A hemispherical goldfish tank with radius 15 cm (as shown in the figure above) was initially filled with water. The tank has a defect and water is leaking at a constant rate of 20 cm^3 per min. The volume of water in the tank is given by $V = \frac{\pi}{3}(45h^2 - h^3),$ where $h$ is the depth of water at the centre of the tank in cm. Show that $r$, the radius of the water surface in cm, is given by $r = \sqrt{30h - h^2}$. Given that the minimum depth of water needed for the goldfish to survive is 5 cm, find, at this instant,

1. Find the derivative of the volume $V$ with respect to the depth $h$.

Given $V = \frac{\pi}{3}(45h^2 - h^3)$,
$\frac{dV}{dh} = \frac{\pi}{3}(90h - 3h^2) = \pi(30h - h^2)$

2. Set up the connected rates of change equation using the chain rule and substitute the given values.

By the chain rule,
$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$
We are given that water is leaking at $20 \text{ cm}^3/\text{min}$, so $\frac{dV}{dt} = -20$.
When $h = 5$,
$\frac{dV}{dh} = \pi(30(5) - 5^2) = \pi(150 - 25) = 125\pi$

3. Solve for the rate of change of depth, $\frac{dh}{dt}$.

$-20 = 125\pi \times \frac{dh}{dt}$
$\frac{dh}{dt} = -\frac{20}{125\pi} = -\frac{4}{25\pi}$
$\frac{dh}{dt} \approx -0.0509 \text{ cm/min}$ (to 3 s.f.)

1. Calculate the rate of change of depth $\frac{dh}{dt}$ at $h = 5$ from part (i)

From part (i), we use the chain rule $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$.
Given $V = \frac{\pi}{3}(45h^2 - h^3)$, differentiating with respect to $h$ gives:
$\frac{dV}{dh} = \frac{\pi}{3}(90h - 3h^2) = \pi(30h - h^2)$
When $h = 5$, $\frac{dV}{dh} = \pi(150 - 25) = 125\pi$.
Given that water leaks at $20\text{ cm}^3\text{/min}$, we have $\frac{dV}{dt} = -20$.
$-20 = 125\pi \frac{dh}{dt} \implies \frac{dh}{dt} = -\frac{4}{25\pi} \text{ cm/min}$

2. Find $\frac{dr}{dh}$ by differentiating the given equation for $r$

Given $r = (30h - h^2)^{\frac{1}{2}}$, we use the chain rule for differentiation:
$\frac{dr}{dh} = \frac{1}{2}(30h - h^2)^{-\frac{1}{2}} \cdot (30 - 2h)$
$\frac{dr}{dh} = \frac{15 - h}{\sqrt{30h - h^2}}$

3. Evaluate $\frac{dr}{dh}$ at the instant $h=5$

Substitute $h = 5$ into the derivative:
$\frac{dr}{dh} = \frac{15 - 5}{\sqrt{30(5) - 5^2}} = \frac{10}{\sqrt{150 - 25}}$
$\frac{dr}{dh} = \frac{10}{\sqrt{125}} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$

4. Calculate the rate of decrease of the radius using connected rates of change

Using the chain rule $\frac{dr}{dt} = \frac{dr}{dh} \times \frac{dh}{dt}$:
$\frac{dr}{dt} = \frac{2}{\sqrt{5}} \times \left(-\frac{4}{25\pi}\right) = -\frac{8}{25\sqrt{5}\pi}$
Rationalising the denominator:
$\frac{dr}{dt} = -\frac{8\sqrt{5}}{125\pi} \approx -0.0456$
Since $\frac{dr}{dt}$ is negative, the radius is decreasing.
The rate of decrease is $\frac{8\sqrt{5}}{125\pi}\text{ cm/min}$.