Question

The volume of a spherical balloon is increasing at the rate of 25 cm³/sec. Find the rate of change of its surface area at the instant when radius is 5 cm.

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{r}}\text{be the radius and}\mathit{\text{V}}\text{be the volume of the sphere at any time}\mathit{\text{t}}\text{.\hspace{0.17em}Then,} \\ \mathit{\text{V}}\text{=}\frac{4}{3}\text{π}{\mathit{\text{r}}}^3 \\ \Rightarrow \frac{dV}{dt}=4\mathrm{\pi }{r}^2\frac{dr}{dt} \\ \Rightarrow \frac{dr}{dt}=\frac{1}{4\mathrm{\pi }{r}^2}\frac{dV}{dt} \\ \Rightarrow \frac{dr}{dt}=\frac{25}{4\mathrm{\pi }{\left(5\right)}^2}\left[\because r=5\mathrm{cm}\mathrm{and}\frac{dV}{dt}=25{\mathrm{cm}}^3/\sec \right] \\ \Rightarrow \frac{dr}{dt}=\frac{1}{4\mathrm{\pi }}\mathrm{cm}/\sec \\ \text{Now, let}\mathit{\text{S}}\text{be the surface area of the sphere at any time}\mathit{\text{t.}}\text{Then,} \\ \mathit{\text{S}}\text{=}\mathrm{4\pi }{r}^2 \\ \text{⇒}\frac{dS}{dt}\text{=8πr}\frac{dr}{dt} \\ \text{⇒}\frac{dS}{dt}\text{=8π}\left(5\right)\text{×}\frac{1}{4\mathrm{\pi }}\left[\because r=5\mathrm{cm}\mathrm{and}\frac{dr}{dt}=\frac{1}{4\mathrm{\pi }}\mathrm{cm}/\sec \right] \\ \text{⇒}\frac{dS}{dt}{\text{=10 cm}}^2\text{/sec} \end{gathered}$$