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 \((63t+27)^{\frac{1}{3}}\)