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

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

\textbf{Question}

A particle is travelling along the curve $y=x^2$, $z=x^3$ and has
constant vertical speed $w=dz/dt=3$. When the particle is at the point
$(2,4,8)$, find both its velocity and acceleration.


\textbf{Answer}

When the $x$-coordinate of the particle is $x$ is has position
$$\underline{r}=x\underline{i}+x^2\underline{j}+ x^3\underline{k},$$
and so has a velocity
$$\underline{v}=\frac{dx}{dt}[\underline{i}+2x\underline{j}+
3x^2\underline{k}].$$

As $w=\frac{dz}{dt}=3x^2\frac{dx}{dt}=3$, when $x=2$
\begin{eqnarray*}
12\frac{dx}{dt} & = & 3\\
\textrm{so\ } \frac{dx}{dt} & = & \frac{1}{4}
\end{eqnarray*}
$\Rightarrow$
$$\underline{v}=\frac{1}{4}\underline{i}+\underline{j}+3
\underline{k}.$$


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