The radius of a circle is increasing uniformly at the rate of $3{\rm{cm/s}}$. Find the rate at which the area of the circle is increasing when the radius is $10{\rm{ cm}}$.

Let us assume at any instant of time is $t$,

the radius of the circle be $r$ and its area be $A$. Then, $A = \pi {r^2}$

$\Rightarrow \cfrac{{dA}}{{dt}} = \pi (2r)\cfrac{{dr}}{{dt}}$ and ${\left( {\cfrac{{dA}}{{at}}} \right)_{r = l0{\rm{cm}}}} = 2\pi (10{\rm{cm}})(3{\rm{cm/sec)}}$

$= 60\pi {\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/sec}}$

Therefore, rate of increase of area of the circle $= 60\pi {\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/sec}}$.