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If the surface area of a sphere of radius $r$ is increasing uniformly at the rate $8 \mathrm{~cm}^2 / \mathrm{s}$, then the rate of change of its volume is:
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Verified Answer
The correct answer is:
proportional to $r$
proportional to $r$
$$
\begin{aligned}
& \mathrm{V}=\frac{4}{3} \pi r^3 \Rightarrow \frac{d \mathrm{~V}}{d t}=4 \pi r^2 \cdot \frac{d r}{d t} \\
& \mathrm{~S}=4 \pi r^2 \Rightarrow \frac{d \mathrm{~S}}{d t}=8 \pi r \cdot \frac{d r}{d t} \\
& \Rightarrow 8=8 \pi r \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{1}{\pi r}
\end{aligned}
$$
Putting the value of $\frac{d r}{d t}$ in (i), we get
$$
\frac{d \mathrm{~V}}{d t}=4 \pi r^2 \times \frac{1}{\pi r}=4 r
$$
$\Rightarrow \frac{d \mathrm{~V}}{d t}$ is proportional to $r$.
\begin{aligned}
& \mathrm{V}=\frac{4}{3} \pi r^3 \Rightarrow \frac{d \mathrm{~V}}{d t}=4 \pi r^2 \cdot \frac{d r}{d t} \\
& \mathrm{~S}=4 \pi r^2 \Rightarrow \frac{d \mathrm{~S}}{d t}=8 \pi r \cdot \frac{d r}{d t} \\
& \Rightarrow 8=8 \pi r \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{1}{\pi r}
\end{aligned}
$$
Putting the value of $\frac{d r}{d t}$ in (i), we get
$$
\frac{d \mathrm{~V}}{d t}=4 \pi r^2 \times \frac{1}{\pi r}=4 r
$$
$\Rightarrow \frac{d \mathrm{~V}}{d t}$ is proportional to $r$.
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