Application of Derivatives — Class 12 Maths Solution

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

Differentiating (i) w.r.t. $r$, we get
$\cfrac{{dV}}{{dt}} = \left( {\cfrac{4}{3}\pi } \right)3{r^2} = 4\pi {r^2} = 4\pi {\left( {10{\rm{cm}}} \right)^2} = 400\pi {\rm{c}}{{\rm{m}}^3}$

Rate of increase of volume with respect to change in radius $= 400\pi {\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/cm}}$