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A truck is dumping sand into a conical pile at the rate of 30 cubic feet per second, and in such a way that the height of the pile is always equal to twice the radius. At what rate is the height increasing when the sand pile has a volume of 300 cubic feet?
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30ft^3/sec = dV/dt h=2r V=1/3 pi r^2 h Use the volume formula to derive dV/dr and dV/dh at V = 300ft^3 300 = (1/3) pi r^2 h = (1/3) pi r^2 * 2r (450/pi)^(1/3)= r 2*(450/pi)^(1/3)=h \[\frac{dV}{dt} = \frac{dV}{dr} * \frac{dr}{dt} + \frac{dv}{dh} * \frac{dh}{dt}\] \[30 = \frac{1}{3}\pi (2r) * h*\frac{dr}{dt}+\frac{1}{3}\pi*r^2*\frac{dh}{dt}\] from here you can use h=2r to solve for dh/dt by taking the time derivative. dh/dt = 2dr/dt
Thanks. I got it.
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