Question

Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min at the same time water is being pumped into the tank at a constant rate If the tank has a height of 6m and the diameter at the top is 4 m and if the water level is rising at a rate of 20 cm/min when the height of the water is 2m, how do you find the rate at which the water is being pumped into the tank?

Answer 1

Let `V` be the volume of water in the tank, in `cm^3`; let `h` be the depth/height of the water, in cm; and let `r` be the radius of the surface of the water (on top), in cm. Since the tank is an inverted cone, so is the mass of water. Since the tank has a height of 6 m and a radius at the top of 2 m, similar triangles implies that `\frac{h}{r}=\frac{6}{2}=3` so that `h=3r`.

The volume of the inverted cone of water is then `V=\frac{1}{3}\pi r^{2}h=\pi r^{3}`.

Now differentiate both sides with respect to time `t` (in minutes) to get `\frac{dV}{dt}=3\pi r^{2}\cdot \frac{dr}{dt}` (the Chain Rule is used in this step).

If `V_{i}` is the volume of water that has been pumped in, then `\frac{dV}{dt}=\frac{dV_{i}}{dt}-10000=3\pi\cdot (\frac{200}{3})^{2}\cdot 20` (when the height/depth of water is 2 meters, the radius of the water is `\frac{200}{3}` cm).

Therefore `\frac{dV_{i}}{dt}=\frac{800000\pi}{3}+10000\approx 847758\ \frac{\mbox{cm}^3}{min}`.