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

\textit{\textbf{Applications}}
\end{center}

\textbf{Question}

Solve the initial value problem
\begin{eqnarray*}
\frac{d \underline{r}}{dt} & = & \underline{k} \times \underline{r}\\
\underline{r}(0) & = & \underline{i} + \underline{k}
\end{eqnarray*}
Describe the curve $\underline{r}=\underline{r}(t)$.


\textbf{Answer}

\begin{eqnarray*}
\frac{d \underline{r}}{dt} & = & \underline{k} \times \underline{r}\\
\underline{r}(0) & = & \underline{i} + \underline{k}
\end{eqnarray*}

If $\un{r}(t)=x(t)\un{i} +y(t)\un{j} +z(t)\un{k}$

Then
\begin{eqnarray*}
x(0) & = & z(0) = 1\\
y(0) & = & 0
\end{eqnarray*}
As $\un{k}\bullet(d\un{r}/dt)=\un{k}\bullet(\un{k}\times\un{r})=0$,
velocity is always perpendicular to $\un{k}$, meaning that $z(t)$ must
be constant.

$z(t)=z(0)=1$, $\forall t$

$\Rightarrow$
$$\frac{dx}{dt}\un{i} + \frac{dy}{dt}\un{j} = \frac{d\un{r}}{dt} =
\un{k} \times \un{r} = xun{j} =y\un{i}$$
In component form this becomes
\begin{eqnarray*}
\frac{dx}{dt} & = & -y\\
\frac{dy}{dt} & = & x\\
\Rightarrow \frac{d^2x}{dt^2} & = & -\frac{dy}{dt}=-x\\
x & = & A\cos t +B\sin t
\end{eqnarray*}
As $x(0)=1$ and $y(0)=0$, $\Rightarrow A=1$ and $B=0$.
\begin{eqnarray*}
\Rightarrow x(t)=\cos t\\
y(t) & = & \sin t
\end{eqnarray*}
So the path has equation
$$\un{r} = \cos t\un{i} +\sin t \un{j} + \un{k}$$

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