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Calculate change in height of fluid in a cylinder based on flow out

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data

    A cylinder with radius five has water in it. The water flows out of the cylinder with a rate of 5π units cubed per minute. At what rate does the height of the fluid in the cylinder change?

    2. Relevant equations

    volume of a cylinder = [itex]πr^{2}h[/itex]

    3. The attempt at a solution

    I know the height of the cylinder is constant. I think using [itex]\int f(t) \mathrm{d} t = \frac{5}{2}πt^{2} + C[/itex] would help me, where [itex]f(t) = 5πt[/itex], or the amount of water released in t minutes.

    Thanks for any help.
     
  2. jcsd
  3. Dec 14, 2011 #2
    So you already have that the volume of the water is
    [tex] V = \pi r^2h [/tex]
    And you're given that the rate of loss of water is [itex]\displaystyle 5 units^3min^{-1}[/itex] so you have
    [tex] \frac{dV}{dt} = 5 [/tex]
    Now what is [itex]\displaystyle \frac{dV}{dt} [/itex] ?
     
  4. Dec 15, 2011 #3
    So that tells me that the change in volume with respect to time is 5π, no? For example, after t minutes, [itex]5πt[/itex] water would have been released.
     
  5. Dec 15, 2011 #4

    HallsofIvy

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    "I know the height of the cylinder is constant." But the height of the column of water is not. In fact, it is dh/dt that you are asked to find. The radius is constant. Yes, [itex]V=\pi r^2h[/itex]. So what is dV/dt as a function of dh/dt?
     
  6. Dec 15, 2011 #5
    Forgive me if I'm completely wrong, been studying for finals all day and night and it's late here.

    From what I've calculated, [itex]\frac{dV}{dt} = 25π\frac{dh}{dt}[/itex].

    Thanks for the help.
     
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