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{\bf Question}

The volume of a cone with base radius $r$ and height $h$ is given
by $V  =  \frac{1}{3}\pi r^2 h$.  If the radius increases by 5\%
and the height decreases by 10\%, find the approximate percentage
change in $V.$

 \vspace{.25in}

{\bf Answer}


\begin{eqnarray*}
V &amp; = &amp; \frac{1}{3}\pi r^2 h \\ \ln V &amp; = &amp; \ln
\left(\frac{\pi}{3} \right) + \ln r^2 + \ln h \\ &amp; = &amp; \ln
\left(\frac{\pi}{3}\right) + 2\ln r + \ln h
\\ {\rm Take\ differentials} \\ \frac{dV}{V} &amp; \approx &amp; 2
\frac{dr}{r} + \frac{dh}{h} \\ {\rm In\ this\ question} \\
\frac{dr}{r} &amp; \approx &amp; 0.05 \hspace{.2in} \frac{dh}{h} = -0.1 \\
\Rightarrow\ \frac{dV}{V} &amp; \approx &amp; 2 \times 0.05 - 0.1 \approx
0
\end{eqnarray*}



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