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Differential Equation of a hemispherical bowl

  1. Feb 8, 2006 #1
    A hemispherical bowl has a radious of R (metres) and at a time t = 0 is full of water. At that moment a circular hole of radius a (centimetres) is opened in the bottom of the bowl. Let y be the vertical height of the water above the hole at time t. Then y is governed by the differential equation

    [tex]\pi(Ry-y^2)\frac{dy}{dt} = -\pi(a10^{-2})^2\sqrt{2gy},[/tex]​

    where [tex]g = 9.8m/s^2[/tex] is gravity.

    Solve this differential equation to find y as an implicit function of t.


    I need help with this. Im unsure of how to start. Do i try gather all the y terms on one side and everything else on the other?
     
  2. jcsd
  3. Feb 8, 2006 #2
    Do you notice the equation is separable, so that f(y)dy = Cdt for some constant C?
     
  4. Feb 8, 2006 #3
    Yes, can i say:

    [tex] \frac{(Ry-y^2)dy}{\sqrt{2gy}} = \frac{-\pi(a10^{-2})^2dt}{\pi} [/tex] ?
     
  5. Feb 8, 2006 #4
    Now you should be able to integrate both sides (don't forget the constant of integration afterwards).
     
  6. Feb 8, 2006 #5

    Hurkyl

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    I've merged the old thread into this one.
     
  7. Feb 13, 2006 #6
    im having difficulty integrating this. how do i get it into a form that is managable or atleast more recognisable??

    thanks
     
  8. Feb 13, 2006 #7
    You have
    (Ry-y^2)dy/sqrt(2gy) = Rydy/sqrt(2gy) - y^2dy/sqrt(2gy). Reduce the fractions, and you should be able to find the antiderivative of each term. Does that help?
     
  9. Feb 14, 2006 #8
    um kinda. Im not 100% sure on integrating them.

    would i get:

    [tex] \frac{2Ry^2}{\sqrt{2gh}} = \frac{3y^2}{\sqrt{2gh}} ?? [/tex]
     
  10. Feb 14, 2006 #9

    HallsofIvy

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    Where did "h" come from?? There was no h in anything you wrote before.

    Before, you had
    [tex]\frac{(Ry-y^2)dy}{\sqrt{2gy}} [/tex]
    [tex]= \frac{1}{\sqrt{2g}}\left(Ry^{\frac{1}{2}}- y^{\frac{3}{2}}\right)dy[/tex]
    That should be easy to integrate.
     
    Last edited: Feb 14, 2006
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