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 $0.5\text{ mm}^2$ per second. When $x = 5$, find the exact value of

1. Express the circular cross-sectional area in terms of $x$ and find its derivative with respect to $x$.

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

2. Apply the chain rule to relate the rates of change and substitute the given values.

By the chain rule:
$\frac{\text{d}A}{\text{d}t} = \frac{\text{d}A}{\text{d}x} \times \frac{\text{d}x}{\text{d}t}$
Since the area is decreasing at $0.5 \text{ mm}^2\text{/s}$, $\frac{\text{d}A}{\text{d}t} = -0.5$.
Substitute $x = 5$ and $\frac{\text{d}A}{\text{d}t} = -0.5$ into the equation:
$-0.5 = 2\pi(5) \times \frac{\text{d}x}{\text{d}t}$
$-0.5 = 10\pi \frac{\text{d}x}{\text{d}t}$

3. Solve for the exact value of $\frac{\text{d}x}{\text{d}t}$.

$\frac{\text{d}x}{\text{d}t} = \frac{-0.5}{10\pi} = -\frac{1}{20\pi}$

1. Calculate the rate of change of the radius $\frac{\text{d}x}{\text{d}t}$ when $x=5$ (Result from part a).

Let $A$ be the cross-sectional area. $A = \pi x^2$, so $\frac{\text{d}A}{\text{d}x} = 2\pi x$.
Using the chain rule: $\frac{\text{d}A}{\text{d}t} = \frac{\text{d}A}{\text{d}x} \times \frac{\text{d}x}{\text{d}t} = 2\pi x \frac{\text{d}x}{\text{d}t}$.
Given $\frac{\text{d}A}{\text{d}t} = -0.5$, when $x = 5$:
$2\pi(5) \frac{\text{d}x}{\text{d}t} = -0.5$
$\frac{\text{d}x}{\text{d}t} = -\frac{0.5}{10\pi} = -\frac{1}{20\pi}$

2. Formulate the volume of the tablet in terms of $x$ and differentiate with respect to $x$.

Let $V$ be the volume of the cylindrical tablet.
$V = \pi r^2 h = \pi x^2 \left(\frac{1}{3}x\right) = \frac{\pi}{3} x^3$
$\frac{\text{d}V}{\text{d}x} = \pi x^2$

3. Apply the chain rule to find the rate of change of the volume when $x = 5$.

$\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}x} \times \frac{\text{d}x}{\text{d}t}$
Substitute $x = 5$ and $\frac{\text{d}x}{\text{d}t} = -\frac{1}{20\pi}$:
$\frac{\text{d}V}{\text{d}t} = \pi (5)^2 \times \left(-\frac{1}{20\pi}\right)$
$\frac{\text{d}V}{\text{d}t} = 25\pi \times \left(-\frac{1}{20\pi}\right) = -\frac{25}{20} = -1.25$