WIP

quarto/README-quarto.md

@@ -9,13 +9,14 @@ quarto publish gh-pages --no-render
 ```
 
 
-But better to 
+But better to
 
 ```
 quarto render
 # commit changes and push
 # fix typos
 quarto render
+julia adjust_plotly.jl
 quarto publish gh-pages --no-render
 ```
 

quarto/derivatives/implicit_differentiation.qmd

@@ -923,6 +923,73 @@ $$
 y(x^2 + a^2) = a^3.
 $$
 
+::: {#fig-witch-agnesi}
+```{julia}
+#| echo: false
+gr()
+let
+
+	function ABP(θ,a=1)
+		# y/x = 2a/x = tan(θ)
+		A = (2a/tan(θ), a)
+		# x = y/tan(theta); x^2 + (y-a)^2 = a^2
+		# y^2/t^2 + y^2 - 2ya + a^2 = a^2
+		# y/t^2 + y - 2a = 0
+		# y = 2a/(1 + 1/t^2)
+		y = 2a/(1 + 1/tan(θ)^2) # = 2a sin(θ)^2
+		x = y/tan(θ)
+		B = (x, y-a)
+		P = (A[1],B[2])
+		(;A,B,P)
+	end
+
+
+    a = 1
+
+	ts = range(0, 2pi, 200)
+	plot(;empty_style..., aspect_ratio=:equal)
+
+	plot!(a*cos.(ts), a*sin.(ts); line=(:black, 1))
+	Δ = 1.5
+	plot!(Δ*[-1,1],[-1,-1], line=(:gray, 1))
+	plot!(Δ*[-1,1],[1,1], line=(:gray, 1))
+	plot!([(0,0), (0,a)]; line=(:gray, 1, :dash))
+
+	witch(θ,a=1) = ABP(θ,a).P
+
+	θs = range(pi/4,pi/2, 100)
+	plot!(witch.(θs); line=(:black, 3))
+
+    # fix a specific angle
+	θ = pi/3
+	A,B,P = ABP(θ)
+	O = (0, -a)
+
+	plot!([O,A]; line=(:black,1))
+	plot!([B,P,A]; line=(:gray,1, :dash))
+	scatter!([A,B,P,(0,0)])
+
+	ts = (range(0, θ, 100))
+	λ = a/5
+	plot!([(λ*cos(t),λ*sin(t)-a) for t in ts]; line=(:gray,1, 0.75),arrow=true)
+
+	annotate!([(A..., text(L"A",:bottom)),
+			  (B..., text(L"B", :right)),
+			   (P..., text(L"P", :top)),
+               (0,0,text(L"O", :right)),
+			  (0,1/2, text(L"a",:right)),
+			  (a/4*cos(θ/2), a/4*sin(θ/2)-a, text(L"\theta",:left))])
+end
+```
+```{julia}
+#| echo: false
+plotly()
+nothing
+```
+
+The Witch of Agnesi can be expressed implicitly or parametrically in terms of $\theta$.
+:::
+
 If $a=1$, numerically find a value of $y$ when $x=2$.
 
 
@@ -950,6 +1017,30 @@ answ = 1
 radioq(choices, answ)
 ```
 
+In @fig-witch-agnesi for a given $\theta$ the point $P = (x,y)$ where $x$ is the $x$ value of the intersection of the drawn line with the line $y=a$ and $y$ is the $y$ value of the intersection of the drawn line with the circle $x^2 + y^2 = a^2$.
+
+Suppose $O=(0,0)$ and $A=(u,v)$. Which of these formulas is true:
+
+```{julia}
+#| echo: false
+choices = [
+L"(v+a)/u = 2a/u = \tan(\theta)",
+L"v/u = a/u = \tan(\theta)"
+],
+radioq(choices, 1)
+```
+
+Suppose $B=(u,v)$. Which of these is true:
+
+```{julia}
+#| echo: false
+choices = [
+L"$(v+a)/u = \tan(\theta)$ and $u^2 + v^2 = a^2$",
+L"$v/u = \tan(\theta)$ and $u^2 + v^2 = a^2$"
+]
+radioq(choices, 1)
+```
+
 ###### Question
 
 

quarto/differentiable_vector_calculus/scalar_functions.qmd

@@ -1828,6 +1828,43 @@ answ = 1
 radioq(choices, answ)
 ```
 
+###### Question
+
+[Durer](https://mathshistory.st-andrews.ac.uk/Curves/Durers/)'s curves are parameterized by $a$ and $b$ and given by:
+
+```{julia}
+durer(a=1, b=1) = (x,y) ->(x^2 + x*y + a*x - b^2)^2 - (b^2 - x^2)*(x-y+a)^2
+```
+
+They can be visualized with a contour plot as follows (a plot of an implicit function)
+
+```{julia}
+xs = ys = range(-5, 5, 500)
+b = 4; a = b/4
+contour(xs, ys, durer(a,b); levels=[0])
+```
+
+The definition of `durer` above creates a closure. Take the values of $b=4$ and $a=b/2$. Is the point $(-2a, -a)$ on the curve?
+
+```{julia}
+#| echo: false
+b = 4
+a = b/2
+yesnoq(durer(a,b)(-2a, -a) == 0)
+```
+
+What about the point $(-2a, a)$?
+
+```{julia}
+#| echo: false
+b = 4
+a = b/2
+yesnoq(durer(a,b)(-2a, a) == 0)
+```
+
+(One is the cusp which is a loop if `a=b/4` and smooths out if `a=b/(3/2)`, say.
+
+
 ###### Question
 
 

quarto/differentiable_vector_calculus/vector_valued_functions.qmd

@@ -1038,8 +1038,7 @@ $$
 ---
 
 
-Kepler's second law can also be derived from vector calculus. This derivation follows that given at [MIT OpenCourseWare](https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-c-parametric-equations-for-curves/session-21-keplers-second-law/MIT18_02SC_MNotes_k.pdf) and [OpenCourseWare](https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/index.htm).
-
+Kepler's second law can also be derived from vector calculus. This derivation follows that given at MIT OpenCourseWare (link defunct); a succinct approach is given by [Strang](https://ocw.mit.edu/courses/res-18-001-calculus-fall-2023/mitres_18_001_f17_ch12.pdf).
 
 The second law states that the area being swept out during a time duration only depends on the duration of time, not the time. Let $\Delta t$ be this duration. Then if $\vec{x}(t)$ is the position vector, as above, we have the area swept out between $t$ and $t + \Delta t$ is visualized along the lines of:
 

quarto/limits/sequences_series.qmd

@@ -254,8 +254,8 @@ p_n &= \sum_{k=0}^n \frac{1}{k!}b_{n,l}\\
 &> \sum_{k=0}^n \frac{1}{k!} \cdot \left(1 - \frac{(k-1)k}{2n}\right)\\
 &= s_n - \sum_{k=0}^n \frac{1}{k!}\frac{(k-1)k}{2n}\\
 &= s_n - \frac{1}{2n} \sum_{k=2}^n \frac{1}{(k-2)!}\\
-&= s_n - \frac{1}{2n} s_{n-2}
-$> s_n - \frac{3}{2n}
+&= s_n - \frac{1}{2n} s_{n-2}\\
+&> s_n - \frac{3}{2n}
 \end{align*}
 $$