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

\textit{\textbf{One variable functions}}
\end{center}

\textbf{Question}

An object travels on the curve of intersection of the cylinders
$y=-x^2$ and $z=x^2$ with increasing $x$. When the particle is at
$(1,-1,1)$, it has a speed of $9cm/s$ which is increasing at a rate of
$3cm/s^2$. Given that all distances are in $cm$, find the velocity and
acceleration of the object at that point.


\textbf{Answer}

\begin{eqnarray*}
\underline{r} & = & x\underline{i} -x^2\underline{j} +x^2
\underline{k}\\
\underline{v} & = & \frac{dx}{dt}(\underline{i}- 2x\underline{j}
+2x\underline{k} )\\
\underline{a} & = & \frac{d^2x}{dt^2}(\underline{i} -2x\underline{j}
+2x\underline{k}) +\left ( \frac{dx}{dt} \right )^2 (-2\underline{j}
+2\underline{k} ).
\end{eqnarray*}
$\Rightarrow$
\begin{eqnarray*}
|\underline{v}| & = & \left | \frac{dx}{dt} \right |
 \sqrt{1+4x^4+4x^4}\\
& = & \sqrt{1+8x^4}\frac{dx}{dt},
\end{eqnarray*}
as $x$ is increasing.

At $(1,-1,1)$, $x=1$ and $|\underline{v}|=9$, $\Rightarrow
\frac{dx}{dt} =3$.

$\Rightarrow$ at that point
$$\underline{v}=3\underline{i} -6\underline{j} +6\underline{k}$$.

Now
$$\frac{d}{dt}|\underline{v}| = \sqrt{1+8x^4}\frac{d^2}{dt^2}
+\frac{16x^3}{\sqrt{1+8x^4}}\left ( \frac{dx}{dt} \right )^2.$$
When $x=1$, the left side is 3.

$\Rightarrow$
\begin{eqnarray*}
3\left ( \frac{d^2x}{dt^2} \right ) +48 & = & 3\\
\textrm{and \ \ \ } \frac{d^2x}{dt^2} =15
\end{eqnarray*}
at that point. 

$\Rightarrow$ Acceleration at that point
\begin{eqnarray*}
\underline{a} & = & -15(\underline{i} -2\underline{j} +2\underline{k}
) +9( -2\underline{j} +2\underline{k} )\\
& = & -15\underline{i} +12\underline{j} -12\underline{k}
\end{eqnarray*}

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