Question

Water is leaking out of an inverted conical tank at a rate of 11,500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate (in cm3/min) at which water is being pumped into the tank. (Round your answer to the nearest integer.)

Answer 1

290,753 cm³/min

Explanation:

You want the rate at which water is being pumped into a 6 m high 4 m diameter inverted conical tank if water is leaking out at 11,500 cm³/min and the level is rising at 20 cm/min.

Volume change

The dimensions of the tank tell us the radius is (2 m)/(6 m) = 1/3 of the height of the tank. In terms of height, the volume of the filled portion of the tank is then ...

V = 1/3π(r²)h

V = (π/3)(h/3)²h = (π/27)h³

The rate of change of volume is then ...

V' = (π/9)h²·h'

V' = (π/9)(200 cm)²·(20 cm/min) ≈ 279,253 cm³/min

Flow

The input flow rate must increase the volume at this rate, and accommodate the leakage from the tank. Therefore, it must be ...

279,253 +11,500 = 290,753 . . . . . cm³/min

Water is pumped in at 290,753 cm³/min.

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