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Help with Differential equation

  1. Feb 2, 2013 #1
    A hemispherical bowl of radius a has its axis vertical and is full of water. At time t=0 water starts running out of a small hole in the bottom of the bowl so that the depth of water in the bowl at t is x. The rate at which the volume of water is decreasing is proportional to x. Given that the volume of water in the bowl when the depth is x is [tex]\pi(ax^2-\frac{1}{3}x^3)[/tex] show that there is a positive constant k such that

    [tex]\pi(2ax-x^2)\dfrac{dx}{dt}=-kx[/tex]
    My method
    [tex]-\dfrac{dV}{dt}\propto x[/tex]
    [tex]\dfrac{dV}{dt}=-kx[/tex]
    [tex]\dfrac{dV}{dt}=\dfrac{dV}{dx} \dfrac{dx}{dt}[/tex]
    [tex]V=\pi(ax^2-\frac{1}{3}x^3)[/tex]
    [tex]\dfrac{dV}{dx}=\pi(2ax-x^2)[/tex]
    [tex]\pi(2ax-x^2)\dfrac{dx}{dt}=-kx[/tex]

    (ii) Given that the bowl is empty after a time T, show that

    [tex]k=\dfrac{3 \pi a^2}{2T}[/tex]

    My method (I'm not sure how to answer this part).

    [tex]\pi(2ax-x^2)\dfrac{dx}{dt}=-kx[/tex]
    [tex]\displaystyle \int (2\pi a-\pi x )dx=\displaystyle \int-k dt[/tex]
    [tex]2\pi ax-\frac{\pi x^2}{2}=-kt + C[/tex]

    x=0 when t=T

    I'm not sure how to go about showing [tex]k=\dfrac{3 \pi a^2}{2T}[/tex]with the information in the question.

    Can you help me?

    Thanks.
     
    Last edited: Feb 2, 2013
  2. jcsd
  3. Feb 2, 2013 #2

    Dick

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    There's a mistake in your integration in part (ii). The integral of 2*pi*a*dx is not 2*pi*a. Fix that first. Then use that at t=0, x=a (the hemisphere was initially full) to find the value of C. Once you've found C put x=0 and solve for t.
     
    Last edited by a moderator: Feb 2, 2013
  4. Feb 2, 2013 #3
    I got [tex]C=2a^2(\pi-\frac{1}{4})[/tex]

    So

    [tex]2 \pi ax-\frac{\pi x^2}{2}=-kt+2a^2(\pi -\frac{1}{4})[/tex]

    when x=0 t=T

    [tex]kT=2a^2(\pi-\frac{1}{4})[/tex]

    [tex]k=\dfrac{2a^2(\pi-\frac{1}{4})}{T}[/tex]

    [tex]k=\dfrac{2a^2 \pi-\frac{1}{2}a^2}{T}[/tex]

    [tex]k=\dfrac{4a^2 \pi-a^2}{2T}[/tex]

    [tex]k=\dfrac{a^2(4 \pi -1)}{2T}[/tex]

    This doesn't equal [tex]k=\dfrac{3\pi a^2}{2T}[/tex]

    Where have I gone wrong?
     
    Last edited: Feb 2, 2013
  5. Feb 2, 2013 #4

    Dick

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    If I put x=a into ##2 \pi ax-\frac{\pi x^2}{2}##, which is what you did I hope, I don't get ##2a^2(\pi-\frac{1}{4})##.
     
  6. Feb 2, 2013 #5
    I simplified it wrong then.

    I'll have another look.
     
  7. Feb 2, 2013 #6
    Oh my god. This is incredibly simple. I over complicated it I guess. So I have

    [tex]C=\dfrac{3\pi a^2}{2}[/tex]

    [tex]2\pi ax- \dfrac{\pi x^2}{2}=-kx+\dfrac{3 \pi a^2}{2}[/tex]

    x=0 when t=T

    [tex]kT=\dfrac{3 \pi a^2}{2}[/tex]

    [tex]k=\dfrac{3 \pi a^2}{2T}[/tex]

    Got it.
     
  8. Feb 2, 2013 #7

    Dick

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    Yeah, you had the right idea all along. Some simple errors were tripping you up.
     
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