Sand is pouring from a pipe at the rate of $\displaystyle 15 \mathrm{~cm}^{3} /$ minute . The falling sand forms a cone on the ground such that the height of the cone is always one- third of the radius of the base. How fast is the height of the sand cone increasing at the instant when the height is $\displaystyle 4$ cm ?

Let ' $\displaystyle V$ ' be the volume, ' $\displaystyle h$ ' and ' $\displaystyle r$ ' be the height and radius of the cone,\begin{aligned} \therefore & h=\frac{r}{3}, \frac{d V}{d t}=15 \mathrm{~cm}^{3} / \mathrm{min} & V=\frac{1}{3} \pi r^{2} h=3 \pi h^{3} \Rightarrow & \left.\left.\frac{d}{d t}\left(\frac{\pi}{3} r^{2} h\right)=15 \Rightarrow 3 \pi \frac{d\left(h^{3}\right)}{d t}=15 \Rightarrow \frac{d h}{d t}\right]_{h=4}=\frac{5}{3 \pi h^{2}}\right]_{h=4}=\frac{5}{48 \pi} \mathrm{~cm} / \mathrm{min} \end{aligned} $$