Water flows into a conical tank at a rate of 2 ft³/min. If the radius of the top of the tank is 4 ft and the height is 6 ft, determine how quickly the water level is rising when the water is 2 ft deep in the tank.

Verified step by step guidance

1. Identify the relationship between the volume of the cone and the height of the water. The volume V of a cone is given by the formula V = (1/3)πr²h, where r is the radius and h is the height.
2. Since the tank is conical, the radius r of the water at height h can be expressed in terms of h using similar triangles. The ratio of the radius to the height of the tank is constant: r/h = 4/6, which simplifies to r = (2/3)h.
3. Substitute r = (2/3)h into the volume formula to express V in terms of h: V = (1/3)π((2/3)h)²h = (4/27)πh³.
4. Differentiate the volume V with respect to time t to find dV/dt: dV/dt = (4/27)π(3h²)(dh/dt). Simplify this to dV/dt = (4/9)πh²(dh/dt).
5. Set dV/dt equal to the rate at which water flows into the tank (2 ft³/min) and solve for dh/dt when h = 2 ft: 2 = (4/9)π(2²)(dh/dt).