Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into the tank at a constant rate.

Question

Water is leaking out of an inverted conical tank at a rate of 10,500 cm³/min at the same time that water is being pumped into the tank at a constant rate. The tank has a height of 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

Given:

d = 4m

∴ r = 2m

h = 6m

Therefore r = h/3

The formula for the volume of the cone is:

V=(1/3)π•r²h

substitute:

V=(1/3)π (h/3)²h

V=(1/27)π h³

Derive both sides with respect to time:

dV/dt = (1/9) π h²• dh/dt

When h= 2 meters = 200 cm when dh/dt = 20

dV/dt = (1/9) π (200)²• (20)

dV/dt = (800,000/9) π ~ 279,252.68

*Meaning the volume of water is increasing at a rate of 279,252.68 cm³/min.

However, it's leaking out at a rate of 10,500 cm³/min. Therefore the rate of the water being pumped into the tank is:

279,252.68 + 10,500 = 289,752.68 ~ 289,753 cm³/min.