Rate of rise of water level in a cylindrical tank

The volume of water in a cylindrical tank of radius $r$ and water depth $h$ is $V = \pi r^2 h.$ Since the radius $r = 2$ m is constant for this tank, differentiating both sides with respect to $t$ using the chain rule gives $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt},$ where $r^2$ stays fixed because $r$ does not change over time.

We are given $\dfrac{dV}{dt} = 4$ m$^3$/min and $r = 2$ m, so $r^2 = 4$. Substituting into $\dfrac{dV}{dt} = \pi r^2 \dfrac{dh}{dt}$ gives $4 = \pi (4) \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{4}{4\pi} = \frac{1}{\pi}.$ So the water level rises at a rate of $\dfrac{1}{\pi}$ m/min, approximately $0.318$ m/min.