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Homework Help: Related rates question conical reservoir

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data
    A conical reservoir is 50m deep and 200m across the top. Water is being pumped in at a rate of 7000m^3/min, and at the same time, water is being drained out at a rate of 9000 m^3/min. What is the change in diameter when the height of the water in the reservoir is 20m?


    2. Relevant equations
    v=(1/3)pi(r^2)(h)


    3. The attempt at a solution
    h=(.5r)
    v=(1/3)(pi)(r^2)(.5r)
    =(1/6)(pi)(r^3)
    v'=(.5)(pi)(r^2)(r')

    7000=.5pi(40^2)(r')
    = 2.785
    -9000= .5pi(40^2)(r')
    =-3.581

    2.785-3.581 = -0.796 m/min

    my answer, however, is supposed to be -1.6m/min
     
  2. jcsd
  3. Mar 8, 2010 #2

    cronxeh

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    Gold Member

    What is (40^2) ?
     
  4. Mar 8, 2010 #3
    I got the 40 from h=.5r and r=20
     
  5. Mar 8, 2010 #4

    cronxeh

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    Gold Member

    Well then, very well I suppose.

    Your question is to find change in diameter, hence multiply it by 2 and you get your -1.592 ~ -1.6
     
  6. Mar 8, 2010 #5
    AHHH silly mistake. thankyou!!
     
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