Question

A cone whose height is always equal to its diameter is increasing in volume at the rate of 40 cm³/sec. At what rate is the radius increasing when its circular base area is 1 m²?
(a) 1 mm/sec
(b) 0.001 cm/sec
(c) 2 mm/sec
(d) 0.002 cm/sec

Answer 1

(d) 0.002 cm/sec

$$\begin{gathered} \text{Let}\mathit{\text{r}}\text{be the radius,}\mathit{\text{h}}\text{be the height and}V\text{be the volume of the cone at any time}\mathit{\text{t}}\text{. Then,} \\ V=\frac{1}{3}\mathrm{\pi }{r}^2\mathit{\text{h}} \\ \Rightarrow V=\frac{2}{3}\mathrm{\pi }{r}^3\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because h=\mathit{2}r\right] \\ \Rightarrow \frac{dV}{dt}=2\times {10}^4\frac{dr}{dt}\left[\because \mathrm{\pi }{r}^2=1{\mathrm{m}}^2\mathrm{or}{10}^4{\mathrm{cm}}^2\right] \\ \Rightarrow \frac{dr}{dt}=\frac{1}{2\times {10}^4}\frac{dV}{dt} \\ \Rightarrow \frac{dr}{dt}=\frac{40}{2\times {10}^4} \\ \Rightarrow \frac{dr}{dt}=0.002\mathrm{cm}/\sec \end{gathered}$$