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,
$$\begin{gathered} S=4\mathrm{\pi }{r}^2 \\ \Rightarrow \frac{dS}{dt}=8\mathrm{\pi }\mathit{}r\frac{dr}{dt}.....\left(1\right) \\ \mathrm{Given}:\hspace{0.17em}\frac{dS}{dt}\mathrm{\alpha }t \\ \Rightarrow \frac{dS}{dt}=kt,\mathrm{where}k\mathrm{is}\mathrm{any}\mathrm{constant} \\ \mathrm{Putting}\frac{dS}{dt}=kt\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \Rightarrow kt=8\mathrm{\pi }\mathit{}r\frac{dr}{dt} \\ kt\mathit{}dt=8\mathrm{\pi }r\mathit{}dr \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int kt\mathit{}dt=\int 8\mathrm{\pi }r\mathit{}dr \\ \Rightarrow \frac{k{t}^2}{2}=8\mathrm{\pi }\times \frac{r^2}{2}+\mathrm{C}.....\left(2\right) \\ \mathrm{At}t=0\mathrm{s},r\mathit{=}1\mathrm{unit}\mathrm{and}\mathrm{at}\mathit{}t=3\mathrm{s},r=2\mathrm{units}\left(\mathrm{Given}\right) \\ \therefore 0=8\mathrm{\pi }\times \frac{1}{2}+\mathrm{C} \\ \Rightarrow \mathrm{C}=-4\mathrm{\pi } \\ \mathrm{And} \\ \frac{9}{2}k=8\mathrm{\pi }\times 2+\mathrm{C} \\ \Rightarrow \frac{9}{2}k=12\mathrm{\pi } \\ \Rightarrow k=\frac{8}{3}\mathrm{\pi } \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{C}\mathrm{and}k\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \frac{8t^2}{6}\mathrm{\pi }=8\mathrm{\pi }\times \frac{r^2}{2}-4\mathrm{\pi } \\ \Rightarrow \frac{4{\mathrm{t}}^2}{3}=4{\mathrm{r}}^2-4 \\ \Rightarrow \frac{{\mathrm{t}}^2}{3}={\mathrm{r}}^2-1 \\ \Rightarrow {\mathrm{r}}^2=1+\frac{{\mathrm{t}}^2}{3} \\ \Rightarrow r=\sqrt{1+\frac{1}{3}{t}^2} \end{gathered}$$