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\title{Math 147A Sample Midterm}
\date{}


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\begin{document}
\maketitle
\textbf{April 28, 2016}
The midterm will be 5 problems, some computational, some conceptual.  Feel free to bring a reasonable sized paper for notes and formulas. 

\begin{enumerate}
\item Compute the curvature of 
\begin{equation*}
\gamma(t) = (\cos^3(t),\sin^3(t)).
\end{equation*}
\item Compute the curvature and torsion (if it exists) of
\begin{equation*}
\gamma(t) = \left(\frac{4}{5}\cos(t),1-\sin(t),-\frac{3}{5}\cos(t)\right)
\end{equation*}
Show that $\gamma$ parametrizes a circle, find its center, radius and the plane in which it lies.
\item Show that a unit speed curve with $\kappa(s) >0$ for each $s\in [a,b]$ is a plane curve if and only if the torsion vanishes everywhere. 
\item State and prove the isoperimetric inequality, assuming Wirtinger's inequality and Green's theorem for area.
\item Show that a reparametrization by arc-length gives a unit-speed curve.
\item Derive the Frenet-Serret equation.
\item Let $\gamma:(a,b) \to \mathbb{R}^2$ be a regular plane curve and let $a \in \mathbb{R}^2$ such that $\gamma(t) \neq a$ for all $t$.  If there exists a $t_0 \in (a,b)$ such that 
\begin{equation*}
\|\gamma(t) - a\| \geq \|\gamma(t_0) -a\|
\end{equation*}
for all $t \in (a,b)$, show that the straight line joining the point $a$ with $\gamma(t_0)$ is the normal line of $\gamma$ at $t_0$.  The same is true if we reverse the inequality.  Draw a situation that illustrates both cases.
\item Let $\gamma:(a,b) \to \mathbb{R}^2$ be a regular plane curve and let $[\alpha,\beta] \subset (a,b)$ be such that $\gamma(\alpha) \neq \gamma(\beta)$.  Prove that there exists some $t_0 \in (\alpha,\beta)$ such that the tangent line of $\gamma$ at $t_0$ is parallel to the segment of the straight line joining $\gamma(\alpha)$ with $\gamma(\beta)$.  This is a curve version of mean value theorem.  Hint: Consider $f(t) = \det(\gamma(t), \gamma(\beta)-\gamma(\alpha))$.
\item Prove that a unit speed curve $\gamma:(a,b) \to \mathbb{R}^2$  is an arc of a circle if and only if all its normal lines pass through a given point.
\item Let $\gamma:(a,b) \to \mathbb{R}^3$ be a unit speed curve with positive curvature.  If $\|\gamma(s)\| =1$ for all $s$, i.e. $\gamma$ is a curve on a sphere, and it has constant torsion $\tau$, prove that there exists $b,c \in \mathbb{R}$ such that
\begin{equation*}
\kappa(s) = \frac{1}{b\cos(\tau s)+c\sin(\tau s)}.
\end{equation*}
\item Let $\gamma$ be a unit speed curve in $\mathbb{R}^3$ with constant curvature and zero torsion.  Show that $\gamma$ is a parametrization of a circle.
\end{enumerate}

\textbf{Solutions}

1. Book problem 2.1.1

2. Book problem 2.3.1

3. Book Prop 2.3.3

4. Book Theorem 3.2.2

5. Let $\gamma(t)$ be a curve and let $s(t) = \int_0^t \|\dot{\gamma}(u)\|du$ be the arc length.  Then 
\begin{align*}
\frac{d}{ds}\gamma(t) = \dot{\gamma}\frac{dt}{ds}= \frac{\dot{\gamma}}{\frac{ds}{dt}} = \frac{\dot{\gamma}}{\|\dot{\gamma}\|}.
\end{align*}
Hence $\|\frac{d\gamma}{ds}\| =1$ (as long as $\gamma$ is regular).

6.  See paragraph before Theorem 2.3.4

7. Let $f(t) = \|\gamma(t) -a\|^2$. Then $t=t_0$ is a minimum therefore
\begin{align*}
0 = f'(t_0) = 2\langle \gamma'(t_0), \gamma(t_0) -a \rangle. 
\end{align*}
The same proof will work for a maximum. 

8. We have $f(a) = f(b) = 0$.  Therefore, by mean value theorem, there is a point such that 
\begin{align*}
0 = f'(t_0) = \det(\gamma'(t_0),\gamma(\beta)-\gamma(\alpha))
\end{align*}
This shows that $\gamma'(t_0)$ and $\gamma(\beta)-\gamma(\alpha)$ are linearly dependent and for $\mathbb{R}^2$, they are parallel.

9. Suppose $\gamma$ is an arc of a circle.  If we center at the origin, then the position vector is the normal vector, hence passes through the center.  Conversely, Suppose $\gamma$ is a unit speed curve such that all its normal lines pass through a given point, say $a$.  Consider $f(t) = \|\gamma(t) -a\|^2$.  Then
\begin{align*}
f'(t) = 2\langle \gamma'(t),\gamma(t)-a\rangle.
\end{align*}
The vector $\gamma(t)-a$ is a normal line by assumption and $\gamma'(t)$ is a tangent vector, hence they are perpendicular so $f'(t) = 0$ for all $t$ where the assumption holds.  Hence $f(t)$ is a constant, i.e. $\|\gamma(t) - a\| = R$ for some $R$.

10. Differentiating $\|\gamma(s)\|^2 =1$ twice, we get
\begin{align*}
&\langle \gamma'(s),\gamma(s)\rangle = 0\\
&\langle \gamma''(s),\gamma(s)\rangle + \langle \gamma'(s),\gamma'(s)\rangle = 0.
\end{align*}
Using the fact that $\gamma' = T$, $\|\gamma'(s)\| =1$ and $\gamma''(s) = \dot{T} = \kappa N$,
\begin{align*}
\langle N(s), \gamma(s)\rangle = -\frac{1}{\kappa}.
\end{align*}
Taking the derivative once more, we have
\begin{equation*}
-\left(\frac{1}{\kappa}\right)' = \langle \dot{N},\gamma\rangle + \langle N,\dot{\gamma}\rangle = \langle \dot{N},\gamma\rangle
\end{equation*}
since $\dot{\gamma}$ and $N$ are perpendicular.  From the Frenet-Serret equations,
\begin{equation*}
\langle \dot{N},\gamma\rangle = -K\langle T,\gamma\rangle + \tau \langle B,\gamma\rangle = -K\langle \dot{\gamma},\gamma\rangle +  \tau \langle B,\gamma\rangle =  \tau \langle B,\gamma\rangle
\end{equation*}
where we used the fact that $\gamma$ lies on a sphere so $\langle \dot{\gamma},\gamma\rangle = 0$. Finally taking the derivative once more
\begin{align*}
-\left(\frac{1}{\kappa}\right)'' = \tau\langle \dot{B},\gamma\rangle + \tau \langle B,\dot{\gamma}\rangle = \tau\langle \dot{B},\gamma\rangle
\end{align*}
and use the Frenet-Serret equations once more to get
\begin{equation*}
-\left(\frac{1}{\kappa}\right)'' = -\tau^2 \langle N,\gamma\rangle = -\tau^2\left(\frac{1}{\kappa}\right).
\end{equation*}
Hence we want to solve an ODE of the form $y''+\tau^2 y =0$.  It has a solution
\begin{equation*}
y(s) = b\cos(\tau s) + c\sin(\tau s) 
\end{equation*}
for some constants $b$ and $c$.  Since $y=\frac{1}{\kappa}$, we obtain the result.

11. Book Prop 2.3.5

\end{document}