The surface area of a spherical balloon is increasing at the rate $2 \mathrm{~cm}^2 / \mathrm{sec}$. Then rate of increase in the volume of the balloon is , when the radius of the balloon is $6 \mathrm{~cm}$.

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Aicharya · Accepted Answer

We have $\frac{\mathrm{dA}}{\mathrm{dt}}=2$ and $\mathrm{A}=4 \pi \mathrm{r}^2$ $\begin{aligned} & 	herefore \frac{\mathrm{dA}}{\mathrm{dt}}=8 \pi \mathrm{r} \frac{\mathrm{dt}}{\mathrm{dt}} \Rightarrow 2=8 \pi(6) \frac{\mathrm{dr}}{\mathrm{dt}} \ & 	herefore \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{24 \pi} \end{aligned}$ Here $\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^3$ $\begin{aligned} & 	herefore \frac{\mathrm{dV}}{\mathrm{dt}}=4 \pi \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}} \ & =4 \pi(6)^2\left(\frac{1}{24 \pi}ight)=6 \mathrm{~cm}^3 / \mathrm{sec} \end{aligned}$

The surface area of a spherical balloon is increasing at the rate $2 \mathrm{~cm}^2 / \mathrm{sec}$. Then…

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