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Find the rate at which water is draining from the tank after the following times.

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data

    If a tank holds 3500 gallons of water, which drains from the bottom of the tank in 50 minutes, then Toricelli's Law gives the volume V of water remaining in the tank after t minutes as the following.

    V = 3500 (1-t/50)^2 0 </=t<=50

    Find the rate at which water is draining from the tank after the following amounts of time.

    (a) 5 min

    (b) 10 min

    (c) 20 min


    2. Relevant equations
    Basic knowledge of derivatives. The quotient rule?


    3. The attempt at a solution

    I attempted to solve for V'(t).

    V = 3500((1-(t/50)2)

    V'(t) = 1(1-2t(2500)-t^2(1))/(2500)2

    I then plugged in 5 for t = 5 and definitely got the wrong answer. The answer should be negative, since the water in the tank is decreasing, and it should be greatest at the smaller times since there is more water in the tank to leave it.
     
    Last edited: Mar 1, 2009
  2. jcsd
  3. Mar 2, 2009 #2
    I factored out the equation so that V=3500(1-t/50)² becomes V=3500(1-2t/50+t²/2500)
    Then I derived that to get: V'(t)=3500(-2/50+2t/2500)
    Much easier than quotient rule, I think.
    Then sub in your values and get negative results as required.
     
  4. Mar 2, 2009 #3

    HallsofIvy

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    There is no need to use the quotient rule- there is no variable in the deominator: just the chain rule: V'= (3500)(2)(1- t/50)(-1/50).

    And, redargon, you didn't "factor out" anything, you multiplied what was already factored.
     
  5. Mar 3, 2009 #4
    ah, indeed. :shy:
     
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