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Related Rates of a water leak

  1. Oct 24, 2004 #1
    Hi guys, I'm completely stuck here when doing related rates questions. Here is the question, and following the setup I have so far.

    Water is leaking out of an inverted conical tank
    at a rate of 10,00cm^3/min at the same time that water is
    being pumped into the tank at a constant rate. The tank has
    height 6m and diameter at top is 4m. IF the water level is
    rising at a rate of 20cm/min when the height of water is
    2m, find teh rate at which water is being pumped into the


    d = 4m
    h = 6m
    2m = 20cm/min.
    v = (1/3) (pi) r^2 h
    dl/dt = 10,000cm^3/min (rate at which water leaves cone)
    dp/dt = ? (rate at which water is being pumped into cone)

    thats really all I have. I don't know where to go after that. Any help is appreciated.
  2. jcsd
  3. Oct 24, 2004 #2


    User Avatar
    Science Advisor

    this belongs under "Calculus" not "Differential Equations".k

    The volume of a cone of radius r and height h is (1/3)πr2h. In this case entire tank has r= 3 and h= 4. Whatever the level of the water, the ratio of height to radius must be the same: r= (3/4)h so the volume is given by
    V= (1/3)π(9/16)h3= (3/16)πh3 (I've written r in terms of h because we are given the rate of change of the height, h).
    Now differentiate with respect to t: dV/dt= (9/16)πh2 (dh/dt).

    You are told that h= 2 and dh/dt= 20 cm/min= .2 m/min so dV/dt= (9/16)π(.04)(20)= 0.4 &pi cubic m /min= 400000 cubic cm/min. Since we are also told that "Water is leaking out of an inverted conical tank at a rate of 10,00cm^3/min", in order to account for the water leaking out AND the increase in volume, the water must be coming in at 400010 cubic cm./min.
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