\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath, amssymb, amsthm}
\usepackage{caption}
\usepackage{enumitem}
\usepackage{geometry}
\usepackage{graphicx}
\usepackage{hyperref}
\hypersetup{colorlinks=true, urlcolor=blue}
\usepackage{url}
\usepackage[export]{adjustbox}
\geometry{margin=0.75in}
\setlength{\parindent}{0pt}

\begin{document}

\begin{center}
MIT OCW 16.940J Computational Geometry Problem Set 1 Solutions \\
AnthonyMakesVideos
\end{center}

\begin{enumerate}
%
% Problem 1
%
\item \textbf{Show that the curvature of a planar curve is independent of the parametrization. Namely, if}
\begin{align}
\textbf{r}(t) = [x(t), y(t)]
\end{align}
\textbf{is the curve then a change of variables}
\begin{align}
t = w(u) \text{ with } w^\prime(u) \neq 0
\end{align}
\textbf{does not affect the curvature.} \\

Curvature: $\displaystyle \kappa(t) = \frac{\sqrt{(\textbf{r}^\prime \cdot \textbf{r}^\prime)(\textbf{r}^{\prime\prime} \cdot \textbf{r}^{\prime\prime}) - (\textbf{r}^\prime \cdot \textbf{r}^{\prime\prime})^2}}{(\textbf{r}^\prime \cdot \textbf{r}^\prime)^{3/2}}$. Note: planar curves have zero torsion, i.e., in this problem $\tau = 0$. \\

First, re-write the general non-arc-length parametrized formulation for curvature in two dimensions in terms of the components $x, \ y$ of $\textbf{r}(t)$ as given.
\begin{align*}
\kappa(t) &= \frac{\sqrt{((x^\prime, y^\prime)\cdot(x^\prime, y^\prime))((x^{\prime\prime}, y^{\prime\prime})\cdot(x^{\prime\prime}, y^{\prime\prime}))-((x^\prime, y^\prime)\cdot(x^{\prime\prime}, y^{\prime\prime}))^2}}{((x^\prime, y^\prime)\cdot(x^{\prime\prime}, y^{\prime\prime}))^{3/2}} \\
&= \frac{\sqrt{(x^{\prime 2)} + y^{\prime 2})(x^{\prime\prime 2)} + y^{\prime\prime 2})-(x^\prime x^{\prime\prime} + y^\prime y^{\prime\prime})^2}}{(x^{\prime 2} + y^{\prime 2})^{3/2}} \\
&= \frac{\sqrt{(x^{\prime 2}x^{\prime\prime 2} + x^{\prime 2}y^{\prime\prime 2} + y^{\prime 2}x^{\prime\prime 2} + y^{\prime 2}y^{\prime\prime 2})-((x^\prime x^{\prime\prime})^2 + 2(x^\prime x^{\prime\prime}y^\prime y^{\prime\prime}) + (y^\prime y^{\prime\prime})^2)}}{(x^{\prime 2} + y^{\prime 2})^{3/2}} \\
&= \frac{\sqrt{(x^\prime y^{\prime\prime})^2 - 2(x^\prime y^{\prime\prime}y^\prime x^{\prime\prime})+(y^\prime x^{\prime\prime})^2}}{(x^{\prime 2} + y^{\prime 2})^{3/2}} \\
\kappa(t) &= \frac{\vert x^\prime y^{\prime\prime} - y^\prime x^{\prime\prime}\vert}{(x^{\prime 2} + y^{\prime 2})^{3/2}} \tag{$\star$}\label{0}
\end{align*}

Denote the change of variables $\textbf{r}(t) \mapsto \textbf{r}(w(u)) \equiv \textbf{r}(u)$. Then next, compute $\textbf{r}^\prime(u)$ and $\textbf{r}^{\prime\prime}(u)$ by the chain rule, use the results to recompute (\ref{0}), and finally compare the resulting curvature expression with (\ref{0}).
\begin{align*}
\textbf{r}^\prime (u) = \frac{d}{du}[x(w(u), y(w(u)] = [w^\prime x^\prime(u), w^\prime x^\prime(u)] = w^\prime \textbf{r}^\prime(u) = w^\prime[x^\prime(u), y^\prime(u)] \\
\textbf{r}^{\prime\prime}(u) = \frac{d}{du}[w^\prime \textbf{r}^\prime(u)] = w^{\prime\prime}\textbf{r}^\prime(u) + (w^\prime)^2\textbf{r}^{\prime\prime}(u) = w^{\prime\prime}[x^\prime(u), y^\prime(u)] + (w^{\prime})^2[x^{\prime\prime}(u), y^{\prime\prime}(u)]\\
= [w^{\prime\prime}x^\prime(u) + (w^\prime)^2 x^{\prime\prime}(u), w^{\prime\prime}y^\prime(u) + (w^\prime)^2y^{\prime\prime}(u)]
\end{align*}
\begin{align*}
\Rightarrow \kappa(u) &= \frac{\vert w^\prime x^\prime(u) (w^{\prime\prime}y^\prime(u) + (w^\prime)^2 y^{\prime\prime}(u)) - w^\prime y^\prime(u)(w^{\prime\prime}x^\prime(u) + (w^\prime)^2 x^{\prime\prime}(u)) \vert}{((w^\prime x^\prime(u))^2 + (w^\prime y^\prime(u))^2)^{3/2}} \\
&= \frac{\vert w^\prime [x^\prime y^\prime w^{\prime\prime} + x^\prime y^{\prime\prime}(w^\prime)^2 - x^\prime y^\prime w^{\prime\prime} - y^\prime x^{\prime\prime}(w^\prime)^2]\vert}{((w^\prime)^2(x^{\prime 2} + y^{\prime 2})^{3/2})} = \frac{\vert (w^\prime)^3 \vert \vert x^\prime y^{\prime\prime} - y^\prime x^{\prime\prime} \vert}{\vert (w^\prime)^3 \vert (x^{\prime 2} + y^{\prime 2})^{3/2}}
\end{align*}
\begin{align*}
\Rightarrow \kappa(u) = \frac{\vert x^\prime y^{\prime\prime} - y^\prime x^{\prime\prime}\vert}{(x^{\prime 2} + y^{\prime 2})^{3/2}} = \kappa(t)
\end{align*}
\newpage
%
% Problem 2
%
\item \textbf{Let a }[vector-parametrized]\textbf{ curve $\mathbf{X}$ be defined by
\begin{align}
\textbf{X}(t) = a\int \textbf{g}(t) \times \textbf{g}^\prime (t) \ \mathrm{d}t, \ a = const. \neq 0,
\end{align} 
where $\textbf{g}(t)$ is a vector function satisfying $\vert \textbf{g}(t)\vert = 1$ and $[\textbf{g}\ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}] \neq 0$. Show that the curvature and the torsion of the curve are $\kappa \neq 0$ and $\tau = 1/a$, respectively.} \\

Curvature: $\displaystyle \kappa(t) = \frac{\sqrt{(\textbf{X}^\prime \cdot \textbf{X}^\prime)(\textbf{X}^{\prime\prime} \cdot \textbf{X}^{\prime\prime}) - (\textbf{X}^\prime \cdot \textbf{X}^{\prime\prime})^2}}{(\textbf{X}^\prime \cdot \textbf{X}^\prime)^{3/2}}$. Torsion: $\displaystyle \tau(t) = \frac{\vert \textbf{X}^\prime \textbf{X}^{\prime\prime} \textbf{X}^{\prime\prime\prime}\vert}{(\textbf{X}^\prime \times \textbf{X}^{\prime\prime})\cdot (\textbf{X}^\prime \times \textbf{X}^{\prime\prime})}$. \\
Compute the terms of the definitions above and then insert the results directly into the same definitions. 
\begin{align*}
\textbf{X}^\prime(t) &= a(\textbf{g}(t)\times\textbf{g}^\prime(t))\\
\text{Identity: } A\times A = 0 \Rightarrow \textbf{X}^{\prime\prime}(t) &= a[\textbf{g}^\prime \times \textbf{g}^\prime + \textbf{g}\times\textbf{g}^{\prime\prime}] = a(\textbf{g}\times\textbf{g}^{\prime\prime})\\
\text{Identity: }A\times B = -(B\times A) \Rightarrow \textbf{X}^{\prime\prime\prime}(t) &= a[\textbf{g}^{\prime\prime}\times\textbf{g}^\prime + \textbf{g}^\prime\times\textbf{g}^{\prime\prime}+\textbf{g}^\prime\times\textbf{g}^{\prime\prime}+\textbf{g}\times\textbf{g}^{\prime\prime\prime}] = a[\textbf{g}^\prime\times\textbf{g}^{\prime\prime}+\textbf{g}\times\textbf{g}^{\prime\prime\prime}].
\end{align*} 
Next, compute the dot products $\textbf{X}^\prime \cdot \textbf{X}^\prime, \ \textbf{X}^{\prime\prime} \cdot \textbf{X}^{\prime\prime}$, and $\textbf{X}^\prime \cdot \textbf{X}^{\prime\prime}$,
\begin{align*}
\text{Identity: }(A\times B)\cdot C = C\cdot (A\times B) = \vert A B C \vert \Rightarrow \textbf{X}^\prime \cdot \textbf{X}^\prime &= a^2(\textbf{g}\times \textbf{g}^\prime)\cdot(\textbf{g}\times\textbf{g}^\prime)=a^2\vert \textbf{g}\times\textbf{g}^\prime\vert^2 \\
\textbf{X}^{\prime\prime} \cdot \textbf{X}^{\prime\prime} &= a^2[\textbf{g}^\prime\times\textbf{g}^\prime + \textbf{g}\times\textbf{g}^{\prime\prime}]\cdot[\textbf{g}^\prime\times\textbf{g}^\prime + \textbf{g}\times\textbf{g}^{\prime\prime}]\\
&= a^2[\textbf{g}\times\textbf{g}^{\prime\prime}]\cdot[\textbf{g}\times\textbf{g}^{\prime\prime}] = a^2\vert \textbf{g}\times\textbf{g}^{\prime\prime}\vert^2 \\
\textbf{X}^\prime \cdot \textbf{X}^{\prime\prime} &= a^2(\textbf{g}\times\textbf{g}^\prime)\cdot[\textbf{g}^\prime\times\textbf{g}^\prime + \textbf{g}\times\textbf{g}^{\prime\prime}] \\
&= a^2(\textbf{g}\times\textbf{g}^\prime)\cdot(\textbf{g}\times\textbf{g}^{\prime\prime}).
\end{align*}
Next, compute the cross-product of $\textbf{X}^\prime$ and $\textbf{X}^{\prime\prime}$,
\begin{align*}
\textbf{X}^\prime\times\textbf{X}^{\prime\prime} &= a(\textbf{g}\times\textbf{g}^\prime)\times a(\textbf{g}\times\textbf{g}^{\prime\prime})\\
\textbf{X}^\prime\times\textbf{X}^{\prime\prime} &= a^2 (\textbf{g}\times\textbf{g}^\prime)\times (\textbf{g}\times\textbf{g}^{\prime\prime})\\
\text{Identity: }(A)\times(B\times C) = (A\cdot C)B - (A\cdot B)C \Rightarrow \textbf{X}^\prime\times\textbf{X}^{\prime\prime} &= a^2[((\textbf{g}\times\textbf{g}^\prime)\cdot \textbf{g}^{\prime\prime})\textbf{g}-(\textbf{g}\times\textbf{g}^\prime)\cdot \textbf{g})\textbf{g}^{\prime\prime}]\\
\textbf{g}\text{ is orthogonal to }\textbf{g}\times\textbf{g}^\prime \Rightarrow (\textbf{g}\times\textbf{g}^\prime)\cdot \textbf{g})\textbf{g}^{\prime\prime} = 0\Rightarrow \textbf{X}^\prime\times\textbf{X}^{\prime\prime} &= a^2((\textbf{g}\times\textbf{g}^\prime)\cdot \textbf{g}^{\prime\prime})\textbf{g}\\
\text{Identity: }(A\times B)\cdot C = C\cdot (A\times B) = \vert A B C \vert \Rightarrow \textbf{X}^\prime\times\textbf{X}^{\prime\prime} &= a^2\vert \textbf{g} \ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}\vert \textbf{g}.
\end{align*}
Next, simplify the numerator of curvature $\kappa$ and compute
\begin{align*}
\text{Identity: }\vert\textbf{a}\times\textbf{b}\vert^2 = \vert\textbf{a}\vert^2\vert\textbf{b}\vert^2 - (\textbf{a}\cdot\textbf{b})^2\Rightarrow \sqrt{(\textbf{X}^\prime \cdot \textbf{X}^\prime)(\textbf{X}^{\prime\prime} \cdot \textbf{X}^{\prime\prime}) - (\textbf{X}^\prime \cdot \textbf{X}^{\prime\prime})^2}  &= \sqrt{\vert\textbf{X}^\prime\times\textbf{X}^{\prime\prime}\vert^2} = \vert\textbf{X}^\prime\times\textbf{X}^{\prime\prime}\vert\end{align*}\begin{align*}
\textbf{X}^\prime\times\textbf{X}^{\prime\prime} = a^2\vert \textbf{g} \ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}\vert \textbf{g} \Rightarrow \vert\textbf{X}^\prime\times\textbf{X}^{\prime\prime}\vert &= \sqrt{a^4\vert\textbf{g} \ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}\vert^2\vert\textbf{g}\vert} \\
\vert \textbf{g}\vert = 1 \Rightarrow \vert\textbf{X}^\prime\times\textbf{X}^{\prime\prime}\vert &= a^2\vert\textbf{g} \ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}\vert.
\end{align*}
It is given that $a$ is a non-zero constant and that $[\textbf{g}\ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}] \neq 0$, therefore $\kappa \neq 0$ if the denominator of curvature $\kappa$ is also non-zero. Thus, simplify the denominator of curvature $\kappa$ and compute
\begin{align*}
(\textbf{X}^\prime \cdot \textbf{X}^\prime)^{3/2} = (\vert\textbf{X}^\prime\vert^2)^{3/2} = \vert\textbf{X}^\prime\vert^3.
\end{align*}
Notice that $\vert \textbf{X}^\prime\vert = a\vert\textbf{g}\times\textbf{g}^\prime\vert = a\sqrt{\vert\textbf{g}\vert^2\vert\textbf{g}^\prime\vert^2 -(\textbf{g}\cdot\textbf{g}^\prime)^2}$. Since it is given that $\vert\textbf{g}\vert=1$ and $\textbf{g}^\prime\neq 0$ necessarily otherwise the curve is constant, $\vert\textbf{X}^\prime\vert$, the denominator of curvature $\kappa$ and the entire expression for curvature $\kappa$ itself are non-zero.
Next, compute the denominator of torsion $\tau$,
\begin{align*}
\textbf{X}^\prime\times\textbf{X}^{\prime\prime} = a^2\vert \textbf{g} \ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}\vert \textbf{g} \Rightarrow (\textbf{X}^\prime\times\textbf{X}^{\prime\prime})\cdot (\textbf{X}^\prime\times\textbf{X}^{\prime\prime}) &= a^4\vert \textbf{g}\ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}\vert^2\vert \textbf{g}\vert^2 \\
\text{From the given }\vert \textbf{g} \vert = 1 \Rightarrow (\textbf{X}^\prime\times\textbf{X}^{\prime\prime})\cdot (\textbf{X}^\prime\times\textbf{X}^{\prime\prime}) = a^4\vert \textbf{g}\ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}\vert^2.
\end{align*}
Next, compute the numerator of torsion $\tau$,
\begin{align*}
\text{Identity: }(A\times B)\cdot C = C\cdot (A\times B) = \vert A B C \vert\Rightarrow \vert \textbf{X}^\prime \ \textbf{X}^{\prime\prime} \ \textbf{X}^{\prime\prime\prime} \vert &= (\textbf{X}^\prime \times \textbf{X}^{\prime\prime})\cdot \textbf{X}^{\prime\prime\prime}\\
\vert \textbf{X}^\prime \ \textbf{X}^{\prime\prime} \ \textbf{X}^{\prime\prime\prime} \vert &= a^3\vert \textbf{g} \ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}\vert[\textbf{g}\cdot(\textbf{g}^\prime \times \textbf{g}^{\prime\prime})+\textbf{g}\cdot(\textbf{g}\times\textbf{g}^{\prime\prime\prime})] \\
\textbf{g}\text{ is orthogonal to }\textbf{g}\times\textbf{g}^{\prime\prime\prime}\text{, and }\textbf{g}\cdot(\textbf{g}^\prime \times \textbf{g}^{\prime\prime})=\vert\textbf{g}\ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}\vert\Rightarrow \vert \textbf{X}^\prime \ \textbf{X}^{\prime\prime} \ \textbf{X}^{\prime\prime\prime} \vert &= a^3\vert\textbf{g}\ \textbf{g}^\prime \ \textbf{g}^{\prime\prime}\vert^2.
\end{align*}
Therefore, $\displaystyle \tau = \frac{a^3\vert\textbf{g}\textbf{g}^\prime\textbf{g}^{\prime\prime}\vert^2}{a^4\vert \textbf{g}\textbf{g}^\prime\textbf{g}^{\prime\prime}\vert^2} = 1/a$.
\newpage
%
% Problem 3
%
\item \textbf{Find the parametric equation of a curve whose curvature }$\kappa$\textbf{ and torsion }$\tau$\textbf{ are respectively}
\begin{align}
\kappa=\frac{a}{a^2+b^2}, \ \tau=\frac{b}{a^2+b^2},
\end{align}
\textbf{where }$a>0$\textbf{ and }$b$\textbf{ are constants.}\\ \\
First, let $c = a^2 + b^2$ and notice that 
\begin{align*}
\kappa^2 + \tau^2 = \frac{a^2}{(a^2+b^2)^2}+\frac{b^2}{(a^2+b^2)^2} = \frac{c}{c^2} = \frac{1}{c}.
\end{align*}
Use the given curvature $\kappa$ and torsion $\tau$ to formulate the Frenet-Serret (differential) equations and solve for the parametric curve $\textbf{r}(s)$ where $s$ denotes arc-length. The Frenet-Serret (hereafter ``FS") equations form a system of linear ordinary differential equations from the first derivatives of $\textbf{t} = \textbf{v}_1$, the unit tangent vector, $\textbf{p} = \textbf{v}_2$, the unit principal normal vector, and $\textbf{b} = \textbf{v}_3$, the unit binormal vector. As unit vectors, $\vert \textbf{v}_j \vert = 1$ for $j = 1, 2, 3$. The FS equations may be written as $\displaystyle \textbf{v}^\prime_j = \sum_{k=1}^3 c_{jk}\textbf{v}_k$ for $j = 1, 2, 3$ such that the $c_{jk}$ form the skew-symmetric matrix $C$, i.e., 
\begin{align*}
C = \begin{bmatrix}
0 & \kappa & 0 \\
-\kappa & 0 & \tau \\
0 & -\tau & 0
\end{bmatrix}.
\end{align*}
Note that a prime (i.e., $\textbf{v}^\prime_j$) here denotes a derivative with respect to the arc length $s$ of the curve $\textbf{r}(s)$ implied by the FS equations. In total the FS system of differential equations are written $\textbf{F}^\prime = C\textbf{F}$ where $\textbf{F}=[\textbf{v}_1 \ \textbf{v}_2 \ \textbf{v}_3]^{\intercal}$. Written in this form, $\displaystyle \textbf{t} = \textbf{v}_1 = \frac{d\textbf{r}}{ds}$. \\

A general exponential solution satisfying the FS equations is sought in the form of $\displaystyle \textbf{F}(s) = \textbf{F}(0)e^{sC}$ where $\textbf{F}(0) = [\textbf{v}_1(0) \ \textbf{v}_2(0) \ \textbf{v}_3(0)]^\intercal$ is the initial FS frame a.k.a. ``tpb frame" at arc-length $s = 0$. In the following, denote $\displaystyle\omega =\frac{1}{\sqrt{c}} = \frac{1}{\sqrt{a^2 + b^2}}$. Begin by calculating the first, second, and third derivatives of $\textbf{v}_1$. 
\begin{align*}
\textbf{v}_1^\prime &= \kappa\textbf{v}_2 \\
\Rightarrow \textbf{v}_1^{\prime\prime} &= \kappa\textbf{v}_2^{\prime} = \kappa(-\kappa\textbf{v}_1 + \tau\textbf{v}_3) = -\kappa^2\textbf{v}_1 + \kappa\tau\textbf{v}_3 \\
\Rightarrow \textbf{v}_1^{\prime\prime\prime} &= -\kappa^2\textbf{v}_1^\prime + \kappa\tau\textbf{v}_3^{\prime}\\
&= -\kappa^2(\kappa\textbf{v}_2) + \kappa\tau(-\tau\textbf{v}_2)\\
&= -\kappa^3\textbf{v}_2 - \kappa\tau^2\textbf{v}_2 \\
&= -\kappa(\kappa^2 + \tau^2)\textbf{v}_2 \\
\kappa^2 + \tau^2 = \frac{1}{c} = \frac{1}{\omega^2} \Rightarrow \textbf{v}_1^{\prime\prime\prime} &= \frac{-\kappa}{\omega^2}\textbf{v}_2 \\
\textbf{v}_1^\prime/\kappa = \textbf{v}_2 \Rightarrow \textbf{v}_1^{\prime\prime\prime} &= \frac{-\kappa}{\omega^2}\frac{1}{\kappa}\textbf{v}_1^\prime = \frac{-1}{\omega^2}\textbf{v}_1^\prime \\
&\Rightarrow \textbf{v}_1^{\prime\prime\prime} + \frac{1}{\omega^2}\textbf{v}_1^\prime = 0
\end{align*}
Therefore, the characteristic equation of the FS system of differential equations for the given curvature and torsion with roots $q$ satisfying
\begin{align*}
q^3 + \frac{1}{\omega^2}q = q(q^2 + \frac{1}{\omega^2}) = 0.
\end{align*}
Thus the roots of this characteristic equation are
\begin{align*}
q &= 0, \ \pm i\omega
\end{align*}
Note that the root $q = 0$ gives a constant vector along the binormal $\textbf{b} \equiv \textbf{v}_3$.
\newpage
The general exponential solution for $\textbf{v}_1$ is such that $\displaystyle \vert\textbf{v}_1\vert = 1$ and $\displaystyle \vert\textbf{v}_1^\prime\vert = \kappa = \frac{a}{c}$. Next, the constant from $\vert\textbf{v}_1^\prime\vert=\kappa\ast\text{constant}$ is considered.
\begin{align*}
\textbf{v}_1(s) &= (k\cos{(\omega s)}, \ k\sin{(\omega s)}, \ h) \\
\vert\textbf{v}_1\vert = 1 \Rightarrow \vert \textbf{v}_1\vert^2 &= k^2\cos^2{(\omega s)} + k^2\sin^2{(\omega s)} + h^2 = k^2 + h^2 = 1
\end{align*}
Due to $\vert\textbf{v}_1^\prime\vert=\kappa\cdot\text{constant}$,
\begin{align*}
\vert\textbf{v}_1^\prime\vert &= k\omega = \kappa = \frac{a}{c} \\
k &= \frac{a}{c\omega} = \frac{a}{c\cdot\frac{1}{\sqrt{c}}} = \frac{a}{\sqrt{c}} \\
\Rightarrow k^2 &= \frac{a^2}{c} \\
k^2 + h^2 = 1 \Rightarrow h^2 &= 1 - \frac{a^2}{c} = \frac{c - a^2}{c} = \frac{b^2}{c}
\end{align*}
Thus, $\displaystyle \textbf{v}_1(s) = \left(\frac{a}{\sqrt{c}}\cos{(\omega s)}, \frac{a}{\sqrt{c}}\sin{(\omega s)}, \frac{b}{\sqrt{c}}\right)$. \\
Integrating the tangent vector with respect to arc length yields the parametrized curve of interest, thus,
\begin{align*}
\textbf{r}(s) = \int{\textbf{v}_1(s) \ ds} &= \int{\left(\frac{a}{\sqrt{c}}\cos{(\omega s)}, \ \frac{a}{\sqrt{c}}\sin{(\omega s)}, \ \frac{b}{\sqrt{c}}\right)} ds \\
&= \left(\frac{a}{\sqrt{c}\omega}\sin{(\omega s)}, \ -\frac{a}{\sqrt{c}\omega}\cos{(\omega s)}, \ \frac{b}{\sqrt{c}}s\right) + \textbf{D}\\
&= \left(a\sin{(\frac{s}{\sqrt{a^2 + b^2}})}, \ -a\cos{(\frac{s}{\sqrt{a^2 + b^2}})}, \ \frac{bs}{\sqrt{a^2 + b^2}}\right) + \textbf{D}
\end{align*}
where $\textbf{D}$ is the integration constant. Choose $\textbf{D} = [0 \ 0 \ 0]^\intercal$ to fix the start of the parametric curve at the origin. \\
Thus, $\displaystyle \textbf{r}(s) = \left(a\sin{\left(\frac{s}{\sqrt{a^2 + b^2}}\right)}, \ -a\cos{\left(\frac{s}{\sqrt{a^2 + b^2}}\right)}, \ \frac{bs}{\sqrt{a^2 + b^2}}\right)$.
\newpage
%
% Problem 4
%
\item \textbf{A curve $\mathbf{C}_1$ is called an \textit{involute} of a given curve $\mathbf{C}$ if tangents of $\mathbf{C}$ are normal to $\mathbf{C}_1$. The curve $\mathbf{C}$ is called an \textit{evolute} of $\mathbf{C}_1$. Show that the curvature $\kappa_1$ of $\mathbf{C}_1$ is given by}
\begin{align}
\kappa_1^2 = \frac{\kappa^2+\tau^2}{\kappa^2(c-s)^2},
\end{align}
\textbf{where $c$ is constant, $s$ is the arc length of $\mathbf{C}$ measured from a fixed point on $\mathbf{C}$, and $\kappa$ and $\tau$ are the curvature and torsion of $\mathbf{C}$.}\\

The position vector $\mathbf{r}(s)$
of the involute $\mathbf{C}_1$ at a point on $\mathbf{C}$ is $\mathbf{r}_1(s)$, namely
\begin{align*}
\mathbf{r}_1(s) = \mathbf{r}(s) + (c-s)\mathbf{T}(s)
\end{align*}
where $\mathbf{T}(S)$ denotes the tangent vector for the displacement implied by $(c-s)$. Note that $\mathbf{T}^\prime(s) = \kappa\cdot\mathbf{P}(s)$ is the relation between this tangent vector and the principal normal vector $\mathbf{P}(s)$. Furthermore, the arc length $s_1$ of $\mathbf{C}_1$ is related to $s$ by $\displaystyle \frac{\mathrm{d}s_1}{\mathrm{d}s}=\vert\mathbf{r}_1^\prime(s)\vert = \vert c-s \vert\kappa$. First, differentiate $\mathbf{r}_1(s)$,
\begin{align*}
\mathbf{r}_1^\prime(s) = \frac{\mathrm{d}}{\mathrm{d}s}[\mathbf{r}(s) + (c-s)\mathbf{T}(s)] &= \mathbf{r}^\prime(s) + (c-s)\mathbf{T}^\prime(s) - \mathbf{T}(s)\\
&= (c-s)\kappa\mathbf{P}(s)
\end{align*}
Similarly, since $\mathbf{P}(s)$ is a unit vector (i.e., $\vert\mathbf{P}(s)\vert = 1$) and $\kappa > 0$, the magnitude of $\mathbf{r}_1^\prime(s)$ is thus
\begin{align*}
\vert\mathbf{r}_1^\prime(s)\vert = \vert (c-s)\kappa\mathbf{P}(s)\vert = \vert c-s \vert\kappa
\end{align*}
Following this, the the unit tangent $\mathbf{T}_1$ of $\mathbf{C}_1$ is
\begin{align*}
\mathbf{T}_1(s)=\frac{\mathbf{r}_1^\prime(s)}{\vert \mathbf{r}_1^\prime(s)\vert}=\frac{(c-s)\kappa\mathbf{P}(s)}{\vert c-s\vert\kappa}=\mathbf{P}(s)
\end{align*}
Next, compute curvature $\displaystyle\kappa_1=\left|\frac{\mathrm{d}\mathbf{T}_1}{\mathrm{d}s}\right|$. By the FS equations,
\begin{align*}
\mathbf{T}_1 = \mathbf{P} \Rightarrow \frac{\mathrm{d}\mathbf{T}_1}{\mathrm{d}s} = \frac{\mathrm{d}\mathbf{P}}{\mathrm{d}s} = -\kappa\mathbf{T} + \tau\mathbf{B}
\end{align*}
where $\mathbf{B}$ denotes the binormal vector. By the chain rule for derivatives,
\begin{align*}
\frac{\mathrm{d}\mathbf{T}_1}{\mathrm{d}s_1} = \frac{\mathrm{d}\mathbf{T}_1}{\mathrm{d}s}\cdot\frac{\mathrm{d}s}{\mathrm{d}s_1} = \frac{\mathbf{P}^\prime}{\mathrm{d}s_1/\mathrm{d}s} = \frac{-\kappa\mathbf{T} + \tau\mathbf{B}}{\vert c-s\vert\kappa}
\end{align*}
Without loss of generality, assume that $c>s$. Next, to compute curvature, calculate the magnitude of $\displaystyle \frac{\mathrm{d}\mathbf{T}_1}{\mathrm{d}s_1}$,
\begin{align*}
\kappa_1 = \left|\frac{-\kappa\mathbf{T} + \tau\mathbf{B}}{(c-s)\kappa}\right| = \frac{\vert -\kappa\mathbf{T} + \tau\mathbf{B}\vert}{(c-s)\kappa}
\end{align*}
$\mathbf{T}$ and $\mathbf{B}$ are orthonormal i.e., $\vert\mathbf{T}\vert = \vert\mathbf{B}\vert = 1$ and $\mathbf{T}\cdot\mathbf{B}=0$, thus
\begin{align*}
\Rightarrow \vert -\kappa\mathbf{T} + \tau\mathbf{B}\vert &= \sqrt{(-\kappa)^2 + \tau^2} = \sqrt{\kappa^2 + \tau^2}\\
\Rightarrow \kappa_1^2 &= \frac{\kappa^2+\tau^2}{\kappa^2(c-s)^2}.
\end{align*}
\newpage
%
% Problem 5
%
\item \textbf{Let $E, F, G$ be the coefficients of the first fundamental form of a regular surface $\mathbf{R} = \mathbf{R}(u, v)$. Let $f(u, v) = c$ and $g(u, v) = d$ be two families of regular curves defined in the parameter space $u - v$ of the surface with images in 3D space obtained for various constants $c$ and $d$. Prove that the 3D images of these two families of curves are orthogonal (i.e., whenever two curves of distinct families meet, their tangents are orthogonal) if and only if}
\begin{align}
Ef_vg_v - F(f_ug_v + f_vg_u) + Gf_ug_u = 0
\end{align}
\textbf{where $E = \mathbf{R}_u\cdot\mathbf{R}_u, F=\mathbf{R}_u\cdot\mathbf{R}_v, G=\mathbf{R}_v\cdot\mathbf{R}_v$, and subscripts $u, v$ denote partial derivatives.}\\

$\nabla f$ is normal to the curve in the $u-v$ plane and is thus perpendicular to $(-f_v, f_u)$ and $(-g_v, g_u)$. Start by parametrizing and differentiating using the chain rule:
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t}f(u(t), v(t)) = f_u u^\prime + f_v v^\prime = 0.
\end{align*}
Thus, $(u^\prime, v^\prime) \propto (-f_v, f_u)$, so for scalar $k$
\begin{align*}
\mathbf{T}_f = \frac{\mathrm{d}}{\mathrm{d}t}\mathbf{R}(u(t), v(t)) &= \mathbf{R}_u u^\prime + \mathbf{R}_v v^\prime\\
&= k(-f_v\mathbf{R}_u + f_u \mathbf{R}_v)
\end{align*}
and similarly $u^\prime g_u + v^\prime g_v = 0$, then for scalar $m$
\begin{align*}
\mathbf{T}_g = \mathbf{R}_u u^\prime + \mathbf{R}_v v^\prime = m(-g_v\mathbf{R}_u + g_u\mathbf{R}_v).
\end{align*}
Orthogonality in 3D requires that $\mathbf{T}_f\cdot\mathbf{T}_g = 0$, thus
\begin{align*}
\mathbf{T}_f\cdot\mathbf{T}_g = km [(-f_v\mathbf{R}_u + f_u\mathbf{R}_v)\cdot(-g_v\mathbf{R}_u + g_u\mathbf{R}_v)]
\end{align*}
The bracketed term in the above expression is equivalent to
\begin{align*}
f_vg_v(\mathbf{R}_u\cdot\mathbf{R}_u) - f_vg_u(\mathbf{R}_u\cdot\mathbf{R}_v) - f_ug_v(\mathbf{R}_u\cdot\mathbf{R}_v) + f_ug_u(\mathbf{R}_v\cdot\mathbf{R}_v)
\end{align*}
Thus, since it is given that $E = \mathbf{R}_u\cdot\mathbf{R}_u, F=\mathbf{R}_u\cdot\mathbf{R}_v, G=\mathbf{R}_v\cdot\mathbf{R}_v$,
\begin{align*}
\mathbf{T}_f\cdot\mathbf{T}_g = km[Ef_vg_v - F(f_vg_u+f_ug_v)+Gf_ug_u].
\end{align*}
Note that the curves are orthogonal if and only if the bracketed term in the above expression equals $0$ since $km \neq0$ assuming nonzero tangent vectors. $\blacksquare$
\newpage
%
% Problem 6
%
\item \textbf{Consider a torus parametrized as follows:}
\begin{align}
\mathbf{r}(u, v) = [(R + a\cos{u})\cos{v}, (R + a\cos{u})\sin{v}, a\sin{u}]
\end{align}
\textbf{where $0\leq u \leq 2\pi, \ 0\leq v\leq2\pi$, and $R$ and $a$ are constants such that $R > a$. Derive formulae for the Gauss, mean, and principal curvatures. Sketch the torus and subdivide it into hyperbolic, parabolic, and elliptic regions. In a follow-up sketch, illustrate the lines of curvature of the torus. Explain the above subdivision and sketches. \footnote{Problem 17 in ``Shape Interrogation for Computer Aided Design and Manufacturing'' by N.~M.~Patrikalakis and T.~Maekawa, 1st ed., Springer, 2002.}}

For a torus embedded in 3D space, say $x_1x_2x_3$ space, there will be $3-1 = 2$ principal directions and corresponding principal curvatures. Denote the components of the first fundamental form $g_{jk}$ for $j,\ k = 1, \ 2$ respectively and the components of the second fundamental form $b_{jk}$ for $j,\ k = 1, \ 2$ respectively. In this case, $j$ or $k$ being $1$ corresponds to $u$ and likewise being $2$ corresponds to $v$.

First, compute the first and second derivatives of $\textbf{r}$ needed for the first and second fundamental forms,
\begin{align*}
\textbf{r}_u &= [(-a\sin{u})\cos{v}, (-a\sin{u})\sin{v}, a\cos{u}], \quad \textbf{r}_v = [-(R + a\cos{u})\sin{v}, (R + a\cos{u})\cos{v}, 0],\\
\textbf{r}_{uu} &= [-a\cos{u}\cos{v}, -a\cos{u}\sin{v}, -a\sin{u}], \ \textbf{r}_{uv} =[a\sin{u}\sin{v}, -a\sin{u}\cos{v}, 0], \\
\textbf{r}_{vv} &= [-(R+a\cos{u})\cos{v}, -(R + a\cos{u})\sin{v}, 0]
\end{align*}
Next, compute the components of the first fundamental form and the discriminant of the first fundamental form,
\begin{align*}
g_{11} &= \textbf{r}_u\cdot\textbf{r}_u = a^2\sin^2{u}(\sin^2{v}+\cos^2{v}) + a^2\cos^2{u} = a^2\\
g_{12} &= \textbf{r}_u\cdot\textbf{r}_v = ((a\sin{u})\cos{v})((R + a\cos{u})\sin{v}) - ((a\sin{u})\sin{v})((R + a\cos{u})\cos{v}) + a\cos{u}\cdot 0 = 0 \\
g_{22} &= \textbf{r}_v\cdot\textbf{r}_v = (-(R + a\cos{u})\sin{v})^2 + ((R + a\cos{u})\cos{v})^2 + 0^2 = (R + a\cos{u})^2\\
g &= g_{11}g_{22} - (g_{12})^2 = a^2(R + a\cos{u})^2.
\end{align*}
Next, compute the normal vector for the torus starting with the appropriate cross product of first derivatives of $\textbf{r}$,
\begin{align*}
\textbf{r}_u\times\textbf{r}_v &= \begin{vmatrix}
i & j & k \\
(-a\sin{u})\cos{v} & (-a\sin{u})\sin{v} & a\cos{u} \\
-(R + a\cos{u})\sin{v} & (R + a\cos{u})\cos{v} & 0
\end{vmatrix} \\
&= [-a\cos{u}\cos{v}(R + a\cos{u}), -a\cos{u}\sin{v}(R + a\cos{u}), -a\sin{u}(R+a\cos{u})] \\
\textbf{n} = \frac{\mathbf{x}_1\times\mathbf{x}_2}{+\sqrt{g}} &= \frac{[-a\cos{u}\cos{v}(R + a\cos{u}), -a\cos{u}\sin{v}(R + a\cos{u}), -a\sin{u}(R+a\cos{u})]}{a(R + a\cos{u})}\\
&= [-\cos{u}\cos{v}, -\cos{u}\sin{v}, -\sin{u}].
\end{align*}
Next, use the previous results to compute the components of the second fundamental form $b_{11}= \textbf{r}_{uu}\cdot\textbf{n}$, $b_{12} = \textbf{r}_{uv}\cdot\textbf{n}$, $b_{22} = \textbf{r}_{vv}\cdot\textbf{n}$, and the discriminant of the second fundamental form $b = b_{11}b_{22} - b_{12}^2$,
\begin{align*}
b_{11} &= a\cos^2{u}\cos^2{v} + a\cos^2{u}\sin^2{v} + a\sin^2{u} = a\\
b_{12} &= a\sin{u}\cos{u}\sin{v}\cos{v} - a\sin{u}\cos{u}\sin{v}\cos{v} = 0\\
b_{22} &= (R + a\cos{u})\cos{u}(\sin^2{v}+\cos^2{v}) = (R + a\cos{u})\cos{u}\\
b &= a(R + a\cos{u})\cos{u}
\end{align*}
Thus, for the given parametrization of the torus the principal, Gauss, and mean curvatures are
\begin{align*}
\kappa_1 &= \frac{b_{11}}{g_{11}} = -\frac{1}{a}, \quad \kappa_2 = \frac{b_{22}}{g_{22}} = -\frac{\cos{u}}{R+a\cos{u}} \\
\Rightarrow K = \frac{b}{g} &= \kappa_1\kappa_2 = \frac{\cos{u}}{a(R+a\cos{u})}, \quad
H = \frac{1}{2}(\kappa_1 + \kappa_2) = -\frac{1}{2}\left(\frac{-\cos{u}}{R+a\cos{u}}-\frac{1}{a}\right).
\end{align*}
\newpage
If $v = \pm \frac{\pi}{2}$ then $K=0$ i.e., circles at the maximum and minimum height $a$ on the torus, say $S$, measured from the torus center perpendicular to R consist of parabolic points only. Points on $S$ whose distance from the center of $S$ to its height extrema along the $x_3$-axis in the figure below is greater than $\sqrt{R^2 + a^2}$ are elliptic while those whose distance from the origin/center of $S$ is smaller than $\sqrt{R^2 + a^2}$ are hyperbolic. The lines of curvature on a torus consist of meridians and parallels. The curves of constant Gaussian curvature are the meridians (i.e, the lines of curvature which as plane curves lie in planes parallel to the $x_1-x_2$ plane). Note that a second circle consisting of only parabolic points on the underside of the torus at its minimum $z$-axis height in the left figure below is not visible.
\begin{figure}[h!]
\centering
\begin{minipage}{\textwidth}
	\includegraphics[width=\linewidth]{problem6TorusPlots.pdf}
	\caption{Problem 6 Torus Curvature Plots. Left: 3D Torus plot with $R=6, \ a = 2$. Right: Torus Cross-Section.}
\end{minipage}
\end{figure}
\begin{figure}[h!]
\centering
\begin{minipage}{0.48\textwidth}
    \centering
    \includegraphics[scale=0.5]{Patrikalakis_torus.jpg}
    \caption*{\textbf{Torus curvature plot by Patrikalakis from the course homepage}\protect\footnotemark}
    \label{fig:PatrikalakisTorus}
\end{minipage}
\hfill
\begin{minipage}{0.48\textwidth}
    \centering
    \includegraphics[scale=2]{fig52Torus.png}
    \caption*{\textbf{Figure 52 from} \textit{Differential Geometry} \textbf{by Erwin Kreyszig} (\S 44. Torus)\protect\footnotemark}
    \label{fig:KreyszigTorus}
\end{minipage}

\caption{Reference figures for comparison}
\label{fig:combinedReferenceFigures}

\end{figure}

\footnotetext[2]{Image adapted from N.~M.~Patrikalakis, ``Computational Geometry,'' MIT OpenCourseWare, Course 2.158J, Spring 2003, \url{https://ocw.mit.edu/courses/2-158j-computational-geometry-spring-2003/}.}

\footnotetext[3]{Image reproduced from Figure 52 in E.~Kreyszig, \emph{Differential Geometry}, Dover Publications, Inc., Mineola, NY, 1991 (unabridged republication of the 1963 edition), p.~136, Section 44.}
%
% Problem 7
%
\item \textbf{Show that the surface area on a Monge patch $\mathbf{X}(u, v) = u\mathbf{e}_1 + v\mathbf{e}_2 + f(u, v)\mathbf{e}_3$ is given by the integral}
\begin{align}
A = \int\int_W{\sqrt{1+f_u^2 + f_v^2} \ \mathrm{d}u \ \mathrm{d}v},\label{9}
\end{align}
\textbf{where $W$ is the parameter domain, and $\mathbf{e}_1, \mathbf{e}_2$, and $\mathbf{e}_3$ are the unit coordinate vectors.}\\

A ``\textit{Monge Patch}" refers to the parametrization of a surface by its height over a flat reference plane, and is sometimes also called a portion of a surface. Denote the components of the first fundamental form as $g_{11}, \ g_{12} = g_{21}$, and $g_{22}$. Furthermore, denote the discriminant of the first fundamental form as $g = g_{11}g_{22} - (g_{12})^2$. Taking the definition of the area of a portion of a surface or Monge Patch,
\begin{align*}
A = \int\int_W{\sqrt{g} \ \mathrm{d}u \ \mathrm{d}v}
\end{align*}
compute the are directly from the first fundamental form for $\mathbf{X}$. Begin by computing the derivatives with respect to $u$ and $v$ of $\mathbf{X}$ and then compute the first fundamental form,
\begin{align*}
\frac{\mathrm{d}\mathbf{X}}{\mathrm{d}u} = \mathbf{e}_1 + f_u\mathbf{e}_3, \quad \frac{\mathrm{d}\mathbf{X}}{\mathrm{d}v} = \mathbf{e}_2 + f_u\mathbf{e}_3 \\
g_{11} = (\mathbf{e}_1 + f_u\mathbf{e}_3)\cdot(\mathbf{e}_1 + f_u\mathbf{e}_3) = 1 + f_u^2\\
g_{22} = (\mathbf{e}_2 + f_u\mathbf{e}_3)\cdot(\mathbf{e}_2 + f_u\mathbf{e}_3) = 1 + f_v^2\\
g_{12} = (\mathbf{e}_1 + f_u\mathbf{e}_3)\cdot(\mathbf{e}_2 + f_u\mathbf{e}_3) = f_u f_v\\
g = (1 + f_u^2)(1+f_v^2) - (f_u f_v)^2 = 1 + f_u^2 + f_v^2.
\end{align*}
Re-inserting the resulting discriminant of the first fundamental form into $A = \int\int_W{\sqrt{g} \ \mathrm{d}u \ \mathrm{d}v}$ recapitulates (\ref{9}).
\newpage
%
% Problem 8
%
\item \textbf{Show that the second fundamental form on a Monge patch $\mathbf{X}(u, v) = u\mathbf{e}_1 + v\mathbf{e}_2 + f(u, v)\mathbf{e}_3$ is}
\begin{align}
II = (f_u^2 + f_v^2 + 1)^{-\frac{1}{2}}\left[f_{uu}du^2 + 2f_{uv}\mathrm{d}u^2 + 2f_{uv}\mathrm{d}u\mathrm{d}v + f_{vv}\mathrm{d}v^2\right],\label{8}
\end{align}
\textbf{where $\mathbf{e}_1, \mathbf{e}_2$, and $\mathbf{e}_3$ are the unit coordinate vectors.}\\

For a given general vector parametrized curve $\mathbf{X}$, the second fundamental form is defined in Einstein notation (i.e., summation is implicit) by $b_{\alpha\beta}\mathrm{d}u^\alpha \ \mathrm{d}u^\beta$ where $\alpha, \beta = 1, 2$ respectively, namely
\begin{align*}
b_{\alpha\beta} &= \mathbf{x}_{\alpha\beta}\cdot\mathbf{n}\text{ where }\mathbf{n}\text{ denotes the normal vector to }\mathbf{X},\\
II &= b_{11}(\mathrm{d}u^1)^2 + 2b_{12}\mathrm{d}u^1\mathrm{d}u^2 + b_{22}(\mathrm{d}u^2)^2 = L(\mathrm{d}u^1)^2 + 2M\mathrm{d}u^1\mathrm{d}u^2 + N(\mathrm{d}u^2)^2.
\end{align*} 
Furthermore, the normal vector is generally defined by
\begin{align*}
\mathbf{n} = \frac{\mathbf{x}_1\times\mathbf{x}_2}{\vert \mathbf{x}_1\times\mathbf{x}_2\vert} = \frac{\mathbf{x}_1\times\mathbf{x}_2}{+\sqrt{g}}
\end{align*}
where $g$ denotes the discriminant of the first fundamental form. First compute the first and second derivatives of the given $\mathbf{X}$, then directly compute the second fundamental form. Derivatives:
\begin{align*}
\frac{\mathrm{d}\mathbf{X}}{\mathrm{d}u} = \mathbf{e}_1 + f_u\mathbf{e}_3, \quad \frac{\mathrm{d}\mathbf{X}}{\mathrm{d}v} = \mathbf{e}_2 + f_v\mathbf{e}_3, \\
\frac{\mathrm{d}^2\mathbf{X}}{\mathrm{d}u^2} = f_{uu}\mathbf{e}_3, \quad \frac{\mathrm{d}^2\mathbf{X}}{\mathrm{d}u\mathrm{d}v} = f_{uv}\mathbf{e}_3, \quad \frac{\mathrm{d}^2\mathbf{X}}{\mathrm{d}v^2} = f_{vv}\mathbf{e}_3.
\end{align*} 
Next, compute the cross product of the first derivatives of $\mathbf{X}$,
\begin{align*}
\mathbf{X}_u\times\mathbf{X}_v &= (\mathbf{e}_1 + f_u\mathbf{e}_3)\times( \mathbf{e}_2 + f_v\mathbf{e}_3)\\
&= \mathbf{e}_1\times\mathbf{e}_2 + \mathbf{e}_1\times f_v\mathbf{e}_3 + f_u\mathbf{e}_3\times\mathbf{e}_2 + f_u\mathbf{e}_3\times f_v\mathbf{e}_3\\
&= \mathbf{e}_3 - f_v\mathbf{e}_2 - f_u\mathbf{e}_1.
\end{align*}
Next, compute the dot products of first derivatives of $\mathbf{X}$,
\begin{align*}
\mathbf{X}^\prime_u\cdot\mathbf{X}^\prime_u &= (\mathbf{e}_1 + f_u\mathbf{e}_3)\cdot(\mathbf{e}_1 + f_u\mathbf{e}_3)\\
&= \mathbf{e}_1\cdot\mathbf{e}_1 + \mathbf{e}_1\cdot f_u\mathbf{e}_3 + f_u\mathbf{e}_3\cdot\mathbf{e}_1 + f_u^2\mathbf{e}_3\cdot\mathbf{e}_3 \\
&= 1 + f_u^2 \\
\mathbf{X}_u\cdot\mathbf{X}_v &= (\mathbf{e}_1 + f_u\mathbf{e}_3)\cdot(\mathbf{e}_2 + f_v\mathbf{e}_3)\\
&= \mathbf{e}_1\cdot\mathbf{e}_2 + \mathbf{e}_1\cdot f_v\mathbf{e}_3 + f_u\mathbf{e}_3\cdot\mathbf{e}_2 + f_u f_v\mathbf{e}_3\cdot\mathbf{e}_3\\
&= f_u f_v \\
\mathbf{X}_v\cdot\mathbf{X}_v &= (\mathbf{e}_2 + f_v\mathbf{e}_3)\cdot(\mathbf{e}_2 + f_v\mathbf{e}_3)\\
&= \mathbf{e}_2\cdot\mathbf{e}_2 + \mathbf{e}_2\cdot f_v\mathbf{e}_3 + f_v\mathbf{e}_3\cdot\mathbf{e}_2 + f_v^2\mathbf{e}_3\cdot\mathbf{e}_3\\
&= 1 + f_v^2.
\end{align*}
Next, compute the discriminant of the first fundamental form,
\begin{align*}
g &= g_{11}g_{22} - (g_{12})^2\\
g &= (1 + f_u^2)(1+f_v^2) - (f_u f_v)^2 \\
&= 1 + f_u^2 + f_v^2 + f_u^2f_v^2 - (f_u^2f_v^2) \\
&= f_u^2 + f_v^2 + 1 \\
\Rightarrow \sqrt{g} &= \sqrt{f_u^2 + f_v^2 + 1}.
\end{align*}
Next, compute the normal vector,
\begin{align*}
\mathbf{n} = \frac{\mathbf{e}_3 - f_v\mathbf{e}_2 - f_u\mathbf{e}_1}{\sqrt{f_u^2 + f_v^2 + 1}}
\end{align*}
Note that
\begin{align*}
\mathbf{e}_3\cdot(\mathbf{e}_3 - f_v\mathbf{e}_2-f_u\mathbf{e}_1) = 1\tag{$\star\star$}\label{2}
\end{align*}
Finally, compute the second fundamental form from the definition and the results above, and compare with (\ref{8})
\begin{align*}
f_{uu}\mathbf{e}_3\cdot\mathbf{n} &= \frac{f_{uu}}{\sqrt{f_u^2 + f_v^2+1}}\cdot(\star\star)\\
f_{uv}\mathbf{e}_3\cdot\mathbf{n} &= \frac{f_{uv}}{\sqrt{f_u^2 + f_v^2 + 1}}\cdot(\star\star)\\
f_{vv}\mathbf{e}_3\cdot\mathbf{n} &= \frac{f_{vv}}{\sqrt{f_u^2 + f_v^2 + 1}}\cdot(\star\star)\\
\Rightarrow II &= (f_u^2 + f_v^2 + 1)^{-\frac{1}{2}}\left[f_{uu}du^2 + 2f_{uv}\mathrm{d}u^2 + 2f_{uv}\mathrm{d}u\mathrm{d}v + f_{vv}\mathrm{d}v^2\right].
\end{align*}
\newpage
%
% Problem 9
%
\item \textbf{Show that the principal curvatures of the surface $f(x, y, z) = x\sin{(z)} -y\cos{(z)} = 0$ are $\pm(x^2 + y^2 + 1)^{-1}$.}

Introduce a change of variables to cylindrical coordinates given by $x = r\cos{\theta}$, $y = r\sin{\theta}$, $\theta=z-n\pi$,
\begin{align*}
f(x, y, z) \mapsto \textbf{r}(r, z) = (r\cos{(z - n\pi)}, r\sin{(z-n\pi)}, z).
\end{align*}
This parametrization satisfies the original $f(x, y, z)$. Without loss of generality, let $n=0$, thus the change of variables for the given surface parametrization may be written
\begin{align*}
\textbf{r}(r, z)=(r\cos{z}, r\sin{z}, z).
\end{align*}
The coefficients of the first fundamental form are thus
\begin{align*}
E=\textbf{r}_r\cdot \textbf{r}_r, \ F = \textbf{r}_r\cdot\textbf{r}_z, \ G = \textbf{r}_z\cdot\textbf{r}_z.
\end{align*}
Proceed to compute the first derivatives in the above expressions, the coefficients of the first fundamental form, and the subsequent first fundamental form:
\begin{align*}
\textbf{r}_r = (\cos{z}, \sin{z}, 0)&, \ \textbf{r}_z = (-r\sin{z}, r\cos{z}, 1)\\
E: \cos^2{z}&+\sin^2{z} + 0 = 1 \\
F: -r\cos{z}\sin{z} &+ r\sin{z}\cos{z} = 0 \\
G: (-r\sin{z})^2 &+ (r\cos{z})^2 + 1^2 = r^2 + 1 \\
\Rightarrow I = & \ \mathrm{d}s^2 = \mathrm{d}r^2 + (r^2 + 1)\mathrm{d}z^2 .
\end{align*}
Next, compute the normal vector:
\begin{align*}
\textbf{r}_r\times\textbf{r}_z &=
\begin{vmatrix}
i & j & k \\
\cos{z} & \sin{z} & 0 \\
-r\sin{z} & r\cos{z} & 1
\end{vmatrix}
= (\sin{z}, -\cos{z}, r) \\
\Rightarrow \vert \textbf{r}_r\times\textbf{r}_z \vert &= \sqrt{\sin^2{z} + \cos^2{z} + r^2} = \sqrt{1 + r^2}\\
\Rightarrow \textbf{n} &= \frac{(\sin{z}, -\cos{z}, r)}{\sqrt{1 + r^2}}.
\end{align*}
Next, compute the second derivatives of $\textbf{r}$, the coefficients of the second fundamental form, and the subsequent second fundamental form
\begin{align*}
\textbf{r}_{rr} &= (0, 0, 0) \\
\textbf{r}_{rz} &= (-\sin{z}, \cos{z}, 0) \\
\textbf{r}_{zz} &= (-r\cos{z}, -r\sin{z}, 0) \\
L: \textbf{n}\cdot \textbf{r}_{rr} &= \textbf{n}\cdot (0, 0, 0) = 0 \\
M: \textbf{n}\cdot \textbf{r}_{rz} &= \frac{(\sin{z})(-\sin{z}) + (-\cos{z})(\cos{z}) + (r)(0)}{\sqrt{1 + r^2}}\\
&= \frac{-\sin^2{z} - \cos^2{z}}{\sqrt{1 + r^2}} = \frac{-1}{\sqrt{1 + r^2}} \\
N: \textbf{n}\cdot \textbf{r}_{zz} &= \frac{(\sin{z})(-r\cos{z}) + (-\cos{z})(-r\sin{z}) + (r)(0)}{\sqrt{1 + r^2}}\\
&= \frac{-r\sin{z}\cos{z} + r\cos{z}\sin{z}}{\sqrt{1 + r^2}} = 0 \\
\Rightarrow II &= \frac{-1}{\sqrt{1 + r^2}}\mathrm{d}r \ \mathrm{d}z.
\end{align*}
\newpage
The principal curvatures can be calculated as the eigenvalues of the shape operator given by the Weingarten matrix $W = I^{-1}II$ where in this case $I$ and $II$ are the matrices of the $1^{\text{st}}$ and $2^{\text{nd}}$ fundamental forms, respectively.
\begin{align*}
I &= \begin{bmatrix}
E & F \\
F & G
\end{bmatrix} = \begin{bmatrix}
1 & 0 \\
0 & r^2 + 1
\end{bmatrix}; \ I^{-1} = {\Large 
\begin{bmatrix}
1 & 0 \\
0 & \frac{1}{r^2 + 1}
\end{bmatrix}}%
\\
II &= \begin{bmatrix}
L & M \\
M & N
\end{bmatrix} = {\Large
\begin{bmatrix}
0 & \frac{-1}{\sqrt{1 + r^2}}\\
\frac{-1}{\sqrt{1 + r^2}} & 0
\end{bmatrix}}%
\\
\Rightarrow W &= I^{-1}II = {\Large
\begin{bmatrix}
1 & 0 \\
0 & \frac{1}{r^2 + 1}
\end{bmatrix}}%
{\Large
\begin{bmatrix}
0 & \frac{-1}{\sqrt{1 + r^2}}\\
\frac{-1}{\sqrt{1 + r^2}} & 0
\end{bmatrix}}%
= {\Large
\begin{bmatrix}
0 & \frac{-1}{\sqrt{1 + r^2}}\\
\frac{-1}{(r^2 + 1)\sqrt{1 + r^2}} & 0
\end{bmatrix}}%
.
\end{align*}

Next, compute the eigenvalues of $W$,
\begin{align*}
\textrm{det}(W - \lambda I) &= \textrm{det}
{\Large
\begin{pmatrix}
-\lambda & \frac{-1}{\sqrt{1 + r^2}}\\
\frac{-1}{(1 + r^2)\sqrt{1 + r^2}} & -\lambda
\end{pmatrix}}%
= \lambda^2 - (\frac{-1}{\sqrt{1 + r^2}})(\frac{-1}{(1 + r^2)\sqrt{1 + r^2}}) = \lambda^2 - \frac{1}{(r^2 + 1)^2}\\
&\Rightarrow\lambda = \pm \frac{1}{r^2 + 1}.
\end{align*}
Finally, perform the inverse change of variables back to rectangular coordinates and confirm that the result recapitulates the given expression for the principal curvatures,
\begin{align*}
\text{Principal curvatures: }\lambda = \pm \frac{1}{r^2 + 1} = \pm \frac{1}{x^2 + y^2 +1}.
\end{align*}
\newpage
%
% Problem 10
%
\item \textbf{Consider the parametrized surface}
\begin{align}
\mathbf{r}(u, v) = \left(u - \frac{u^3}{3}+uv^2, v-\frac{v^3}{3}+vu^2, u^2 - v^2\right).
\end{align}
\textbf{Show that}
\begin{enumerate}
\item \textbf{The coefficients of the first fundamental form are}
\begin{align}
E = G = (1 + u^2 + v^2)^2, \quad F = 0.
\end{align}
\item \textbf{The coefficients of the second fundamental form are}
\begin{align}
L = 2, \ M = -2, \ N = 0.
\end{align}
\item \textbf{The principal curvatures are}
\begin{align}
\kappa_1 = \frac{2}{(1 + u^2 + v^2)^2}, \quad \kappa_2 = -\frac{2}{(1 + u^2 + v^2)^2}.
\end{align}
\end{enumerate}
\end{enumerate}
\begin{enumerate}[label=(\alph*)]
\item Compute the first derivatives of $\textbf{r}$ and their various dot products to subsequently compute the coefficients of the first fundamental form,
\begin{align*}
\textbf{r}_u &= (1 - u^2 + v^2, 2vu, 2u) \\
\textbf{r}_v &= (2vu, 1-v^2 + u^2, -2v) \\
\textbf{r}_u\cdot\textbf{r}_u &= (1-u^2 + v^2)^2 + (2vu)^2 + (2u)^2\\
&= 1 - u^2 + v^2 - u^2 + u^4 -u^2v^2+v^2-u^2v^2 + v^4 + 4u^2v^2 + 4u^2\\
&= 1 + 2u^2 + 2v^2 + 2u^2v^2 + u^4 + v^4\\
&= (1 + u^2+v^2)^2 = E \\
\textbf{r}_u\cdot\textbf{r}_v &= (2vu)^2 + (1 - v^2 + u^2)^2 + (-2v)^2 \\
&= (1 + u^2+v^2)^2 = G \\
\textbf{r}_u\cdot\textbf{r}_v &= (1 - u^2 + v^2)(2vu) + (1 - v^2 + u^2)(2vu) + (2u)(-2v)\\
&= 2vu(1 - u^2 + v^2 + 1 - v^2 + u^2)-4uv \\
&= 4vu - 4uv = 0 = F. 
\end{align*}
\item Compute the cross product $\textbf{r}_u\times\textbf{r}_v$,
\begin{align*}
\textbf{r}_u\times\textbf{r}_v &= \begin{vmatrix}
\textbf{i} & \textbf{j} & \textbf{k}\\
1 - u^2 + v^2 & 2vu & 2u\\
2vu & 1-v^2 + u^2 & -2v
\end{vmatrix}\\
= &\textbf{i}[(2vu)(-2v)-(2u)(1 - v^2 + u^2)]\\
&-\textbf{j}[(1 - u^2 + v^2)(-2v)-(2u)(2vu)]\\
&+\textbf{k}[(1-u^2+v^2)(1-v^2 + u^2)-(2vu)(2vu)]\\
= &\textbf{i}[-4v^2u - 2u + 2uv^2-2u^3]\\
&-\textbf{j}[-2v(1-u^2+v^2)-4u^2v]\\
&+\textbf{k}[1-(v^2 - u^2)^2 - 4u^2v^2]\\
= &\textbf{i}[-2u(1 + u^2+v^2)]\\
&+\textbf{j}[2v(1+u^2 + v^2)]\\
&+\textbf{k}[1 - u^4 - v^4 - 2u^2v^2].
\end{align*}
Let $w = u^2 + v^2 + 1$, thus 
\begin{align*}
\Rightarrow \textbf{r}_u \times \textbf{r}_v &= (-2uw, 2vw, 1- (u^2 + v^2)^2)\\
\Rightarrow \vert \textbf{r}_u \times \textbf{r}_v \vert^2 &= (-2uw)^2 + (2vw)^2 + (1-(u^2 + v^2))^2 = w^4\\
\Rightarrow \vert \textbf{r}_u \times \textbf{r}_v \vert &= \sqrt{(-2uw)^2 + (2vw)^2 + (1-(u^2 + v^2))^2} = w^2.
\end{align*}
Thus, the normal vector is
\begin{align*}
\textbf{n} = \frac{\textbf{r}_u\times\textbf{r}_v}{\vert \textbf{r}_u\times\textbf{r}_v\vert} = \frac{(-2uw, 2vw, 1- (u^2 + v^2)^2)}{w^2} = (\frac{-2u}{w}, \frac{2v}{w}, \frac{2}{w}-1).
\end{align*}
Next, compute the second derivatives of $\textbf{r}$,
\begin{align*}
\textbf{r}_{uu} &= (-2u, 2v, 2)\\
\textbf{r}_{vv} &= (2u, -2v, -2)\\
\textbf{r}_{uv} &= (2v, 2u, 0).
\end{align*}
Next, compute the coefficients of the second fundamental form $L= \textbf{r}_{uu}\cdot\textbf{n}$, $M = \textbf{r}_{uv}\cdot\textbf{n}$, $N = \textbf{r}_{vv}\cdot\textbf{n}$,
\begin{align*}
L: (2u, 2v, 2)\cdot\textbf{n} &= (-2u)(\frac{-2u}{w}) + (2v)(\frac{2v}{w}) +2(\frac{2}{w}-1) \\
&= \frac{4u^2}{w} + \frac{4v^2}{w} + \frac{4}{w} - 2 = \frac{4w}{w} - 2 = 2\\
M: (2v, 2u, 0)\cdot\textbf{n} &= (2v)(\frac{-2u}{w}) + (2u)(\frac{2v}{w}) + 0(\frac{2}{w}-1) \\
&= \frac{-4uv + 4uv}{w} = 0 \\
N: (2u, -2v, -2)\cdot\textbf{n} &= (2u)(\frac{-2u}{w}) + (-2v)(\frac{2v}{w}) + (-2)(\frac{2}{w}-1)\\
&= \frac{-4u^2}{w} - \frac{4v^2}{w} + 2(1 - \frac{2}{w}) = \frac{-4(w-1)-4}{w}+2 = \frac{-4w + 4 - 4}{w} + 2 = -4 + 2 = -2.
\end{align*}
\item Using the above results, compute the principal curvatures as the eigenvalues of the shape operator given by the Weingarten matrix $W = I^{-1}II$,
\begin{align*}
I &= \begin{bmatrix}
E & F \\
F & G
\end{bmatrix} = \begin{bmatrix}
(1 + u^2 + v^2)^2 & 0 \\
0 & (1 + u^2 + v^2)^2
\end{bmatrix}; \ I^{-1} = 
{\Large
\begin{bmatrix}
\frac{1}{(1 + u^2 + v^2)^2} & 0 \\
0 & \frac{1}{(1 + u^2 + v^2)^2}
\end{bmatrix}}%
\\
II &= \begin{bmatrix}
L & M \\
M & N
\end{bmatrix} = \begin{bmatrix}
2 & 0 \\
0 & -2
\end{bmatrix}\\
\Rightarrow W &= I^{-1}II =
{\Large
\begin{bmatrix}
\frac{1}{(1 + u^2 + v^2)^2}
& 0 \\
0 & \frac{1}{(1 + u^2 + v^2)^2}
\end{bmatrix}}%
\begin{bmatrix}
2 & 0 \\
0 & -2
\end{bmatrix} =
{\Large
\begin{bmatrix}
\frac{2}{(1 + u^2 + v^2)^2} & 0 \\
0 & \frac{-2}{(1 + u^2 + v^2)^2}
\end{bmatrix}}%
\end{align*}
Finally, compute the eigenvalues i.e., the principal curvatures, via the determinant of $W - \lambda I$,
\begin{align*}
\textrm{det}(W - \lambda I) &= \textrm{det}
{\Large
\begin{pmatrix}
\frac{2}{(1 + u^2 + v^2)^2} - \lambda & 0 \\
0 & \frac{-2}{(1 + u^2 + v^2)^2} - \lambda
\end{pmatrix}}%
\\
&\Rightarrow\lambda = \pm \frac{2}{(1 + u^2 + v^2)^2} \\
&\Rightarrow \kappa_1 = \frac{2}{(1 + u^2 + v^2)^2}, \quad \kappa_2 = -\frac{2}{(1 + u^2 + v^2)^2}.
\end{align*}
\end{enumerate}
\end{document}
