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The rate of change of the surface area of a sphere of radius $r$, when the radius is increasing at the rate of $2 \mathrm{~cm} / \mathrm{s}$ is proportional to
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$r$
Surface area of sphere,
$\begin{array}{l}
\mathrm{S}=4 \pi \mathrm{r}^{2} \\
\text { and } \frac{\mathrm{dr}}{\mathrm{dt}}=2 \\
\therefore \frac{\mathrm{ds}}{\mathrm{dt}}=4 \pi \times 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=8 \pi \mathrm{r} \times 2=16 \pi \mathrm{r} \\
\Rightarrow \frac{\mathrm{ds}}{\mathrm{dt}} \propto \mathrm{r}
\end{array}$
$\begin{array}{l}
\mathrm{S}=4 \pi \mathrm{r}^{2} \\
\text { and } \frac{\mathrm{dr}}{\mathrm{dt}}=2 \\
\therefore \frac{\mathrm{ds}}{\mathrm{dt}}=4 \pi \times 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=8 \pi \mathrm{r} \times 2=16 \pi \mathrm{r} \\
\Rightarrow \frac{\mathrm{ds}}{\mathrm{dt}} \propto \mathrm{r}
\end{array}$
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