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\newcommand{\un}{\underline}
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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 given by the parameterization
$\underline{r}=3u\underline{i}+ 3u^2\underline{j}+ 2u^3\underline{k}$.
Given that the object has a
constant speed of 6 and that $u$ is increasing, find the velocity and
acceleration of the object at the point $(3,3,2)$.


\textbf{Answer}

\begin{eqnarray*}
\underline{r} &amp; = &amp; 3u\underline{i}+3u^2\underline{j}
+2u^3\underline{k}\\
\underline{v} &amp; = &amp; \frac{du}{dt}(3\underline{i} +6u\underline{j}+
6u^2\underline{k})\\
\underline{a} &amp; = &amp; \frac{d^2u}{dt^2}(3\underline{i} +6u\underline{j}+
6u^2\underline{k})\\
&amp; &amp; + \left ( \frac{du}{dt} \right )^2(6\underline{j}
+12u\underline{k})
\end{eqnarray*}
As the speed of the object of 6 and $u$ is increasing,
\begin{eqnarray*}
6=|v| &amp; = &amp; 3\frac{du}{dt}\sqrt{1+4u^2+4u^4}\\
 &amp; = &amp; 3(1+2u^2)\frac{du}{dt}
\end{eqnarray*}
$\Rightarrow$
$$\frac{du}{dt} = \frac{2}{1+2u^2}, \ \textrm{and}$$
\begin{eqnarray*}
\frac{d^2u}{dt^2} &amp; = &amp; \frac{-2}{(1+2u^2)^2}4u\frac{du}{dt}\\
&amp; = &amp; \frac{-16u}{(1+2u^2)^3}
\end{eqnarray*}

It can be seen that the object is as $(3,3,2)$ when $u=1$. 

At this point $\frac{du}{dt} =2/3$ and $\frac{d^2u}{dt^2}=-16/27$.

$\Rightarrow$
\begin{eqnarray*}
\underline{v} &amp; = &amp; \frac{2}{3}(3\underline{i}+ 6u\underline{j} +6u^2
\underline{k} )\\
&amp; = &amp; 2\underline{i} +4\underline{j} +4\underline{k}\\
\underline{a} &amp; = &amp; -\frac{16}{27}(3\underline{i} +6\underline{j}
+6\underline{k})+ \left ( \frac{2}{3} \right )^2 (6\underline{j} +12
\underline{k})\\
&amp; = &amp; \frac{8}{9}(-2\underline{i} -\underline{j} +2\underline{k})
\end{eqnarray*}

\end{document}
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