泡打粉以 $2\pi\text{ cm}^3\text{s}^{-1}$ 的恒定速率倒在水平面上，形成一个直圆锥。圆锥的半径始终是其高度的 $\frac{1}{18}$。求倒入 3 秒后圆锥半径的变化率。 $\left[\text{圆锥体积} = \frac{1}{3}\pi r^2 h\right]$

1. Express the volume of the cone in terms of radius $r$ only.

Given $r = \frac{1}{18}h$, we have $h = 18r$.
$V = \frac{1}{3}\pi r^2 h$
$V = \frac{1}{3}\pi r^2 (18r)$
$V = 6\pi r^3$

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

$\frac{\text{d}V}{\text{d}r} = 18\pi r^2$

3. Find the volume of the cone after 3 seconds.

Rate of pouring, $\frac{\text{d}V}{\text{d}t} = 2\pi\text{ cm}^3\text{s}^{-1}$.
After 3 seconds, $V = 2\pi \times 3 = 6\pi\text{ cm}^3$.

4. Find the radius $r$ after 3 seconds.

$6\pi r^3 = 6\pi$
$r^3 = 1$
$r = 1\text{ cm}$

5. Use the chain rule to find the rate of change of the radius.

$\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}r} \times \frac{\text{d}r}{\text{d}t}$
$2\pi = (18\pi (1)^2) \times \frac{\text{d}r}{\text{d}t}$
$2\pi = 18\pi \frac{\text{d}r}{\text{d}t}$
$\frac{\text{d}r}{\text{d}t} = \frac{2\pi}{18\pi} = \frac{1}{9}\text{ cm s}^{-1}$