Mathematics

The key is connecting the volume formula for a cylinder to how fast the height changes. Since $V = \pi r^2 h$ and the radius stays constant at 2 m, you can differentiate both sides with respect to time to relate the rate water is being added ($\frac{dV}{dt}$) to the rate the height rises ($\frac{dh}{dt}$). This is a classic related rates problem where a constant input rate translates into a constant height-change rate.

The key is connecting the volume formula for a cylinder to how fast the height changes. Since $V = \pi r^2 h$ and the radius stays constant at 2 m, you can differentiate both sides with respect to time to relate the rate water is being added ($\frac{dV}{dt}$) to the rate the height rises ($\frac{dh}{dt}$). This is a classic related rates problem where a constant input rate translates into a constant height-change rate.