Water is being poured, at a constant rate of 42 m^3/s, into an empty container. After t seconds, the depth of water is h m and the volume, V m^3, of the water in the container is given by $V = \frac{(5+2h)^3}{3\pi}.$

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

$V = \frac{(5+2h)^3}{3\pi}$

$\frac{dV}{dh} = \frac{1}{3\pi} \times 3(5+2h)^2 \times 2$

$\frac{dV}{dh} = \frac{2(5+2h)^2}{\pi}$

2. Set up the chain rule for connected rates of change.

$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$

3. Substitute the given rate of change of volume and solve for the rate of change of depth.

Given $\frac{dV}{dt} = 42$:

$42 = \frac{2(5+2h)^2}{\pi} \times \frac{dh}{dt}$

$\frac{dh}{dt} = \frac{42\pi}{2(5+2h)^2}$

$\frac{dh}{dt} = \frac{21\pi}{(5+2h)^2}$

1. Recall the expression for the rate of change of depth from part (i)

From part (i), $\frac{dh}{dt} = \frac{21\pi}{(5+2h)^2}$.

2. Determine the behaviour of the rate of change as $h$ increases

As $h$ increases, the value of $(5+2h)^2$ increases. Since the numerator $21\pi$ is a positive constant and the denominator increases, the overall fraction $\frac{21\pi}{(5+2h)^2}$ decreases. Therefore, the rate of change of $h$ decreases.