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If the rate of increase of the radius of a circle is $5 \mathrm{~cm} / \mathrm{sec}$, then the rate of increase of its area, when the radius is $20 \mathrm{~cm}$, will be
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The correct answer is:
$200 \pi$
Hints : $\mathrm{A}=\pi \mathrm{r}^2 \quad \frac{d r}{d t}=5$
$$
\begin{aligned}
& \frac{\mathrm{dA}}{\mathrm{dt}}=2 \pi r \frac{\mathrm{dr}}{\mathrm{dt}}=2 \pi 20(5) \\
& =200 \pi
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\mathrm{dA}}{\mathrm{dt}}=2 \pi r \frac{\mathrm{dr}}{\mathrm{dt}}=2 \pi 20(5) \\
& =200 \pi
\end{aligned}
$$
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