Water is poured at a rate of 0.1 $\text{m}^3$ per minute into a container in the form of an open cone. The semi-vertical angle of the cone is $30^\circ$. At time $t$ minutes after the start, the radius of the water surface is $r$ m (see diagram). Find the rate of increase of the depth of water when the volume of the water is 3 $\text{m}^3$. [The volume of a cone of base radius $r$ and height $h$ is given by $V = \frac{1}{3}\pi r^2 h$]

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 $h^2 = \frac{9}{\pi^{2/3}}$):

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