Question

The radius $r$ of a right circular cone is decreasing at a rate of $3$ cm per minute and the height $h$ is increasing at a rate of $2$ cm per minute. When $r=9cm$ and $h=6cm$ find the rate of change of its volume.

Answer 1

Solution-

Volume of right circular cone = $\frac{\pi r^{2}h}{3}$

$\frac{\partial V}{\partial t}=\frac{\pi }{3}(2rh\frac{\partial r}{\partial t}+r^2\frac{\partial h}{\partial t})$

$\frac{\partial v}{\partial t}=\frac{\pi }{3}(108\times -3+81\times 2)$

$\frac{\partial v}{\partial t}=\frac{\pi }{3}(-162)=-54\pi c{m}^2/min$

Volume is decreasing at rate $54\pi c{m}^2/min$