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 $V = \frac{\pi}{3}\left(45h^2 - h^3\right),$ where $h$ is the depth of water at the centre of the tank in cm.

1. Apply Pythagoras' theorem to the right-angled triangle formed by the water surface radius, the vertical distance to the tank's centre, and the tank's radius.

$r^2 + (15 - h)^2 = 15^2$

2. Expand the squared expression and make $r$ the subject.

$r^2 + 225 - 30h + h^2 = 225$
$r^2 = 30h - h^2$

Since $r$ represents a length, $r > 0$, therefore:
$r = \sqrt{30h - h^2} \quad \text{(shown)}$

1. Differentiate the given volume $V$ with respect to the depth $h$.

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

2. Apply the chain rule to find the rate of change of depth when $h=5$.

By the chain rule, $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$.
Since water is leaking at $20 \text{ cm}^3/\text{min}$, $\frac{dV}{dt} = -20$.
When $h = 5$,
$\frac{dV}{dh} = \pi(30(5) - 5^2) = 125\pi$
Substitute the values into the chain rule:
$-20 = 125\pi \times \frac{dh}{dt}$
$\frac{dh}{dt} = -\frac{20}{125\pi} = -\frac{4}{25\pi} \approx -0.0509 \text{ cm/min}$

1. Calculate $\frac{dh}{dt}$ at $h = 5$ (dependency from part (b)(i))

$V = \frac{\pi}{3}(45h^2 - h^3)$
$\frac{dV}{dh} = \pi(30h - h^2)$
When $h = 5$, $\frac{dV}{dh} = \pi(30(5) - 5^2) = 125\pi$.
Given $\frac{dV}{dt} = -20$, using the chain rule $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$:
$-20 = 125\pi \times \frac{dh}{dt} \implies \frac{dh}{dt} = -\frac{4}{25\pi}$

2. Differentiate the radius $r$ with respect to $h$ and evaluate at $h = 5$

From part (a), $r = (30h - h^2)^{\frac{1}{2}}$.
$\frac{dr}{dh} = \frac{1}{2}(30h - h^2)^{-\frac{1}{2}}(30 - 2h) = \frac{15 - h}{\sqrt{30h - h^2}}$
When $h = 5$:
$\frac{dr}{dh} = \frac{15 - 5}{\sqrt{30(5) - 5^2}} = \frac{10}{\sqrt{125}} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$

3. Calculate the rate of decrease of the radius $r$

Using the chain rule for connected rates of change:
$\frac{dr}{dt} = \frac{dr}{dh} \times \frac{dh}{dt}$
$\frac{dr}{dt} = \left(\frac{2}{\sqrt{5}}\right) \left(-\frac{4}{25\pi}\right) = -\frac{8}{25\sqrt{5}\pi}$
Rationalising the denominator gives $-\frac{8\sqrt{5}}{125\pi}$.
Since the question asks for the rate of decrease, we drop the negative sign.
Rate of decrease $= \frac{8\sqrt{5}}{125\pi} \approx 0.0456 \text{ cm/min}$