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water is leaking from an inverted conical tank at a rate of . water is also being pumped into the tank at a constant rate. the tank has height and the diameter at the top is . if the water level is rising at a rate of when the height of the water is , find the rate at which water is being pumped into the tank. hint: let denote the rate (in ) at which the tank is being filled, and let denote the volume (in ) of water in the tank. then rate in - rate out , so it suffic

Respuesta :

The rate of the water being pumped into the tank is 289,753 cm3/min

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

Learn more about volume here:

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