Problem Statement
Water is leaking out of an inverted conical tank at a constant rate of $0.2\text{ m}^3\text{/min}$. The cone has height $6\text{ m}$ and base radius $2\text{ m}$. Find the rate at which the water level is falling when the water depth is $3\text{ m}$.
Verified Solution & Marking Scheme
Geometric relationship using similar triangles
Let $r$ be the radius of the water surface and $h$ be the water depth.
By similar triangles: $\frac{r}{h} = \frac{2}{6} = \frac{1}{3} \implies r = \frac{h}{3}$.
Express Volume strictly in terms of depth h
Volume of cone: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left( \frac{h}{3} \right)^2 h = \frac{1}{27}\pi h^3$.
Apply Chain Rule to find dh/dt
Differentiating with respect to time $t$:
$$\frac{dV}{dt} = \frac{d}{dh}\left( \frac{\pi h^3}{27} \right) \frac{dh}{dt} = \frac{\pi h^2}{9} \frac{dh}{dt}$$
Given $\frac{dV}{dt} = -0.2 = -\frac{1}{5}\text{ m}^3\text{/min}$. At $h = 3\text{ m}$:
$$-\frac{1}{5} = \frac{\pi (3)^2}{9} \frac{dh}{dt} = \frac{9\pi}{9} \frac{dh}{dt} = \pi \frac{dh}{dt}$$
$$\frac{dh}{dt} = -\frac{1}{5\pi}\text{ m/min}$$
Therefore, the water level is falling at a rate of $\frac{1}{5\pi}\text{ m/min} \approx 0.0637\text{ m/min}$.