The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Solution: Here, Volume is changing at a constant rate.
So, $\frac{dV}{dt}=k$, where $k$ is a constant.
Now, Volume of a sphere, $V=\frac{4}{3}\pi r^3$
So, $\frac{dV}{dt}=\frac{4}{3}\ast 3\pi r^2\frac{dr}{dt}$
$\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$
As,$\frac{dV}{dt}=k$.So,
$\Rightarrow kdt=4\pi r^{2}dr$
Now, integrating both sides,
$\Rightarrow kt+c=\frac{4}{3}\pi r^3+c_1$->(1)
$\Rightarrow kt+C=\frac{4}{3}\pi r^3$, where $C=c-c_1$
So, at $\left(r,t\right)=\left(3,0\right)$
$\Rightarrow C=\frac{4}{3}\pi \ast {\left(3\right)}^3\Rightarrow C=36\pi$
At$\left(r,t\right)=\left(6,3\right)$
$\Rightarrow k\left(3\right)+36\pi =\frac{4}{3}\pi {\left(6\right)}^3\Rightarrow 3k=288\pi -36\pi \Rightarrow k=84\pi$
Putting values of $C$ and $k$ in (1),
$84\pi t+36\pi =\frac{4}{3}\pi r^3\Rightarrow 84t+36=\frac{4}{3}{r}^3$
$\Rightarrow r^3=\left(63t+27\right)\Rightarrow r={\left(63t+27\right)}^{\frac{1}{3}}$
So, at any time $t$ radius of the baloon will be ${\left(63t+27\right)}^{\frac{1}{3}}$.