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Math Help - Rate of change

  1. #1
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    Rate of change

    An inverted cone of base radius 4cm and height 8cm is initially filled with water. Water drips out from the vertex at a constant rate of 2pi cm^3/s. Find the rate of decrease in the depth of the water in the cone 16seconds after dripping starts.

    From what i understand:

    dv/dt = -2pi cm^3/s
    dh/dr = ?

    Vol.of cone = (1/3)(pi)(r^2)(h)
    dv/dh = 16pi/3

    dv/dt = dh/dt * dv/dh
    ....
    dh/dt = -2pi * (3/16pi) = -0.375cm/s
    16seconds after dripping starts = 16 * -0.375

    My ans was wrong, any idea what went wrong? Thanks in advance!
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  2. #2
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    Where did you get dv/dh? I get 4\pi\sqrt[3]{4} for 16 seconds in.

    Thank you for showing some of your work, but you didn't show this important piece.

    Did you notice that r = h/2 is a constant relationship?
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  3. #3
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    Re: Rate of change

    Using r = h/2,
    Vol.of cone = (1/3)pi(r^2)h = (1/3)pi[(h/2)^2]h = (pi/12)h^3
    Therefore, dv/dh = (pi/12)(3h^2) = (pi/4)h^2
    How do I bring in the '16seconds after dripping starts' from here on?
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  4. #4
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    Re: Rate of change

    Start with the volume formula. How much have you lost after 16 seconds? Solve for 'h' at that moment.
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  5. #5
    Newbie sugarT's Avatar
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    Re: Rate of change

    Rate of increase is constant therefore rate of decrease is constant so shouldn't they be the same?
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  6. #6
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    Re: Rate of change

    What rate of increase? Shouldn't what be the same as what?
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