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.

: Volume of spherical balloon $= V = \cfrac{4}{3}\pi {r^3}$

…(1)
Differentiating (1) w.r.t. $t$,

we get $\cfrac{{dV}}{{dt}} = 4\pi {r^2}\cfrac{{dr}}{{dt}}$

Now, $\cfrac{{dV}}{{dt}} = a \Rightarrow 4\pi {r^2}\cfrac{{dr}}{{dt}} = a$

…(2)
Integrating (2) both sides,

we get
$\int 4 \pi {r^2}dr = \int a dt\; \Rightarrow 4\pi \cfrac{{{r^3}}}{3} = at + C$

When $t = 0,r = 3$
$\therefore \cfrac{{4\pi {{(3)}^3}}}{3} = C\; \Rightarrow C = 36\pi \;\,\,\therefore \,\,\,\cfrac{{4\pi {r^3}}}{3} = at + 36\pi$

When $t = 3,r = 6$
$\therefore \cfrac{{4\pi {{(6)}^3}}}{3} = 3a + 36\pi \Rightarrow \cfrac{{4\pi \times 36 \times 6}}{3} = 3a + 36\pi$

$\Rightarrow 288\pi = 3a + 36\pi \Rightarrow 96\pi = a + 12\pi \Rightarrow a = 84\pi$

Hence, $\cfrac{{4\pi {r^3}}}{3} = 84\pi t + 36\pi \Rightarrow \cfrac{{4{r^3}}}{3} = 84t + 36$

$\Rightarrow \cfrac{{{r^3}}}{3} = 21t + 9 \Rightarrow {r^3} = 63t + 27 \Rightarrow r = {(63t + 27)^{1/3}}$