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\begin{center}
\textbf{Vector Functions and Curves}

\textit{\textbf{One variable functions}}
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\textbf{Question}


Show that $\underline{r}=\underline{r}_0\cos(\omega t)+ 
(\underline{v}_0/\omega)\sin(\omega t)$ satisfies the initial-value problem.
\begin{eqnarray*}
\frac{d^2 \underline{r}}{dt^2} & = & -\omega^2\underline{r}\\
\underline{r}'(0) & = & \underline{v}_0\\
\underline{r}(0) & = & \underline{r}_0
\end{eqnarray*}  
In this case, describe the path $\underline{r}(t)$ and determine what happens
to the paths if $\underline{r}_0$ is perpendicular to $\underline{v}_0$. 


\textbf{Answer}

\begin{eqnarray*}
\underline{r} & = & \underline{r}_0 \cos\omega t + \left (
\frac{\underline{v}_0}{\omega} \right ) \sin\omega t\\
\Rightarrow \frac{d\underline{r}}{dt} & = & -\omega \underline{r}_0
\sin\omega t + \underline{v}_0\cos\omega t\\
\Rightarrow \frac{d^2\underline{r}}{dt^2} & = & -\omega^2
\underline{r}_0 \cos\omega t - \omega \underline{v}_0 \sin \omega t\\
& = & -\omega^2 \underline{r}
\end{eqnarray*}
$$\underline{r}(0)=\underline{r}_0, \ \ \ \ \
\left. \frac{d\underline{r}}{dt} \right |_{t=0} =\underline{v}_0$$

See that $\underline{r}\bullet (\underline{r}_0\times \underline{v}_0
)$ for all $t$.

So the path lies in a plane through the origin with normal
$$\underline{N}=\underline{r}_0 \times \underline{v}_0.$$

So choose the system of coordinates so that
\begin{eqnarray*}
\underline{r}_0 = a\underline{i} & \ & (a>0)\\
\underline{v}_0 = \omega b \underline{i} + \omega c \underline{j} \ \
& \ & (c>0)
\end{eqnarray*}
$\Rightarrow$ $\underline{N}$ is in the direction of $\underline{k}$.

Parameterization of the path gives
\begin{eqnarray*}
x & = & a \cos \omega t + b\sin \omega t\\
y & = & c \sin \omega t
\end{eqnarray*}

The curve has the quadratic equation
$$ \frac{1}{a^2} \left ( x-\frac{by}{c} \right )^2 + \frac{y^2}{c^2}
=1 $$
so it is a conic section. As the path is bounded by
$$|\underline{r}(t)|\le |\underline{r}_0| +
(|\underline{v}_0|/\omega)$$
it must be an ellipse.

If $\underline{r}_0$ is perpendicular to $\underline{v}_0$, then
$b=0$, making the path an ellipse with equation
$$(x/a)^2+(y/c)^2=1$$
and semi-axes $a=|\underline{r}_0|$ and $c=|\underline{v}_0|/\omega$.


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