Question

The surface area of a spherical bubble is increasing at the rate of $2{\text{cm}}^2/s.$ When the radius of the bubble is $6$ cm, at what rate is the volume of the bubble increasing?

Answer 1

We have,
$S=4\pi r^2\Rightarrow \frac{dS}{dt}=8\pi r\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{2}{8\pi r}$ $[\because \frac{ds}{dt}=2]$
Now,
$V=\frac{4}{3}\pi r^3$
$\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$
$\Rightarrow \frac{dV}{dt}=4\pi r^2\times \frac{2}{8\pi r}$
$\Rightarrow \frac{dV}{dt}=r$
Hence,$\frac{dV}{dt}=6$ when $r=6$