Question

The volume of a sphere is increasing at the rate of 4π cm³/sec. The rate of increase of the radius when the volume is 288 π cm³, is
(a) 1/4
(b) 1/12
(c) 1/36
(d) 1/9

Answer 1

(c) 1/36

$$\begin{gathered} \text{Let}\mathit{\text{r}}\text{be the radius and}\mathit{\text{V}}\text{be the volume of the sphere at any time}\mathit{\text{t}}\text{. Then,} \\ V\text{=}\frac{4}{3}\text{π}{\mathit{\text{r}}}^3 \\ \Rightarrow \frac{4}{3}\text{π}{\mathit{\text{r}}}^3\text{=288π} \\ \Rightarrow r^3=\frac{288\times 3}{4} \\ \Rightarrow r^3=216 \\ \Rightarrow r=6 \\ \Rightarrow \frac{dV}{dt}=4\mathrm{\pi }{r}^2\frac{dr}{dt} \\ \Rightarrow \frac{dV}{dt}=4\mathrm{\pi }{\left(6\right)}^2\frac{dr}{dt} \\ \Rightarrow 4\mathrm{\pi }=144\mathrm{\pi }\frac{dr}{dt} \\ \Rightarrow \frac{dr}{dt}=\frac{1}{36} \end{gathered}$$