Problem Statement
Water is dripping out from a conical funnel at a uniform rate of $4\text{ cm}^3/\text{s}$ through a tiny hole at the vertex in the bottom. When the slant height of water is $3\text{ cm}$, find the rate of decrease of the slant height of water, given that the semi-vertical angle of the funnel is $30^\circ$.
Verified Solution & Marking Scheme
Express radius and height in terms of slant height l
Let $l$ be the slant height, $r$ be the base radius, and $h$ be the depth of water at time $t$.
Given semi-vertical angle $\alpha = 30^\circ$:
$$r = l \sin 30^\circ = \frac{l}{2}$$
$$h = l \cos 30^\circ = \frac{l\sqrt{3}}{2}$$
Given $\frac{dV}{dt} = -4\text{ cm}^3/\text{s}$.
Express Volume V in terms of l
$$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{l}{2}\right)^2 \left(\frac{l\sqrt{3}}{2}\right) = \frac{\pi\sqrt{3}}{24} l^3$$
Differentiate with respect to time t and solve for dl/dt
$$\frac{dV}{dt} = \frac{\pi\sqrt{3}}{24} \cdot 3l^2 \frac{dl}{dt} = \frac{\pi\sqrt{3}}{8} l^2 \frac{dl}{dt}$$
When $l = 3\text{ cm}$ and $\frac{dV}{dt} = -4$:
$$-4 = \frac{\pi\sqrt{3}}{8} (3)^2 \frac{dl}{dt} = \frac{9\pi\sqrt{3}}{8} \frac{dl}{dt}$$
$$\frac{dl}{dt} = -\frac{32}{9\pi\sqrt{3}} = -\frac{32\sqrt{3}}{27\pi}\text{ cm/s}$$
Thus, the rate of decrease of slant height is $\frac{32\sqrt{3}}{27\pi}\text{ cm/s}$.