The volume of a ball is increasing at the rate of $4 \pi \mathrm{cc} / \mathrm{sec}$. The rate of increase of the radius, when the volume is $288 \pi \mathrm{cc}$, is

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Aicharya · Accepted Answer

$\begin{aligned} & V=\frac{4}{3} \pi r^3 \ & \Rightarrow 288 \pi=\frac{4}{3} \pi r^3 \ & \Rightarrow r=6 \mathrm{~cm}\end{aligned}$ $\begin{aligned} & \quad \mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3 \ & 	herefore \quad \frac{\mathrm{dV}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \cdot \frac{\mathrm{dr}}{\mathrm{dt}} \ & \Rightarrow 4 \pi=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \ & 	herefore \quad \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{\mathrm{r}^2}=\frac{1}{36} \mathrm{~cm} / \mathrm{sec}\end{aligned}$

The volume of a ball is increasing at the rate of $4 \pi \mathrm{cc} / \mathrm{sec}$. The rate of increase…

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