Question

The volume of a 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 the balloon after `t` seconds.

Answer 1

Let r be the radius and V be the volume of the balloon at any time 't'.
Then, we have,
\[V = \frac{4}{3} \pi r^3 \]
Given :- 
\[\frac{dV}{dt} = - k ...............\left(\text{where }k > 0 \right)\]
\[ \Rightarrow \frac{d}{dt}\left( \frac{4}{3}\pi r^3 \right) = - k\]
\[ \Rightarrow 4 \pi r^2 \frac{dr}{dt} = - k\]
\[ \Rightarrow 4\pi r^2 dr = - k\ dt \]
Integrating both sides, we get
\[\int4\pi r^2 dr = - \int k\ dt \]
\[\frac{4}{3}\pi r^3 = - kt + C ............(1)\]
It is given that at t = 0, r = 3 . 
\[\text{ Substituting }t = 0\text{ and }r = 3\text{ in }(1), \text{ we get }\]
\[C = 36\pi\]
\[\text{ Putting }C = 36\pi\text{ in }(1),\text{ we get }\]
\[\frac{4}{3}\pi r^3 = - kt + 36\pi .............(2)\]
It is also given that at t = 3, r = 6 . 
\[\text{ Putting }t = 3\text{ and }r = 6\text{ in }(1), \text{ we get }\]
\[288 \pi = - 3k + 36\pi\]
\[ \Rightarrow k = - 84\pi\]
\[\text{ Putting }k = - 84 \pi\text{ in }(2),\text{ we get }\]
\[\frac{4}{3}\pi r^3 = 84\pi t + 36 \pi\]
\[ \Rightarrow r^3 = 63 t + 27\]
\[ \Rightarrow r = \left( 63 t + 27 \right)^\frac{1}{3} \]