Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 9 cubic feet per minute. if the pool has radius 3 feet and height 8 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 5 feet?

Question

Mathematics

Ayaan Randolph

28 August, 06:41

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 9 cubic feet per minute. if the pool has radius 3 feet and height 8 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 5 feet?

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Jaydin Hendricks · Answer 1

Since the cross section is uniform, the depth of the water is irrelevant. The rate of change of height is the rate of change of volume divided by the area.

dh/dt = (dV/dt) / (π·r²)

= (9 ft³/min) / (π· (3 ft) ²)

= 1/π ft/min ≈ 0.3813 ft/min