Problem Statement
Water is poured into an inverted right circular cone of base radius $6\text{ m}$ and height $12\text{ m}$ at a constant rate of $2\text{ m}^3\text{ min}^{-1}$.
Find the rate at which the water level is rising when the depth of the water is $4\text{ m}$. [7 Marks]
Verified Solution & Marking Scheme
Establish Geometric Similarity Ratio
For an inverted cone of height $H = 12$ and radius $R = 6$, at water height $h$ with surface radius $r$:
$\frac{r}{h} = \frac{6}{12} = \frac{1}{2} \implies r = \frac{h}{2}$
Express Volume Exclusively in terms of h
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi \left(\frac{h^2}{4}\right) h = \frac{\pi}{12} h^3$
Differentiate with respect to time t
$\frac{dV}{dt} = \frac{d}{dh}\left(\frac{\pi}{12} h^3\right) \cdot \frac{dh}{dt} = \frac{3\pi}{12} h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$
Substitute Known Rates and Height h = 4
Given $\frac{dV}{dt} = 2\text{ m}^3\text{ min}^{-1}$ and $h = 4\text{ m}$:
$2 = \frac{\pi}{4}(4)^2 \frac{dh}{dt} = \frac{\pi}{4}(16) \frac{dh}{dt} = 4\pi \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi} \approx 0.159\text{ m min}^{-1}$