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Integration help

  1. Oct 31, 2006 #1
    I am having difficulty in integrating this one. My objective is to find height as a function of time; however each time I actually do the integration, I end up with being able to find height as a function of both time and height. I have attempted using Maple to find a unique solution but it produces a (Lambar delta) (possible spelling error). Here is my integration.

    int(t)=int([8(mu)(h)]/[2(sigma)(g)(r)-(rho)(g)(h)(r^2)] dh

    I keep getting non real answers when I plug in values. Sigma is a surface tension but it keeps getting caught into natural logs with a negative value(impossible)

    Any help would be quite appreciated.
     
  2. jcsd
  3. Oct 31, 2006 #2

    arildno

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    What are you trying to do here??
    Is it to find the anti-derivative:
    [tex]I=\int\frac{8\mu{h}}{2\sigma{gr}-\rho{gh}r^{2}}dh[/tex]
    If so, what are constants, and what is variable quantities with respect to which variables?
     
  4. Oct 31, 2006 #3
    the only variables are time and height. The others are constants.
     
  5. Oct 31, 2006 #4

    arildno

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    What is r?
     
  6. Oct 31, 2006 #5
    a non changing radius.
    I will define all variables
    t=time
    h=height
    r=radius
    mu= fluid viscocity
    sigma=surface tension
    g=gravitational force
    rho= density
     
  7. Oct 31, 2006 #6

    arildno

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    Well, and how does t enter the picture.

    Whatever do you mean by int(t)??
     
  8. Oct 31, 2006 #7
    the integral of t from t=0 to t so basically the left hand side of the equation reduces to t.
     
  9. Oct 31, 2006 #8

    arildno

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    The expression [itex]\int_{0}^{t}tdt[/itex]????
    First of all, your notation is a meaningless conjunction of integration variable and limit of integration, nor would it in this abuse of notation equal t.

    Are you sure your left-hand side shouldn't be [itex]\int_{0}^{t}d\tau=t[/itex]?
     
  10. Oct 31, 2006 #9
    right I forgot to say with what respect I was integrating the left hand side, but I am sure that the left hand side reduces to t.

    I agree with the first equation that you posted except the left hand side should be t instead of I. the limits of integration from the right side are h= 0 to h=h(final)
     
  11. Oct 31, 2006 #10

    arildno

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    Okay, so what your ACTUAL equation should be is:
    [tex]t=\int_{0}^{h(t)}\frac{8\mu{H}}{2\sigma{gr}-\rho{gH}r^{2}}dH[/tex]

    That, at the very least, is a mathematically meaningful expression.
    Is that what you meant?
     
  12. Oct 31, 2006 #11
    yes. I'm sorry, but I don't know how to use the equation editor
     
  13. Oct 31, 2006 #12

    arildno

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    Now, you may rewrite your right-hand side as:
    [tex]t=K\int_{0}^{h(t)}\frac{h}{1-CH}dH[/tex]
    for suitable constants K, C.
    Can you perform that integration?
     
  14. Oct 31, 2006 #13
    I'm sorry I don't guess I follow you. That isn't something that I have seen in my undergraduate mathematics.
     
  15. Oct 31, 2006 #14

    arildno

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    Sorry, it should have been:

    [tex]t=K\int_{0}^{h(t)}\frac{H}{1-CH}dH[/tex]
    for suitable constants K, C.

    That's just algebra.
     
  16. Oct 31, 2006 #15
    okay, now I see. Thanks. I can do that integration quite easily.
     
  17. Oct 31, 2006 #16

    arildno

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    You'll get an implicit expression for h(t).
     
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