A spherical snow ball is forming so that its volume is increasing at the rate of $8 \mathrm{~cm}^3 / \mathrm{sec}$. Find the rate of increase of radius when radius is $2 \mathrm{~cm}$

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Aicharya · Accepted Answer

$\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3$ $	herefore \frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3} \pi\left(3 \mathrm{r}^2ight) \frac{\mathrm{dr}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}$ $	herefore 8=4 \pi(2)^2 \frac{\mathrm{dr}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{2 \pi}$

A spherical snow ball is forming so that its volume is increasing at the rate of $8 \mathrm{~cm}^3 /…

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