Water flows into a tank at a rate of $v=1.2$ liters per second. At what speed does the water level rise when the water has just reached half the height of the tank, if the tank is in the shape of a right circular cone, its axis is vertical, and it is placed with its vertex downward; its height is $2.4 \mathrm{~m}$, and the radius of the tank's opening is $1 \mathrm{~m} ?$
The speed of the rise is the time derivative of the height with respect to time. To determine this, we need to know how the height of the water in the container depends on time. In time $t$, $v t$ liters of water flow into the container. This amount of water fills the cone to a height of $m$. In this height, the radius of the cross-section is $\varrho$. Therefore, the volume of the cone of water is $\frac{\varrho^{2} \pi m}{3}$. The ratio of the radius to the height of the cone is constant. $\frac{\varrho}{m}=\frac{1}{2.4}$

$$
\text { Therefore } \frac{1.2 \cdot t}{10^{3}}=\frac{\varrho^{2} \pi m}{3}=\frac{m^{3} \pi}{2.4^{2} 3} \text {, }
$$

thus

$$
\frac{m^{3}}{3}=\frac{1.2 \cdot 2.4^{2} t}{10^{3} \pi}
$$

If we differentiate both sides with respect to $t$, then

$$
\frac{3 m^{2}}{3} \frac{d m}{d t}=\frac{1.2 \cdot 2.4^{2} \cdot 3}{10^{3} \pi}
$$

from the equation

$$
\frac{d m}{d t}=\frac{1.2 \cdot 2.4^{2}}{10^{3} \pi m^{2}}
$$

At half the height of the cone $\frac{2.4}{m}=\frac{2.4}{1.2}=2$

$$
\frac{d m}{d t}=\frac{1.2}{\pi \cdot 10^{3}} \cdot 4 \sim 1.528 \text{ mm} / \text{sec}
$$

Remark: It can be seen that it was unnecessary to specify the height of the cone. The speed of the rise depends only on the area of the water surface at half the height.