A conical tank has a radius of 2 ft at the top and a height of 3 ft. If a liquid flows in a rate of 2 ft^3/min, how fast is the water level rising when it is 2 ft high?

0.358 ft/min

Step-by-step explanation:

Let's say r and h are the radius and height of the water, and R and H are the radius and height of the tank.

The volume of the water is:

V = π/3 r² h

Using similar triangles:

r / h = R / H

r = h R/H

Substitute:

V = π/3 (h R/H) ² h

V = π/3 (R/H) ² h³

Take derivative with respect to time:

dV/dt = π (R/H) ² h² dh/dt

Plug in values:

2 = π (2/3) ² (2) ² dh/dt

2 = 16π/9 dh/dt

1 = 8π/9 dh/dt

dh/dt = 9 / (8π)

dh/dt ≈ 0.358

The water is rising at approximately 0.358 ft/min.