2. Problem: A cube with sides $1 \mathrm{~m}$ in length is filled with water, and has a tiny hole through which the water drains into a cylinder of radius $1 \mathrm{~m}$. If the water level in the cube is falling at a rate of $1 \mathrm{~cm} / \mathrm{s}$, at what rate is the water level in the cylinder rising?
Solution: The magnitude of the change in volume per unit time of the two solids is the same. The change in volume per unit time of the cube is $1 \mathrm{~cm} \cdot \mathrm{m}^{2} / \mathrm{s}$. The change in volume per unit time of the cylinder is $\pi \cdot \frac{d h}{d t} \cdot m^{2}$, where $\frac{d h}{d t}$ is the rate at which the water level in the cylinder is rising.
Solving the equation $\pi \cdot \frac{d h}{d t} \cdot m^{2}=1 \mathrm{~cm} \cdot m^{2} / \mathrm{s}$ yields $\frac{1}{\pi} \mathrm{cm} / \mathrm{s}$.