Question

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.

Answer 1

Let the rate of change of the volume of the balloon be k(where k is a constant).
⇒\frac{dv}{dt}=k$$⇒\frac{d}{dt}(\frac{4\pi}{3}r^3)=k [Volume of sphere$=\frac{4}{3}\pi r^3$]
$⇒\frac{4π}{3}.3r^2.\frac{dr}{dt}=k$
$⇒4πr^2dr=k\, dt$
Integrating both sides,we get:
$4π∫r^2dr=k∫1dt$
$⇒4π.\frac{r^3}{3}=kt+C$
$⇒4πr^3=(kt+C)...(1)$
Now,at t=0,r=3:
$⇒\frac{4π\times27}{3}=C$
$⇒C=36π$
At t=3,r=6:
$⇒4π\times6^3=3k+C$
$⇒3k=-288π-36π=252π$
$⇒k=84π$
Substituting the values of k and C in equation(1),we get:
$4πr^3=84πt+36π$
$⇒r^3=63t+27$
$⇒r=(63t+27)^{\frac{1}{3}}$
Thus the radius of the balloon after t seconds is $(63t+27)^{\frac{1}{3}}$