A tank containing water is in the form of a cone with vertex $C$. The axis is vertical and the semi-vertical angle is $60^{\circ}$, as shown in the diagram. At time $t=0$, the tank is filled with $94 \pi \mathrm{cm}^3$ of water. At this instant, a tap at $C$ is turned on and water begins to flow out at a constant rate of $2 \pi \mathrm{cm}^3 \mathrm{~s}^{-1}$. Denoting $h \mathrm{~cm}$ as the depth of water at time $t \mathrm{~s}$, find the rate of decrease of $h$ when $t=15$, leaving your answer in exact form.

[The volume $V$ of a cone of vertical height $h$ and base radius $r$ is given by $V=\frac{1}{3} \pi r^2 h$.]

Rate at which $h$ is decreasing at the instant when $t=15$ is $\frac{1}{24} \mathrm{cms}^{-1} \text {. }$

$\tan 60^{\circ}=\frac{r}{h} \Rightarrow r=h \sqrt{3}$

Therefore

$V=\pi h^{3}$

$\Rightarrow \quad \frac{\mathrm{d} V}{\mathrm{~d} t}=3 \pi h^{2} \frac{\mathrm{d} h}{\mathrm{~d} t}$

When $t=15, V=94 \pi-(2 \pi)(15)=64 \pi, \pi h^{3}=64 \pi \Rightarrow h=4$

$\therefore \frac{\mathrm{d} h}{\mathrm{~d} t}=-2 \pi \div 3 \pi(4)^{2}=-\frac{1}{24}$

$\therefore$ Rate at which $h$ is decreasing at the instant when $t=15$ is $\frac{1}{24} \mathrm{cms}^{-1} \text {. }$