An upright cone, with a closed circular base, is shown in the diagram above. It has a circular base radius of $4\text{ cm}$ and height $12\text{ cm}$ and is initially full of water. Water is leaking from the circular base of the cone at a rate of $2\pi\text{ cm}^3\text{ min}^{-1}$. If $h$ is the depth of water in the cone at time $t$ minutes, show that the volume $V$, of water remaining, $V$, in the cone at time $t$ minutes is given by $V = 64\pi - \frac{\pi}{27}(12-h)^3$. Hence find the rate of change of the depth of water when the depth of water is $6\text{ cm}$.

1. Calculate the total volume of the cone

V_{\text{total}} = \frac{1}{3}\pi r^2 H = \frac{1}{3}\pi (4)^2 (12) = 64\pi

2. Find the dimensions and volume of the empty space at the top of the cone

\text{The empty space is a cone of height } (12-h).
\text{Let its base radius be } x.
\text{By similar triangles, } \frac{x}{4} = \frac{12-h}{12} \implies x = \frac{12-h}{3}.

V_{\text{empty}} = \frac{1}{3}\pi x^2 (12-h) = \frac{1}{3}\pi \left(\frac{12-h}{3}\right)^2 (12-h) = \frac{\pi}{27}(12-h)^3

3. Formulate the expression for the volume of water remaining

V = V_{\text{total}} - V_{\text{empty}}
V = 64\pi - \frac{\pi}{27}(12-h)^3 \text{ (shown)}

4. Differentiate the volume $V$ with respect to the depth $h$

\frac{\text{d}V}{\text{d}h} = -\frac{\pi}{27} \times 3(12-h)^2 \times (-1) = \frac{\pi}{9}(12-h)^2

5. Apply the chain rule to relate the rates of change

\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}h} \times \frac{\text{d}h}{\text{d}t}
\text{Since water is leaking, } \frac{\text{d}V}{\text{d}t} = -2\pi.
\text{Substitute } h = 6:
-2\pi = \frac{\pi}{9}(12-6)^2 \times \frac{\text{d}h}{\text{d}t}

6. Solve for the rate of change of depth

-2\pi = \frac{\pi}{9}(36) \times \frac{\text{d}h}{\text{d}t}
-2\pi = 4\pi \frac{\text{d}h}{\text{d}t}
\frac{\text{d}h}{\text{d}t} = -0.5