A water tank has the shape of an inverted circular cone with base radius 2m and height 4m. If water is being pumped into the tank at a rate of 2 m3 / min, find the rate at which the water level is rising when the water is 3 m deep.

The volume of the tank is a circular cone, therefore it is given by:

V = πr²h/3

V = volume, r = base radius, h = height or water level

Differentiate both sides with respect to time t. The base radius r doesn't change over time so treat it as a constant:

dV/dt = πr² (dh/dt) / 3

Given values:

dV/dt = 2m³/min

r = 2m

Plug in and solve for dh/dt:

2 = π (2) ² (dh/dt) / 3

dh/dt = 0.48m/min

The water level is increasing at a rate of 0.48m/min