Question

The volume of a sphere is increasing at 3 cubic centimeter per second. Find the rate of increase of the radius, when the radius is 2 cms.

Answer 1

$$\begin{gathered} \text{Let}\mathit{\text{r}}\text{be the radius and}\mathit{\text{V}}\text{be the volume of the sphere at any time}\mathit{\text{t.}}\text{Then}\mathit{\text{,}} \\ V=\frac{4}{3}\text{π}{\mathit{\text{r}}}^3 \\ \Rightarrow \frac{dV}{dt}=4\text{π}{\mathit{\text{r}}}^2\frac{dr}{dt} \\ \Rightarrow \frac{dr}{dt}=\frac{1}{4\text{π}{\mathit{\text{r}}}^2}\frac{dV}{dt} \\ \Rightarrow \frac{dr}{dt}=\frac{3}{4\text{π}{\left(\text{2}\right)}^2}\left[\because r=2\mathrm{cm}\mathrm{and}\frac{dV}{dt}=3{\mathrm{cm}}^3/\sec \right] \\ \Rightarrow \frac{dr}{dt}=\frac{3}{16\mathrm{\pi }}\mathrm{cm}/\sec \end{gathered}$$