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Solvng a differential equation

  1. Apr 8, 2004 #1
    Problem:

    A cylindrical bucket of cross-sectional area A has water in it up to an initial depth of d at t=0. The water has density p, an the gravitational acceleration is g. The water leaks out the bucket through a hole in the bottom with the rate of change of he volume of the water in the bucket proportional to the pressure in the bottom of he bucket, dV/dt=-kP, with k postive constant. Find the volume of the water in the bucket as a function of time.
    I tried doing some of it but I'm not sure if I'm doing this right. Can someone help?
    dV/dt=-kP = dV/dt= -k(d-g-p) = dV/(d-g-p)=-k dt = integral(dv/(d-g-p))=-kt+C
     
  2. jcsd
  3. Apr 9, 2004 #2
    dV/dt is the derivative of V(t) function. So if:

    [tex]V'(t) = -kP[/tex]

    You need to integrate it to find the actualy function V(t):

    [tex]V(t) = -kPt + C[/tex]

    But what is C? You can see that the function V(t) gets the value of C at t = 0. What is the volume of the water in the bucket right before it starts to leak?
     
  4. Apr 10, 2004 #3
    How it can be true pressure is not constant
     
  5. Apr 10, 2004 #4
    Yes, it does depend on the volume.

    [tex]P = d\rho g = \frac{\rho g}{A}V[/tex]

    [tex]V'(t) = -kP = -\frac{K\rho g}{A}V = -cV[/tex]

    [tex]V(t) = V_0e^{-ct} = V_0e^{-\frac{K\rho g}{A}t}[/tex]

    Where V0 is d0A. Is that more like it?
     
  6. Apr 10, 2004 #5
    [tex] V=\pi r^2 h[/tex]
    [tex] P= \rho g h [/tex]
    So u can set up an equation from given condition that
    [tex] \frac{dh}{dt}= - \frac{k\rho g }{\pi r^2} h[/tex]
     
  7. Apr 10, 2004 #6
    They ask for the function V(t) though, not h(t).
     
  8. Apr 10, 2004 #7
    Oh Yes I believe thats right
     
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