
\section{Torsion}\label{sec:torsion}

\begin{enumerate}[1.]


\item
{\it 
Let $C$ be the curve $x=e^{2t}$, $y=2e^t $, $z=t$ for $t \in\mathbb{R}$. 
Find all points 
on the curve where the 
osculating plane is parallel to the plane $x-6y+18z=0$. (If there are no such points, prove it.)}

\medskip


We have $\bfr'(t)=2e^{2t}\ii + 2e^t \jj + \kk$, so that $\bfr''(t)=4e^{2t}\ii + 2e^t \jj $.
The vector normal to the osculating plane is ${\bf B}$, which is parallel to
$$\bfr' \times \bfr''=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
2e^{2t} & 2e^t & 1\\
4e^{2t} & 2e^t & 0\\
            \end{matrix}\right|
 =-2e^t {\bf i} + 4e^{2t}\jj -4e^{3t}\kk= -2e^t(\ii -2e^t \jj + 2e^{2t} \kk).$$
For the osculating plane to be parallel to $x-6y+18z=0$, this vector needs to be parallel to
$\ii-6\jj + 18\kk$, so that we need to have
$$ 2e^t=6,\ \ 2e^{2t}= 18.$$
From the first equation we get $e^t=3,\ t=\ln 3$. Then $2e^{2t}=2\cdot 9 = 18$, so that the second equation is also satisfied. The point where $t=\ln 3$ has coordinates $(x,y,z) = (9,6,\ln 3)$.

\bigskip

\item
{\it 
Prove that if ${\bf r}(t)$ is a smooth curve such that $\kappa\neq 0$ and ${\bf r}+\kappa^{-1}{\bf N}$ (the center of the osculating circle) is constant, then
${\bf r}(t)$ is a circle. 
(Hint: By the uniqueness theorem for curves, it suffices to prove that $\kappa$ is constant and the torsion $\tau$ is 0.) }


\medskip

We use the arclength parametrization.
By the Frenet-Serret formulas, we have 
\begin{align*}
\frac{d{\bf N}}{ds}&= \tau {\bf B} - \kappa {\bf T}.
\end{align*}
Therefore
\begin{align*}
\frac{d}{ds}({\bf r}+\kappa^{-1}{\bf N}) &= 
\frac{d{\bf r}}{ds} + (\frac{d}{ds}\kappa^{-1}){\bf N}+ \kappa^{-1} \frac{d{\bf N}}{ds}
\\
&= {\bf T}+  (\frac{d}{ds}\kappa^{-1}){\bf N}+ \kappa^{-1}(-\kappa {\bf T} + \tau {\bf B})
\\
&=  (\frac{d}{ds}\kappa^{-1}){\bf N}+ \kappa^{-1} \tau {\bf B}.
\end{align*}
By assumption, this is equal to ${\bf 0}$ for all $s$.
Since ${\bf N}$ and ${\bf B}$ are linearly independent, it follows that both coefficients must be zero. Thus, $\frac{d}{ds}\kappa^{-1}\equiv 0$,
so that $\kappa^{-1}$ is constant and therefore $\kappa$ is constant. Also, $\kappa^{-1} \tau=0$, so that $\tau\equiv 0$. 
By the uniqueness theorem for curves, ${\bf r}(t)$ is a circle. 


\bigskip

\item{\it 
Let $C$ be a smooth curve parametrized by $\bfr(t):\mathbb{R}\to\mathbb{R}^3$. Assume that (0,0,0) belongs to the osculating plane at some point $\bfr(t_0)$. Prove that at $t=t_0$,}
$$
(\bfr\times \bfr')\cdot \bfr''=0.
$$


\medskip
The normal vector to the osculating plane is 
${\bf B}
=\frac{\bfr'\times\bfr''}{|\bfr'\times\bfr''|}$, parallel to $\bfr'\times\bfr''$. Therefore the equation of the osculating plane at $t=t_0$ is
$$
(\langle x,y,z\rangle  -\bfr(t_0) )\cdot (\bfr'(t_0)\times\bfr''(t_0))=0.
$$
If $(x,y,z)=(0,0,0)$ satisfies this equation, we have
$$
\bfr(t_0) )\cdot (\bfr'(t_0)\times\bfr''(t_0))=0.
$$
But this is the triple product:
$$
\bfr(t)\cdot( \bfr'(t)\times \bfr''(t))= \left|\begin{matrix}
x(t) & y(t) & z (t)\\
x'(t)& y' (t)& z' (t) \\
x'' (t)& y'' (t)& z'' (t) \\
            \end{matrix}\right|
= (\bfr(t)\times \bfr'(t))\cdot \bfr''(t),
$$
so the last expression must be 0 at $t=t_0$.

 








\end{enumerate}

