
\section{Velocity and Acceleration}

\begin{enumerate}[1.]






\item 
{\it The position of an object is given by a smooth vector-valued function ${\bf r}(t)$. Let ${\bf v}(t)=\bfr'(t)$, and ${\bf a}(t)=\bfr''(t)$. Assume that $
{\bf a}(t) =\kk \times {\bf v}(t)$ for all $t$.

\medskip

(a) Prove that the speed $v$ is constant.

\medskip

(b) Assuming that $v$ is nonzero, prove that the angle $\theta(t)$ between ${\bf v}(t)$ and $\kk$ is constant.

}

\medskip

(a) We have
$$
\frac{d}{dt} ({\bf v}\cdot {\bf v})={\bf a}\cdot {\bf v}+ {\bf v}\cdot {\bf a}=0,
$$
since ${\bf a}$ is perpendicular to ${\bf v}$. Therefore ${\bf v}\cdot {\bf v}= |{\bf v}|^2=v^2$ is constant, so that $v$ is constant.

\medskip

(b) Since ${\bf a}$ is perpendicular to ${\bf u}$, we have
$$
\frac{d}{dt} ({\bf v}\cdot {\bf k})={\bf a}\cdot {\bf k}+ {\bf v}\cdot {\bf 0}=0.
$$
Therefore ${\bf v}\cdot {\bf k}$ is constant.
We also have $|\kk|=1$, and $|{\bf v}|=v$ is constant by (a) and nonzero by assumption. It follows that
$$
\cos\theta = \frac{{\bf v}\cdot {\bf k}}{v}
$$
is constant, as required.


\bigskip


\item 
{\it Assume that ${\bf a}(t)=c\,\bfr(t)$ for some constant $c$. (This describes the motion of an object under a central force.)

\medskip

(a) Prove that ${\bf w}:=\bfr\times{\bf v}$ is constant. 

\medskip

(b) Assuming that ${\bf w}\neq {\bf 0}$, prove that the curve $\bfr(t)$ lies in the plane passing through the origin and perpendicular to ${\bf w}$.}


\medskip

(a) We first prove that ${\bf w}$ is constant:
$$
{\bf w}'(t)=\frac{d}{dt} ({\bf r}\times {\bf v})={\bf v}\times {\bf v}+ {\bf r}\times {\bf a}={\bf 0}+ \bfr\times (c \bfr) ={\bf 0}.
$$

(b) It follows that ${\bf r}$ is always perpendicular to the fixed vector ${\bf w}$, which we assume to be nonzero. Therefore the curve lies in the plane ${\bf r}\cdot {\bf w}=0$.

\bigskip


\item 
{\it Suppose that a curve $\mathcal{C}$ in $\rr^3$ has a smooth parametrization $\bfr(t)$ for all $t\in\rr$,
the curvature $\kappa$ is a positive constant, and the torsion $\tau$ is zero. Prove, without using the Fundamental Theorem of Calculus for curves, that $\mathcal{C}$ is a circle. (Hint: how would you find the center of the circle? You may also want to use the Frenet-Serret formulas.)}

\medskip

Reverse-engineering from the conclusion, if $\mathcal{C}$ were a circle, then $\mathcal{C}$ would be its own osculating circle at all points. In particular, the center of the osculating circle should remain fixed as we travel the curve.

Now, we do not actually know that $\mathcal{C}$ is a circle -- we need to prove this. As a first step, we will try to prove that the center $Q$ of the osculating circle remains at a fixed point. Let ${\bf w}=\vec{OQ}=\bfr + \rho {\bf N}$, where $\rho=1/\kappa$. Since $\kappa$ is constant, so is $\rho$. We want $Q$ to be independent of $t$. To prove this, we check the derivative with respect to $s$ (this is easier here than differentiating in $t$):
\begin{align*}
\frac{d}{ds} ({\bf r}+\rho {\bf N})& = {\bf T}+ \frac{d}{ds} (\rho {\bf N})
\\
&= {\bf T}+ \kappa^{-1} \frac{d}{ds} {\bf N}
\\
&= {\bf T}+ \kappa^{-1} (-\kappa {\bf T}) ={\bf 0},
\end{align*}
where at the last step we used Frenet-Serret. So $Q$ is constant as claimed. 

Based on this, we need to prove that $\mathcal{C}$ is a circle centered at $Q$. In Section \ref{sec:torsion} of this manual, this was done using the Fundamental Theorem of Calculus for curves. Here, you are asked to find an alternative proof that does not rely on the theorem. It suffices to prove the following two facts:

\medskip

\begin{itemize} 
\item The distance from $Q$ to ${\bf r}(t)$ is constant: this follows from
$$
|\bfr-\vec{OQ}|=|\bfr-(\bfr+ \rho {\bf N})|=|\rho {\bf N}|=\rho.
$$
\item The curve lies in a fixed 2-dimensional plane through $Q$. Indeed, we claim that the binormal vector ${\bf B}$ is constant, and that the curve lies in the plane $(\bfr-\vec{OQ})\cdot {\bf B}=0$. To see this, we first verify that ${\bf B}$ is constant:
$$
\frac{d}{ds} {\bf B}=-\tau {\bf N}={\bf 0},
$$
using Frenet-Serret again. Finally, we check the equation of the plane:
$$
(\bfr-\vec{OQ})\cdot {\bf B}= (-\rho {\bf N})\cdot {\bf B}=0.
$$
\end{itemize}
The two bullet points together imply that the curve is a circle.





\bigskip




\item 
{\it Let $C$ be a curve on the sphere $\mathcal{S}$ of radius 1 centered at ${\bf 0}$. Suppose that $C$ has a smooth parametrization ${\bf r}(t)$ such that the velocity ${\bf r}'(t)$ is never zero and the acceleration $\bfr''(t)$ is always perpendicular to $\mathcal{S}$ at $\bfr(t)$. Prove that the curve is contained in a great circle of $\mathcal{S}$ (that is, a circle where $\mathcal{S}$ intersects a plane through ${\bf 0}$).}


\medskip 


Let ${\bf b}=\bfr'\times \bfr$. Then:

\medskip

\begin{itemize}
\item ${\bf b}$ is constant since
$$
\frac{d}{dt}(\bfr'\times \bfr)=\bfr''\times \bfr + \bfr'\times\bfr'= \bfr''\times \bfr.
$$
Since $\bfr''$ is perpendicular to $\mathcal{S}$, it is parallel to the normal vector to $\mathcal{S}$, therefore parallel to $\bfr$. It follows that the last cross-product is zero, and so ${\bf b}$ is constant.

\medskip

\item ${\bf b}$ is nonzero: since $|{\bf r}|=1$ for all $t$, $\bfr'$ is perpendicular to $\bfr$, so that 
$|{\bf b}|=|\bfr'\times \bfr|=|\bfr'||\bfr|=|\bfr'|\neq 0$ by assumption.

\medskip


\item $\bfr\cdot {\bf b}=0$, since ${\bf b}=\bfr'\times \bfr$ is perpendicular to $\bfr$.
\end{itemize}

\medskip


Therefore $\bfr$ lies in the plane through the origin perpendicular to ${\bf b}$, as claimed.


\bigskip


\item {\it
Let $\bfr(s):\R\to\R^3$ be a smooth curve, parametrized by arc length.
Assume that $\frac{d^2\bfr}{ds^2}=\bfr\times {\bf w}$ for all $s$, where ${\bf w}$ is a fixed nonzero vector. Prove that this is only possible when $\frac{d^2\bfr}{ds^2}=\bfr\times {\bf w}={\bf 0}$ for all $s$.


\medskip\noindent
(Hint: suppose that there is a value of $s$ where $\bfr\times {\bf w}\neq {\bf 0}$. Prove first that there are scalar functions $\lambda(s),\mu(s)$ such that $\frac{d\bfr}{ds}=\lambda\bfr + \mu {\bf w}$ near that point. What does this imply about $\frac{d^2\bfr}{ds^2}$?)}

\medskip

As per hint, assume towards contradiction that $\bfr\times {\bf w}\neq{\bf 0}$  for some $s$. By continuity, this is also true on some neighbourhood of that point. In particular, we may find a part of the curve where $\bfr\times {\bf w}\neq{\bf 0}$.

We always have $|\frac{d\bfr}{ds}|=|{\bf T}|=1$, so that $\frac{d^2\bfr}{ds^2}$ is perpendicular to $\frac{d\bfr}{ds}$. But we know that $\frac{d^2\bfr}{ds^2}$ is also perpendicular to $\bfr$ and ${\bf w}$, and these two vectors are nonzero and not parallel to each other. Therefore $\frac{d\bfr}{ds}$ must belong to the span of these two vectors, so that $\frac{d\bfr}{ds}=\lambda\bfr + \mu {\bf w}$
for some scalar functions $\lambda(s),\mu(s)$.

Next, differentiating this in $s$ and using that ${\bf w}$ is constant, we get
\begin{align*}
\frac{d^2\bfr}{ds^2} &= \frac{d}{ds}(\lambda\bfr + \mu {\bf w})
\\
&= \frac{d\lambda}{ds}\bfr + \lambda \frac{d\bfr}{ds} +  \frac{d\mu}{ds}{\bf w}
\\
&= \frac{d\lambda}{ds}\bfr + \lambda (\lambda\bfr + \mu {\bf w} ) +  \frac{d\mu}{ds}{\bf w}.
\end{align*}
This is a linear combination of $\bfr$ and ${\bf w}$. But $\frac{d^2\bfr}{ds^2}$ was also supposed to be perpendicular to both of these vectors. Therefore it is also perpendicular to any linear combination of its vectors, including itself. Hence
$$\frac{d^2\bfr}{ds^2}= \bfr\times {\bf w}={\bf 0},
$$
contradicting our assumption.







\end{enumerate}



