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 $k\sqrt{h}\ \mathrm{m^{3}\ per\ minute}$, where $k$ is a positive constant. At time $t$ minutes, the height of the water surface from the hole is $h\ \mathrm{m}$ (see diagram).

1. Determine the radius of the water surface in terms of $h$.

Let the total distance from the imaginary apex to the water surface be $H$. Since the hole is $10\text{ m}$ from the apex, $H = h + 10$.
Let the radius of the water surface be $r$.
Using the semi-vertical angle $\alpha$, we have:
$\tan \alpha = \frac{r}{H} \implies 0.5 = \frac{r}{h + 10}$
$r = 0.5(h + 10)$

2. Express the volume of the cone from the imaginary apex up to the water surface.

$V_{\text{large}} = \frac{1}{3}\pi r^2 H$
$V_{\text{large}} = \frac{1}{3}\pi \left[0.5(h+10)\right]^2 (h+10)$
$V_{\text{large}} = \frac{1}{3}\pi (0.25)(h+10)^2(h+10)$
$V_{\text{large}} = \frac{1}{12}\pi(h+10)^3$

3. Calculate the volume of the removed smaller cone.

The smaller cone has a base radius of $5\text{ m}$ and a height of $10\text{ m}$.
$V_{\text{small}} = \frac{1}{3}\pi (5)^2 (10)$
$V_{\text{small}} = \frac{250}{3}\pi$

4. Subtract to find the volume of the water and simplify to the required form.

$V = V_{\text{large}} - V_{\text{small}}$
$V = \frac{1}{12}\pi(h+10)^3 - \frac{250}{3}\pi$
Expressing with a common denominator of $12$:
$V = \frac{1}{12}\pi(h+10)^3 - \frac{1000}{12}\pi$
$V = \frac{1}{12}\pi\bigl[(h+10)^3 - 1000\bigr]$ (Shown)

1. Find the depth of the water $h$ when the volume is $120\pi$.

$\frac{1}{12}\pi\left[(h+10)^3 - 1000\right] = 120\pi$
$(h+10)^3 - 1000 = 1440$
$(h+10)^3 = 2440$
$h+10 = \sqrt[3]{2440} = 2\sqrt[3]{305}$
$h = 2\sqrt[3]{305} - 10 \approx 3.4632$

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

$V = \frac{1}{12}\pi\left[(h+10)^3 - 1000\right]$
$\frac{dV}{dh} = \frac{1}{12}\pi \left[3(h+10)^2\right] = \frac{\pi}{4}(h+10)^2$
Substitute $h+10 = 2\sqrt[3]{305}$:
$\frac{dV}{dh} = \frac{\pi}{4}\left(2\sqrt[3]{305}\right)^2 = \pi(305)^{2/3} \approx 142.36$

3. Relate the rates of change using the chain rule to find $\frac{dh}{dt}$.

Since water flows out, the volume is decreasing, meaning the rate of change of volume is negative: $\frac{dV}{dt} = -k\sqrt{h}$.
Using the chain rule, $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$:
$-k\sqrt{2\sqrt[3]{305} - 10} = \pi(305)^{2/3} \times \frac{dh}{dt}$
$\frac{dh}{dt} = -\frac{\sqrt{2\sqrt[3]{305}-10}}{\pi(305)^{2/3}} k \approx -0.0131 k$