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The radius of a circle is increasing at the rate $2 \mathrm{~cm} / \mathrm{sec}$. The rate at which its area is
increasing when the radius of the circle is 5 decimeters is
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increasing when the radius of the circle is 5 decimeters is
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Verified Answer
The correct answer is:
$200 \pi \mathrm{cm}^{2} / \mathrm{sec}$
Here $\frac{\mathrm{dr}}{\mathrm{dt}}=2$ and $\mathrm{r}=5$ decimeter $=50 \mathrm{~cm}$
Now Area of circle $=\mathrm{A}=\pi \mathrm{r}^{2}$.
$\begin{aligned}
\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}} &=2 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \\
&=2 \times \pi \times 50 \times 2=200 \pi \mathrm{cm}^{2} / \mathrm{sec}
\end{aligned}$
Now Area of circle $=\mathrm{A}=\pi \mathrm{r}^{2}$.
$\begin{aligned}
\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}} &=2 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \\
&=2 \times \pi \times 50 \times 2=200 \pi \mathrm{cm}^{2} / \mathrm{sec}
\end{aligned}$
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