The radius of an air bubble is increasing at the rate of $\cfrac{1}{2}{\rm{cm/s}}$. At what rate is the volume of the bubble increasing when the radius is $1{\rm{cm}}$?

Let us assume that at any instant of time $t$, the radius of the bubble be $r$ and its volume be $V$, then
$V = \cfrac{4}{3}\pi {r^3}$ …(i)

Differentiating (i), w.r.t. $t$, we get
$\cfrac{{dV}}{{dt}} = \left( {\cfrac{4}{3}\pi } \right)\left( {3{r^2}\cfrac{{dr}}{{dt}}} \right) = 4\pi {r^2}\cfrac{{dr}}{{dt}}$

$= 4\pi {\left( {1{\rm{cm}}} \right)^2}\left( {\cfrac{1}{2}{\rm{cm/sec}}} \right) = 2\pi {\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/sec}}$

Hence, the rate of increase of the volume of the bubble$= 2\pi {\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/sec}}$.