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The question:
Gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 19 feet high? Recall that the volume of a right circular cone with height h and base radius r is given by V=(1/3)(pi)(r^2)(h)
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I've attempted this problem four times and I'm really just confusing myself at this point. Can anyone tell me what I'm doing wrong?
Calculations so far:
diameter = height
radius = height/2
V = (1/3)pi(h/2)^2(h)
V = (1/3)pi((h^2)/4)(h)
V = (1/3)pi((h^3)/4)
V = (pi*h^3)/12
Implicit differentiation:
dV/dt = (3pi/12)h^2(dh/dt) = (pi/4)h²(dh/dt)
dh/dt = (dV/dt)/(pi/4)h^2
dV/dt = 40 cubic feet per minute
h = 19 feet
dV/dt = (40)/(pi/4)*(19^2)
dV/dt = 40/(361*pi)/4
dV/dt = 40/22.5625
but that answer's apparently incorrect.
What do I need to do?