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@@ -1828,6 +1828,43 @@ answ = 1
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radioq(choices, answ)
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```
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###### Question
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[Durer](https://mathshistory.st-andrews.ac.uk/Curves/Durers/)'s curves are parameterized by $a$ and $b$ and given by:
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```{julia}
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durer(a=1, b=1) = (x,y) ->(x^2 + x*y + a*x - b^2)^2 - (b^2 - x^2)*(x-y+a)^2
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```
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They can be visualized with a contour plot as follows (a plot of an implicit function)
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```{julia}
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xs = ys = range(-5, 5, 500)
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b = 4; a = b/4
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contour(xs, ys, durer(a,b); levels=[0])
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```
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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?
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```{julia}
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#| echo: false
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b = 4
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a = b/2
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yesnoq(durer(a,b)(-2a, -a) == 0)
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```
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What about the point $(-2a, a)$?
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```{julia}
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#| echo: false
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b = 4
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a = b/2
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yesnoq(durer(a,b)(-2a, a) == 0)
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```
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(One is the cusp which is a loop if `a=b/4` and smooths out if `a=b/(3/2)`, say.
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###### Question
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@@ -1038,8 +1038,7 @@ $$
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---
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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).
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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).
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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:
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