A cup has the shape of a right circular cone. The height of the cup is 12 cm, and the radius of the opening is 3 cm. Water is poured into the cup at a constant rate of 32 cm / sec. What is the rate at which the water level is rising when the depth of water in the cup is 5cm?

about 6.52 cm/s

Step-by-step explanation:

The radius of the cone is 3/12 = 1/4 of its height, so the radius of the water surface at the time of interest is (5 cm) / 4 = 1.25 cm.

The area of the water's surface is ...

A = πr² = π (5/4 cm) ² = 25π/16 cm²

The rate of change of depth multiplied by this area will give the rate of change of volume.

dV/dt = (25π/16 cm²) (dh/dt)

dh/dt = (32 cm³/s) / (25π/16 cm²) = 512 / (25π) cm/s ≈ 6.52 cm/s

The water is rising at the rate of about 6.52 cm/s.