Question

The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec. Find the rate of increase of its surface area, when the radius is 7 cm.

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{r}}\text{be the radius and}\mathit{\text{S}}\text{be the surface area of the spherical ball at any time}\mathit{\text{t.}}\text{Then}\mathit{\text{,}} \\ S=4\mathrm{\pi }{r}^2 \\ \Rightarrow \frac{dS}{dt}=8\mathrm{\pi }r\frac{dr}{dt} \\ \Rightarrow \frac{dS}{dt}=8\mathrm{\pi }\times 7\times 0.2\left[\because r=7\mathrm{cm}\mathrm{and}\frac{dr}{dt}=0.2\mathrm{cm}/\sec \right] \\ \Rightarrow \frac{dS}{dt}=11.2\mathrm{\pi }{\mathrm{cm}}^2/\sec \end{gathered}$$