
\section{The Principal Normal and The Osculating Plane}

\begin{enumerate}[1.]
\item
{\it  Let $${\bf r}(t)=\langle\sqrt{2}t^{3/2},(t+2)^{3/2},(t-2)^{3/2}\rangle, \textrm{ for } t\in(2,\infty).$$
\noindent Find the unit tangent vector ${\bf T}(t)$, the principal normal vector ${\bf N}(t)$, and 
the curvature $\kappa(t)$. Find the limit of ${\bf N}(t)$ as $t\to 2^+$. (No need for an $\epsilon-\delta$ 
proof here, it suffices to use the general rules for limits.)}

\medskip

We calculate ${\bf T}$ and ${\bf N}$ as follows:
\begin{align*}
{\bf r}'(t)& =\left\langle \frac{3\sqrt{2}}{2}t^{1/2},
\frac{3}{2}(t+2)^{1/2},
\frac{3}{2}(t-2)^{1/2}\right\rangle,
\\
|{\bf r}'(t)|&=\frac{3}{2}\Big(2t+(t+2)+(t-2)\Big)^{1/2}
=\frac{3}{2}\sqrt{4t}=3\sqrt{t},
\\
{\bf T}(t)& =\frac{\bfr'(t)}{|\bfr'(t)|}
=\frac{3}{2\cdot 3\sqrt{t}}\left\langle \sqrt{2t},\sqrt{t+2},\sqrt{t-2}\right\rangle
=\frac{1}{2}\left\langle \sqrt{2},\sqrt{1+\frac{2}{t}},\sqrt{1-\frac{2}{t}}\right\rangle,
\\
{\bf T'}(t)
&=\frac{1}{2}\left\langle 0,\frac{1}{2\sqrt{1+\frac{2}{t}}}\cdot\frac{-2}{t^2}
,\frac{1}{2\sqrt{1-\frac{2}{t}}}\cdot\frac{2}{t^2}\right\rangle 
=\frac{1}{2t^2}\left\langle 0,-\sqrt{\frac{t}{t+2}}, \sqrt{\frac{t}{t-2}}\right\rangle,
\end{align*}
\begin{align*}
|{\bf T}'(t)|
&=\frac{1}{2t^2}\left( \frac{t}{t+2}+\frac{t}{t-2}\right) ^{1/2}
=\frac{1}{2t^2}\left( \frac{t^2-2t+t^2+2t}{t^2-4}\right)^{1/2}
\\
&=\frac{1}{2t^2}\left(\frac{2t^2}{t^2-4}\right)^{1/2}
=\frac{1}{t\sqrt{2(t^2-4)}},
\\
{\bf N}(t)&=\frac{{\bf T}'(t)}{|{\bf T}'(t)|}
=\frac{t\sqrt{2(t^2-4)}}{2t^2}\cdot
\left\langle 0,\sqrt{\frac{t}{t+2}}, -\sqrt{\frac{t}{t-2}}\right\rangle
\\
&=\sqrt{\frac{(t+2)(t-2)}{2t^2}}
\left\langle 0,-\sqrt{\frac{t}{t+2}}, \sqrt{\frac{t}{t-2}}\right\rangle 
=\left\langle 0,-\sqrt{\frac{t-2}{2t}}, \sqrt{\frac{t+2}{2t}}\right\rangle .
\end{align*}
From the last formula, we have
$$
\lim_{t\to 2^+}{\bf N}(t)
=(0,0, 1).$$
With the work done so far, the easiest way to compute the curvature
is
$$\kappa(t)=\frac{|{\bf T}'(t)|}{|\bfr'(t)|}
=\frac{1}{t\sqrt{2(t^2-4)}}\cdot\frac{1}{3\sqrt{t}}
=\frac{1}{3t^{3/2}\sqrt{2(t^2-4)}}.$$


\bigskip





\item {\it
Let $\bfr(t)=t\ii + t^2 \jj +t^3\kk$ for $t\in\rr$.


\medskip

(a) 
Find the equation of the osculating plane at the point $t=0$. }

\medskip




We have
$$
\bfr '=  \ii +2t  \jj +3t^2\kk,\ \ 
\bfr''= 2\jj+6t\kk.
$$
At $t=0$, we have $\bfr(0)={\bf 0}$, $\bfr'(0)=\ii$, $\bfr''(0)=2\jj$. The osculating plane is parallel to both $\bfr'(0)$ and $\bfr''(0)$, therefore to $\ii$ and $\jj$. It also passes through $(0,0,0)$. Therefore the osculating plane is the plane $z=0$.

\medskip


{\it (b) For $s,t\in\rr$ such that $s,t,0$ are all distinct, find the unit vector ${\bf n}(t,s)$ normal to the plane that passes through the points $\bfr(t)$, $\bfr(s)$, $\bfr(0)$. (There are two such vectors, pointing in opposite directions; choose the one that has a positive $\kk$-component.) Prove that 
$$
\lim_{s,t\to 0,\ s\neq t} {\bf n}(t,s) 
$$
is perpendicular to the osculating plane found in (a).}


\medskip
The vector ${\bf n}$ is perpendicular to $t\ii + t^2 \jj +t^3\kk$ and $s\ii + s^2 \jj +s^3\kk$, therefore to $\ii + t \jj +t^2\kk$ and $\ii + s \jj +s^2\kk$. It must therefore be parallel to their cross product: 
$$
\left|\begin{matrix}
\ii& \jj& \kk\\
            1 & t & t^2\\
            1& s & s^2\\
            \end{matrix}\right|
= (ts^2-st^2)\ii - (s^2-t^2)\jj + (s-t)\kk
$$
$$
= (s-t)\Big( st\ii -(s+t)\jj + k\Big),
$$
so that 
$$
{\bf n}(s,t)=\frac{1}{g(s,t)}\Big( st\ii -(s+t)\jj + k\Big).\hbox{ where } g(s,t)=\sqrt{(st)^2+(s+t)^2 +1}.
$$
As $(s,t)\to(0,0)$, we have $g(s,t)\to 1$, $st\to 0$, and $s+t\to 0$, so that ${\bf n}(s,t)\to \kk$. This is perpendicular to the osculating plane from (a) as claimed.





\bigskip





\item (12 marks)
{\it Let $\bfr(t)=t\ii + at^2 \jj $ for some $a>0$.

(a) 
Find the center $Q$ and radius $\rho$ of the osculating circle at the point $(0,0)$. }


We have
\begin{align*}
\bfr '&=  \ii +2at  \jj ,
\\
|{\bf r}'|&=\sqrt{1+4a^2t^2},
\\
\bfr''&= 2a\jj,
\end{align*}
so that
$$
\bfr' \times \bfr''= 
\left|\begin{matrix}
\ii& \jj& \kk\\
            1 & 2at & 0\\
            0& 2a  & 0\\
            \end{matrix}\right|
= 2a \kk,$$
$$
\kappa(t) = \frac{ | \bfr' \times \bfr''|}{|\bfr'|^3} = \frac{2a}{(1+4a^2t^2)^{3/2}}.
$$
At $t=0$, we have $\kappa=2a$, so that $\rho=(2a)^{-1}$. The center $Q$ of the osculating circle lies on the positive $y$ half-axis at distance $\rho$ from $(0,0)$, so that $Q=(0,(2a)^{-1})$.

\smallskip
(Alternatively, we could solve this using the calculations from Worked Problem 3. You should think about how to do this!)


\medskip



{\it (b) For $t>0$, find the center $Q(t)$ and radius $R(t)$ of the circle in the $xy$-plane that passes through the points $\bfr(t)$, $\bfr(-t)$, $\bfr(0)$. Prove that $Q(t)\to Q$ and $R(t)\to \rho$ as $t\to 0$, where $Q$ and $\rho$ are the answers in (a).)}

\medskip

By symmetry, $Q(t)$ lies on the positive $y$ half-axis, so that $Q(t)=(0,b)$ for some $b=b(t)$. The distance from $Q(t)$ to all of the points $\bfr(t)$, $\bfr(-t)$, $\bfr(0)$ has to be the same (and equal to $R(t)$). Let $O=(0,0)$, $P=(t,at^2)$, and $P'=(-t,at^2)$, then
\begin{align*}
|OQ(t)|^2&=b^2, 
\\
|PQ(t)|^2& =|P'Q(t)|^2= (0-t)^2+ (b-at^2)^2
\\
&= t^2+b^2 -2ab t^2 + a^2t^4.
\end{align*}
For these to be equal, we solve for $b$ as a function of $t$ with :
\begin{align*}
b^2&= t^2+b^2 -2ab t^2 + a^2t^4
\\
0&= t^2 -2ab t^2 + a^2t^4= t^2(1-2ab+a^2t^2).
\end{align*}
For $t\neq 0$, this implies that $1-2ab+a^2t^2=0$, so that $b=\frac{{a^2t^2+1}}{2a}$ and
$$
Q(t)=\left(0, \frac{{a^2t^2+1}}{2a}\right),\ \ R(t)=\frac{{a^2t^2+1}}{2a}.
$$
As $t\to 0$, we have $Q(t)\to (0,(2a)^{-1})=Q$ and $R(t)\to (2a)^{-1}=\rho$.













\bigskip






\item 
{\it Let $\bfr$ be a parametrized curve in $\rr^3$ such that $\bfr$ and its derivatives of all orders
are continuous, $\bfr'(t)\neq 0$ for $t$ in some interval $I$, and $|\bfr (t)|=r$ for
all $t\in I$ (i.e. the curve lies on the sphere $S$ of radius $r$ centered
at the origin $O$). Let $P=\bfr(t)$ be a point on
the curve.  
\medskip


(a) Prove that $\rho=r \cos\theta$, where $\theta$ is the angle between 
$\vec{PO}$ and ${\bf N}$. Use this to prove that $\kappa\geq 1/r$.

\smallskip

(Hint: Use arclength parametrization. We have $\bfr(s)\cdot \frac{d\bfr}{ds}=0$ (why?).
If we differentiate this in $s$ and use that $|\frac{d\bfr}{ds}|=1$, what does it say about 
the dot product $\vec{PO}\cdot {\bf N}$?)}


\medskip

Let $s$ be the arclength parameter. Since $|\bfr|=r$ is constant, we know from class that
$0=\bfr(s)\cdot \bfr'(s)$.
Differentiating this in $s$, and then using that
$|\bfr'(s)|=1$, we get
$$0=\bfr'(s)\cdot\bfr'(s)+\bfr(s)\cdot\bfr''(s)=1+\bfr(s)\cdot\bfr''(s). \eqno(1)$$
Hence 
$\bfr(s)\cdot\bfr''(s)=-1$.  But $\bfr'(s)={\bf T}$, so 
$\bfr''(s)=\frac{d{\bf T}}{ds}=\kappa {\bf N}$, therefore we get from (1) that
$$\bfr \cdot \kappa{\bf N}=-1.\eqno(2)$$
Therefore $\kappa \neq 0$. 
Let $\theta$ be the angle between $\vec{PO}=-\bfr$
and ${\bf N}$, then, on one hand, from (2) we have
$$\vec{PO}\cdot {\bf N} = -\bfr \cdot  {\bf N} =\frac{1}{\kappa}=\rho.$$
On the other hand, $\vec{PO}\cdot {\bf N} = |{PO}|\, |{\bf N}| \cos\theta
= r  \cos\theta$, so that $\rho=r\cos\theta$.
(This in particular implies that $\cos\theta>0$.) 
Since $0<\cos\theta\leq 1$, we get that
$$
\kappa = \frac{1}{r\cos\theta}\geq \frac{1}{r}.
$$

 
{\it (b)
Prove that for any point $P=\bfr(t)$ on the curve, the osculating circle for the curve at that point
lies on the sphere $S$. }

\medskip

The osculating circle at $P$ is 
the unique circle that
lies in the osculating plane at $P$, has radius $\rho=1/\kappa$,
and its center $Q$ satisfies $\vec{PQ}=\rho {\bf N}$.
If  $Q$ coincides with
the origin $O$, then the osculating circle is centered at $O$ and has radius $\rho=|OP|=r$, so it
lies on the sphere. Otherwise, we claim that the vector $\vec{OQ}$ is
perpendicular to both ${\bf N}$ and ${\bf T}$, therefore to the osculating plane.
To prove this:

\begin{itemize}
\item Consider the triangle $OPQ$. We have $|OP|=r$, $|PQ|=\rho$, and we proved in (a)
that $\rho=r\cos\theta$, where $\theta$ is the angle at $P$. Therefore $OPQ$ is a right triangle
with a right angle at $Q$, so that $\vec{OQ}\perp \vec{PQ}$. Since $\vec{PQ}=\rho {\bf N}$,
$\vec{OQ}$ is perpendicular to ${\bf N}$.

\item We also have $\vec{OQ}\cdot{\bf T}= (\vec{OP}+ \vec{PQ})\cdot{\bf T}
=(\bfr +\rho {\bf N})\cdot {\bf T} = \bfr \cdot {\bf T} +\rho {\bf N}\cdot {\bf T}
= 0+0=0.$ At the last step, we used that $\bfr\cdot\bfr'=0$, so that 
$\bfr\cdot {\bf T}=0$.
\end{itemize}

It follows that the osculating circle is centered at $Q$, lies in the plane perpendicular to
$\vec{OQ}$, and passes through $P$. Therefore the entire circle lies on the sphere.


\begin{figure}[htbp]
\centering\includegraphics[width=3in]{figures/sphere.pdf}%\caption{The 4-corner set, 2nd iteration.}
\end{figure}



\bigskip


\item {\it 
Let $\bfr$ be a closed, simple parametrized curve in $\rr^3$.
We use a ``wrap-around" parametrization: if ${\bf r}(t)$ for $t\in[0,a]$ parametrizes one full loop around $C$ returning to the starting point, we extend ${\bf r}$ to a periodic function $\bfr:\rr\to\rr^3$ so that $\bfr(t+a)=\bfr(t)$ for all $t$. (This is similar to parametrizing the unit circle by $\bfr(t)=\cos(t)\ii + \sin(t)\jj$, with period $2\pi$.)

Assume that $\bfr$ and its derivatives of all orders
are continuous, $\bfr'(t)\neq 0$, and $|\bfr (t)|\leq 1$ for
all $t$ (i.e. the curve lies in the closed ball of radius $1$ centered
at the origin $O$). 
Prove that there is a point $\bfr(t_0)$ on the curve where $\kappa(t_0)\geq 1$.  

(Hint: Let $f(t)=|\bfr(t)|^2$, then $f(t)$
attains its maximum at some point $t=t_0$, What does this say about $f'(t_0)$ and $f''(t_0)$?
It may be helpful to use arclength parametrization, starting from $t=t_0$ so that $s(t_0)=0$.)}





\medskip

Let $s$ be the arclength parameter, measured from $\bfr(t_0)$ so that $s(t_0)=0$. Let $f(s)=|\bfr(s)|^2$, then
\begin{align*}
f'(s)&=2\bfr(s)\cdot \bfr'(s)=2\bfr(s)\cdot {\bf T}(s),
\\
f''(s)&=2\bfr'(s)\cdot {\bf T}(s)+2\bfr(s)\cdot {\bf T}'(s)= 2{\bf T}(s)\cdot {\bf T}(s)+2\bfr(s)\cdot \kappa{\bf N}(s)\\
&=2+ 2\kappa \bfr(s)\cdot {\bf N}(s)
\end{align*}
Since $f$ has a local maximum at $s=0$, we have $f'(0)=0$ and $f''(0)\leq 0$. This means that
\begin{align*}
&\bfr(0)\cdot {\bf T}(0)=0,
\\
&2+ 2\kappa \bfr(0)\cdot {\bf N}(0)\leq 0.
\end{align*}
Since $|{\bf N}(0)|=1$, we have $\bfr(0)\cdot {\bf N}(0)=|{\bf r}(0)|\cos\theta$, where $\theta$ is the angle between $\bfr(0)$ and ${\bf N}(0)$. Therefore
$$
2+ 2\kappa |{\bf r}(0)|\cos\theta \leq 0,
$$
so that $\kappa|{\bf r}(0)|\cos\theta \leq -1$. Since $|{\bf r}(0)|\leq 1$ and $|\cos\theta|\leq 1$, it follows that $\kappa(0)\geq 1$.






\end{enumerate}






