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Sand is being dropped at the rate of 10 cubic meter per minute onto a conical pile. If the height of the pile is always twice the base radius, at what rate is the height increasing when the pile is 8 meters high?

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    Volume = V = (1/3)π r^2 h
    but h = 2r ---> r = h/2

    V = (1/3)π (h^2/4)(h)
    = (1/12)π h^3

    dV/dt = (1/12)π h^2 dh/dt
    10 = 2π (64) dr/dt
    dr/dt = 10/(128π) m/min
    = 5π/64 m/min

    check my arithmetic

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    Thank you for answering my question. I just can't understand how it come up to the equation 10=2Π(64)dr/dt. I will be very thankful if you elaborate. :)

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    I can see why you are puzzled, since I have two typo errors.

    from V = (1/12) π h^3, I should have had ...

    dV/dt = (1/4)π h^2 dh/dt
    now subbing in our given...
    10 = (1/4)π(64) dh/dt
    10 =16π dh/dt
    dh/dt =10/(16π) = 5/(8π) m/min


    I apologize for those blatant errors.

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    I'm very thankful for being kind to me. Last one thing, 8 meters is the height of the conical pile. Does it also mean that it is the same as the height of the sand?

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    Since we assume that the pile of sand forms a cone,
    the height of the cone is equal to height of the sand pile

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    thank you so much. :) you really help me a lot. Godbless!

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    hii... am in the solution with the V = (1/3)π (h^2/4)(h) still have h in 2/4??

    isn't the derivative of r = h/2 is just 2/4??

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