Merge branch 'master' of https://github.com/jverzani/CalculusWithJuli…

…a.jl

.github/workflows/ci.yml

@@ -44,3 +44,25 @@ jobs:
       - uses: codecov/codecov-action@v1
         with:
           file: lcov.info
+  docs:
+    name: Documentation
+    runs-on: ubuntu-latest
+    steps:
+      - uses: actions/checkout@v2
+      - uses: julia-actions/setup-julia@v1
+        with:
+          version: '1'
+      - run: |
+          julia --project=docs -e '
+            using Pkg
+            Pkg.develop(PackageSpec(path=pwd()))
+            Pkg.instantiate()'
+      - run: |
+          julia --project=docs -e '
+            using Documenter: doctest
+            using CalculusWithJulia
+            doctest(CalculusWithJulia)'
+      - run: julia --project=docs docs/make.jl
+        env:
+          GITHUB_TOKEN: ${{ secrets.GITHUB_TOKEN }}
+          DOCUMENTER_KEY: ${{ secrets.DOCUMENTER_KEY }}

CwJ/TODO/ladder-questions.md

@@ -0,0 +1,115 @@
+###### Question (Ladder [questions](http://www.mathematische-basteleien.de/ladder.htm))
+
+
+A ``7``meter ladder leans against wall with the base ``1.5``meters from wall at its base. At which height does the ladder touch the wall?
+
+```julia; hold=true; echo=false
+l = 7
+adj = 1.5
+opp = sqrt(l^2 - adj^2)
+numericq(opp, 1e-3)
+```
+
+
+----
+
+A ``7``meter ladder leans against the wall. Between the ladder and the wall is a ``1``m cube box. The ladder touches the wall, the box and the ground. There are two such positions, what is the height of the ladder of the more upright position?
+
+You might find this code of help:
+
+```julia; eval=false
+@syms x y
+l, b = 7, 1
+eq = (b+x)^2 + (b+y)^2
+eq = subs(eq, x=> b*(b/y)) # x/b = b/y
+solve(eq ~ l^2, y)
+```
+
+What is the value `b+y` in the above?
+
+```julia; echo=false
+radioq(("The height of the ladder",
+        "The height of the box plus ladder",
+        "The distance from the base of the ladder to the box,"
+        "The distance from the base of the ladder to the base of the wall"
+        ),1)
+```
+
+
+What is the height of the ladder
+
+```julia; hold=true; echo=false
+numericq(6.90162289514212, 1e-3)
+```
+
+
+----
+
+A ladder of length ``c`` is to moved through a 2-dimensional hallway of width ``b`` which has a right angled bend. If ``4b=c``, when will the ladder get stuck?
+
+Consider this picture
+
+```julia; hold=true; echo=false
+p = plot(; axis=nothing, legend=false, aspect_ratio=:equal)
+x,y=1,2
+b = sqrt(x*y)
+plot!(p, [0,0,b+x], [b+y,0,0], linestyle=:dot)
+plot!(p, [0,b+x],[b,b], color=:black, linestyle=:dash)
+plot!(p, [b,b],[0,b+y], color=:black, linestyle=:dash)
+plot!(p, [b+x,0], [0, b+y], color=:black)
+```
+
+
+Suppose ``b=5``, then with ``b+x`` and ``b+y``  being the lengths on the walls where it is stuck *and* by similar triangles ``b/x = y/b`` we can solve for ``x``. (In the case take the largest positive value. The answer would be the angle ``\theta`` with ``\tan(\theta) = (b+y)/(b+x)``.
+
+```julia; hold=true; echo=false
+b = 5
+l = 4*b
+@syms x y
+eq = (b+x)^2 + (b+y)^2
+eq =subs(eq, y=> b^2/x)
+x₀ = N(maximum(filter(>(0), solve(eq ~ l^2, x))))
+y₀ = b^2/x₀
+θ₀ = Float64(atan((b+y₀)/(b+x₀)))
+numericq(θ₀, 1e-2)
+```
+
+
+-----
+
+Two ladders of length ``a`` and ``b`` criss-cross between two walls of width ``x``. They meet at a height of ``c``.
+
+```julia; hold=true; echo=false
+p = plot(; legend=false, axis=nothing, aspect_ratio=:equal)
+ya,yb,x = 2,3,1
+plot!(p, [0,x],[ya,0], color=:black)
+plot!(p, [0,x],[0, yb], color=:black)
+plot!(p, [0,0], [0,yb], color=:blue, linewidth=5)
+plot!(p, [x,x], [0,yb], color=:blue, linewidth=5)
+plot!(p, [0,x], [0,0], color=:blue, linewidth=5)
+xc = ya/(ya+yb)
+c = yb*xc
+plot!(p, [xc,xc],[0,c])
+p
+```
+
+Suppose ``c=1``, ``b=3``, and ``a=5``. Find ``x``.
+
+Introduce ``x = z + y``, and ``h`` and ``k`` the heights of the ladders along the left wall and the right wall.
+
+The ``z/c = x/k`` and ``y/c = x/h`` by similar triangles. As ``z + y`` is ``x`` we can solve to get
+
+```math
+x = z + y = \frac{xc}{k} + \frac{xc}{h}
+ = \frac{xc}{\sqrt{b^2 - x^2}} + \frac{xc}{\sqrt{a^2 - x^2}}
+```
+
+With ``a,b,c`` as given, this can be solved with
+
+```julia; hold=true; echo=false
+a,b,c =  5, 3, 1
+f(x) = x*c/sqrt(b^2 - x^2) + x*c/sqrt(a^2 - x^2) - x
+find_zero(f, (0, b))
+```
+
+The answer is ``2.69\dots``.

CwJ/derivatives/derivatives.jmd

@@ -1392,7 +1392,8 @@ Consider this picture of composition:
 ```julia; hold=true; echo=false
 f(x) = sin(x)
 g(x) = exp(x)
-a,b = 0, 1.6
+a,b = 0, 1.55
+
 xs = range(a, b, length=100)
 ys = g.(xs)
 us = range(extrema(ys)..., length=100)
@@ -1425,59 +1426,24 @@ Assuming the approximation gets better for ``h`` close to ``0``, as it visually
 &= \lim_{h\rightarrow 0} \frac{f(g(1) + g'(1)h)-f(g(1))}{h}\\
 &= \lim_{h\rightarrow 0} \frac{f(g(1) + g'(1)h)-f(g(1))}{g'(1)h} \cdot g'(1)\\
 &= \lim_{h\rightarrow 0} (f\circ g)'(1) \cdot g'(1).
-\end{align*
+\end{align*}
 ```
 
 Whta limit law (described below assuming all limits exist) allows the last equals sign?
 
 ```julia; hold=true; echo=false
 choices = [
-"""
-The limit of a sum is the sum of the limits:
-``\lim_{x\rightarrow c}(af(x)+bg(x)) = a\lim_{x\rightarrow c}f(x) + b\lim_{x\rightarrow c}g(x)``
-""",
-"""
-The limit of a product is the product of the limits:
-``\lim_{x\rightarrow c}(f(x)\cdot g(x)) = \lim_{x\rightarrow c}f(x) \cdot \lim_{x\rightarrow c}g(x)``
-""",
-"""
-The limit of a composition (``f`` is continuous):
-``\lim_{x \rightarrow c}f(g(x)) = \lim_{v \rightarrow \lim_{x \rightarrow c}g(x)} f(v)``.
-"""
-]
-radioq(choices, 3, keep_order=true)
-```
-
-
-Consider the function ``k(x) = f(a + bx)``. We can directly find the ``k'(0)`` using a limit (as an alternative to using the chain rule):
-
-```math
-\begin{align*}
-g'(x) &= \lim_{h\rightarrow 0} \frac{k(h) - k(0)}{h}\\
-&= \lim_{h\rightarrow 0} \frac{f(a + b\cdot h) - f(a)}{h}\\
-&= \lim_{h\rightarrow 0} \frac{f(a+ bh)) - f(a)}{bh} \cdot b\\
-&= \lim_{v\rightarrow 0} \frac{f(a+ v)) - f(a)}{v} \cdot b \quad\text{where } v = bh\\
-&= f'(a) \cdot b
-\end{align*
-```
-
-Which, with ``a = g(x)`` and ``b=g'(x)`` shows heuristically that ``(f\circ g)'(x) = f'(g(x)) \cdot g'(x)``.
-
-What limit law (described below assuming all limits exist) allows the second to last equals sign?
-
-```julia; hold=true; echo=false
-choices = [
-"""
+raw"""
 The limit of a sum is the sum of the limits:
-``\lim_{x\rightarrow c}(af(x)+bg(x)) = a\lim_{x\rightarrow c}f(x) + b\lim_{x\rightarrow c}g(x)``
+``\lim_{x\rightarrow c}(au(x)+bv(x)) = a\lim_{x\rightarrow c}u(x) + b\lim_{x\rightarrow c}v(x)``
 """,
-"""
+raw"""
 The limit of a product is the product of the limits:
-``\lim_{x\rightarrow c}(f(x)\cdot g(x)) = \lim_{x\rightarrow c}f(x) \cdot \lim_{x\rightarrow c}g(x)``
+``\lim_{x\rightarrow c}(u(x)\cdot v(x)) = \lim_{x\rightarrow c}u(x) \cdot \lim_{x\rightarrow c}v(x)``
 """,
-"""
-The limit of a composition (``f`` is continuous):
-``\lim_{x \rightarrow c}f(g(x)) = \lim_{v \rightarrow \lim_{x \rightarrow c}g(x)} f(v)``.
+raw"""
+The limit of a composition (under assumptions on ``v``):
+``\lim_{x \rightarrow c}u(v(x)) = \lim_{w \rightarrow \lim_{x \rightarrow c}v(x)} u(w)``.
 """
 ]
 radioq(choices, 3, keep_order=true)

CwJ/derivatives/implicit_differentiation.jmd

@@ -936,11 +936,10 @@ imgfile = "figures/fcarc-may2016-fig35-350.gif"
 caption = """
 Image number 35 from L'Hospitals calculus book (the first). Given a description of the curve, identify the point ``E`` which maximizes the height.
 """
-#ImageFile(:derivatives, imgfile, caption)
+ImageFile(:derivatives, imgfile, caption)
 nothing
 ```
 
-![Image number 35 from L'Hospitals calculus book (the first). Given a description of the curve, identify the point ``E`` which maximizes the height.](https://raw.githubusercontent.com/jverzani/CalculusWithJulia.jl/master/CwJ/derivatives/figures/fcarc-may2016-fig35-350.gif)
 
 The figure above shows a problem appearing in L'Hospital's first calculus book. Given a function defined implicitly by ``x^3 + y^3 = axy``  (with ``AP=x``, ``AM=y`` and ``AB=a``) find the point ``E`` that maximizes the height. In the [AMS feature column](http://www.ams.org/samplings/feature-column/fc-2016-05) this problem is illustrated and solved in the historical manner, with the comment that the concept of implicit differentiation wouldn't have occurred to L'Hospital.
 

CwJ/derivatives/mean_value_theorem.jmd

@@ -394,15 +394,15 @@ That is the function $f(x)$, minus the secant line between $(a,f(a))$ and $(b, f
 
 ```julia; hold=true; echo=false
 # Need to bring jsxgraph into PLUTO
-caption = """
-Illustration of the mean value theorem from
-[jsxgraph](https://jsxgraph.uni-bayreuth.de/).
-The polynomial function interpolates  the points ``A``,``B``,``C``, and ``D``.
-Adjusting these creates different functions. Regardless of the
-function -- which as a polynomial will always be continuous and
-differentiable -- the slope of the secant line between ``A`` and ``B`` is always matched by **some** tangent line between the points ``A`` and ``B``.
-"""
-JSXGraph(:derivatives, "mean-value.js", caption)
+#caption = """
+#Illustration of the mean value theorem from
+#[jsxgraph](https://jsxgraph.uni-bayreuth.de/).
+#The polynomial function interpolates  the points ``A``,``B``,``C``, and ``D``.
+#Adjusting these creates different functions. Regardless of the
+#function -- which as a polynomial will always be continuous and
+#differentiable -- the slope of the secant line between ``A`` and ``B`` is alway#s matched by **some** tangent line between the points ``A`` and ``B``.
+#"""
+#JSXGraph(:derivatives, "mean-value.js", caption)
 nothing
 ```
 

CwJ/derivatives/taylor_series_polynomials.jmd

@@ -1014,7 +1014,7 @@ radioq(choices, ans)
 
 ###### Question
 
-	The ``5``th order Taylor polynomial for $\sin(x)$ about $c=0$ is: $x - x^3/3! + x^5/5!$. Use this to find the first ``3`` terms of the Taylor polynomial of $\sin(x^2)$ about $c=0$.
+The ``5``th order Taylor polynomial for $\sin(x)$ about $c=0$ is: $x - x^3/3! + x^5/5!$. Use this to find the first ``3`` terms of the Taylor polynomial of $\sin(x^2)$ about $c=0$.
 
 They are: