Question

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

Answer 1

Let the rate of change of the volume of the balloon be K (where k, is constant)
$\frac{d}{dt}\left(volume\right)=constant\Rightarrow \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=k[\because Volumeofsphere=\frac{4}{3}\pi r^3]\Rightarrow \left(\frac{4}{3}\pi \right)\left(3r^2\frac{dr}{dt}\right)=k$
On separating the variables, we get $4\pi r^{2}dr=kdt...\left(i\right)$
On integrating bothe sides, we get $4\pi \int r^{2}dr=k\int dt$
$\Rightarrow 4\pi \frac{r^3}{3}=kt+c\Rightarrow 4\pi r^3=3(kt+C)$ ...(ii)
Now, at t=0, r=3, initially;
$\therefore 4\pi (3{)}^3=3(k\times 0+C)\Rightarrow 108\pi =3C\Rightarrow C=36\pi$
Also, when t=3, r=6, then from Eq. (ii)
$4\pi (6{)}^3=3(k\times 0+C)\Rightarrow 864\pi =3(3k+36\pi )\Rightarrow 3k=288\pi -36\pi =252\pi \Rightarrow k=84\pi$
On substituting the values of k and C in Eq. (ii), we get
$4\pi r^3=3[84\pi t+36\pi ]\Rightarrow 4\pi r^3=4\pi (63t+27)\Rightarrow r^3=63t+27\Rightarrow r=(63t+27{)}^{1/3}$
which is the required radius of the balloon at time t.