When the hemispherical bowl above contains water to a depth of $x$ cm, the volume, $V\text{ cm}^3$, of the water is given by $V = \frac{1}{3}\pi x^2 (18 - x)$. The bowl is initially empty. After water has been poured into the bowl at a constant rate for 9 seconds, the depth of water is 4.5 cm.

1. Calculate the volume of water when the depth is 4.5 cm.

When $x = 4.5$,
$V = \frac{1}{3}\pi (4.5)^2 (18 - 4.5)$
$V = \frac{1}{3}\pi (20.25)(13.5)$
$V = 91.125\pi \text{ cm}^3$

2. Calculate the constant rate of change of volume.

Since the water is poured at a constant rate and the bowl was initially empty, the rate of change of volume is $\frac{\Delta V}{\Delta t}$.
$\frac{dV}{dt} = \frac{91.125\pi - 0}{9 - 0}$
$\frac{dV}{dt} = 10.125\pi \text{ cm}^3/\text{s}$

1. Find the constant rate of change of volume, $\frac{dV}{dt}$ (from part a)

When $x = 4.5$, $V = \frac{1}{3}\pi (4.5)^2 (18 - 4.5) = 91.125\pi$.
Since the rate is constant and the bowl was initially empty, $\frac{dV}{dt} = \frac{91.125\pi}{9} = 10.125\pi$.

2. Differentiate $V$ with respect to $x$

$V = 6\pi x^2 - \frac{1}{3}\pi x^3$
$\frac{dV}{dx} = 12\pi x - \pi x^2$

3. Evaluate $\frac{dV}{dx}$ at $x = 4.5$

When $x = 4.5$,
$\frac{dV}{dx} = 12\pi (4.5) - \pi (4.5)^2 = 54\pi - 20.25\pi = 33.75\pi$

4. Use the chain rule to find $\frac{dx}{dt}$

$\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$
$10.125\pi = 33.75\pi \times \frac{dx}{dt}$
$\frac{dx}{dt} = \frac{10.125\pi}{33.75\pi} = 0.3$