Problem Statement
A conical water tank of height $12\text{ m}$ and base radius $4\text{ m}$ is positioned with its vertex pointing downwards.
Water is leaking out of the vertex at a constant rate of $0.2\text{ m}^3\text{min}^{-1}$.
**(a)** [3 marks] Show that the volume of water when the depth is $h$ meters is given by $V = \frac{\pi}{27} h^3$.
**(b)** [4 marks] Find the rate at which the water level is dropping when the depth of water is $3\text{ meters}$.
Verified Solution & Marking Scheme
Part (a) [M1]
By similar triangles: $\frac{r}{h} = \frac{4}{12} = \frac{1}{3} \implies r = \frac{h}{3}$
Part (a) [M1]
Volume of cone: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h$
Part (a) [A1]
$V = \frac{1}{3}\pi \frac{h^2}{9} h = \frac{\pi}{27} h^3$ (AG)
Part (b) [M1]
Differentiate with respect to time $t$: $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$
Part (b) [A1]
$\frac{dV}{dh} = \frac{3\pi}{27}h^2 = \frac{\pi}{9}h^2$
Part (b) [M1]
At $h = 3$: $\frac{dV}{dh} = \frac{\pi}{9}(3^2) = \pi$. Given $\frac{dV}{dt} = -0.2$
Part (b) [A1]
$-0.2 = \pi \frac{dh}{dt} \implies \frac{dh}{dt} = -\frac{0.2}{\pi} \approx -0.0637\text{ m min}^{-1}$. The water level is dropping at $0.0637\text{ m min}^{-1}$