calculusrate problem
posted by beth .
A conical tank (with vertex down) is 10 feet acros the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Let h be the depth.
Then the base radius (at the top) at depth h is 5*h/12. So volume of water is
V= 1/3 PI (5h/12)^2 h
take the derivative, set it equal to 10, solve for dh/dt when h=8
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