + + f(x) = x^7+2*x^3-5*x^2+2 + p=plot(f, (x, -5, 5), ymin=-15, ymax=15, color='blue', thickness=3) + p+=text('Graph C', (0, 8), fontsize=18, color='black') + p + + + + + f(x) = -2*x^4+3*x^2-5 + p=plot(f, (x, -5, 5), ymin=-15, ymax=15, color='blue', thickness=3) + p+=text('Graph D', (0, 8), fontsize=18, color='black') + p + + + + + + + +

How would you describe the behavior of Graph A as you approach the ends? +

    +
  1. Graph A rises on the left and on the right.
    2. +
  2. Graph A rises on the left, but falls on the right.
    3. +
  3. Graph A rises on the right, but falls on the left.
    4. +
  4. Graph A falls on the left and on the right.

+ + + + +

How would you describe the behavior of Graph B as you approach the ends? +

    +
  1. Graph B rises on the left and on the right.
    2. +
  2. Graph B rises on the left, but falls on the right.
    3. +
  3. Graph B rises on the right, but falls on the left.
    4. +
  4. Graph B falls on the left and on the right.

+ + + + +

How would you describe the behavior of Graph C as you approach the ends? +

    +
  1. Graph C rises on the left and on the right.
    2. +
  2. Graph C rises on the left, but falls on the right.
    3. +
  3. Graph C rises on the right, but falls on the left.
    4. +
  4. Graph C falls on the left and on the right.

+ + + + +

How would you describe the behavior of Graph D as you approach the ends? +

    +
  1. Graph D rises on the left and on the right.
    2. +
  2. Graph D rises on the left, but falls on the right.
    3. +
  3. Graph D rises on the right, but falls on the left.
    4. +
  4. Graph D falls on the left and on the right.

+ + + + + + +

Typically, when given an equation of a polynomial function, we look at the degree and leading coefficient to help us determine the behavior of the ends. The degree is the highest exponential power in the polynomial. The leading coefficient is the number written in front of the variable with the highest exponential power. +

+ + + + + +Let's refer back to the graphs in and look at the equations of those polynomial functions. Let's apply to see if we can determine how the degree and leading coefficients of those graphs affect their end behavior. +
  • Graph A: f(x)=-11x^3+32+42x^2+x^4-64x
    • +
  • Graph B: g(x)=2x^2+3-4x^3
    • +
  • Graph C: h(x)=x^7+2x^3-5x^2+2
    • +
  • Graph D: j(x)=3x^2-2x^4-5
    • +
+ + + + + f(x) = x^4-11*x^3+42*x^2-64*x+32 + p=plot(f, (x, -5, 5), ymin=-15, ymax=15, color='blue', thickness=3) + p+=text('Graph A', (0, 8), fontsize=18, color='black') + p + + + + + f(x) = -4*x^3+2*x^2+3 + p=plot(f, (x, -5, 5), ymin=-15, ymax=15, color='blue', thickness=3) + p+=text('Graph B', (0, 8), fontsize=18, color='black') + p + + + + + + + f(x) = x^7+2*x^3-5*x^2+2 + p=plot(f, (x, -5, 5), ymin=-15, ymax=15, color='blue', thickness=3) + p+=text('Graph C', (0, 8), fontsize=18, color='black') + p + + + + + f(x) = -2*x^4+3*x^2-5 + p=plot(f, (x, -5, 5), ymin=-15, ymax=15, color='blue', thickness=3) + p+=text('Graph D', (0, 8), fontsize=18, color='black') + p + + + + + + + +

What is the degree and leading coefficient of Graph A? +

    +
  1. Degree: -64; Leading Coefficient: 4
    2. +
  2. Degree: 4; Leading Coefficient: 0
    3. +
  3. Degree: 1; Leading Coefficient: -64
    4. +
  4. Degree: 4; Leading Coefficient: 1

+ + + + +

What is the degree and leading coefficient of Graph B? +

    +
  1. Degree: 3; Leading Coefficient: -4
    2. +
  2. Degree: -4; Leading Coefficient: 3
    3. +
  3. Degree: 2; Leading Coefficient: 3
    4. +
  4. Degree: 3; Leading Coefficient: 4

+ + + + +

What is the degree and leading coefficient of Graph C? +

    +
  1. Degree: -5; Leading Coefficient: 2
    2. +
  2. Degree: 0; Leading Coefficient: 7
    3. +
  3. Degree: -5; Leading Coefficient: 3
    4. +
  4. Degree: 7; Leading Coefficient: 1

+ + + + +

What is the degree and leading coefficient of Graph D? +

    +
  1. Degree: -2; Leading Coefficient: 4
    2. +
  2. Degree: 3; Leading Coefficient: 2
    3. +
  3. Degree: -2; Leading Coefficient: 4
    4. +
  4. Degree: -5; Leading Coefficient: 4

+ + + + +

Notice that Graph A and Graph D have their ends going in the same direction. What conjectures can you make about the relationship between their degrees and leading coefficients with the behavior of their graphs? +

+ + + + +

Notice that Graph B and Graph C have their ends going in opposite directions. What conjectures can you make about the relationship between their degrees and leading coefficients with the behavior of their graphs? +

+ + + + + + +

From , we saw that the degree and leading coefficient of a polynomial function can give us more clues about the behavior of the function. In summary, we know: +

  • If the degree is even, the ends of the polynomial function will be going in the same direction. If the leading coefficient is positive, both ends will be pointing up. If the leading coefficient is negative, both ends will be pointing down.
    • +
  • If the degree is odd, the ends of the polynomial function will be going in opposite directions. If the leading coefficient is positive, the left end will fall and the right end will rise. If the leading coefficient is negative, the left end will rise and the right end will fall.
+

+ + + + + +

When describing end behavior, mathematicians typically use arrow notation. Just as the name suggests, arrows are used to indicate the behavior of certain values on a graph. +

+

For end behavior, students are often asked to determine the behavior of y-values as x-values either increase or decrease. The statement "As x\to \infty, f(x)\to -\infty" can be translated to "As x approaches infinity (or as x increases), f(x) (or the y-values) go to negative infinity (i.e., it decreases)."

+ + + + + +

Use the graph of f(x) to answer the questions below.

+
+ + A polynomial function with one even and one odd multiplicity + + f(x) = (x-5)*(x+1)^2 + p=plot(f, (x, -10, 10), ymin=-40, ymax=20, color='blue', thickness=3) + + +
+ + + +

How would you describe the end behavior of f(x) ? +

    +
  1. f(x) rises on the left and on the right.
    2. +
  2. f(x) rises on the left, but falls on the right.
    3. +
  3. f(x) rises on the right, but falls on the left.
    4. +
  4. f(x) falls on the left and on the right.

+ + + + +

How would you describe the end behavior of f(x) using arrow notation? +

    +

  1. As x\to -\infty, f(x)\to -\infty

    +

    As x\to \infty, f(x)\to -\infty

    2. +

  2. As x\to -\infty, f(x)\to -\infty

    +

    As x\to \infty, f(x)\to \infty

    3. +

  3. As x\to -\infty, f(x)\to \infty

    +

    As x\to \infty, f(x)\to -\infty

    4. +

  4. As x\to -\infty, f(x)\to \infty

    +

    As x\to \infty, f(x)\to \infty

+

+ + + + + + +

When graphing polynomial functions, you may notice that these functions have some "hills" and "valleys." These characteristics of the graph are known as the local maxima and local minima of the graph - similar to what we've already seen with quadratic functions. Unlike quadratic functions, however, a polynomial graph can have many local maxima/minima (quadratic functions only have one).

+ + + + + +

Sketch the function, f(x)=(x-2)(x+1)(x-3)^2 , by first finding the given characteristics. +

+ + + +

Find the zeros of f(x).

+ + + + +

Find the multiplicities at each zero.

+ + + + +

Find the y-intercept of f(x).

+ + + + +

Describe the end behavior of f(x).

+ + + + +

Estimate where any local maximums and minimums may occur.

+ + + + + + +

Sketch the graph of a function f(x) that meets all of the following criteria. Be sure to scale your axes and label any important features of your graph.

+ +
  • The x-intercepts of f(x) are 0, 2, and 5 .
    • +
  • f(x) has one maximum at 0. f(x) has one minimum at -5 and another at -16 .
    • +
  • The end behavior of f(x) is given as: As x\to \infty, f(x)\to\infty and As x\to -\infty, f(x)\to-\infty
    • +
+ + + + +

Now that we know all the different characterisitics of polynomials, we should also be able to identify them from a graph. Use the graph below to find the given charactertistics.

+
+ + Polynomial Function + + f(x) = (1*x^3-3*x-2) + p=plot(f, (x, -5, 5), ymin=-10, ymax=10, color='blue', thickness=3) + + +
+ + + +

What are the x-intercept(s) of the polynomial function? Select all that apply. +

    +
  1. (1, 0)
    2. +
  2. (-1, 0)
    3. +
  3. (2, 0)
    4. +
  4. (0, -2)

+ + + + +

What are the y-intercept(s) of the polynomial function? +

    +
  1. (1, 0)
    2. +
  2. (-1, 0)
    3. +
  3. (2, 0)
    4. +
  4. (0, -2)

+ + + + +

How many zeros does this polynomial function have? +

    +
  1. 0
    2. +
  2. 1
    3. +
  3. 2
    4. +
  4. 3
+

+ + + + +

At what point is the local minimum located? +

    +
  1. (2, -4)
    2. +
  2. (-1, 0)
    3. +
  3. (-2, 0)
    4. +
  4. (1, -4)
    5. +
  5. (2, 0)
    6. +

+ + + + +

At what point is the local maximum located? +

    +
  1. (2, -4)
    2. +
  2. (-1, 0)
    3. +
  3. (-2, 0)
    4. +
  4. (1, -4)
    5. +
  5. (2, 0)
    6. +

+ + + + +

How do you describe the behavior of the polynomial function as x\to \infty? +

    +
  1. the y-values go to negative infinity
    2. +
  2. f(x) \to \infty
    3. +
  3. the y-values go to positive infinity
    4. +
  4. f(x) \to -\infty
    5. +

+ + + + +

How do you describe the behavior of the polynomial function as x\to -\infty? +

    +
  1. the y-values go to negative infinity
    2. +
  2. f(x) \to \infty
    3. +
  3. the y-values go to positive infinity
    4. +
  4. f(x) \to -\infty
    5. +

+ + + + + + +

Use the given function, f(x)=(x+1)^2(x-5) , to answer the following questions. +

+ + + +

What are the zeros of f(x) ? +

    +
  1. -1, -5
    2. +
  2. -1, 5
    3. +
  3. 1, -5
    4. +
  4. 1, 5

+ + + + +

What are the multiplicities at each zero? +

    +

  1. At x=-1, the mulitplicity is even.

    +

    At x=5, the mulitplicity is even.

    2. +

  2. At x=-1, the mulitplicity is even.

    +

    At x=5, the mulitplicity is odd.

    3. +

  3. At x=-1, the mulitplicity is odd.

    +

    At x=5, the mulitplicity is even.

    4. +

  4. At x=-1, the mulitplicity is odd.

    +

    At x=5, the mulitplicity is odd.

+

+ + + + +

What is the end behavior of f(x) ? +

    +
  1. f(x) rises on the left and on the right.
    2. +
  2. f(x) rises on the left, but falls on the right.
    3. +
  3. f(x) rises on the right, but falls on the left.
    4. +
  4. f(x) falls on the left and on the right.

+ + + + +

+ Using what you know about the zeros, multiplicities, and end behavior, where on the graph can we estimate the local maxima and minima to be? +

+ + + + +

+ Now look at the graph of f(x). At which zero does a local maximum or local minimum occur? Explain how you know. +

+ + + + + + +

We can estimate where these local maxima and minima occur by looking at other characteristics, such as multiplicities and end behavior. +

+ +

From , we saw that when the function touches the x-axis at a zero, then that zero could be either a local maximum or a local minimum of the graph. When the function crosses the x-axis, however, the local maximum or local minimum occurs between the zeros. +

+ + + + + +

Exercises available at .

+ + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/04-PR/06.ptx b/precalculus/source/04-PR/06.ptx new file mode 100644 index 00000000..c73e8285 --- /dev/null +++ b/precalculus/source/04-PR/06.ptx @@ -0,0 +1,851 @@ + + +
+ Properties and Graphs of Rational Functions (PR6) + + + + + Activities + + + + +

A function r is rationalrational function provided that it is possible to write r as the ratio of two polynomials, p and q. That is, r is rational provided that for some polynomial functions p and q, we have + r(x) = \frac{p(x)}{q(x)}\text{.} + +

+ + + + + +

+ Rational functions occur in many applications, so our goal in this lesson is to learn about their properties and be able to graph them. In particular we want to investigate the domain, end behavior, and zeros of rational functions. +

+ + + + + +

Consider the rational function + r(x) = \frac{x^2-3x+2}{x^2-4x+3}\text{.}

+ + + + +

Find r(1) , r(2) , r(3), and r(4).

+ + + + +

Label each of these four points as giving us information about the DOMAIN of r(x), information about the ZEROES of r(x), or NEITHER. +

+ + + + + + + + + +

+ Let p and q be polynomial functions so that r(x) = \frac{p(x)}{q(x)} is a rational function. The domainrational functiondomain of r is the set of all real numbers except those for which q(x) = 0. +

+ + + + + + + +

Let's investigate the domain of r(x) more closely. We will be using the same function from the previous activity:

+ r(x) = \frac{x^2-3x+2}{x^2-4x+3}\text{.}

+ + + + +

Rewrite r(x) by factoring the numerator and denominator, but do not try to simplify any further. What do you notice about the relationship between the values that are not in the domain and how the function is now written?

+ + + + + +

The function was not defined for x=3. Make a table for values of r(x) near x=3. + + + + x + r(x) + + + 2 + + + 2.9 + + + 2.99 + + + 2.999 + + + 3 + undefined + + + 3.001 + + + 3.01 + + + 3.1 + + +

+ + + + +

Which of the following describe the behavior of the graph near x=3? +

    +
  1. As x\to 3, r(x) approaches a finite number
    2. +
  2. As x\to 3 from the left, r(x)\to\infty
    3. +
  3. As x \to 3 from the left, r(x)\to -\infty
    4. +
  4. As x \to 3 from the right, r(x)\to \infty
    5. +
  5. As x \to 3 from the right, r(x)\to -\infty
+

+ + + + +

The function was also not defined for x=1. Make a table for values of r(x) near x=1. + + + + x + r(x) + + + 0 + + + 0.9 + + + 0.99 + + + 0.999 + + + 1 + undefined + + + 1.001 + + + 1.01 + + + 1.1 + + +

+ + + + +

Which of the following describe the behavior of the graph near x=1? +

    +
  1. As x\to 1, r(x) approaches a finite number
    2. +
  2. As x\to 1 from the left, r(x)\to\infty
    3. +
  3. As x \to 1 from the left, r(x)\to -\infty
    4. +
  4. As x \to 1 from the right, r(x)\to \infty
    5. +
  5. As x \to 1 from the right, r(x)\to -\infty
+

+ + + + +

The function is behaving differently near x=1 than it is near x=3. Can you see anything in the factored form of r(x) that may help you account for the difference?

+ + + + + + + + + + + Features of a rational function +

+ Let r(x) = \frac{p(x)}{q(x)} be a rational function. +

+ +

+

    +
  • +

    + If p(a) = 0 and q(a) \ne 0, then r(a) = 0, so r has a zerorational functionzerorational functionzero at x = a. +

    +
    • +
  • +

    + If q(a) = 0 and p(a) \ne 0, then r(a) is undefined and r has a vertical asymptoterational functionvertical asymptoterational functionvertical asymptote at x = a. +

    +
    • +
  • +

    + If p(a) = 0 and q(a) = 0 and we can show that there is a finite number L such that + + r(x) \to L + , + then r(a) is not defined and r has a holerational functionholerational functionhole at the point (a,L). +

    +
    • +
+

+ + + + +

Another property of rational functions we want to explore is the end behavior. This means we want to explore what happens to a given rational function r(x) when x goes toward positive infinity or negative infinity. + +

+ + + + + + +

+ Consider the rational function \displaystyle r(x)=\frac{1}{x^3}. +

+ + + +

Plug in some very large positive numbers for x to see what r(x) is tending toward. Which of the following best describes the behavior of the graph as x approaches positive infinity? +

    +
  1. As x\to \infty, r(x)\to \infty.
    2. +
  2. As x\to \infty, r(x)\to -\infty.
    3. +
  3. As x\to \infty, r(x)\to 0.
    4. +
  4. As x\to \infty, r(x)\to 1.
    5. +
+

+ + + + + + + +

Now let's look at r(x) as x tends toward negative infinity. Plug in some very large negative numbers for x to see what r(x) is tending toward. Which of the following best describes the behavior of the graph as x approaches negative infinity? +

    +
  1. As x\to -\infty, r(x)\to \infty.
    2. +
  2. As x\to -\infty, r(x)\to -\infty.
    3. +
  3. As x\to -\infty, r(x)\to 0.
    4. +
  4. As x\to -\infty, r(x)\to 1.
    5. +
+

+ + + + + + + + +

+ We can generalize what we have just found to any function of the form \frac{1}{x^n}, where n>0. Since x^n increases without bound as x \to \infty, we find that \frac{1}{x^n} will tend to 0. In fact, the numerator can be any constant and the function will still tend to 0! +

+

+ Similarly, as x \to -\infty, we find that \frac{1}{x^n} will tend to 0 too. +

+ + + + + + + +

+ Consider the rational function \displaystyle r(x) = \frac{3x^2 - 5x + 1}{7x^2 + 2x - 11}. +

+

+ Observe that the largest power of x that's present in r(x) is x^2. In addition, because of the dominant terms of 3x^2 in the numerator and 7x^2 in the denominator, both the numerator and denominator of r increase without bound as x increases without bound. +

+ + + + +

+ In order to understand the end behavior of r, we will start by writing the function in a different algebraic form. +

+

+ Multiply the numerator and denominator of r by \frac{1}{x^2}. Then distribute and simplify as much as possible in both the numerator and denominator to write r in a different algebraic form. Which of the following is that new form? + +

    +
  1. + \frac{3x^4 - 5x^3 +x^2}{7x^4+2x^3-11x^2} + +
    2. +
  2. + \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{7 + \frac{2}{x} - \frac{11}{x^2}} + +
    3. +
  3. + \frac{\frac{3x^2}{x^2} - \frac{5x}{x^2} + \frac{1}{x^2}}{\frac{7x^2}{x^2} + \frac{2x}{x^2} - \frac{11}{x^2}} + +
    4. +
  4. + + \frac{3x^2 - 5x +1}{7x^4+2x^3-11x^2} + +
    5. + +
+

+ + + + + +

+ Now determine the end behavior of each piece of the numerator and each piece of the denominator. +

+ + +

+ Use to help! +

+ + + + + +

+ Simplify your work from the previous step. Which of the following best describes the end behavior of r(x)? +

    +
  1. As x \to \pm \infty, r(x) goes to 0. +
    2. +
  2. As x \to \pm \infty, r(x) goes to \frac{3}{7}. +
    3. +
  3. As x \to \pm \infty, r(x) goes to \infty. +
    4. +
  4. As x \to \pm \infty, r(x) goes to -\infty. +
    5. + +
+ +

+ + + + + + +

+ If the end behavior of a function tends toward a specific value a, then we say that the function has a horizontal asymptote at y=a. +

+ + + + + + + +

+ Find the horizontal asymptote (if one exists) of the following rational functions. Follow the same method we used in . +

+ + + + +

+ f(x)=\dfrac{4x^3-3x^2+6}{9x^3+7x-5} +

    +
  1. y=0 +
    2. +
  2. y=\frac{4}{9} +
    3. +
  3. y=-\frac{3}{7} +
    4. +
  4. y=-\frac{6}{5} +
    5. +
  5. There is no horizontal asymptote. +
    6. +
+

+ + + + +

+ g(x)=\dfrac{4x^3-3x^2+6}{9x^5+7x-5} +

+
    +
  1. y=0 +
    2. +
  2. y=\frac{4}{9} +
    3. +
  3. y=-\frac{3}{7} +
    4. +
  4. y=-\frac{6}{5} +
    5. +
  5. There is no horizontal asymptote. +
    6. +
+ + + + +

+ h(x)=\dfrac{4x^5-3x^2+6}{9x^3+7x-5} +

    +
  1. y=0 +
    2. +
  2. y=\frac{4}{9} +
    3. +
  3. y=-\frac{3}{7} +
    4. +
  4. y=-\frac{6}{5} +
    5. +
  5. There is no horizontal asymptote. +
    6. +
+

+ + + + + + + +

+ Some patterns have emerged from the previous problem. Fill in the rest of the sentences below to describe how to find horizontal asymptotes of rational functions. +

+ + + + +

+ If the degree of the numerator is the same as the degree of the denominator, then... +

+ + + + +

+ If the degree of the numerator is less than the degree of the denominator, then... +

+ + + + +

+ If the degree of the numerator is greater than the degree of the denominator, then... +

+ + + + + + +

Consider the following six graphs of rational functions:

+ + + + + f(x) = 2*x/(x^2-2*x-8) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((-2,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((4,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,0),(x,-7,7),color='red',linestyle="--") + p+=text('Graph A', (0, 8), fontsize=18, color='black') + p + + + + + f(x) = (x^2+3)/(2*x^2-8)-1 + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((-2,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((2,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,-1),(x,-7,7),color='red',linestyle="--") + p+=text('Graph B', (0, 8), fontsize=18, color='black') + p + + + + + f(x) = 1-(x^2+3)/(2*x^2-8) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((-2,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((2,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,1),(x,-7,7),color='red',linestyle="--") + p+=text('Graph C', (0, 8), fontsize=18, color='black') + p + + + + + + + f(x) = 2*x/(x^2-2*x-8)-3 + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((-2,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((4,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,-3),(x,-7,7),color='red',linestyle="--") + p+=text('Graph D', (0, 8), fontsize=18, color='black') + p + + + + + f(x) = (x^2+3)/(2*x^2-8) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((-2,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((2,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,0),(x,-7,7),color='red',linestyle="--") + p+=text('Graph E', (0, 8), fontsize=18, color='black') + p + + + + + f(x) = -2*x/(x^2-2*x-8) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((-2,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((4,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,0),(x,-7,7),color='red',linestyle="--") + p+=text('Graph F', (0, 8), fontsize=18, color='black') + p + + + + + + + +

Which of the graphs above represents the function \displaystyle f(x)=\frac{2x}{x^2-2x-8}?

+ + + + +

Which of the graphs above represents the function \displaystyle g(x)=\frac{x^2+3}{2x^2-8}?

+ + + + + + + + +

Consider the following six graphs of rational functions:

+ + + + + f(x) = ((x+2)*(x+3))/((x+2)*(x-5)*(x+4)) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((5,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((-4,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,0),(x,-7,7),color='red',linestyle="--") + p+=text('Graph A', (0, 8), fontsize=18, color='black') + p+=point((-2, -1/14), pointsize=50, markeredgecolor='blue',color='white',zorder=7) + p + + + + + + f(x) = ((x+2)*(x+3))/((x+2)*(x-5)*(x+4)) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((5,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((-4,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,0),(x,-7,7),color='red',linestyle="--") + p+=text('Graph B', (0, 8), fontsize=18, color='black') + p+=point((-3, 0), pointsize=50, markeredgecolor='blue',color='white',zorder=7) + p + + + + + + f(x) = ((x+2))/((x+2)*(x-5)*(x+4)) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((-4,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((5,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,0),(x,-7,7),color='red',linestyle="--") + p+=text('Graph C', (0, 8), fontsize=18, color='black') + p+=point((-2, 1/14), pointsize=50, markeredgecolor='blue',color='white',zorder=7) + p + + + + + + + + f(x) = ((x-5))/((x+2)*(x-5)*(x+4)) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((-4,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((-2,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,0),(x,-7,7),color='red',linestyle="--") + p+=text('Graph D', (0, 8), fontsize=18, color='black') + p+=point((5, 1/63), pointsize=50, markeredgecolor='blue',color='white',zorder=7) + p + + + + + f(x) = ((x+2)*(x-5))/((x+2)*(x-5)*(x+4)) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((-4,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,0),(x,-7,7),color='red',linestyle="--") + p+=text('Graph E', (0, 8), fontsize=18, color='black') + p+=point((-2, 1/2), pointsize=50, markeredgecolor='blue',color='white',zorder=7) + p+=point((5, 1/9), pointsize=50, markeredgecolor='blue',color='white',zorder=7) + p + + + + + f(x) = ((x+2)*(x+3))/((x+2)*(x-5)*(x+4)) + p=plot(f, (x, -7, 7), ymin=-7, ymax=7, color='blue', thickness=3, detect_poles=True) + p+=parametric_plot((5,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((-4,x),(x,-8,8),color='red',linestyle="--") + p+=parametric_plot((x,0),(x,-7,7),color='red',linestyle="--") + p+=text('Graph F', (0, 8), fontsize=18, color='black') + p + + + + + + + + + +

Which of the graphs above represents the function \displaystyle f(x)=\frac{x^2+5x+6}{(x^2-3x-10)(x+4)}?

+ + + + +

Which of the graphs above represents the function \displaystyle g(x)=\frac{x^2-3x-10}{(x+2)(x^2-x-20)}?

+ + + + + + +

+ Let f(x)=\displaystyle \frac{ -{\left(x - 1\right)} {\left(x - 4\right)} }{ 2 \, {\left(x + 3\right)}^{2} {\left(x - 1\right)} }. +

+ + + +

Find the roots of f(x).

+ + + + + +

Find the y-intercept of the graph of f(x).

+ + + + + +

Find any horizontal asymptotes on the graph of f(x).

+ + + + + + +

Find any vertical asymptotes on the graph of f(x).

+ + + + + +

Find any holes on the graph of f(x).

+ + + + + +

Sketch the graph of f(x).

+ + + + + + + + + + +

For each of the following rational functions, identify the location of any potential hole in the graph. Then, create a table of function values for input values near where the hole should be located. Use your work to decide whether or not the graph indeed has a hole, with written justification.

+ + + + +

+ \displaystyle r(x) = \frac{x^2-16}{x+4} +

+ + + + +

+ \displaystyle s(x) = \frac{(x-2)^2(x+3)}{x^2 - 5x - 6} +

+ + + + +

+ \displaystyle u(x) = \frac{(x-2)^3(x+3)}{(x^2 - 5x - 6)(x-7)} +

+ + + + +

+ \displaystyle w(x) = \frac{x^2 + x - 6}{(x^2 + 5x + 6)(x+3)} +

+ + + + + + +

+ Suppose you are given a function r(x) = \frac{p(x)}{q(x)}, and you know that p(3) = 0 and q(3) = 0. + What can you conlude about the function r(x) at x = 3? +

+
    +
  1. r(x) has a hole at x=3.
    2. +
  2. r(x) has an asymptote at x=3.
    3. +
  3. r(x) has either a hole or an asymptote at x=3.
    4. +
  4. r(x) has neither a hole nor an asymptote at x=3.
    5. +
+ + + + + + + + + + +

Exercises available at .

+ + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/04-PR/main.ptx b/precalculus/source/04-PR/main.ptx new file mode 100644 index 00000000..ba7f3965 --- /dev/null +++ b/precalculus/source/04-PR/main.ptx @@ -0,0 +1,18 @@ + + + + Polynomial and Rational Functions (PR) + + + + + + + + + + + + + + \ No newline at end of file diff --git a/precalculus/source/04-PR/outcomes/01.ptx b/precalculus/source/04-PR/outcomes/01.ptx new file mode 100644 index 00000000..e50ff027 --- /dev/null +++ b/precalculus/source/04-PR/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+Graph quadratic functions and identify their axis of symmetry, and maximum or minimum point. +

\ No newline at end of file diff --git a/precalculus/source/04-PR/outcomes/02.ptx b/precalculus/source/04-PR/outcomes/02.ptx new file mode 100644 index 00000000..6499552c --- /dev/null +++ b/precalculus/source/04-PR/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+Use quadratic models to solve an application problem and establish conclusions. +

\ No newline at end of file diff --git a/precalculus/source/04-PR/outcomes/03.ptx b/precalculus/source/04-PR/outcomes/03.ptx new file mode 100644 index 00000000..526338c8 --- /dev/null +++ b/precalculus/source/04-PR/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+Rewrite a rational function as a polynomial plus a proper rational function. +

diff --git a/precalculus/source/04-PR/outcomes/04.ptx b/precalculus/source/04-PR/outcomes/04.ptx new file mode 100644 index 00000000..b0a82c16 --- /dev/null +++ b/precalculus/source/04-PR/outcomes/04.ptx @@ -0,0 +1,6 @@ + +

+Determine the zeros of a real polynomial function, write a polynomial function given +information about its zeros and their multiplicities, and apply the Factor Theorem +and the Fundamental Theorem of Algebra. +

\ No newline at end of file diff --git a/precalculus/source/04-PR/outcomes/05.ptx b/precalculus/source/04-PR/outcomes/05.ptx new file mode 100644 index 00000000..670f5796 --- /dev/null +++ b/precalculus/source/04-PR/outcomes/05.ptx @@ -0,0 +1,5 @@ + +

+Find the intercepts, estimated locations of maxima and minima, and end behavior of a polynomial function, and use this +information to sketch the graph. +

\ No newline at end of file diff --git a/precalculus/source/04-PR/outcomes/06.ptx b/precalculus/source/04-PR/outcomes/06.ptx new file mode 100644 index 00000000..f7c23d52 --- /dev/null +++ b/precalculus/source/04-PR/outcomes/06.ptx @@ -0,0 +1,7 @@ + +

+Find the domain and range, +vertical and horizontal asymptotes, +and intercepts of a rational function +and use this information to sketch the graph. +

\ No newline at end of file diff --git a/precalculus/source/04-PR/outcomes/main.ptx b/precalculus/source/04-PR/outcomes/main.ptx new file mode 100644 index 00000000..45c3190f --- /dev/null +++ b/precalculus/source/04-PR/outcomes/main.ptx @@ -0,0 +1,31 @@ + +> + +

+ BIG IDEA for the chapter goes here, in outcomes/main.ptx +

+

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
+ \ No newline at end of file diff --git a/precalculus/source/04-PR/readiness.ptx b/precalculus/source/04-PR/readiness.ptx new file mode 100644 index 00000000..abb43153 --- /dev/null +++ b/precalculus/source/04-PR/readiness.ptx @@ -0,0 +1,121 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Evaluate functions from an equation.

    +
      +

    • Review and Practice: Khan Academy

      +
      • +

    • Review and Practice: Section 2.2

      +
      • +
    +
    2. +
  2. +

    Find the intercepts of a line.

    +
      +

    • Review and Practice: Khan Academy

      +
      • + +
    +
    3. +
  3. +

    Find the intercepts of a graph of a function.

    +
      +

    • Review and Practice: Section 2.3

      +
      • +
    +
    4. + + +
  4. +

    Find the degree and leading coefficient of a polynomial.

    +
      + +

    • Review and Practice: Khan Academy

      +
      • +
    +
    5. +
  5. +

    Determine if a graph is a polynomial function.

    +
      + +

    • Review: YouTube (Math Tutorials)

      +
      • +
    +
    6. + +
  6. +

    Determine where a function is increasing and decreasing.

    +
      + +

    • Review and Practice: Khan Academy

      +
      • +
    +
    7. + + + +
  7. +

    Find the equation of a line given two points.

    +
      + +

    • Review and Practice: Khan Academy

      +
      • +

    • Review and Practice: Section 3.2

      +
      • +
    +
    8. +
  8. +

    Factor quadratics and polynomials.

    +
      + +

    • Review and Practice (Quadratics): Khan Academy

      +
      • +

    • Review and Practice (Polynomials): Khan Academy

      +
      • +
    +
    9. +
  9. +

    Perform long division of two whole numbers.

    +
      + +

    • Review and Practice: Khan Academy

      +
      • +
    +
    10. +
  10. +

    Solve quadratic equations with complex roots.

    +
      + +

    • Review and Practice: Khan Academy

      +
      • +

    • Review and Practice: Section 1.5

      +
      • +
    +
    11. + + + +
  11. +

    Simplify rational expressions.

    +
      + +

    • Review and Practice: Khan Academy

      +
      • +
    +
    12. + + + + + + +
+ + diff --git a/precalculus/source/05-EL/01.ptx b/precalculus/source/05-EL/01.ptx new file mode 100644 index 00000000..9205cbce --- /dev/null +++ b/precalculus/source/05-EL/01.ptx @@ -0,0 +1,962 @@ + + +
+ Introduction to Exponentials (EL1) + + + + + Activities + + + Linear functions have a constant rate of change - that is a constant change in output for every change in input. Let's consider functions which do not fit this model - those which grow more rapidly and change by a varying amount for every change in input. + + + +

+ You have two job offers on the horizon. One has offered to pay you \$10{,}000 per month while the other is offering \$0.01 the first month, \$0.02 the second month, \$0.04 the third month and doubles every month. Which job would you rather take? +

+ + + +

+ Make a table representing how much money you will be paid each month for the first two years from the first job - paying \$10{,}000 per month. +

+ + +

+ + + month + Job 1 + + + 1 + 10{,}000 + + + 2 + 10{,}000 + + + 3 + 10{,}000 + + + 4 + 10{,}000 + + + 5 + 10{,}000 + + + 6 + 10{,}000 + + + 7 + 10{,}000 + + + 8 + 10{,}000 + + + 9 + 10{,}000 + + + 10 + 10{,}000 + + + 11 + 10{,}000 + + + 12 + 10{,}000 + + + 13 + 10{,}000 + + + 14 + 10{,}000 + + + 15 + 10{,}000 + + + 16 + 10{,}000 + + + 17 + 10{,}000 + + + 18 + 10{,}000 + + + 19 + 10{,}000 + + + 20 + 10{,}000 + + + 21 + 10{,}000 + + + 22 + 10{,}000 + + + 23 + 10{,}000 + + + 24 + 10{,}000 + + +

+ + + + +

+ Make a table representing how much money you will be paid each month for the first two years from the second job - paying \$0.01 the first month and doubling every month after. +

+ + +

+ + + month + Job 2 + + + 1 + 0.01 + + + 2 + 0.02 + + + 3 + 0.04 + + + 4 + 0.08 + + + 5 + 0.16 + + + 6 + 0.32 + + + 7 + 0.64 + + + 8 + 1.28 + + + 9 + 2.56 + + + 10 + 5.12 + + + 11 + 10.24 + + + 12 + 20.48 + + + 13 + 40.96 + + + 14 + 81.92 + + + 15 + 163.84 + + + 16 + 327.68 + + + 17 + 655.36 + + + 18 + 1{,}310.72 + + + 19 + 2{,}621.24 + + + 20 + 5{,}242.88 + + + 21 + 10{,}485.76 + + + 22 + 20{,}971.52 + + + 23 + 41{,}943.04 + + + 24 + 83{,}886.08 + + +

+ + + + +

+ Which job is earning more money per month after one year? +

+ + +

+ Job 1 is earning \$10{,}000 per month. Job 2 is earning \$20.48 per month. +

+ + + + +

+ Which job is earning more money per month after 18 months? +

+ + +

+ Job 1 is earning \$10{,}000 per month. Job 2 is earning \$1{,}310.72 per month. +

+ + + + +

+ According to your tables, does the second job ever earn more money per month than the first job? +

+ + +

+ Yes! After 21 months, Job 1 is earning \$10{,}000 per month. Job 2 is earning \$10{,}485.76 per month. +

+ + + + + This idea of a function that grows very rapidly by a factor, ratio, or percent each time, like the second job in , is considered exponential growth. + + + +

+ Let a be a non-zero real number and b \neq 1 a positive real number. An exponential function takes the form + + f(x)=ab^{x} + + a is the initial value and b is the base. + +

+ + + + +

+ Evaluate the following exponential functions. +

+ + + +

+ f(x)=4^{x} for f(3) +

+ + +

+ f(3)=64 +

+ + + +

+ f(x)=\left( \frac{1}{3} \right)^{x} for f(3) +

+ + +

+ f(3)=\frac{1}{27} +

+ + + + +

+ f(x)=3\left( 5 \right)^{x} for f(-2) +

+ + +

+ f(-2)=\frac{3}{25} +

+ + + + +

+ f(x)=-2^{3x-4} for f(4) +

+ + +

+ f(4)=-256 +

+ + + + + +

+ Notice that in part (a) the ouput value is larger than the base, + while in part (b) the output value is smaller than the base. This is similar to the difference + between a positive and negative slope for linear functions. +

+ + + + +

+ Consider two exponential functions f(x)=100(2)^{x} and g(x)=100 \left( \frac{1}{2} \right)^{x}. +

+ + + +

+ Fill in the table of values for f(x). + + + x + f(x) + + + 0 + + + 1 + + + 2 + + + 3 + + + 4 + + +

+ + +

+ + + x + f(x) + + + 0 + 100 + + + 1 + 200 + + + 2 + 400 + + + 3 + 800 + + + 4 + 1600 + + +

+ + + + +

+ Fill in the table of values for g(x). + + + x + g(x) + + + 0 + + + 1 + + + 2 + + + 3 + + + 4 + + +

+ + +

+ + + x + f(x) + + + 0 + 100 + + + 1 + 50 + + + 2 + 25 + + + 3 + 12.5 + + + 4 + 6.25 + + +

+ + + + +

+ How do the values in the tables compare? +

+ + +

+ The values of f(x) are getting larger while those of g(x) are getting smaller. +

+ + + + +

+ In , the only difference between the two exponential functions was the base. + f(x) has a base of 2, while g(x) has a base of \frac{1}{2}. + Let's use this fact to update . +

+ + + +

+ An exponential function of the form f(x)=ab^{x} will grow (or increase) if b \gt 1 and decay (or decrease) if 0\lt b \lt 1. +

+ + + + +

+ For each year t, the population of a certain type of tree in a forest is represented by the function F(t)=856(0.93)^t. +

+ + + +

+ How many of that certain type of tree are in the forest initially? +

+ + +

+ 856 trees +

+ + + + +

+ Is the number of trees of that type growing or decaying? +

+ + +

+ Decaying b=0.93 \lt 1 +

+ + + + + +

+ To begin creating equations for exponential functions using a and b, let's compare a linear function and an exponential function. The tables show outputs for two different functions r and s that correspond to equally spaced input. + + + + + + x + r(x) + + + 0 + 12 + + + 3 + 10 + + + 6 + 8 + + + 9 + 6 + + + + + + + + x + s(x) + + + 0 + 12 + + + 3 + 9 + + + 6 + 6.75 + + + 9 + 5.0625 + + + + +

+ + + +

+ Which function is linear? +

+ + +

+ r(x) since the outputs decrease by 2 every time. +

+ + + + +

+ What is the initial value of the linear function? +

+ + +

+ 12 +

+ + + + +

+ What is the slope of the linear function? +

+ + +

+ -\frac{2}{3} +

+ + + + +

+ What is the initial value of the exponential function? +

+ + +

+ 12 +

+ + + + +

+ What is the ratio of consecutive outputs in the exponential function? +

    +
  1. \frac{4}{3}
    2. +
  2. \frac{3}{4}
    3. +
  3. -\frac{4}{3}
    4. +
  4. -\frac{3}{4}
    5. +
+

+ + +

+ B +

+ + + + + + + In a linear function the differences are constant, while in an exponential function the ratios are constant. + + + + +

+ Find an equation for an exponential function passing through the points (0,4) and (1,6). +

+ + + +

+ Find the initial value. +

    +
  1. 0
    2. +
  2. 4
    3. +
  3. 1
    4. +
  4. 6
    5. +
+

+ + +

+ B +

+ + + + +

+ Find the common ratio. +

    +
  1. \frac{3}{2}
    2. +
  2. 6
    3. +
  3. \frac{2}{3}
    4. +
  4. \frac{1}{6}
    5. +
+

+ + +

+ A +

+ + + + +

+ Find the equation. +

    +
  1. f(x)=6\left( \frac{3}{2} \right)^x
    2. +
  2. f(x)=4\left( 6 \right)^x
    3. +
  3. f(x)=4\left( \frac{2}{3} \right)^x
    4. +
  4. f(x)=4\left( \frac{3}{2} \right)^x
    5. +
+

+ + +

+ D: f(x)=4\left( \frac{3}{2} \right)^x +

+ + + + + + + + +

+ The letter e represents an irrational number and is used as a base for many real-world exponential models. To work with base e, we use the approximation, e \approx 2.718282. It can also be found on most calculators. +

+ + + + + + +

+ Use a calculator to evaluate the following exponentials involving the base e. +

+ + + +

+ f(x)=-2e^{x}-2 for f(-2) +

+
    +
  1. -0.0366
    2. +
  2. -2.2707
    3. +
  3. -1.7293
    4. +
  4. -16.778
    5. +
+ + +

+ B +

+ + + + +

+ f(x)=\frac{1}{3}e^{x+1} for f(-1) +

    +
  1. 1
    2. +
  2. 0
    3. +
  3. 1.122
    4. +
  4. \frac{1}{3}
    5. +
+

+ + +

+ D +

+ + + + + Recall the negative rule of exponents which states that for any nonzero real number a and natural number n + + a^{-n}=\frac{1}{a^n} + + + + + +

+ Let's consider the two exponential functions f(x)=2^{-x} and g(x)=\left(\frac{1}{2}\right)^{x}. +

+ + + +

+ Fill in the table of values for f(x). + + + x + f(x) + + + -2 + + + -1 + + + 0 + + + 1 + + + 2 + + +

+ + +

+ + + x + f(x) + + + -2 + 4 + + + -1 + 2 + + + 0 + 1 + + + 1 + \frac{1}{2} + + + 2 + \frac{1}{4} + + +

+ + + + +

+ Fill in the table of values for g(x). + + + x + g(x) + + + -2 + + + -1 + + + 0 + + + 1 + + + 2 + + +

+ + +

+ + + x + g(x) + + + -2 + 4 + + + -1 + 2 + + + 0 + 1 + + + 1 + \frac{1}{2} + + + 2 + \frac{1}{4} + + +

+ + + + +

+ What do you notice about the two functions? +

+ + +

+ Their values are the same. +

+ + + + +

+ Use and other properties of exponents to try and rewrite f(x) as g(x). +

+ + +

+ + f(x)=2^{-x}=\frac{1}{2^x}=\left(\frac{1}{2}\right)^x=g(x) + +

+ + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/05-EL/02.ptx b/precalculus/source/05-EL/02.ptx new file mode 100644 index 00000000..5cc87522 --- /dev/null +++ b/precalculus/source/05-EL/02.ptx @@ -0,0 +1,1082 @@ + + +
+ Graphs of Exponential Functions (EL2) + + + + + Activities + + + +

Consider the function f(x)=2^x. +

+ + + + +

+ Fill in the table of values for f(x). Then plot the points on a graph. + + + x + f(x) + + + -2 + + + -1 + + + 0 + + + 1 + + + 2 + + +

+ + + + + + + x + f(x) + + + -2 + \frac{1}{4} + + + -1 + \frac{1}{2} + + + 0 + 1 + + + 1 + 2 + + + 2 + 4 + + + + + + p=point([(-2,.25),(-1,.5),(0,1),(1,2),(2,4)],pointsize=50,color='blue',ymin=-2, ymax=8, xmin=-8,xmax=8, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + + + + + + + + + +

+ What seems to be happening with the graph as x goes toward infinity? Plug in large positive values of x to test your guess, then describe the end behavior. + +

    +

  1. As x \to \infty, f(x) \to -\infty.

    2. +

  2. As x \to \infty, f(x) \to -2.

    3. +

  3. As x \to \infty, f(x) \to 0.

    4. +

  4. As x \to \infty, f(x) \to 2.

    5. +

  5. As x \to \infty, f(x) \to \infty.

    6. + + +
+

+ + + E. + + + + + +

+ What seems to be happening with the graph as x goes toward negative infinity? Plug in large negative values of x to test your guess, then describe the end behavior. + +

    +

  1. As x \to-\infty, f(x) \to -\infty.

    2. +

  2. As x \to -\infty, f(x) \to -2.

    3. +

  3. As x \to -\infty, f(x) \to 0.

    4. +

  4. As x \to -\infty, f(x) \to 2.

    5. +

  5. As x \to -\infty, f(x) \to \infty.

    6. +
+

+ + + C. + + + + + +

+ Complete the graph you started in , connecting the points and including the end behavior you've just determined. +

+ + +

+ + + f(x) = 2^x + p=point([(-2,.25),(-1,.5),(0,1),(1,2),(2,4)],pointsize=50,color='blue') + p+=plot(f, (x, -9, 8), ymin=-2, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + +

+ + + + + +

+ Does your graph seem to have any asymptotes? + +

    + +

  1. No. There are no asymptotes.

    2. + +

  2. There is a vertical asymptote but no horizontal one.

    3. + +

  3. There is a horizontal asymptote but no vertical one.

    4. + +

  4. The graph has both a horizontal and vertical asymptote.

    5. + + +
+ +

+ + +

+ C. +

+ + + + + +

+ What the equation for each asymptote of f(x)? Select all that apply. + +

    +

  1. There are no asymptotes.

    2. +

  2. x=0

    3. +

  3. x=3

    4. +

  4. y=0

    5. +

  5. y=3

    6. +
+ +

+ + +

+ D. +

+ + + + + +

+ Find the domain and range of f(x). Write your answers using interval notation. +

+ + +

+ Domain: (-\infty,\infty), Range: (0, \infty) +

+ + + + + +

+ Find the interval(s) where f(x) is increasing and the interval(s) where f(x) is decreasing. Write your answers using interval notation. +

+ + +

+ Increasing: (-\infty,\infty), Decreasing: nowhere +

+ + + + + + +

+ The graph of an exponential function f(x)=b^x where b>1 has the following characteristics: + +

    +

  • Its domain is (-\infty,\infty) and its range is (0,\infty).

    • +

  • It is an exponential growth function; that is it is increasing on (-\infty,\infty).

    • +

  • There is a horizontal asymptote at y=0. There is no vertical asymptote.

    • +

  • There is a y-intercept at (0,1). There is no x-intercept.

    • +
+

+ + + + +

Consider the function g(x)=\left(\frac{1}{2}\right)^x. +

+ + + + +

+ Fill in the table of values for g(x). Then plot the points on a graph. + + + x + g(x) + + + -2 + + + -1 + + + 0 + + + 1 + + + 2 + + +

+ + + + + + + x + g(x) + + + -2 + 4 + + + -1 + 2 + + + 0 + 1 + + + 1 + \frac{1}{2} + + + 2 + \frac{1}{4} + + + + + + p=point([(-2,4),(-1,2),(0,1),(1,.5),(2,.25)],pointsize=50,color='blue',ymin=-2, ymax=8, xmin=-8,xmax=8, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + + + + + + + + + +

+ What seems to be happening with the graph as x goes toward infinity? Plug in large positive values of x to test your guess, then describe the end behavior. + +

    +

  1. As x \to \infty, g(x) \to -\infty.

    2. +

  2. As x \to \infty, g(x) \to -2.

    3. +

  3. As x \to \infty, g(x) \to 0.

    4. +

  4. As x \to \infty, g(x) \to 2.

    5. +

  5. As x \to \infty, g(x) \to \infty.

    6. + + +
+

+ + + C. + + + + + +

+ What seems to be happening with the graph as x goes toward negative infinity? Plug in large negative values of x to test your guess, then describe the end behavior. + +

    +

  1. As x \to-\infty, g(x) \to -\infty.

    2. +

  2. As x \to -\infty, g(x) \to -2.

    3. +

  3. As x \to -\infty, g(x) \to 0.

    4. +

  4. As x \to -\infty, g(x) \to 2.

    5. +

  5. As x \to -\infty, g(x) \to \infty.

    6. +
+

+ + + E. + + + + + +

+ Complete the graph you started in , connecting the points and including the end behavior you've just determined. +

+ + +

+ + + f(x) = (.5)^x + p=point([(-2,4),(-1,2),(0,1),(1,.5),(2,.25)],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + +

+ + + + + +

+ What are the equations of the asymptote(s) of the graph? +

+ + +

+ y=0 +

+ + + + + +

+ Find the domain and range of f(x). Write your answers using interval notation. +

+ + +

+ Domain: (-\infty,\infty), Range: (0, \infty) +

+ + + + + +

+ Find the interval(s) where f(x) is increasing and the interval(s) where f(x) is decreasing. Write your answers using interval notation. +

+ + +

+ Increasing: nowhere, Decreasing: (-\infty,\infty) +

+ + + + + + + +

Consider the two exponential functions we've just graphed: f(x)=2^x and g(x)=\left(\frac{1}{2}\right)^x. +

+ + + + +

+ How are the graphs of f(x) and g(x) similar? +

+ + +

+ Answers could include basic shape, asymptote, domain, range. +

+ + + + + +

+ How are the graphs of f(x) and g(x) different? +

+ + +

+ Answers could include reflection over y-axis, one is increasing, one is decreasing. +

+ + + + + + +

+ We can now update so that it includes all values of the base of an exponential function. +

+

+ The graph of an exponential function f(x)=b^x has the following characteristics: + +

    +

  • Its domain is (-\infty,\infty) and its range is (0,\infty).

    • +
  • +

    If b>1, f(x) is increasing on (-\infty,\infty) and is an exponential growth function. If 0 < b < 1, f(x) is decreasing on (-\infty,\infty) and is an exponential decay function.

    +
    • +

  • There is a horizontal asymptote at y=0. There is no vertical asymptote.

    • +

  • There is a y-intercept at (0,1). There is no x-intercept.

    • +
+

+ + + + + +

Let's look at a third exponential function, h(x)=2^{-x}. +

+ + + + +

+ Before plotting any points or graphing, what do you think the graph might look like? What sort of characteristics might it have? +

+ + +

+ With the 2 as the base, students may assume exponential growth. But the negative in the exponent may change their mind if they remember transformations! +

+ + + + + +

+ Fill in the table of values for h(x). Then plot the points on a graph. + + + x + h(x) + + + -2 + + + -1 + + + 0 + + + 1 + + + 2 + + +

+ + + + + + + x + h(x) + + + -2 + 4 + + + -1 + 2 + + + 0 + 1 + + + 1 + \frac{1}{2} + + + 2 + \frac{1}{4} + + + + + + p=point([(-2,4),(-1,2),(0,1),(1,.5),(2,.25)],pointsize=50,color='blue',ymin=-2, ymax=8, xmin=-8,xmax=8, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + + + + + + + + + +

+ This function h(x) looks to be the same as a function we looked at previously. Use properties of exponents to rewrite h(x) in a different way. +

+ + +

+ 2^{-x} = (2^{-1})^x = \left(\frac{1}{2} \right)^x +

+ + + + + +

+ In addition to plotting points, we can use transformations to graph. If we consider f(x)=2^x to be the parent function, what transformation is needed to graph h(x)=2^{-x}? + +

    +

  1. A vertical stretch.

    2. +

  2. A horizontal stretch.

    3. +

  3. A reflection over the x-axis.

    4. +

  4. A reflection over the y-axis.

    5. + +
+ +

+ + +

+ D. +

+ + + + + + +

For a reminder of transformations, see and the following definitions: +

    +

    • +

    • +

    • +

    • +

    • +

    • +
+

+ + + + +

Let f(x)=4^{x}. +

+ + + + +

+ Graph f(x). +

+ + +

+ + + f(x) = (4)^x + p=point([(-2,1/8),(-1,1/4),(0,1),(1,4),(2,16)],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=20, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,20]]) + p + + +

+ + + + + +

+ Match the following functions to their graphs. +

    +

  • g(x)= -4^x

    • +

  • h(x)= 4^{-x}

    • +

  • j(x)= 4^{x+1}

    • +

  • k(x)= 4^x +1

    • +
+ +
    +

  1. + + f(x) = (4)^x +1 + p=point([(-2,9/8),(-1,5/4),(0,2),(1,5),(2,17)],pointsize=50,color='blue') + p+=line([(-9,1),(9,1)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=20, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,20]]) + p + +

    2. + +

  2. + + f(x) = (4)^(-x) + p=point([(2,1/8),(1,1/4),(0,1),(-1,4),(-2,16)],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=20, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,20]]) + p + +

    3. + +

  3. + + f(x) = (4)^(x+1) + p=point([(-3,1/8),(-2,1/4),(-1,1),(0,4),(1,16)],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=20, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,20]]) + p + +

    4. + +

  4. + + f(x) = -4^(x) + p=point([(-2,-1/8),(-1,-1/4),(0,-1),(1,-4),(2,-16)],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-20, ymax=2, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-20,-19,..,2]]) + p + +

    5. + + +
+

+ + +

+

    +

  • g(x)= -4^x is D

    • +

  • h(x)= 4^{-x} is B

    • +

  • j(x)= 4^{x+1} is C

    • +

  • k(x)= 4^x +1 is A

    • +
+

+ + + + + +

+ Find the domain, range, and equation of the asymptote for the parent function \left(f(x)\right) and each of the four transformations \left(g(x), h(x), j(x), \text{ and } k(x)\right). +

+ + +

+ f(x): +

    +

  • Domain: (-\infty,\infty)

    • +

  • Range: (0,\infty)

    • +

  • Asymptote: y=0

    • +
+ + g(x): +
    +

  • Domain: (-\infty,\infty)

    • +

  • Range: (-\infty,0)

    • +

  • Asymptote: y=0

    • +
+ + h(x): +
    +

  • Domain: (-\infty,\infty)

    • +

  • Range: (0,\infty)

    • +

  • Asymptote: y=0

    • +
+ + j(x): +
    +

  • Domain: (-\infty,\infty)

    • +

  • Range: (0,\infty)

    • +

  • Asymptote: y=0

    • +
+ + k(x): +
    +

  • Domain: (-\infty,\infty)

    • +

  • Range: (1,\infty)

    • +

  • Asymptote: y=1

    • +
+

+ + + + + +

+ Which of the transformations affected the domain of the exponential function? Select all that apply. + +

    +

  1. A vertical shift.

    2. +

  2. A horizontal shift.

    3. +

  3. A reflection over the x-axis.

    4. +

  4. A reflection over the y-axis.

    5. +

  5. None of these.

    6. +
+

+ + +

+ E. +

+ + + + + +

+ Which of the transformations affected the range of the exponential function? Select all that apply. + +

    +

  1. A vertical shift.

    2. +

  2. A horizontal shift.

    3. +

  3. A reflection over the x-axis.

    4. +

  4. A reflection over the y-axis.

    5. +

  5. None of these.

    6. +
+

+ + +

+ A., C. +

+ + + + + +

+ Which of the transformations affected the asymptote of the exponential function? Select all that apply. + +

    +

  1. A vertical shift.

    2. +

  2. A horizontal shift.

    3. +

  3. A reflection over the x-axis.

    4. +

  4. A reflection over the y-axis.

    5. +

  5. None of these.

    6. +
+

+ + +

+ A. +

+ + + + + + +

+ Consider the function f(x)=e^{x}. +

+ + + + +

+ Graph f(x)=e^{x}. First find f(0) and f(1). Then use what you know about the characteristics of exponential graphs to sketch the rest. Then state the domain, range, and equation of the asymptote. (Recall that e \approx 2.72 to help estimate where to put your points.) +

+ + + + + f(x) = e^(x) + p=point([(0,1),(1,e)],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+

    +

  • Domain: (-\infty,\infty)

    • +

  • Range: (0,\infty)

    • +

  • Asymptote: y=0

    • +
+

+ + + + + + +

+ Sketch the graph of g(x)=e^{x-2} using transformations. State the transformation(s) used, the domain, the range, and the equation of the asymptote. +

+ + + + + + f(x) = e^(x-2) + p=point([(2,1),(3,e)],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + +

+

    +

  • Transformation: shift right 2

    • +

  • Domain: (-\infty,\infty)

    • +

  • Range: (0,\infty)

    • +

  • Asymptote: y=0

    • +
+

+ + + + + + +

+ Sketch the graph of h(x)=-3e^x using transformations. State the transformation(s) used, the domain, the range, and the equation of the asymptote. +

+ + + + + f(x) = -1*3*e^x + p=point([(0,f(x=0)),(1,f(x=1))],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-10, ymax=2, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-10,-9,..,2]]) + p + + +

+

    +

  • Transformations: vertical stretch of 3, reflection over x-axis

    • +

  • Domain: (-\infty,\infty)

    • +

  • Range: (-\infty,0)

    • +

  • Asymptote: y=0

    • +
+

+ + + + + +

+ Sketch the graph of g(x)=e^{-x}-4 using transformations. State the transformation(s) used, the domain, the range, and the equation of the asymptote. +

+ + + + + + f(x) = e^(-x)-4 + p=point([(0,f(x=0)),(-1,f(x=-1))],pointsize=50,color='blue') + p+=line([(-9,-4),(9,-4)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-6, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + +

+

    +

  • Transformations: reflection over y-axis, shift down 4

    • +

  • Domain: (-\infty,\infty)

    • +

  • Range: (-4,\infty)

    • +

  • Asymptote: y=-4

    • +
+

+ + + + + + + + + +

Graph each of the following exponential functions. Include any asymptotes with a dotted line. State the domain, the range, the equation of the asymptote, and whether it is growth or decay. +

+ + + + +

+ f(x)=3^x +

+ + + + + + f(x) = (3)^x + p=point([(0,1),(1,3)],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + +

+

    +

  • Exponential growth

    • +

  • Domain: (-\infty,\infty)

    • +

  • Range: (0,\infty)

    • +

  • Asymptote: y=0

    • +
+

+ + + + + + +

+ f(x)=6^{-x} +

+ + + + + + f(x) = (6)^(-x) + p=point([(0,1),(-1,6)],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + +

+

    +

  • Exponential decay

    • +

  • Domain: (-\infty,\infty)

    • +

  • Range: (0,\infty)

    • +

  • Asymptote: y=0

    • +
+

+ + + + + +

+ f(x)=\frac{1}{5}^{x-2} +

+ + + + + + f(x) = (1/5)^(x-2) + p=point([(1,5),(2,1)],pointsize=50,color='blue') + p+=line([(-9,0),(9,0)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=8, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + +

+

    +

  • Exponential decay

    • +

  • Domain: (-\infty,\infty)

    • +

  • Range: (0,\infty)

    • +

  • Asymptote: y=0

    • +
+

+ + + + + +

+ f(x)=\frac{1}{3}^{x}+4 +

+ + + + + + f(x) = (1/3)^(x)+4 + p=point([(0,5),(-1,7)],pointsize=50,color='blue') + p+=line([(-9,4),(9,4)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=12, thickness=2) + p+=plot(f, (x, -9, 8), ymin=-2, ymax=12, color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,12]]) + p + + + +

+

    +

  • Exponential decay

    • +

  • Domain: (-\infty,\infty)

    • +

  • Range: (4,\infty)

    • +

  • Asymptote: y=4

    • +
+

+ + + + + + + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/05-EL/03.ptx b/precalculus/source/05-EL/03.ptx new file mode 100644 index 00000000..e92cd6d3 --- /dev/null +++ b/precalculus/source/05-EL/03.ptx @@ -0,0 +1,657 @@ + + +
+ Introduction to Logarithms (EL3) + + + + + Activities + +

+ In , we introduced the idea of an inverse function. The fundamental idea is that f has an inverse function if and only if there exists another function g such that f and g “undo” one another's respective processes. In other words, the process of the function f is reversible, and reversing f generates a related function g. +

+ + + +

+ Let P(t) be the function given by P(t)=10^t. +

+ + + +

+ Fill in the table of values for P(t). + + + t + y=P(t) + + + -3 + + + -2 + + + -1 + + + 0 + + + 1 + + + 2 + + + 3 + + +

+ + +

+ + + t + y=P(t) + + + -3 + 0.001=10^{-3} + + + -2 + 0.01=10^{-2} + + + -1 + 0.1=10^{-1} + + + 0 + 1=10^{0} + + + 1 + 10=10^{1} + + + 2 + 100=10^{2} + + + 3 + 1000=10^{3} + + +

+ + + + +

+ Do you think P will have an inverse function? Why or why not? +

+ + +

+ Students may say, each input has a distinct output, the function is one-to-one. +

+ + + + +

+ Since P has an inverse function, we know there exists some other function, say L, such that y=P(t) represent the same relationship between t and y as t=L(y). In words, this means that L reverses the process of raising to the power of 10, telling us the power to which we need to raise 10 to produce a desired result. + + Fill in the table of values for L(y). + + + y + L(y) + + + 10^{-3} + + + 10^{-2} + + + 10^{-1} + + + 10^{0} + + + 10^{1} + + + 10^{2} + + + 10^{3} + + +

+ + +

+ + + y + L(y) + + + 10^{-3} + -3 + + + 10^{-2} + -2 + + + 10^{-1} + -1 + + + 10^{0} + 0 + + + 10^{1} + 1 + + + 10^{2} + 2 + + + 10^{3} + 3 + + +

+ + + + +

+ What are the domain and range of P? +

+ + +

+ Domain (-\infty, \infty) Range (0,\infty) +

+ + + + +

+ What are the domain and range of L? +

+ + +

+ Domain (0,\infty) Range (-\infty, \infty) +

+ + + + + The powers of 10 function P(t) has an inverse L. This new function L is called the base 10 logarithm. But, we could have done a similar procedure with any base, which leads to the following definition. + + + +

+ The base b logarithm of a number is the exponent we must raise b to get that number. We represent this function as y=\log_b(x). +

+

+ We read the logarithmic expression as "The logarithm with base b of x is equal to y," or "log base b of x is y." +

+ + + + We can use to express the relationship between logarithmic form and exponential form as follows: + + \log_{b}(x)=y \iff b^{y}=x + + whenever b \gt 0, b \neq 1 + + + +

+ Write the following logarithmic equations in exponential form. +

+ + + +

+ \log_{7} (\sqrt{7})=\dfrac{1}{2} +

    +
  1. 7^{\frac{1}{2}}=\sqrt{7}
    2. +
  2. 7^{\sqrt{7}}=\dfrac{1}{2}
    3. +
  3. \sqrt{7}^{\frac{1}{2}}=7
    4. +
  4. \frac{1}{2}^{7}=\sqrt{7}
    5. +
+

+ + +

+ A +

+ + + + +

+ \log_{3} (m)=r +

    +
  1. 3^{m}=r
    2. +
  2. r^{3}=m
    3. +
  3. 3^{r}=m
    4. +
  4. m^{r}=3
    5. +
+

+ + +

+ C +

+ + + + +

+ \log_{2} (x)=6 +

    +
  1. 2^{x}=6
    2. +
  2. 6^{2}=x
    3. +
  3. x^{x}=6
    4. +
  4. 2^{6}=x
    5. +
+

+ + +

+ D +

+ + + + + + +

+ Write the following exponential equations in logarithmic form. +

+ + + +

+ 5^{2}=25 +

    +
  1. \log_{5} (2)=25
    2. +
  2. \log_{5} (25)=2
    3. +
  3. \log_{25} (5)=2
    4. +
  4. \log_{2} (25)=5
    5. +
+

+ + +

+ B +

+ + + + +

+ 3^{-1}=\frac{1}{3} +

    +
  1. \log_{3} (-1)=\frac{1}{3}
    2. +
  2. \log_{\frac{1}{3}} (-1)=3
    3. +
  3. \log_{3} \left(\frac{1}{3}\right)=-1
    4. +
  4. \log_{-1} \left(\frac{1}{3}\right)=3
    5. +
+

+ + +

+ C +

+ + + + +

+ 10^{a}=n +

    +
  1. \log_{10} (n)=a
    2. +
  2. \log_{10} (a)=n
    3. +
  3. \log_{n} (10)=a
    4. +
  4. \log_{a} (n)=10
    5. +
+

+ + +

+ A +

+ + + + + + +

+ We can use the idea of converting a logarithm to an exponential to evaluate logarithms. +

+ + + +

+ Consider the logarithm \log_{3}(9). If we want to evaluate this, which question should you try and solve? +

    +
  1. To what exponent must 9 be raised in order to get 3?
    2. +
  2. What exponent must be raised to the third in order to get 9?
    3. +
  3. To what exponent must 3 be raised in order to get 9?
    4. +
  4. What exponent must be raised to the ninth in order to get 3?
    5. +
+

+ + +

+ C +

+ + + + +

+ Evaluate the logarithm, \log_{3}(9), by answering the question from part (a). +

+ + +

+ 2 +

+ + + + + + +

+ Evaluate the following logarithms. +

+ + + +

+ \log_2(8) +

    +
  1. 4
    2. +
  2. \frac{1}{4}
    3. +
  3. -3
    4. +
  4. 3
    5. +
+

+ + +

+ D +

+ + + + +

+ \log_{144}(12) +

    +
  1. \frac{1}{2}
    2. +
  2. -2
    3. +
  3. 2
    4. +
  4. -\frac{1}{2}
    5. +
+

+ + +

+ A +

+ + + + +

+ \log_{10}\left(\frac{1}{1000}\right) +

    +
  1. \frac{1}{3}
    2. +
  2. -3
    3. +
  3. 3
    4. +
  4. -\frac{1}{3}
    5. +
+

+ + +

+ B +

+ + + + +

+ \log_{e}\left(e^{3}\right) +

    +
  1. 3
    2. +
  2. e^{3}
    3. +
  3. -3
    4. +
  4. \frac{1}{3}
    5. +
+

+ + +

+ A +

+ + + + +

+ \log_{7}(1) +

    +
  1. 7
    2. +
  2. \frac{1}{7}
    3. +
  3. 0
    4. +
  4. 1
    5. +
+

+ + +

+ C +

+ + + + + Consider the results of part (d) and (e). Using the rules of exponents and the fact that exponents and logarithms are inverses, these properties hold for any base: + + \log_{b}(b^{x})=x + + + \log_b(1)=0 + + + + + + +

There are some logarithms that occur so often, we sometimes write them without noting the base. They are the common logarithm and the natural logarithm. +

+

+ The common logarithm is a logarithm with base 10 and is written without a base. + + \log_{10}(x)=\log (x) + +

+

+ The natural logarithm is a logarithm with base e and has its own notation.

+ + \log_{e}(x)=\ln (x) + + +

+ + + + + + +

+ Evaluate the following logarithms. Some may be done by inspection and others may require a calculator. +

+ + + +

+ \log_4 \left( \frac{1}{64}\right) +

+ + +

+ -3 +

+ + + + +

+ \ln \left( 1\right) +

+ + +

+ 0 +

+ + + + +

+ \ln \left( 12 \right) +

+ + +

+ 2.485 +

+ + + + +

+ \log \left( 100\right) +

+ + +

+ 2 +

+ + + + +

+ \log_5 \left( 32 \right) +

+ + +

+ 2.153 +

+ + + + + +

+ \log_5 \left( \sqrt{5}\right) +

+ + +

+ \frac{1}{2} +

+ + + + +

+ \log \left( -10 \right) +

+ + +

+ Does not exist +

+ + + + + Notice that in part (g) you were unable to evaluate the logarithm. Given that exponentials and logarithms are inverses, their domain and range are related. The range of an expnential function is (0,\infty) which becomes the domain of a logarithmic function. This means that the argument of any logarithmic function must be greater than zero. + + + + +

+ Find the domain of the function \log_3(2x-4). +

+ + + +

+ Set up an inquality that you must solve to find the domain. +

+ + +

+ 2x-4 \gt 0 +

+ + + + +

+ Solve the inequality to find the domain. Write your answer in interval notation. +

+ + +

+ x \gt 2 or (2, \infty) +

+ + + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/05-EL/04.ptx b/precalculus/source/05-EL/04.ptx new file mode 100644 index 00000000..f9a10caf --- /dev/null +++ b/precalculus/source/05-EL/04.ptx @@ -0,0 +1,825 @@ + + +
+ Graphs of Logarithmic Functions (EL4) + + + + + Activities + + + +

Consider the function g(x)=\log_2 x. +

+ + + + +

+ Since we are familiar with graphing exponential functions, we'll use that to help us graph logarithmic ones. Rewrite g(x) in exponential form, replacing g(x) with y. +

    +

  1. x^y=2

    2. +

  2. 2^y=x

    3. +

  3. y^2=x

    4. +

  4. 2^x=y

    5. +

  5. x^2=y

    6. +
+

+ + +

+ B +

+ + + + + +

+ Fill in the table of values. Notice you are given y-values, not x-values to plug in since those are easier in the equivalent exponential form. Then plot the points on a graph. + + + x + y + + + + -2 + + + + -1 + + + + 0 + + + + 1 + + + + 2 + + +

+ + + + + + + x + y + + + \frac{1}{4} + -2 + + + + \frac{1}{2} + -1 + + + + 1 + 0 + + + + 2 + 1 + + + + 4 + 2 + + + + + + +p=point([(.25,-2),(.5,-1),(1,0),(2,1),(4,2)],pointsize=50,color='blue',ymin=-8, ymax=8, xmin=-8,xmax=8, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) +p + + + + + + + + + + +

+ What seems to be happening with the graph as x goes toward infinity? Plug in large positive values of x to test your guess, then describe the end behavior. + +

    +

  1. As x \to \infty, y \to -\infty.

    2. +

  2. As x \to \infty, y \to 0.

    3. +

  3. As x \to \infty, y \to 6.

    4. +

  4. As x \to \infty, y \to \infty.

    5. +

  5. The graph isn't defined as x \to \infty.

    6. + + +
+

+ + + D + + + + + +

+ What seems to be happening with the graph as x goes toward negative infinity? Plug in large negative values of x to test your guess, then describe the end behavior. + +

    +

  1. As x \to -\infty, y \to -\infty.

    2. +

  2. As x \to -\infty, y \to 0.

    3. +

  3. As x \to -\infty, y \to 6.

    4. +

  4. As x \to -\infty, y \to \infty.

    5. +

  5. The graph isn't defined as x \to -\infty.

    6. +
+

+ + + E + + + + + +

+ What seems to be happening with the graph as we approach x-values closer and closer to zero from the positive direction? + +

    +

  1. As x \to 0 from the positive direction, y \to -\infty.

    2. +

  2. As x \to 0 from the positive direction, y \to 0.

    3. +

  3. As x \to 0 from the positive direction, y \to \infty.

    4. +

  4. As x \to 0 from the positive direction, the graph isn't defined.

    5. +
+

+ + + A + + + + + +

+ What seems to be happening with the graph as we approach x-values closer and closer to zero from the negative direction? + +

    +

  1. As x \to 0 from the negative direction, y \to -\infty.

    2. +

  2. As x \to 0 from the negative direction, y \to 0.

    3. +

  3. As x \to 0 from the negative direction, y \to \infty.

    4. +

  4. As x \to 0 from the negative direction, the graph isn't defined.

    5. +
+

+ + + D + + + + + +

+ Complete the graph you started in , connecting the points and including the end behavior and behavior near zero that you've just determined. +

+ + +

+ + + f(x) = log(x)/log(2) + p=point([(.25,-2),(.5,-1),(1,0),(2,1),(4,2)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, .00001, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + + +

+ + + + + + + +

+ Does your graph seem to have any asymptotes? + +

    + +

  1. No. There are no asymptotes.

    2. + +

  2. There is a vertical asymptote but no horizontal one.

    3. + +

  3. There is a horizontal asymptote but no vertical one.

    4. + +

  4. The graph has both a horizontal and vertical asymptote.

    5. + + +
+ +

+ + +

+ B +

+ + + + + +

+ What the equation for each asymptote of f(x)? Select all that apply. + +

    +

  1. There are no asymptotes.

    2. +

  2. x=0

    3. +

  3. x=6

    4. +

  4. y=0

    5. +

  5. y=6

    6. +
+ +

+ + +

+ B +

+ + + + + +

+ Find the domain and range of g(x). Write your answers using interval notation. +

+ + +

+ Domain: (0,\infty), Range: (-\infty, \infty) +

+ + + + + +

+ Find the interval(s) where g(x) is increasing and the interval(s) where g(x) is decreasing. Write your answers using interval notation. +

+ + +

+ Increasing: (0,\infty), Decreasing: nowhere +

+ + + + + + + +

The function we've just graphed, g(x)=\log_2 x, and the function f(x)=2^x (which we graphed in ) are inverse functions. +

+ + + + +

+ How are the graphs of f(x) and g(x) similar? +

+ + +

+ Answers could include basic shape (though mirrored), both are increasing. +

+ + + + + +

+ How are the graphs of f(x) and g(x) different? +

+ + +

+ Answers could include flipped x and y values, flipped asymptote, one has x-intercept and one has y-intercept, domain and range are swapped. +

+ + + + + + +

+ The graph of a logarithmic function g(x)=\log_b x where b>0 and b \neq 1 has the following characteristics: + +

+

+ +

    +

  • Its domain is (0,\infty) and its range is (-\infty,\infty).

    • + +

  • There is a vertical asymptote at x=0. There is no horizontal asymptote.

    • +

  • There is an x-intercept at (1,0). There is no y-intercept.

    • +
+

+ + + + +

Just as with other types of functions, we can use transformations to graph logarithmic functions. For a reminder of these transformations, see and the following definitions: +

    +

    • +

    • +

    • +

    • +

    • +

    • +
+

+ + + + +

Let f(x)=\log_4 x. +

+ + + + +

+ Graph f(x). +

+ + +

+ + + f(x) = log(x)/log(4) + p=point([(1,0),(2,1/2),(4,1),(8,3/2)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, .00001, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+ + + + + +

+ Match the following functions to their graphs. +

    +

  • g(x)= -\log_4 x

    • +

  • h(x)= \log_4 (-x)

    • +

  • j(x)= \log_4 (x+1)

    • +

  • k(x)= \log_4 (x) + 1

    • +
+ +
    +

  1. + + + f(x) = -log(x)/log(4) + p=point([(1,0),(2,-1/2),(4,-1),(8,-3/2)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, .00001, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

    2. + +
  2. +

    + + + f(x) = log(x)/log(4) +1 + p=point([(1,1),(2,3/2),(4,2),(8,5/2)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, .00001, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

    +
    3. + +
  3. +

    + + + f(x) = log(-x)/log(4) + p=point([(-1,0),(-2,1/2),(-4,1),(-8,3/2)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, -.00001), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

    +
    4. + +
  4. +

    + + + f(x) = log(x+1)/log(4) + p=point([(0,0),(1,1/2),(3,1),(7,3/2)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(-1,-9),(-1,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -.99999, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

    +
    5. + + +
+

+ + +

+

    +

  • g(x)= -\log_4 x is A.

    • +

  • h(x)= \log_4 (-x) is C.

    • +

  • j(x)= \log_4 (x+1) is D.

    • +

  • k(x)= \log_4 (x) + 1 is B.

    • +
+

+ + + + +

+ Find the domain, range, and equation of the asymptote for the parent function \left(f(x)\right) and each of the four transformations \left(g(x), h(x), j(x), \text{ and } k(x)\right). +

+ + +

+ f(x): +

    +

  • Domain: (0,\infty)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=0

    • +
+ + g(x): +
    +

  • Domain: (0,\infty)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=0

    • +
+ + h(x): +
    +

  • Domain: (-\infty,0)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=0

    • +
+ + j(x): +
    +

  • Domain: (-1,\infty)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=-1

    • +
+ + k(x): +
    +

  • Domain: (0,\infty)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=0

    • +
+

+ + + + + +

+ Which of the transformations affected the domain of the logarithmic function? Select all that apply. + +

    +

  1. A vertical shift.

    2. +

  2. A horizontal shift.

    3. +

  3. A reflection over the x-axis.

    4. +

  4. A reflection over the y-axis.

    5. +

  5. None of these.

    6. +
+

+ + +

+ B,C +

+ + + + + +

+ Which of the transformations affected the range of the logarithmic function? Select all that apply. + +

    +

  1. A vertical shift.

    2. +

  2. A horizontal shift.

    3. +

  3. A reflection over the x-axis.

    4. +

  4. A reflection over the y-axis.

    5. +

  5. None of these.

    6. +
+

+ + +

+ E +

+ + + + + +

+ Which of the transformations affected the asymptote of the logarithmic function? Select all that apply. + +

    +

  1. A vertical shift.

    2. +

  2. A horizontal shift.

    3. +

  3. A reflection over the x-axis.

    4. +

  4. A reflection over the y-axis.

    5. +

  5. None of these.

    6. +
+

+ + +

+ B +

+ + + + + + + +

+ Consider the function f(x)=\ln(x). +

+ + + +

+ Graph f(x)=\ln(x). First find f(1) and f(e). Then use what you know about the characteristics of logarithmic graphs to sketch the rest. Then state the domain, range, and equation of the asymptote. (Recall that e \approx 2.72 to help estimate where to put your points.) +

+ + + + + f(x) = log(x)/log(e) + p=point([(1,0),(e,1)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, .00001, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+

    +

  • Domain: (0,\infty)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=0

    • +
+

+ + + + + +

+ Sketch the graph of g(x)=\ln(x-3) using transformations. State the transformation(s) used, the domain, the range, and the equation of the asymptote. +

+ + + + + + f(x) = log(x-3)/log(e) + p=point([(4,0),(e+3,1)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(3,-9),(3,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, 3.0000001, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+

    +

  • Transformation: shift right 3

    • +

  • Domain: (3,\infty)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=3

    • +
+

+ + + + + +

+ Sketch the graph of h(x)=3\ln(x) using transformations. State the transformation(s) used, the domain, the range, and the equation of the asymptote. +

+ + + + + f(x) = 3*log(x)/log(e) + p=point([(1,0),(e,3)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, .00001, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+

    +

  • Transformation: vertical stretch with factor of three.

    • +

  • Domain: (0,\infty)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=0

    • +
+

+ + + + + + + + +

Graph each of the following logarithmic functions. Include any asymptotes with a dotted line. State the domain, the range, and the equation of the asymptote. +

+ + + + +

+ f(x)=\log_3 x +

+ + + + + f(x) = log(x)/log(3) + p=point([(1,0),(3,1),(9,2)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, .00001, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+

    +

  • Domain: (0,\infty)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=0

    • +
+

+ + + + + +

+ f(x)=\log_6 (-x) +

+ + + + + f(x) = log(-x)/log(6) + p=point([(-1,0),(-6,1)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, -9, -.000001), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+

    +

  • Domain: (-\infty,0)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=0

    • +
+

+ + + + + +

+ f(x)=\log_{\frac{1}{5}} x +

+ + + + + f(x) = log(x)/log(1/5) + p=point([(1,0),(5,-1)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, .000001, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+

    +

  • Domain: (0,\infty)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=0

    • +
+

+ + + + + +

+ f(x)=\log_{\frac{1}{3}} x +2 +

+ + + + + f(x) = log(x)/log(1/3) +2 + p=point([(1,2),(3,1)],pointsize=50,color='blue', xmin=-8, xmax=8, ymin=-8, ymax=8) + p+=line([(0,-9),(0,9)], linestyle="--", color='red',xmin=-8, xmax=8, ymin=-8, ymax=8, thickness=3) + p+=plot(f, (x, .000001, 9), color='blue', thickness=3, gridlines=[[-8,-7,..,8],[-8,-7,..,8]]) + p + + +

+

    +

  • Domain: (0,\infty)

    • +

  • Range: (-\infty,\infty)

    • +

  • Asymptote: x=0

    • +
+

+ + + + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
diff --git a/precalculus/source/05-EL/05.ptx b/precalculus/source/05-EL/05.ptx new file mode 100644 index 00000000..b45fbedc --- /dev/null +++ b/precalculus/source/05-EL/05.ptx @@ -0,0 +1,612 @@ + + +
+ Properties of logarithms (EL5) + + + + + Activities + + In this section, we will explore the properties of logarithms and learn how to manipulate them that will be helpful when we are ready to solve logarithmic equations. + + + + Recall that \log_b M=\log_b N if and only if M=N. In addition, because exponentials and logarithms are inverses, we also know that \log_b(b^{k})=k. + In addition, according to the law of exponents, we know that: x^a \cdot x^b=x^{a+b} + \frac{x^a}{x^b}=x^{a-b} + \left(x^a\right)^b=x^{a \cdot b} + + Consider all these as you move through the activities in this section. + + + + +

+ Let's begin with the law of exponents to see if we can understand the product property of logs. According to the law of exponents, we know that 10^x \cdot 10^y=10^{x+y}. Start with this equation as you move through this activity. +

+ + + +

+ Let a=10^x and b=10^y. How could you rewrite the left side of the equation 10^x \cdot 10^y? +

    +
  1. a+b
    2. +
  2. a-b
    3. +
  3. 10^{x+y}
    4. +
  4. a \cdot b
+

+ + +

+ D +

+ + + + + +

+ Recall from that \log_b M=\log_b N if and only if M=N. Use this property to apply the logarithm to both sides of the rewritten equation from part a. What is that equation? +

    +
  1. \log_{10}(a+b)=\log_{10}\left(10^{x+y}\right)
    2. +
  2. \log_{10}(a \cdot b)=\log_{10}\left(10^{x+y}\right)
    3. +
  3. \log_{10}(a \cdot b)=\log_{10}\left(10^{a+b}\right)
    4. +
  4. \log_{10}(a+b)=\log_{10}\left(10^{a+b}\right)
+

+ + +

+ B +

+ + + + +

+ Knowing that \log_b(b^k)=k, how could you simplify the right side of the equation? +

+
    +
  1. \log_{10}(a+b)
    2. +
  2. \log_{10}(x+y)
    3. +
  3. a+b
    4. +
  4. x+y
+

+ + +

+ D +

+ + + + + +

+ Recall in part a, we defined 10^x=a and 10^y=b. What would these look like in logarithmic form? +

    +
  1. \log_{10}a=x
    2. +
  2. \log_{x}a=10
    3. +
  3. \log_{10}b=y
    4. +
  4. \log_{y}b=10
+

+ + +

+ A and C +

+ + + + + +

+ Using your solutions in part d, how can we rewrite the right side of the equation? +

    +
  1. 10^{a+b}
    2. +
  2. \log_{10}a-\log_{10}b
    3. +
  3. \log_{10}a+\log_{10}b
    4. +
  4. 10^x+10^y
+

+ + +

+ C +

+ + + + +

Combining parts a and d, which equation represents 10^x \cdot 10^y=10^{x+y} in terms of logarithms? +

    +
  1. \log_{10}(a+b)=10^{a+b}
    2. +
  2. \log_{10}(a \cdot b)=\log_{10}a-\log_{10}b
    3. +
  3. \log_{10}(a \cdot b)=\log_{10}a+\log_{10}b
    4. +
  4. \log_{10}(a \cdot b)=10^x+10^y
+

+ + +

+ C +

+ + + + + + +

+ According to the law of exponents, we know that \frac{10^x}{10^y}=10^{x-y}. Start with this equation as you move through this activity. +

+ + + +

+ Let a=10^x and b=10^y. How could you rewrite the left side of the equation \frac{10^x}{10^y}? +

    +
  1. a+b
    2. +
  2. a-b
    3. +
  3. 10^{x+y}
    4. +
  4. a \cdot b
+

+ + +

+ B +

+ + + + + +

+ Recall from that \log_b M=\log_b N if and only if M=N. Use this property to apply the logarithm to both sides of the rewritten equation from part a. What is that equation? +

    +
  1. \log_{10}(a-b)=\log_{10}\left(10^{x-y}\right)
    2. +
  2. \log_{10}\left(\frac{a}{b}\right)=\log_{10}\left(10^{x-y}\right)
    3. +
  3. \log_{10}\left(\frac{a}{b}\right)=\log_{10}\left(10^{a+b}\right)
    4. +
  4. \log_{10}(a-b)=\log_{10}\left(10^{a-b}\right)
+

+ + +

+ B +

+ + + + +

+ Knowing that \log_b(b^k)=k, how could you simplify the right side of the equation? +

+
    +
  1. \log_{10}(a-b)
    2. +
  2. \log_{10}(x-y)
    3. +
  3. x-y
    4. +
  4. a-b
+

+ + +

+ C +

+ + + + + +

+ Recall in part a, we defined 10^x=a and 10^y=b. What would these look like in logarithmic form? +

    +
  1. \log_{10}a=x
    2. +
  2. \log_{x}a=10
    3. +
  3. \log_{10}b=y
    4. +
  4. \log_{y}b=10
+

+ + +

+ A and C +

+ + + + + +

+ Using your solutions in part d, how can we rewrite the right side of the equation? +

    +
  1. 10^{a+b}
    2. +
  2. \log_{10}a-\log_{10}b
    3. +
  3. \log_{10}a-\log_{10}b
    4. +
  4. 10^{x-y}
+

+ + +

+ B +

+ + + + +

Combining parts a and d, which equation represents \frac{10^x}{10^y}=10^{x-y} in terms of logarithms? +

    +
  1. \log_{10}(a-b)=10^{a+b}
    2. +
  2. \log_{10}(a-b)=\log_{10}a-\log_{10}b
    3. +
  3. \log_{10}\left(\frac{a}{b}\right)=\log_{10}a-\log_{10}b
    4. +
  4. \log_{10}\left(\frac{a}{b}\right)=10^{x-y}
+

+ + +

+ C +

+ + + + + + In and , you explored an example of two common properties of logarithms! + + + + +

+ The product property of logarithms states \log_{a}(m \cdot n)=\log_{a}m+\log_{a}n. + The quotient property of logarithms states \log_{a}\left(\frac{m}{n}\right)=\log_{a}m-\log_{a}n. + Notice that for each of these properties, the base has to be the same. +

+ + + + + +

+ There is still one more property to consider. This activity will investigate the power property. Suppose you are given \log(x^3)=\log(x \cdot x \cdot x). +

+ + + + +

+ By applying , how could you rewrite the right-hand side of the equation? +

    +
  1. \log(3x)
    2. +
  2. \log(x)-\log(x)-\log(x)
    3. +
  3. \log(x^3)
    4. +
  4. \log(x)+\log(x)+\log(x)
+

+ + +

+ D +

+ + + + + +

+ By combining "like" terms, you can simplify the right-hand side of the equation further. What equation do you have after simplifying the right-hand side? +

    +
  1. \log(x^3)=\log(3x)
    2. +
  2. \log(x^3)=-3\log(x)
    3. +
  3. \log(x^3)=\log(x^3)
    4. +
  4. \log(x^3)=3\log(x)
+

+ + +

+ D +

+ + + + + + +

+ The power property of logarithms states \log_a(m^n)=n \cdot \log_a(m). +

+ + + + + +

+ Apply and to expand the following. (Note: When you are asked to expand logarithmic expressions, your goal is to express a single logarithmic expression into many individual parts or components.) +

+ + + +

+ \log_3\left(\frac{6}{19}\right) +

    +
  1. \log_36+\log_319
    2. +
  2. \log_3(6-19)
    3. +
  3. \log_36-\log_319
    4. +
  4. \log_3(6+19)
+

+ + +

+ C +

+ + + + + +

+ \log(a \cdot b)^2 +

    +
  1. 2(\log a+\log b)
    2. +
  2. 2\log a-\log b
    3. +
  3. \log a+2\log b
    4. +
  4. 2\log a+2\log b
+

+ + +

+ A and D +

+ + + + + +

+ \ln\left(\frac{x^3}{y}\right) +

    +
  1. 3\ln x+\ln y
    2. +
  2. 3\ln x-\ln y
    3. +
  3. 3(\ln x-\ln y)
    4. +
  4. \ln x^3-\ln y
+

+ + +

+ B. Note that D is also correct, but it is not expanded fully. +

+ + + + +

+ \log(x \cdot y \cdot z^3) +

    +
  1. \log x+\log y+3\log z
    2. +
  2. \log x+\log y+\log z^3
    3. +
  3. \log x-\log y-3\log z
    4. +
  4. 3(\log x+y+z)
+

+ + +

+ A +

+ + + + + + +

+ Apply and to condense into a single logarithm. +

+ + + +

+ 6\log_6 a+3\log_6 b +

    +
  1. 6(\log_6 a)+3(\log_6 b)
    2. +
  2. \log_6 a^6+\log_6 b^3
    3. +
  3. (\log_6 a)^6+(\log_6 b)^3
    4. +
  4. \log_6\left(a^6 \cdot b^3\right)
+

+ + +

+ D. B is also correct, but it is not condensed fully. +

+ + + + + +

+ \ln x-4\ln y +

    +
  1. \ln x-\ln y^4
    2. +
  2. \ln\left(\frac{x}{y^4}\right)
    3. +
  3. \ln\left(\frac{x}{y}\right)^4
    4. +
  4. \ln(x \cdot y^4)
+

+ + +

+ B +

+ + + + + +

+ 2(\log(2x)-\log y) +

    +
  1. \log\left(\frac{4x^2}{y^2}\right)
    2. +
  2. \log\left(\frac{2x^2}{y^2}\right)
    3. +
  3. 2 \cdot \log\left(\frac{2x}{y}\right)
    4. +
  4. \log\left(\frac{4x^2}{y}\right)
+

+ + +

+ A +

+ + + + + +

+ \log 3+2\log 5 +

    +
  1. \log(3 \cdot 5^2)
    2. +
  2. 2 \cdot \log 75
    3. +
  3. \log 75
    4. +
  4. \log 225
+

+ + +

+ C. Note that A is also correct, but not condensed completely. +

+ + + + + + You might have noticed that a scientific calculator has only "log" and "ln" buttons (because those are the most common bases we use), but not all logs have base 10 or e as their bases. + + + + +

+ Suppose you wanted to find the value of \log_53 in your calculator but you do not know how to input a base other than 10 or e (i.e., you only have the "log" and "ln" buttons on your calculator). Let's explore another helpful tool that can help us find the value of \log_53. +

+ + + +

+ Let's start with the general statement, \log_ba=x. How can we rewrite this logarithmic equation into an exponential equation? +

+ + +

+ b^x=a +

+ + + + +

+ Now take the log of both sides of your equation and apply the power property of logarithms to bring the exponent down. What equation do you have now? +

+ + +

+ x \cdot \log b=\log a +

+ + + + +

+ Solve for x. What does x equal? +

+ + +

+ x=\frac{\log a}{\log b} +

+ + + + +

+ Recall that when we started, we defined x=\log_ba. Substitute \log_ba into your equation you got in part c for x. What is the resulting equation? +

+ + +

+ \log_ba=\frac{\log a}{\log b} +

+ + + + +

+ Apply what you got in part d to find the value of \log_53. What is the approximate value of \log_53? +

+ + +

+ \log_53 is approximately 0.683 +

+ + + + + + Notice in , we were able to calculate \log_53 using logs of base 10. You should now be able to find the value of a logarithm of any base! + + + + +

+ The change of base formula is used to write a logarithm of a number with a given base as the ratio of two logarithms each with the same base that is different from the base of the original logarithm. + \log_{b}a=\frac{\log a}{\log b} +

+ + + + + +

+ Apply and a calculator to approximate the value of each logarithm. +

+ + + +

+ \log_2 30 +

    +
  1. 0.204
    2. +
  2. 4.907
    3. +
  3. \frac{\log30}{\log2}
    4. +
  4. \frac{\log2}{\log30}
+

+ + +

+ B and C +

+ + + + +

+ \ln 183 +

    +
  1. \frac{\ln183}{\ln e}
    2. +
  2. 67.32
    3. +
  3. \frac{\ln e}{\ln183}
    4. +
  4. 5.209
+

+ + +

+ A and D +

+ + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/05-EL/06.ptx b/precalculus/source/05-EL/06.ptx new file mode 100644 index 00000000..8752b296 --- /dev/null +++ b/precalculus/source/05-EL/06.ptx @@ -0,0 +1,824 @@ + + +
+ Solving Exponential and Logarithmic Equations (EL6) + + + + + Activities + + + Recall that we can convert between exponential and logarithmic forms. \log_bx=y is equivalent to b^y=x + + + + +

+ This activity will investigate ways we can solve logarithmic equations using properties and definitions from previous sections. +

+ + + +

+ Given that \log_39=x, how can we rewrite this into an exponential equation? +

    +
  1. 9^x=3
    2. +
  2. \log\left(\frac{9}{3}\right)
    3. +
  3. 3^x=9
    4. +
  4. x^3=9
+

+ + +

+ C +

+ + + + +

+ Now that \log_39=x is rewritten as an exponential equation, what is the value of x? +

    +
  1. -2
    2. +
  2. 2
    3. +
  3. -\frac{1}{2}
    4. +
  4. \frac{1}{2}
+

+ + +

+ B +

+ + + + + + Notice in , you were able to solve a logarithmic equation by converting it into an exponential equation. This is one method in solving logarithmic equations. + + + + +

+ For each of the following, solve the logarithmic equations by first converting them to exponential equations. +

+ + + +

+ \log_{10}(1{,}000{,}000)=x +

    +
  1. -6
    2. +
  2. 100{,}000
    3. +
  3. 6
    4. +
  4. 0.00001
+

+ + +

+ C +

+ + + + +

+ \log_3(x+3)=0 +

    +
  1. 0
    2. +
  2. -2
    3. +
  3. 2
    4. +
  4. -1
+

+ + +

+ B +

+ + + + +

+ \log_5(2x+4)=2 +

    +
  1. 14
    2. +
  2. \frac{29}{2}
    3. +
  3. 16
    4. +
  4. \frac{21}{2}
+

+ + +

+ D +

+ + + + + + +

+ Not all logarithmic equations can be solved by converting to exponential equations. In this activity, we will explore another way to solve logarithmic equations. +

+ + + +

+ Suppose you are given the equation, \log(4x-5)=\log(2x-1), and you + brought all the logs to one side to get \log(4x-5)-\log(2x-1)=0. Using the quotient property of logs, how could you condense the left side of the equation? +

    +
  1. \log\left(\frac{4x-5}{2x-1}\right)
    2. +
  2. \log\left(\frac{2x-1}{4x-5}\right)
    3. +
  3. \frac{\log(4x-5)}{\log(2x-1)}
    4. +
  4. \frac{\log(2x-1)}{\log(4x-5)}
+

+ + +

+ A +

+ + + + +

+ Now that you have a single logarithm, convert this logarithmic form into exponential form. What does this new equation look like? +

    +
  1. \frac{4x-5}{2x-1}=10^0
    2. +
  2. \frac{2x-1}{4x-5}=1
    3. +
  3. \frac{2x-1}{4x-5}=10^0
    4. +
  4. \frac{4x-5}{2x-1}=1
+

+ + +

+ A and D +

+ + + + +

+ Notice that the "log" has disappeared and you now have an equation with just the variable x. Which of the following is equivalent to the equation you got in part b? +

    +
  1. 4x-5=2x-1
    2. +
  2. 4x-5=0
    3. +
  3. 2x-1=0
+

+ + +

+ A +

+ + + + +

+ Compare the answer you got in part c to the original equation given \log(4x-5)=\log(2x-1). What do you notice? +

+ + +

+ Students should notice that the equation they have thus far is 4x-5=2x-1, which is very similar to the original equation (only without the logs). Students may make a conjecture about "dropping" the logs to solve. +

+ + + + +

+ Solve the equation you got in part d to find the value of x. +

    +
  1. -3
    2. +
  2. \frac{5}{4}
    3. +
  3. \frac{1}{2}
    4. +
  4. 2
+

+ + +

+ D +

+ + + + + + Notice in , that you did not have to convert the logarithmic equation into an exponential equation. A faster method, when you have a log on both sides of the equals sign, is to "drop" the logs and set the arguments equal to one another. Be careful though - you can only have one log on each side before you can "drop" them! + + + + +

+ The one-to-one property of logarithms states that if both sides of an equation can be rewritten as a single logarithm with the same base, then the arguments can be set equal to each other (and then solved algebraically). +

+ + + + + +

+ Apply and other properties of logarithms (i.e., product, quotient, and power) to solve the following logarithmic equations. +

+ + + +

+ \log(-2a+9)=\log(7-4a) +

+ + +

+ -1 +

+ + + + +

+ \log_9(x+6)-\log_9x=\log_92 +

+ + +

+ 6 +

+ + + + +

+ \log_82+\log_8(4x^2)=1 +

+ + +

+ 1 and -1 +

+ + + + +

+ \ln(4x+1)-\ln3=5 +

+ + +

+ \frac{3e^5-1}{4} or 111.06 +

+ + + + + + +

+ In some cases, you will get equations with logs of different bases. Apply properties of logarithms to solve the following logarithmic equations. +

+ + + +

+ \log_3(x-6)=\log_9x +

+ + +

+ Use the change of base formula to rewrite \log_9x so that it has a base of 3. +

+ + +

+ x=9 +

+ + + + +

+ \log_2x=\log_8(4x) +

+ + +

+ x=0 and x=2 +

+ + + + + +

+ Now that we've looked at how to solve logarithmic equations, let's see how we can apply similar methods to solving exponential equations. +

+ + + +

+ Suppose you are given the equation 2^x=48. There is no whole number value we can raise 2 to to get 48. What two whole numbers must x be between? +

+ + +

+ Between 5 and 6. +

+ + + + +

+ We'll use logarithms to isolate the variable in the exponent. How can we convert 2^x=48 into a logarithmic equation? +

    +
  1. x=\log_{48}2
    2. +
  2. x=\log_248
    3. +
  3. 48=\log_2x
    4. +
  4. 2=\log_{48}x
+

+ + +

+ B +

+ + + + +

+ Notice that the answer you got in part b is an exact answer for x. There will be times, though, that it will be helpful to also have an approximation for x. Which of the following is a good approximation for x? +

    +
  1. x \approx 5.585
    2. +
  2. x \approx 0.179
    3. +
  3. x \approx 24
    4. +
  4. x \approx \frac{1}{24}
+

+ + +

+ A +

+ + + + + + Notice in we started with an exponential equation and then solved by converting the equation into a logarithmic equation. Logarithms can help us get the variable out of the exponent. + + + + +

+ Although rewriting an exponential equation into a logarithmic equation is helpful at times, it is not the only method in solving exponential equations. In this activity, we will explore what happens when we take the log of both sides of an exponential equation and use the properties of logarithms to solve in another way. +

+ + + +

+ Suppose you are given the equation: 3^x=7. Take the log of both sides. What equation do you now have? +

+ + +

+ \log{3^x}=\log7 +

+ + + + +

+ Apply the power property of logarithms () to bring down the exponent. What equation do you now have? +

+ + +

+ x \cdot \log3=\log7 +

+ + + + +

+ Solve for x. What does x equal? +

+ + +

+ x=\frac{\log7}{\log3} +

+ + + + +

+ Using the change-of-base formula (), rewrite your answer from part c so that x is written as a single logarithm. What is the exact value of x? +

+ + +

+ x=\log_37 +

+ + + + +

+ If you were to solve 3^x=7 by converting it into a logarithmic equation, what would it look like? +

+ + +

+ \log_37=x +

+ + + + +

+ What do you notice about your answer from parts d and e? +

+ + +

+ Students should see that the two answers they got for parts d and e are the same. This might be a good time to discuss why taking the log of both sides is a valid method in solving an exponential equation. +

+ + + + + + +

+ Use the method of taking the log of both sides (as you saw in ) to solve 5^{2x+3}=8. +

+ + + +

+ Take the log of both sides and use the power property of logarithms to bring down the exponent. What equation do you have now? +

    +
  1. 2x+3\log5=\log8
    2. +
  2. 2x+(3\log5)=\log8
    3. +
  3. (2x+3) \cdot \log5=\log8
    4. +
  4. 2x\log5+3=\log8
+

+ + +

+ C +

+ + + + +

+ Solve for x. +

    +
  1. x=\frac{\log_58-3}{2}
    2. +
  2. x=\frac{\log8-3\log5}{2}
    3. +
  3. x=\frac{\log8}{3\log5}-2
    4. +
  4. x=\frac{\log8-3}{2\log5}
+

+ + +

+ A +

+ + + + +

+ What is the approximate value of x? +

    +
  1. x \approx -0.85
    2. +
  2. x \approx -1.5
    3. +
  3. x \approx -0.60
    4. +
  4. x \approx -1.57
+

+ + +

+ A +

+ + + + + + +

+ In this activity, we will explore other types of exponential equations, which will require other methods of solving. +

+ + + +

+ Suppose you are given the equation + 5^{3x}=5^{7x-2} and you decide to take the log of both sides as your first step to get \log5^{3x}=\log5^{7x-2} What would you use next to solve this equation? +

    +
  1. Quotient property of logarithms
    2. +
  2. Power property of logarithms
    3. +
  3. Product property of logarithms
    4. +
  4. Change of base formula
    5. +
  5. One-to-one property of logarithms
+

+ + +

+ B +

+ + + + +

+ Applying the property you chose in part a, what would the resulting equation be? +

    +
  1. 3x\log5=7x-2\log5
    2. +
  2. 3x\log5=7x\log5-2\log5
    3. +
  3. \log5^{3x}=\log5^{7x-2}
    4. +
  4. (3x)\log5=(7x-2)\log5
+

+ + +

+ B and D +

+ + + + +

+ Now that you have a logarithmic equation, divide both sides by \log5 to begin to isolate the variable x. After dividing by \log5, what equation do you now have? +

    +
  1. 3x=\frac{7x}{\log5}-2
    2. +
  2. 3x=7x-2
    3. +
  3. 5^{3x}=5^{7x-2}
    4. +
  4. \frac{3x\log5}{\log5}=\frac{(7x-2)\log5}{\log5}
+

+ + +

+ B +

+ + + + +

+ Compare the equation you got in part c to the original equation given. What do you notice? +

+ + +

+ Students will probably notice that their resulting equation are just the exponents of the original equation set equal to one another. +

+ + + + +

+ Solve for x. +

    +
  1. 2
    2. +
  2. \frac{1}{2}
    3. +
  3. -2
    4. +
  4. \frac{1}{2}
+

+ + +

+ D +

+ + + + + + Notice in , it is much faster to set the exponents equal to one another. Make sure to check that the bases are equal before you set the exponents equal! And if the bases are not equal, you might have to use properties of exponents to help you get the bases to be the same. + + + + +

+ When you are given an exponential equation with the same bases on both sides, you can simply set the exponents equal to one another and solve. This is known as the one-to-one property of exponentials. +

+ + + + + +

+ When an exponential equation has the same base on each side, the exponents can be set equal to one another. If the bases aren't the same, we can rewrite them using properties of exponents and use the one-to-one property of exponentials. +

+ + + +

+ Suppose you are given 5^x=625, how could you rewrite this equation so that both sides have a base of 5? +

+ + +

+ 5^x=5^4 +

+ + + + +

+ Suppose you are given 4^x=32, how could you rewrite this equation so that both sides have a base of 2? +

+ + +

+ 2^{2x}=2^5 +

+ + + + +

+ Suppose you are given 3^{1-x}=\frac{1}{27}, how could you rewrite this equation so that both sides have a base of 3? (Hint: you many need to revisit properties of exponents) +

+ + +

+ 3^{1-x}=3^{-3} +

+ + + + +

+ Suppose you are given 6^{\frac{x-3}{4}}=\sqrt{6}, how could you rewrite this equation so that both sides have a base of 6? (Hint: you many need to revisit properties of exponents) +

+ + +

+ 6^{\frac{x-3}{4}}=6^{\frac{1}{2}} +

+ + + + + + +

+ For each of the following, use properties of exponentials and logarithms to solve. +

+ + + +

+ 6^{-2x}=6^{2-3x} +

    +
  1. 2
    2. +
  2. -\frac{2}{5}
    3. +
  3. -2
    4. +
  4. -\frac{5}{2}
+

+ + +

+ A +

+ + + + +

+ \log_2256=x +

    +
  1. -8
    2. +
  2. 254
    3. +
  3. 8
    4. +
  4. 128
+

+ + +

+ C +

+ + + + +

+ 5\ln(9x)=20 +

    +
  1. e^4
    2. +
  2. \frac{e^4}{9}
    3. +
  3. e^{\left(\frac{4}{9}\right)}
    4. +
  4. \frac{4}{\ln9}
+

+ + +

+ B +

+ + + + +

+ 10^x=4.23 +

    +
  1. \log4.23
    2. +
  2. 42.3
    3. +
  3. 1.44
    4. +
  4. 0.63
+

+ + +

+ A and D +

+ + + + +

+ \log_6(5x-5)=\log_6(3x+7) +

    +
  1. 2
    2. +
  2. 0
    3. +
  3. 6
    4. +
  4. 1
+

+ + +

+ C +

+ + + + + + +

+ For each of the following, use properties of exponentials and logarithms to solve. +

+ + + +

+ \log_82+\log_8{4x^2}=1 +

+ + +

+ x=1,-1 +

+ + + + +

+ 5^{x+7}=3 +

+ + +

+ x=\frac{log3}{log5}-7 +

+ + + + +

+ 8^{\frac{x-6}{6}}=\sqrt8 +

+ + +

+ x=9 +

+ + + + +

+ \log_6(x+1)-\log_6x=\log_6{29} +

+ + +

+ x=\frac{1}{28} +

+ + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
\ No newline at end of file diff --git a/precalculus/source/05-EL/07.ptx b/precalculus/source/05-EL/07.ptx new file mode 100644 index 00000000..6aa6b884 --- /dev/null +++ b/precalculus/source/05-EL/07.ptx @@ -0,0 +1,534 @@ + + +
+ Applications of Exponential and Logarithmic Functions (EL7) + + + + + Activities + + + Now that we have explored multiple methods for solving exponential and logarithmic equations, let's put those in to practice using some real-world application problems. + + + + +

+ A coffee is sitting on Mr. Abacus's desk cooling. It cools according to the function T = 70(0.80)^x + 20, where x is the time elapsed in minutes and T is the temperature in degrees Celsius. +

+ + + + + +

+ What is the initial temperature of Mr. Abacus's coffee? +

    +
  1. 20C
    2. +
  2. 70C
    3. +
  3. 0.80C
    4. +
  4. 90C
+

+ + +

+ D +

+ + + + +

+ What is the temperature of Mr. Abacus's coffee after 10 minutes? +

    +
  1. 7.5C
    2. +
  2. 20C
    3. +
  3. 27.5C
    4. +
  4. 76C
+

+ + +

+ C +

+ + + + +

+ According to the function given, if Mr. Abacus leaves his coffee on his desk all day, what will the coffee eventually cool to? +

    +
  1. 90C
    2. +
  2. 70C
    3. +
  3. 20C
    4. +
  4. 0C
+

+ +

Think about what happens to the function as x\to \infty.

 + +

+ C +

+ + + + + + +

+ A video posted on YouTube initially had 80 views as soon as it was posted. The total number of views to date has been increasing exponentially according to the exponential growth function y=80e^{0.2t}, where t represents time measured in days since the video was posted. +

+ + + +

+ How many views will the video have after 3 days? Round to the nearest whole number. +

    +
  1. 293 views
    2. +
  2. 146 views
    3. +
  3. 98 views
    4. +
  4. 82 views
+

+ + +

+ B +

+ + + + +

+ How many days does it take until 2{,}500 people have viewed this video? +

    +
  1. 18 days
    2. +
  2. 156 days
    3. +
  3. 39 days
    4. +
  4. 17 days
+

+ + +

+ A +

+ + + + + + +

+ In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Recall that the general form of an exponential equation is f(x)=a \cdot b^x, where a is the initial value, b is the growth/decay factor, and t is time. We want to write a function N(t) to represent the deer population after t years. +

+ + + +

+ What is the initial value for the deer population? +

+ + +

+ 80 deer +

+ + + + +

+ We are not given the growth factor, so we must solve for it. Write an exponential equation using the initial population, the 2012 population, and the time elapsed. +

+ + +

+ 180=80 \cdot b^6 +

+ + + + +

+ Take your equation in part b and solve for b using logs. +

+ + +

+ b is approximately 1.1447 +

+ + + + +

+ Now that you have found the growth factor (from part b), what is the equation, in terms of N(t) and t that represents the deer population? +

+ + +

+ N(t)=80(1.1447)^t +

+ + + + +

+ If the growth continues according to this exponential function, when will the population reach 250? +

+ + +

+ It will take 8.43 years, which means the deer population will reach 250 in the year 2015. +

+ + + + + + +

+ The concentration of salt in ocean water, called salinity, varies as you go deeper in the ocean. Suppose f(x) = 28.9 + 1.3\log (x + 1) models salinity of ocean water to depths of 1000 meters at a certain latitude, where x is the depth in meters and f(x) is in grams of salt per kilogram of seawater. (Note that salinity is expressed in the unit g/kg, which is often written as ppt (part per thousand) or ‰ (permil).) +

+ + + +

+ At 50 meters, what is the salinity of the seawater? +

    +
  1. 31.1 ppt
    2. +
  2. 32.1 ppt
    3. +
  3. 51.6 ppt
    4. +
  4. 30.6 ppt
+

+ + +

+ A +

+ + + + +

+ Approximate the depth (to the nearest tenth of a meter) where the salinity equals 33 ppt. +

    +
  1. -1.0 meters
    2. +
  2. 1426.1 meters
    3. +
  3. -0.9 meters
    4. +
  4. 1424.1 meters
+

+ + +

+ D +

+ + + + + + +

+ The first key on a piano keyboard (called A_0) corresponds to a pitch with a frequency of 27.5 cycles per second. With every successive key, going up the black and white keys, the pitch multiplies by a constant. The formula for the frequency, f of the pitch sounded when the nth note up the keyboard is played is given by + n=1+12\log_2\frac{f}{27.5} + +

+ + + + +

+ A note has a frequency of 220 cycles per second. How many notes up the piano keyboard is this? +

    +
  1. 37 notes
    2. +
  2. 39 notes
    3. +
  3. 12 notes
    4. +
  4. 96 notes
+

+ + +

+ A. This note is called A_3. +

+ + + + +

+ What frequency does the 12th note have? Round to the nearest tenth. +

    +
  1. 0.1 cycles per second
    2. +
  2. 227.0 cycles per second
    3. +
  3. 51.9 cycles per second
    4. +
  4. 49.4 cycles per second
+

+ + +

+ C +

+ + + + + + Another application of exponential equations is compound interest. Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account. +

+ Compound interest can be calcuated by using the formula A(t)=P\left(1+\frac{r}{n}\right)^{nt} , where A(t) is the account value, t is measured in years, P is the starting amount of the account (also known as the principal), r is the annual percentage rate (APR) written as a decimal, and n is the number of compounding periods in one year. +

+ + + + +

+ Before we can apply the compound interest formula, we need to understand what "compounding" means. Recall that compounding refers to interest earned not only on the original value, but on the accumulated value of the account. This amount is calculated a certain number of times in a given year. +

+ + + +

+ Suppose an account is compounded quarterly, what value of n would we use in the formula? +

    +
  1. 1
    2. +
  2. 2
    3. +
  3. 4
    4. +
  4. 12
    5. +
  5. 365
+

+ + +

+ C +

+ + + + +

+ Suppose an account is compounded daily, what value of n would we use in the formula? +

    +
  1. 1
    2. +
  2. 2
    3. +
  3. 4
    4. +
  4. 12
    5. +
  5. 365
+

+ + +

+ E +

+ + + + +

+ Suppose an account is compounded semi-annually, what value of n would we use in the formula? +

    +
  1. 1
    2. +
  2. 2
    3. +
  3. 4
    4. +
  4. 12
    5. +
  5. 365
+

+ + +

+ B +

+ + + + +

+ Suppose an account is compounded monthly, what value of n would we use in the formula? +

    +
  1. 1
    2. +
  2. 2
    3. +
  3. 4
    4. +
  4. 12
    5. +
  5. 365
+

+ + +

+ D +

+ + + + +

+ Suppose an account is compounded annually, what value of n would we use in the formula? +

    +
  1. 1
    2. +
  2. 2
    3. +
  3. 4
    4. +
  4. 12
    5. +
  5. 365
+

+ + +

+ A +

+ + + + + + +

+ A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child's future college tuition; the account grows tax-free. Lily currently has $10{,}000 and opens a 529 account that will earn 6 \% compounded semi-annually. +

+ + + +

+ Which equation could we use to determine how much money Lily will have for her granddaughter after t years? +

    +
  1. A(t)=10{,}000\left(1+\frac{6}{2}\right)^{2t}
    2. +
  2. A(t)=10{,}000\left(1+\frac{0.06}{2}\right)^{2t}
    3. +
  3. A(t)=10{,}000\left(1+\frac{6}{\frac{1}{2}}\right)^{\frac{1}{2}t}
    4. +
  4. A(t)=10{,}000\left(1+\frac{0.06}{2}\right)^{18}
+

+ + +

+ B +

+ + + + +

+ To the nearest dollar, how much will Lily have in the account in 10 years? +

    +
  1. $106{,}090
    2. +
  2. $103{,}000
    3. +
  3. $13{,}439
    4. +
  4. $18{,}061
+

+ + +

+ D +

+ + + + +

+ How many years will it take Lily to have $40{,}000 in the account for her granddaughter? Round to the nearest tenth. +

    +
  1. 23.4 years
    2. +
  2. 46.9 years
    3. +
  3. 52.1 years
    4. +
  4. 25.6 years
+

+ + +

+ A +

+ + + + + + +

+ For each of the following, determine the appropriate equation to use to solve the problem. +

+ + + +

+ Kathy plans to purchase a car that depreciates (loses value) at a rate of 14 \% per year. The initial cost of the car is $21{,}000. Which equation represents the value, v, of the car after 3 years? +

    +
  1. v=21{,}000(0.14)^3
    2. +
  2. v=21{,}000(0.86)^3
    3. +
  3. v=21{,}000(1.14)^3
    4. +
  4. v=21{,}000(0.86)(3)
+

+ + +

+ B +

+ + + + +

+ Mr. Smith invested $2{,}500 in a savings account that earns 3 \% interest compounded annually. He made no additional deposits or withdrawals. Which expression can be used to determine the number of dollars in this account at the end of 4 years? +

    +
  1. 2{,}500(1+0.03)^4
    2. +
  2. 2{,}500(1+0.3)^4
    3. +
  3. 2{,}500(1+0.04)^3
    4. +
  4. 2{,}500(1+0.4)^3
+

+ + +

+ A +

+ + + + + + For many real-world phenomena, e is used as the base for exponential functions. Exponential models that use e as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics. +

For all real numbers t, and all positive numbers a and r, continuous growth or decay is represented by the formula A(t)=ae^{rt}, where a is the initial value, r is the continuous growth rate per of unit time, and t is the elapsed time. If r\gt 0, then the formula represents continuous growth. If r\lt 0, then the formula represents continuous decay.

+ +

For business applications, the continuous growth formula is called the continuous compounding formula and takes the form A(t)=Pe^{rt}, where P is the principal or the initial invested, r is the growth or interest rate per of unit time, and t is the period or term of the investment. +

+ + + + +

+ Use the continuous formulas to answer the following questions. +

+ + + +

+ A person invested $1{,}000 in an account earning 10 \% per year compounded continuously. How much was in the account at the end of one year? +

+ + +

+ $1{,}105.17 +

+ + + + +

+ Radon-222 decays at a continuous rate of 17.3 \% per day. How much will 100 mg of Radon-222 decay to in 3 days? +

+ + +

+ 59.51 mg +

+ + + + + + + + Videos +

It would be great to include videos down here, like in the Calculus book!

+ +
diff --git a/precalculus/source/05-EL/main.ptx b/precalculus/source/05-EL/main.ptx new file mode 100644 index 00000000..0473cdf1 --- /dev/null +++ b/precalculus/source/05-EL/main.ptx @@ -0,0 +1,19 @@ + + + + Exponential and Logarithmic Functions (EL) + + + + + + + + + + + + + + + \ No newline at end of file diff --git a/precalculus/source/05-EL/outcomes/01.ptx b/precalculus/source/05-EL/outcomes/01.ptx new file mode 100644 index 00000000..e54ea20f --- /dev/null +++ b/precalculus/source/05-EL/outcomes/01.ptx @@ -0,0 +1,4 @@ + +

+ Determine if a given function is exponential. Find an equation of an exponential function. Evaluate exponential functions (including base e). +

\ No newline at end of file diff --git a/precalculus/source/05-EL/outcomes/02.ptx b/precalculus/source/05-EL/outcomes/02.ptx new file mode 100644 index 00000000..db165ef6 --- /dev/null +++ b/precalculus/source/05-EL/outcomes/02.ptx @@ -0,0 +1,4 @@ + +

+ Graph exponential functions and determine the domain, range, and asymptotes. +

\ No newline at end of file diff --git a/precalculus/source/05-EL/outcomes/03.ptx b/precalculus/source/05-EL/outcomes/03.ptx new file mode 100644 index 00000000..1b40006c --- /dev/null +++ b/precalculus/source/05-EL/outcomes/03.ptx @@ -0,0 +1,4 @@ + +

+ Convert between exponential and logarithmic form. Evaluate a logarithmic function, including common and natural logarithms. +

\ No newline at end of file diff --git a/precalculus/source/05-EL/outcomes/04.ptx b/precalculus/source/05-EL/outcomes/04.ptx new file mode 100644 index 00000000..2a6be94b --- /dev/null +++ b/precalculus/source/05-EL/outcomes/04.ptx @@ -0,0 +1,4 @@ + +

+ Graphs of Logarithmic Functions: Graph logarithmic functions and determine the domain, range, and asymptotes. +

\ No newline at end of file diff --git a/precalculus/source/05-EL/outcomes/05.ptx b/precalculus/source/05-EL/outcomes/05.ptx new file mode 100644 index 00000000..835c7871 --- /dev/null +++ b/precalculus/source/05-EL/outcomes/05.ptx @@ -0,0 +1,4 @@ + +

+ Use properties of logarithms to condense or expand logarithmic expressions. +

\ No newline at end of file diff --git a/precalculus/source/05-EL/outcomes/06.ptx b/precalculus/source/05-EL/outcomes/06.ptx new file mode 100644 index 00000000..41825df4 --- /dev/null +++ b/precalculus/source/05-EL/outcomes/06.ptx @@ -0,0 +1,4 @@ + +

+ Solve exponential and logarithmic equations. +

\ No newline at end of file diff --git a/precalculus/source/05-EL/outcomes/07.ptx b/precalculus/source/05-EL/outcomes/07.ptx new file mode 100644 index 00000000..98d3163f --- /dev/null +++ b/precalculus/source/05-EL/outcomes/07.ptx @@ -0,0 +1,4 @@ + +

+ Solve application problems using exponential and logarithmic equations. +

\ No newline at end of file diff --git a/precalculus/source/05-EL/outcomes/main.ptx b/precalculus/source/05-EL/outcomes/main.ptx new file mode 100644 index 00000000..32316009 --- /dev/null +++ b/precalculus/source/05-EL/outcomes/main.ptx @@ -0,0 +1,34 @@ + +> + +

+ BIG IDEA for the chapter goes here, in outcomes/main.ptx +

+

+By the end of this chapter, you should be able to... +

+ +
    +
  1. + +
    2. +
  2. + +
    3. +
  3. + +
    4. +
  4. + +
    5. +
  5. + +
    6. +
  6. + +
    7. +
  7. + +
    8. +
+ \ No newline at end of file diff --git a/precalculus/source/05-EL/readiness.ptx b/precalculus/source/05-EL/readiness.ptx new file mode 100644 index 00000000..dd2aba88 --- /dev/null +++ b/precalculus/source/05-EL/readiness.ptx @@ -0,0 +1,57 @@ + + + + + Readiness Assurance +

+ Before beginning this chapter, you should be able to... +

+
    +
  1. +

    Simplify expressions using properties of exponents.

    +
      +

    • Review and Practice: Khan Academy

      +
      • +
    +
    2. +
  2. +

    Simplify expressions that involve negative exponents.

    +
      +

    • Review and Practice: Khan Academy

      +
      • +
    +
    3. +
  3. +

    Graph functions using transformations.

    +
      +

    • Review and Practice:

      +
      • +
    +
    4. +
  4. +

    Solve linear inequalities.

    +
      +

    • Review and Practice:

      +
      • +
    +
    5. +
  5. +

    Find the domain and range of a function from its graph.

    +
      +

    • Review and Practice:

      +
      • +
    +
    6. +
  6. +

    Determine horizontal and vertical asymptotes from a graph.

    +
      +

    • Review: YouTube (Mathispower4u)

      +
      • +

    • Practice: IXL

      +
      • +
    +
    7. + +
+ + \ No newline at end of file diff --git a/source/docinfo.ptx b/precalculus/source/docinfo.ptx similarity index 100% rename from source/docinfo.ptx rename to precalculus/source/docinfo.ptx diff --git a/precalculus/source/main.ptx b/precalculus/source/main.ptx new file mode 100644 index 00000000..21acb05e --- /dev/null +++ b/precalculus/source/main.ptx @@ -0,0 +1,27 @@ + + + + + + + + Precalculus for Team-Based Inquiry Learning + 2024 Development Edition + 2024 Development Edition — Instructor Version + + + + + + + + + + + + + + + + + diff --git a/source/backmatter.ptx b/precalculus/source/meta/backmatter.ptx similarity index 100% rename from source/backmatter.ptx rename to precalculus/source/meta/backmatter.ptx diff --git a/precalculus/source/meta/copyright.ptx b/precalculus/source/meta/copyright.ptx new file mode 100644 index 00000000..97170ade --- /dev/null +++ b/precalculus/source/meta/copyright.ptx @@ -0,0 +1,13 @@ + + + + 2023 + Steven Clontz and Drew Lewis + +This work is freely available for noncommerical, educational purposes. +For specific licensing information, including the terms for licensing of derivative works, please visit + + GitHub.com/TeamBasedInquiryLearning +. + + \ No newline at end of file diff --git a/precalculus/source/meta/frontmatter.ptx b/precalculus/source/meta/frontmatter.ptx new file mode 100644 index 00000000..90d53b7f --- /dev/null +++ b/precalculus/source/meta/frontmatter.ptx @@ -0,0 +1,90 @@ + + + + + + + + + TBIL Fellows + + + + Editors + + Steven Clontz + Department of Mathematics and Statistics + University of South Alabama + sclontz@southalabama.edu + + + Drew Lewis + drew.lewis@gmail.com + + + + Contributing Authors + + Tonya DeGeorge + Department of Mathematics + Georgia Gwinnett College + tdegeorge@ggc.edu + + + Abby Noble + Department of Mathematics and Statistics + Middle Georgia State University + abby.noble@mga.edu + + + Kathy Pinzon + Department of Mathematics + Georgia Gwinnett College + kpinzon@ggc.edu + + + + + + + + + + Precalculus for Team-Based Inquiry Learning + + +

+ This work includes materials used under license from the following works: +

+
    +
  • +

    Active Prelude to Calculus

    +
      +
    • +

      +
      • +
    • +

      CC BY-SA 4.0

      +
      • +
    +
    • +
+ +

The authors would like to acknowledge the contributions of Wendy Sheppard to drafts of , , and .

+

This work is supported in part by a grant from Affordable Learning Georgia.

+

This material is based upon work supported by the National Science Foundation under Grant No. DUE-2011807. + Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the National Science Foundation. +

+ + + + diff --git a/project.ptx b/project.ptx index 2415e700..1e3aa6ce 100644 --- a/project.ptx +++ b/project.ptx @@ -1,8 +1,111 @@ - + - - + + + + + + + + + + + + + + + diff --git a/publication/instructor.ptx b/publication/instructor.ptx new file mode 100644 index 00000000..6a281c2d --- /dev/null +++ b/publication/instructor.ptx @@ -0,0 +1,132 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + letterpaper,margin=1in + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/publication/publication.ptx b/publication/publication.ptx index e6ea9a59..08ea1fe7 100644 --- a/publication/publication.ptx +++ b/publication/publication.ptx @@ -18,6 +18,7 @@ + @@ -34,6 +35,7 @@ + @@ -60,16 +62,12 @@ - - - - - - - + + + letterpaper,margin=1in + - - + @@ -101,16 +99,10 @@ - - - + + + + diff --git a/publication/slides.ptx b/publication/slides.ptx new file mode 100644 index 00000000..66fcdadd --- /dev/null +++ b/publication/slides.ptx @@ -0,0 +1,81 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + letterpaper,margin=1in + + + + + + + + + + + + + + + + + + + + + + + diff --git a/site/description.html b/site/description.html index 421ff255..2cf7f322 100644 --- a/site/description.html +++ b/site/description.html @@ -1,7 +1,24 @@ -Team-Based Inquiry Learning (TBIL) is an implementation of Team-Based Learning as -Inquiry-Based Learning for mathematics. Team-Based Learning is popular in other -disciplines such as health sciences, where it is most often implemented as Flipped -Learning and/or Case-Based Learning. In mathematics, however, we view it as a way -to more easily bring Inquiry-Based Learning into lower division courses such as -Precalculus and Linear Algebra. -Learn more! \ No newline at end of file +
+

+I feel like I naturally gained knowledge in the topic, but the structure of the course made it +a lot easier to feel the gains and gain from others as well. +

+
+—Student survey response +
+
+

+Team-Based Learning (TBL) +was first popularized for use in management and health sciences education. This +highly active and structured pedagogy provides scaffolding that allows +mathematics students to engage in +Inquiry-Based Learning +(IBL) in courses as early as precalculus. +

+

+We developed Team-Based Inquiry Learning (TBIL) to combine +the scaffolding of TBL with the rich inquiry of IBL. This site includes several +activity books, exercise banks, research findings, and more, made available +to you as free and open-source resources for your classroom. +

+Learn more! diff --git a/site/pages/about.md b/site/pages/about.md index 9bb05c24..b4669627 100644 --- a/site/pages/about.md +++ b/site/pages/about.md @@ -3,16 +3,16 @@ title: About TBIL slug: about --- -## What is TBIL? +### What is TBIL? Team-Based Inquiry Learning (TBIL) is an implementation of Team-Based Learning as Inquiry-Based Learning for mathematics. Team-Based Learning is popular in other disciplines such as health sciences, where it is most often implemented as Flipped Learning and/or Case-Based Learning. In mathematics, however, we view it as a way to more easily bring Inquiry-Based Learning into lower division courses such as Calculus and Linear Algebra. -## Join the Community +### Join the Community If you're planning to use TBIL in your classroom, you can get a link to join our Slack community by completing this survey. Joining our Slack allows you to chat with other instructors and the TBIL team. -## Publications +### Publications Lewis, D., Clontz, S., and Estis, J. 2021. Team-Based Inquiry Learning. PRIMUS 31 (2), 223-238. https://doi.org/10.1080/10511970.2019.1666440 @@ -28,7 +28,7 @@ Lewis, D., Clontz, S., Parrish, C., Estis, J., and Chaudhury, S.R. To Appear. Su Abstract: Team-Based Inquiry Learning (TBIL) is a novel active learning pedagogy designed to facilitate the use of inquiry-based learning in lower division courses. This preliminary report examines supports provided by the TBIL project to instructors, as well as the fidelity of implementation of TBIL by participants of the project. Initial findings suggest that classroom-ready materials and ongoing support, both synchronous and asynchronous, were most helpful to faculty in their TBIL implementations. -## Presentations +### Presentations Kate Owens* has created these excellent slides overviewing TBIL and resources to implement it in your classroom. diff --git a/source/ch-chapter-title.ptx b/source/ch-chapter-title.ptx deleted file mode 100644 index f8246678..00000000 --- a/source/ch-chapter-title.ptx +++ /dev/null @@ -1,13 +0,0 @@ - - - - Chapter Title - - -

Text before the first section.

- - - - - - \ No newline at end of file diff --git a/source/frontmatter.ptx b/source/frontmatter.ptx deleted file mode 100644 index b81fe274..00000000 --- a/source/frontmatter.ptx +++ /dev/null @@ -1,32 +0,0 @@ - - - - - - - - - You - Your department - Your institution - - - - - - - - - - My Website - - - - 20202023 - You - This work is licensed under the Creative Commons Attribution-ShareAlike 4.0 International License. To view a copy of this license, visit CreativeCommons.org - - - - - diff --git a/source/main.ptx b/source/main.ptx deleted file mode 100644 index dfcbaf1a..00000000 --- a/source/main.ptx +++ /dev/null @@ -1,22 +0,0 @@ - - - - - - - - My Great Book - An example to get you started - - - - - - - - - - - - - diff --git a/source/sec-section-name.ptx b/source/sec-section-name.ptx deleted file mode 100644 index 924b3fee..00000000 --- a/source/sec-section-name.ptx +++ /dev/null @@ -1,8 +0,0 @@ - - -
- Section Title - -

Text of section.

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