\section{Unit Tangent and Curvature}

\begin{enumerate}[1.]
\item
{\it Let $\mathcal{C}$ be the curve $x=\frac{t^3}{3}-2t$, $y=\frac{t^3}{3}+2t  $, $z=\sqrt{2}t^2$ for $t \in\mathbb{R}$. 

\medskip\noindent
(a)  Find the arclength of the curve from $t=0$ to $t=1$.}

\medskip

We have $\bfr'(t)=(t^2-2)\ii + (t^2+2) \jj + 2\sqrt{2}t \kk$, so that
\begin{align*}
|\bfr'(t)|^2& = (t^2-2)^2+(t^2+2)^2+ 8t^2 = t^4-4t^2+4+t^4+4t^2+4+8t^2 
\\
&= 2t^4+8t^2+8 = 2(t^2+2)^2,
\\
|\bfr'(t)|&=\sqrt{2}(t^2+2).
\end{align*}
The arclength is 
$$\int_0^1 \sqrt{2}(t^2+2)dt = \sqrt{2}\Big( \frac{t^3}{3}+2t\Big) \Big|_0^1 =\frac{7 \sqrt{2}}{3}.
$$


\medskip\noindent
{\it (b) Find the curvature at the point where $t=0$.}
\medskip



We have $\bfr'(t)= (t^2-2)\ii + (t^2+2) \jj + 2\sqrt{2}t \kk$
as above, so that $\bfr''(t)=2t \ii + 2t \jj +2\sqrt{2} \kk$.
At $t=0$, we have 
$|\bfr'|=2\sqrt{2}$ (from part (a)), and
$$\bfr' \times \bfr''=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
-2 & 2 & 0\\
0 & 0 & 2\sqrt{2} \\
            \end{matrix}\right|
 = 4\sqrt{2}\, {\bf i} + 4\sqrt{2}\, \jj ,$$
 $$|\bfr' \times \bfr''|= \sqrt{ (4\sqrt{2})^2 + (4\sqrt{2})^2 } =\sqrt{ 32 + 32} =\sqrt{64}=8.$$ 
 For the curvature, we have
$$
\kappa = \frac{|\bfr' \times \bfr''|}{|\bfr'|^3} = \frac{8}{(2\sqrt{2})^3}=\frac{1}{\sqrt{8}}.
$$

\bigskip

\item
{\it 
Let $\bfr(t)=t\ii+\frac{t^2}{2}\jj $, $t\in\rr$ (a parabola in $\rr^2$), and let ${\bf T}(t)$ be the unit tangent vector to
$\bfr(t)$. Suppose that $\bfr(u)$, $u\in I$ (for some interval $I$), is a smooth reparametrization of the parabola such that
$$
\left| \frac{d{\bf T}}{du} \right| =1.
$$
Find $\displaystyle \frac{du}{dt}$ as a function of $t$. Prove that the interval $I$ is finite, and find the length of it.}

\medskip

We have  $r(t) = (t, t^{2}/2)$, and $r'(t) = (1, t)$. Therefore,
$$
{\bf T}(t) = \left\langle \frac{1}{\sqrt{1+t^{2}}}, \frac{t}{\sqrt{1+t^{2}}}\right\rangle,\ \ 
d{\bf T}/dt = \left\langle \frac{-t}{(1+t^{2})^{3/2}}, \frac{1}{(1+t^{2})^{3/2}}\right\rangle.
$$
By a short calculation (you should do it!), we obtain $| d{\bf T}/dt | = \frac{1}{1+t^{2}}$. Applying the Chain Rule, we get
$$
\left| \frac{d{\bf T}}{du} \right| = \left| \frac{d{\bf T}}{dt} \frac{dt}{du} \right| = 1 
$$

Therefore 
$$\Big| \frac{dt}{du} \Big| = \frac{1}{| \frac{d{\bf T}}{dt} |}
= \frac{1}{1/(1+t^{2})},
$$
so that $| \frac{du}{dt} | = | \frac{d{\bf T}}{dt} | = \frac{1}{1+t^{2}}$. This is always positive, so $\frac{du}{dt} = \frac{1}{1+t^{2}}$.

To find the length of the interval $I$, we note that $I$ is the image of $t \mapsto u(t)$ for $t \in \rr$. We have that $\frac{du}{dt} = \frac{1}{1+t^{2}}$, and so we can solve for $u = \arctan(t) + c$ for some constant $c$. The range of $\arctan(t)$ for $t \in \rr$ is the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, so that $u$ has range $ I = (-\frac{\pi}{2} + c , \frac{\pi}{2} + c )$, a finite interval of length $\pi$.

\bigskip







\item {\it 
Let $\bfr:\rr\to \rr^3$ be given by 
$$
\bfr(t)=\cos(t)\ii + \sin(t)\jj + f(t)\kk,
$$
where $f:\rr\to\rr$ satisfies $f(t)=f(t+2\pi)$. Assume that $f$ and all its derivatives are continuous. 
Thus $\bfr(t)$ parametrizes a closed smooth curve that lies on the cylinder $x^2+y^2=1$, with each $t$-interval of length $2\pi$ covering the entire curve. Prove that there is a point $\bfr(t_0)$ on the curve where $\kappa\leq 1$.  }


\medskip

With $\bfr(t)$ as above, we have
$$
{\bf r}'(t)=-\sin(t)\ii + \cos(t)\jj+f'(t)\kk,\
 \ {\bf r}''(t)=-\cos (t)\ii - \sin (t)\jj+f''(t)\kk,
$$
so that 
$$|\bfr'|^2={\sin^2t+\cos^2t+(f'(t))^2}={1+(f'(t))^2},
$$
\begin{align*}
\bfr'\times \bfr''&=\left|\begin{matrix}
\ii& \jj& \kk\\
            -\sin t  & \cos t & f'(t)\\
            -\cos t & -\sin t  & f''(t)\\
            \end{matrix}\right|
            \\
&= (\cos(t)f''+\sin(t)f')\ii+(\sin(t)f''-\cos(t)f')\jj +\kk,
\end{align*}
\begin{align*}
|\bfr'\times \bfr''|^2&= (\cos(t)f''+\sin(t)f')^2+(\sin(t)f''-\cos(t)f')^2+1\\
&= (f'')^2+ (f')^2+1.
\end{align*}
This gives us a formula for curvature,
$$
\kappa=\frac{((f'')^2+ (f')^2+1)^{1/2} }{(1+(f')^2)^{3/2}}.
$$
We need to prove that there is a point on the curve where $\kappa\leq 1$. This is equivalent to $\kappa^2\leq 1$, or
$$
(f'')^2+ (f')^2+1\leq (1+(f')^2)^3 = 1+3(f')^2 + 3 (f')^4 + (f')^6.
$$
If we can find a point $t_0$ where $f''(t_0)=0$, then the inequality holds at that point. 
To find such $t_0$, we observe that since $f$ is periodic with $f(t+2\pi)=f(t)$, we also have that $f'$ is periodic with $f'(t)=f'(t+2\pi)$. In particular, $f'(0)=f'(2\pi)$. By the Mean Value Theorem, there is a $t_0\in(0,2\pi)$ such that $f''(t_0)=0$. That gives us the point we want.





\bigskip





\item 
{\it Let ${\bf r}(t)$ be a smooth vector-valued function,
and suppose that there is a point $P$ that lies on every normal plane to ${\bf r}(t)$. Prove
that the curve lies on the surface of a sphere.}

\medskip

We may assume that $P=(0,0,0)$. The equation of the normal plane to the curve at
${\bf r}(t_0)=(x_0,y_0,z_0)$ is
$${\bf r}'(t_0)\cdot (x-x_0,y-y_0,z-z_0)=0.
$$
If $P$ lies on every such plane, it means that for all $t_0$,
$${\bf r}'(t_0)\cdot {\bf r}(t_0)=0.$$
But this characterizes curves that lie on the surface of a sphere $|{\bf r}(t)|=c$, as proved in Worked Problems, Section 1.1.

\bigskip


\end{enumerate}


