NJC JC1 Promo 9758/2019/Q6 A vessel is formed by removing a smaller cone of radius 5 m from a bigger cone whose semi-vertical angle is $\alpha$, where $\tan \alpha = 0.5$. Water flows out of the vessel at a rate of $k\sqrt{h}\text{ m}^3$ per minute, where $k$ is a positive constant. At time $t$ minutes, the height of the water surface from the hole is $h\text{ m}$ (see diagram).

1. Determine the height of the smaller, removed cone.

Let $H_0$ be the height of the removed cone at the bottom of the vessel.
Using the right-angled triangle formed by the semi-vertical angle $\alpha$, the height $H_0$, and the radius of the hole:
$\tan \alpha = \frac{5}{H_0}$
Given $\tan \alpha = 0.5$,
$0.5 = \frac{5}{H_0} \implies H_0 = 10\text{ m}$

2. Express the radius of the water surface in terms of the water height $h$.

Let $r$ be the radius of the water surface. The total distance from the cone's vertex to the water surface is $h + 10$.
Using the semi-vertical angle again:
$\tan \alpha = \frac{r}{h + 10}$
$0.5 = \frac{r}{h + 10} \implies r = \frac{1}{2}(h + 10)$

3. Formulate the equation for the volume of the water.

The volume of water $V$ is the volume of the cone containing the water (from the vertex to the water surface) minus the volume of the removed small cone.
$V = \frac{1}{3}\pi r^2 (h+10) - \frac{1}{3}\pi (5)^2 (10)$
Substitute $r = \frac{1}{2}(h+10)$ into the equation:
$V = \frac{1}{3}\pi \left[ \frac{1}{2}(h+10) \right]^2 (h+10) - \frac{1}{3}\pi (25)(10)$
$V = \frac{1}{3}\pi \left( \frac{1}{4} \right)(h+10)^3 - \frac{250}{3}\pi$
$V = \frac{1}{12}\pi (h+10)^3 - \frac{1000}{12}\pi$
$V = \frac{1}{12}\pi \left[ (h+10)^3 - 1000 \right] \text{ (shown)}$

1. Find the height $h$ when the volume $V = 120\pi$.

$V = \frac{1}{12}\pi[(h+10)^3 - 1000]$
$120\pi = \frac{1}{12}\pi[(h+10)^3 - 1000]$
$1440 = (h+10)^3 - 1000$
$(h+10)^3 = 2440$
$h+10 = \sqrt[3]{2440} \approx 13.4627$
$h = \sqrt[3]{2440} - 10 \approx 3.4627 \text{ m}$

2. Differentiate the given volume equation with respect to $h$.

$\frac{dV}{dh} = \frac{1}{12}\pi \cdot 3(h+10)^2$
$\frac{dV}{dh} = \frac{\pi}{4}(h+10)^2$

3. Relate the rates of change using the chain rule.

By the chain rule:
$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$
Since water flows out of the vessel, the volume is decreasing, so the rate of change of volume is negative:
$\frac{dV}{dt} = -k\sqrt{h}$
$-k\sqrt{h} = \frac{\pi}{4}(h+10)^2 \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{-4k\sqrt{h}}{\pi(h+10)^2}$

4. Substitute the calculated values to find the rate of change of $h$.

Substitute $h = \sqrt[3]{2440} - 10$ and $(h+10)^2 = 2440^{2/3}$ into the equation for $\frac{dh}{dt}$:
$\frac{dh}{dt} = \frac{-4k\sqrt{\sqrt[3]{2440}-10}}{\pi(2440^{2/3})}$
$\frac{dh}{dt} = \frac{-4k\sqrt{3.4627}}{\pi(181.24)}$
$\frac{dh}{dt} \approx -0.0131 k \text{ m/min} \quad \text{(to 3 s.f.)}$