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\title{Math 147A Midterm}
%\date{April 28, 2016}


\theoremstyle{definition} 
\newtheorem{definition}{Definition}[section]
\newtheorem{conjecture}{Conjecture}[section]
\newtheorem{example}{Example}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{remark}{Remark}[section]
\newtheorem{question}{Question}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem*{notation}{Notation}
\newtheorem{problem}{Problem}[section]

\begin{document}
\maketitle
Choose 5 of the 6.  Each problem is worth 10 points for a total of 50 points.  No extra credit for doing more problems.  Please circle your choice of 5.

\begin{enumerate}
\item Compute the curvature and torsion of
\begin{equation*}
\gamma(t) = \left(2\cos\left(\frac{t}{\sqrt{5}}\right), 2\sin\left(\frac{t}{\sqrt{5}}\right),\frac{t}{\sqrt{5}}\right).
\end{equation*}
What curve is this.
\item Compute the curvature and torsion (if it exists) of
\begin{equation*}
\gamma(t) = \left(\alpha\cos(t),1-\sin(t),\beta\cos(t)\right),
\end{equation*}
where $\alpha^2 + \beta^2 = 1$.  Show that $\gamma$ parametrizes a circle, find its center, radius and the plane in which it lies.
\item Show that a reparametrization by arc-length of a regular curve gives a unit-speed curve.  Show this for the curve $\gamma(t) = (\cos(3t),\sin(3t))$.
\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\| \leq \|\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 $t_0$.  
\item Let $\gamma:(a,b) \to \mathbb{R}^3$ be a unit speed-curve such that $\kappa > 0$ and $0\in (a,b)$. Consider the Frenet-Serret frame $\{T,N,B\}$ centered at $\gamma(0)$ and use coordinates $\gamma(t) = (x(t),y(t),z(t))$ so that $T = (1,0,0)$, $N=(0,1,0)$ and $B= (0,0,1)$, and $\gamma(0) = (0,0,0)$.  Prove that
\begin{equation*}
\tau(t_0) = \lim_{s\to 0}\frac{6z(s)}{\kappa(t_0)s^3}.
\end{equation*}
\item Let $\gamma: (a,b) \to \mathbb{R}^3$ be a unit-speed curve whose curvature $\kappa$ and torsion $\tau$ do not vanish.  Prove that if $\gamma$ lies on a unit-sphere, then
\begin{equation*}
\frac{1}{\kappa^2} + \left(\frac{\kappa'}{\kappa^2\tau}\right)^2 = 1
\end{equation*}
Hint: For an orthonormal frame $\{T,N,B\}$, $\gamma = (\gamma\cdot T)T + (\gamma\cdot N)N + (\gamma\cdot B)B$.

\noindent The converse is true if we assume that $\kappa' \neq 0$.  Try to prove this for fun.
\end{enumerate}

\section*{Solution}

1. The tangent vector is
\begin{equation*}
\dot{\gamma}(t) = \left(-\frac{2}{\sqrt{5}}\sin\left(\frac{t}{\sqrt{5}}\right), \frac{2}{\sqrt{5}}\cos\left(\frac{t}{\sqrt{5}}\right),\frac{1}{\sqrt{5}}\right).
\end{equation*}
Since $\|\dot{\gamma}\| = 1$, $\gamma(t)$ is unit speed and $\dot{\gamma}(t) = T$, the unit tangent vector.  Hence the curvature can be computed by $\|\ddot{\gamma}\|$, which is
\begin{equation*}
\ddot{\gamma}(t) = \left(-\frac{2}{5}\cos\left(\frac{t}{\sqrt{5}}\right), -\frac{2}{5}\sin\left(\frac{t}{\sqrt{5}}\right),0\right)
\end{equation*}
and so $\kappa = \|\ddot{\gamma}\| = \frac{2}{5}$.  The principal normal is given by $N = \left(-\cos\left(\frac{t}{\sqrt{5}}\right),-\sin\left(\frac{t}{\sqrt{5}}\right),0\right)$.  Computing the binormal, we have
\begin{align*}
B = T\times N = \left(\frac{1}{\sqrt{5}}\sin\left(\frac{t}{\sqrt{5}}\right), -\frac{1}{\sqrt{5}}\cos\left(\frac{t}{\sqrt{5}}\right),\frac{2}{\sqrt{5}}\right)
\end{align*}
and
\begin{align*}
\dot{B} = \left(\frac{1}{5}\cos\left(\frac{t}{\sqrt{5}}\right), \frac{1}{5}\sin\left(\frac{t}{\sqrt{5}}\right),0\right) = -\frac{1}{5}\left(-\cos\left(\frac{t}{\sqrt{5}}\right),-\sin\left(\frac{t}{\sqrt{5}}\right),0\right) = -\tau N.
\end{align*}
so that $\tau = \frac{1}{5}$.   The curve is a helix and hence the curvature and torsion can be computed using the formula provided in the book.

\noindent 2. First we compute the tangent vector
\begin{equation*}
\dot{\gamma}(t) = (-\alpha\sin(t),-\cos(t),-\beta\sin(t)).
\end{equation*}
Then $\|\dot{\gamma}\|^2 = 1$, hence is unit speed so $\dot{\gamma} = T$.  The curvature is
\begin{align*}
\kappa = \|\ddot{\gamma}\| = \|(-\alpha\cos(t),\sin(t),-\beta\cos(t))\| = 1.
\end{align*}
So we see that the principal normal is $\dot{T} = \kappa N = N$.  The binormal is given by
\begin{equation*}
B = T\times N = (\beta,0,-\alpha).
\end{equation*}
So $\dot{B} = 0$, hence $\tau = 0$. Since the torsion vanishes and is constant curvature, $\gamma$ parametrizes a circle, with radius $R = \frac{1}{\kappa} = 1$.  The plane that it lies on is given by the binormal vector so $\beta x - \alpha z = 0$ and its center is given by 
\begin{equation*}
a = \gamma - \frac{1}{k}N = (0,1,0).
\end{equation*}

\noindent 3. See solution to sample midterm for the first part.  The arc-length formula is given by
\begin{align*}
s(t) = \int_0^t \|\dot{\gamma}\|du = \int_0^t3 du = 3t
\end{align*}
and so its inverse is given by $t(s) = \frac{s}{3}$.  Plugging this into the original curve, we get $\tilde{\gamma}(s) = \gamma(t(s)) = (\cos(s),\sin(s))$.  Then $\frac{d}{ds}\tilde{\gamma} = (-\sin(s),\cos(s))$ hence is a unit speed curve.

\noindent 4. See solution to sample midterm.

\noindent 5. Consider the third order Taylor expansion
\begin{equation*}
\gamma(s) = \gamma(0) + \gamma'(0)s + \gamma''(0)\frac{s^2}{2} + \gamma'''(0)\frac{s^3}{6} + O(s^4).
\end{equation*}
Now since $\gamma$ is unit-speed, $\gamma'(0) = T(s)$, $\gamma''(s) = \kappa(s) N(s)$ and using the Frenet-Serret equation,
\begin{align*}
\gamma'''(s) &= \kappa'(s)N(s) + \kappa(s) \dot{N}(s) \\
&=\kappa'(s)N(s) + \kappa(s)(-\kappa(s)T(s) + \tau B(s)).
\end{align*}
Hence
\begin{align*}
\gamma(s) &= (x(s),y(s),z(s)) \\
&=Ts + \kappa N\frac{s^2}{2} + \kappa'N\frac{s^3}{6} -\kappa^2T\frac{s^3}{6} + \tau\kappa B\frac{s^3}{6} + O(s^4) \\
&= \left(s -\kappa^2\frac{s^3}{6} + O(s^4),\ \kappa \frac{s^2}{2}+ \kappa' \frac{s^3}{6} + O(s^4),\ \tau \kappa \frac{s^3}{6} +O(s^4)\right)
\end{align*}
so
\begin{equation*}
\tau = \lim\limits_{s\to 0} \frac{6z(s)}{\kappa s^3}.
\end{equation*}

\noindent 6.  Since $\gamma \in S^2$, we have (assuming $S^2$ is centered at $(0,0,0)$) $\|\gamma\| =1$ therefore, taking the derivative we have
\begin{equation*}
\gamma' \cdot \gamma = T \cdot \gamma = 0
\end{equation*}
Taking the derivative once more, we have
\begin{equation*}
\gamma''\cdot \gamma + \gamma'\cdot \gamma' = 0
\end{equation*}
and since $\gamma$ is unit speed, we have
\begin{equation*}
N\cdot\gamma = -\frac{1}{\kappa}.
\end{equation*}
Taking the derivative once more,
\begin{align*}
\dot{N}\cdot\gamma + N\cdot \gamma' = \frac{\kappa'}{\kappa^2}
\end{align*}
and using the Frenet-Serret equation and the fact that $\{T,N,B\}$ form an orthonormal basis,
\begin{equation*}
\frac{\kappa'}{\kappa^2} = -\kappa T\cdot\gamma + \tau B\cdot \gamma = \tau B\cdot \gamma.
\end{equation*}
Now, writing $\gamma$ as a linear combination of $\{T,N,B\}$, we have
\begin{align*}
\gamma &= (\gamma\cdot T)T + (\gamma\cdot N)N + (\gamma\cdot B)B \\
&= -\frac{1}{\kappa}N + \frac{\kappa'}{\kappa^2\tau}B.
\end{align*}
Taking the dot product with $\gamma$, we have
\begin{align*}
1 &= \gamma \cdot \gamma \\
&=  -\frac{1}{\kappa}N\cdot \gamma + \frac{\kappa'}{\kappa^2\tau}B\cdot \gamma\\
&= \frac{1}{\kappa^2} + \left(\frac{\kappa'}{\kappa^2\tau}\right)^2.
\end{align*}
\end{document}