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Rate water has to be added to Leaking cone

  1. Jul 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Water is leaking out of an inverted conical tank at a rate of 6500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.

    2. Relevant equations

    V=1/3 pi(r)^2*h

    relationships:

    4/2r=6/h
    therefore,
    r=1/3h


    Substituting h for r:
    V=1/3 pi (1/3h)^2(h)
    V=pi/27(h)^3
    V'=dv/dt= pi/27 (3h)^2dh/dt
    dv/dt=pi/27 (3*200)^2(20)

    since the cone is leaking we have to find:

    dv/dt-6500=pi/27(3*200)^2(20)
     
  2. jcsd
  3. Jul 13, 2010 #2
    Others may want to confirm, but this looks good to me.
     
  4. Jul 14, 2010 #3
    That answer is incorrect..that is why I posted it...I can't find the correct answer..I was just showing my work so someone could tell me where I went wrong...


    Does anyone know??

    .....


    I got it..I shouldn't have the parenthesis around the 3*200, since only the 200 is squared...
     
    Last edited: Jul 14, 2010
  5. Jul 14, 2010 #4
    Yes, that's right. Sorry I missed the error before. A tip for next time: if you know your answer is wrong, let us know and we (or at least I) will look more carefully for an error.
     
  6. Jul 14, 2010 #5
    Thank you..and I will do that.
     
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