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\textbf{Vector Functions and Curves}

\textit{\textbf{One variable functions}}
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\textbf{Question}

A particle moves along the curve $y=3/x$, travelling to the right. At
the point $(2, \frac{3}{2})$ its speed is 10, what is its velocity?


\textbf{Answer}

When its $x$-coordinate is $x$, the object is at position
$$\underline{r}=x\underline{i}+(3/x)\underline{j}$$,
and it and velocity and speed
$$\underline{v}=\frac{d\underline{r}}{dt}=\frac{dx}{dt}\underline{i}
-\frac{3}{x^2} \frac{dx}{dt}\underline{j}$$
$$\it{v}=\left | \frac{dx}{dt} \right | \sqrt{ 1 +\frac{9}{x^4}}$$
It is known that dx/dt>0 since the particle is moving to the right.

When $x=2$, we have
\begin{eqnarray*}
10=\it{v} & = & (dx/dt)\sqrt{1+(9/16)}\\
 & = & (5/4)(dx/dt)
\end{eqnarray*}
$\Rightarrow$ $dx/dt =8$


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