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 ?

0.002 cm/sec

Let h be the height, r be the radius of the base and V be the volume of the cone at time t. Then,

$V =\frac{1}{3} \pi r^2 h$

$\Rightarrow V =\frac{2}{3} \pi r^3$             [∵ h = 2r]

$\Rightarrow \frac{d V}{d t} =2 \pi r^2 \frac{d r}{d t}$

$\Rightarrow 40 =2\left(10^4\right) \frac{d r}{d t}$       $\left[\begin{array}{r}∵ \pi r^2=1 \mathrm{~m}^2 \text { (given) }=10^4 \mathrm{~cm}^2 \\ \text { and } \frac{d V}{d t}=40 \mathrm{~cm}^3 / \mathrm{sec}\end{array}\right]$

$\Rightarrow \frac{d r}{d t}=\frac{2}{1000} \mathrm{~cm} / \mathrm{sec}=0.002 \mathrm{~cm} / \mathrm{sec}$