Related Rates: Inverted Conical Tank Problem

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SUMMARY

The discussion centers on solving the inverted conical tank problem involving related rates. Water is leaking from the tank at a rate of 10,000 cm³/min while being pumped in at an unknown rate. The tank's dimensions are 6 meters in height and 4 meters in diameter. When the water level is at 2 meters and rising at 20 cm/min, the goal is to determine the rate at which water is being pumped into the tank using the volume formula V = (1/3)πr²h.

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Water is leaking out of an inverted conical tank at a rate of 10000cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank is 6m high and the diameter at the top is 4m. If the water level is rising at a rate of 20cm/min when the height of the water is 2m, find the rate at which the water is being pumped into the tank.

Its not the math I am having trouble with, its setting up the question, ca anyone help?

Thanks
 
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I used [tex]V=1/3 \\(pi) r^2h[/tex]..and then proceeded to find [tex]V= \\(pi) r^3[/tex]..I don't really know what to do now
 
Last edited:
What is v as a function of h?
dh/dt = 20
h = 200
dv/dt = ?

Then algebra will get you to the rate at which the water is being pumped in.
 

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