The volume of a spherical balloon is increasing at the rate of 30 cc per minute. Find the rate of change of…

$\Rightarrow v=\text{volume of sphere of radius} r$

$\Rightarrow v=\frac{4}{3}\mathrm{\pi } {\mathrm{r}}^3$

$\Rightarrow \frac{dv}{dt}=30 cc/min$

$\Rightarrow \frac{d}{dt}\left(\frac{4}{3}\mathrm{\pi } {\mathrm{r}}^3\right)=30$

$\Rightarrow \frac{4}{3}\mathrm{\pi } \left(3 {\mathrm{r}}^2\right)\frac{\mathrm{dr}}{\mathrm{dt}}=30$

$\Rightarrow \frac{dr}{dt}=\frac{30}{4\mathrm{\pi } {\mathrm{r}}^2} \left(1\right)$

$\Rightarrow s=\text{surface area of balloon}$

$\Rightarrow s=4\mathrm{\pi } {\mathrm{r}}^2$

$\text{⇒rate of change of surface area of the balloon}=\frac{ds}{dt}$

$\Rightarrow \frac{ds}{dt}=\frac{d}{dt}\left(4\mathrm{\pi } {\mathrm{r}}^2\right)$

$\Rightarrow \frac{ds}{dt}=8\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}$

$\Rightarrow \frac{ds}{dt}=8\mathrm{\pi r}\frac{30}{4\mathrm{\pi } {\mathrm{r}}^2} \left(\text{from eq. 1}\right)$

$\Rightarrow \frac{ds}{dt}=\frac{60}{r}$

$\Rightarrow \frac{ds}{dt}=\frac{60}{6}=10 {\mathrm{cm}}^2/\min$