<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\un}{\underline}
\usepackage{epsfig}
\begin{document}
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\begin{center}
\textbf{Vector Functions and Curves}

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

\textbf{Question}

A particle travels along the curve of intersection of the plane
$x+y=2$ and the cylinder $z=x^2$ in the direction of increasing
$y$. The particle has constant speed $v=3$, what is its velocity at
the point $(1,1,1)$.

\textbf{Answer}

As the particle moves with increasing $y$ on $x+y=2$, $z=x^2$.

$\Rightarrow$ at time t
$$\underline{r}=(2-g(t))\underline{i} +g(t)\underline{j}+ (2-g(t))^2
\underline{k}$$
where $g(t)$ is an increasing function of time $t$.

$\Rightarrow$
\begin{eqnarray*}
\underline{v} &amp; = &amp;\frac{dg}{dt}[-\underline{i}+\underline{j} -2(2-g(t))
\underline{k}]\\
\it{v} &amp; = &amp; \frac{dg}{dt}\sqrt{1+1+4(2-g(t))^2}=3
\end{eqnarray*}
As the speed is 3.

When $g(t)=1$, we have
$$\frac{dg}{dt} = 3\sqrt{6} =\sqrt{3/2}$$
$\Rightarrow$
$$\underline{v}=\sqrt{\frac{3}{2}}(-\underline{i}+\underline{j}
-2\underline{k}).$$

\end{document}
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