Question

The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.

Answer 1

Let r be the radius and S be the surface area of the balloon at any time t. Then,
\[S = 4\pi r^2 \]
\[ \Rightarrow \frac{dS}{dt} = 8\pi r \frac{dr}{dt} . . . . . \left( 1 \right)\]
\[\text{ Given: }\hspace{0.167em} \frac{dS}{dt}\alpha t\]
\[ \Rightarrow \frac{dS}{dt} = kt,\text{ where k is any constant }\]
\[\text{ Putting }\frac{dS}{dt} = kt\text{ in }(1), \text{ we get }\]
\[ \Rightarrow kt = 8\pi r \frac{dr}{dt}\]
\[kt dt = 8\pi r dr\]
Integrating both sides, we get
\[\int kt dt = \int8\pi r dr\]
\[ \Rightarrow \frac{k t^2}{2} = 8\pi \times \frac{r^2}{2} + C . . . . . (2)\]
\[\text{ At }t = 0 s, r = 1 \text{ unit and at }t = 3 s, r = 2\text{ units }..............\left(\text{Given} \right)\]
\[ \therefore 0 = 8\pi \times \frac{1}{2} + C\]
\[ \Rightarrow C = - 4\pi\]
And
\[\frac{9}{2}k = 8\pi \times 2 + C\]
\[ \Rightarrow \frac{9}{2}k = 12 \pi\]
\[ \Rightarrow k = \frac{8}{3}\pi\]
Substituting the values of C and k in (2), we get 
\[\frac{8 t^2}{6}\pi = 8\pi \times \frac{r^2}{2} - 4\pi\]
\[ \Rightarrow \frac{4 t^2}{3} = 4 r^2 - 4\]
\[ \Rightarrow \frac{t^2}{3} = r^2 - 1\]
\[ \Rightarrow r^2 = 1 + \frac{t^2}{3}\]
\[ \Rightarrow r = \sqrt{1 + \frac{1}{3} t^2}\]