Question

The surface area of a spherical bubble is increasing at the rate of 2 cm²/s. When the radius of the bubble is 6 cm, at what rate is the volume of the bubble increasing?

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{r}}\text{be the radius,}\mathit{\text{S}}\text{be the surface area and}\mathit{\text{V}}\text{be the volume of the sphere at any time}\mathit{\text{t}}\text{. Then,} \\ S=4\mathrm{\pi }{r}^2 \\ \Rightarrow \frac{dS}{dt}=8\mathrm{\pi }r\frac{dr}{dt} \\ \Rightarrow \frac{dr}{dt}=\frac{1}{8\mathrm{\pi }r}\frac{dS}{dt} \\ \Rightarrow \frac{dr}{dt}=\frac{2}{8\mathrm{\pi }\times 6} \\ \Rightarrow \frac{dr}{dt}=\frac{1}{24\mathrm{\pi }}\mathrm{cm}/\sec \\ \mathrm{Now}, \\ \text{Volume of sphere}=\frac{4}{3}\mathrm{\pi }{r}^3 \\ \Rightarrow \frac{dV}{dt}=4\mathrm{\pi }{r}^2\frac{dr}{dt} \\ \Rightarrow \frac{dV}{dt}=\frac{4\mathrm{\pi }{\left(6\right)}^2}{24\mathrm{\pi }} \\ \Rightarrow \frac{dV}{dt}=6{\mathrm{cm}}^3/\sec \end{gathered}$$