A cone whose height is always equal to its diameter is increasing in volume at the rate of $40{\mathrm{cm}}^3/\sec$ At what rate is the  radius increasing when its circular base area is $1{\mathrm{m}}^2?$

Let $h$ be the height, $r$ be the radius of the base and

$V$be the volume of the cone at time t. Then, 

$\begin{array}{l} V=\frac{1}{3}\pi r^{2}h\\ \Rightarrow V=\frac{2}{3}\pi r^3 [\because h=2r] \end{array}$

$\begin{array}{l}\Rightarrow \frac{dV}{dt}=2\pi r^2\frac{dr}{dt}\\ \Rightarrow 40=2\left({10}^4\right)\frac{dr}{dt}\left[\begin{array}{c}\left.\because \pi r^2=1{\mathrm{m}}^2\text{ (given }\right)={10}^4{\mathrm{cm}}^2\\ \text{ and }\frac{dV}{dt}=40{\mathrm{cm}}^3/\sec \end{array}\right]\end{array}$

$\Rightarrow \frac{dr}{dt}=\frac{2}{1000}\mathrm{cm}/\sec =0.002\mathrm{cm}/\sec$