水以每分钟 0.1 $\text{m}^3$ 的速率注入一个敞口圆锥形容器中。圆锥的半顶角为 $30^\circ$。在开始后 $t$ 分钟时，水面的半径为 $r$ m（见图）。求当水的体积为 3 $\text{m}^3$ 时，水深增加的速率。 [底面半径为 $r$ 且高为 $h$ 的圆锥体积由 $V = \frac{1}{3}\pi r^2 h$ 给出]

1. Relate the surface radius $r$ and the depth of water $h$ using the semi-vertical angle.

$\tan(30^\circ) = \frac{r}{h} \implies r = h \tan(30^\circ) = \frac{h}{\sqrt{3}}$

2. Express the volume of the water $V$ entirely in terms of its depth $h$.

$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{\sqrt{3}}\right)^2 h = \frac{1}{3}\pi \left(\frac{h^2}{3}\right) h = \frac{1}{9}\pi h^3$

3. Differentiate $V$ with respect to $h$.

$\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{1}{9}\pi (3h^2) = \frac{1}{3}\pi h^2$

4. Find the depth $h$ at the instant when the volume of water is $3\text{ m}^3$.

$3 = \frac{1}{9}\pi h^3 \implies h^3 = \frac{27}{\pi} \implies h = \frac{3}{\pi^{1/3}}$

5. Apply the chain rule to find the rate of increase of the depth, $\frac{\mathrm{d}h}{\mathrm{d}t}$.

$\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\frac{\mathrm{d}V}{\mathrm{d}t}}{\frac{\mathrm{d}V}{\mathrm{d}h}} = \frac{0.1}{\frac{1}{3}\pi h^2}$

Substituting $h = \frac{3}{\pi^{1/3}}$ (or $h^2 = \frac{9}{\pi^{2/3}}$):

$\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{0.1}{\frac{1}{3}\pi \left(\frac{9}{\pi^{2/3}}\right)} = \frac{0.1}{3\pi^{1/3}} = \frac{1}{30\pi^{1/3}} \approx 0.0228\text{ m/min}$