Fig. 1 A tablet is dissolving in water and is modelled as a cylinder, as shown in Fig. 1. At t seconds after being dropped into the water, the radius of the tablet is $x$ mm, and its thickness is $\frac{1}{3}x$ mm. The circular cross-sectional area of the tablet is decreasing at a constant rate of 0.5 mm^2 per second. When $x = 5$, find the exact value of

1. Express the cross-sectional area in terms of $x$ and differentiate with respect to $x$.

Let $A$ be the circular cross-sectional area.
$A = \pi x^2$
$\frac{dA}{dx} = 2\pi x$

2. Apply the chain rule using the given rate of change to find $\frac{dx}{dt}$.

Using the chain rule for connected rates of change:
$\frac{dA}{dt} = \frac{dA}{dx} \times \frac{dx}{dt}$
Since the area is decreasing at $0.5 \text{ mm}^2\text{s}^{-1}$, $\frac{dA}{dt} = -0.5 = -\frac{1}{2}$.
Substituting $x = 5$:
$-\frac{1}{2} = 2\pi(5) \times \frac{dx}{dt}$
$10\pi \frac{dx}{dt} = -\frac{1}{2}$
$\frac{dx}{dt} = -\frac{1}{20\pi}$

1. Express the volume of the tablet in terms of $x$ and differentiate it with respect to $x$.

The volume of a cylinder is $V = \pi r^2 h$.
Given $r = x$ and $h = \frac{1}{3}x$:
$V = \pi x^2 \left(\frac{1}{3}x\right) = \frac{1}{3}\pi x^3$
$\frac{dV}{dx} = \pi x^2$

2. Use the chain rule to evaluate the rate of change of the volume when $x = 5$.

By the chain rule:
$\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$
From part (a), when $x = 5$, $\frac{dx}{dt} = -\frac{1}{20\pi}$. Substituting $x = 5$ into $\frac{dV}{dx}$:
$\frac{dV}{dx} = \pi (5)^2 = 25\pi$
$\frac{dV}{dt} = 25\pi \times \left(-\frac{1}{20\pi}\right) = -\frac{25}{20} = -\frac{5}{4}$