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Messed up somewhere in my integration

  1. Mar 6, 2006 #1
    [tex]\frac{dv}{dt}= -x^{-3}[/tex]
    when t=0, the particle is at rest with x=1
    Therefore by integrating i get
    [tex]v = \sqrt(x^-2 - 1)[/tex]

    [tex]\frac{dx}{dt}= \sqrt(\frac{1 - x^2}{x^2})[/tex]

    [tex]dx\frac{x}{(/sqrt(1 - x^2))} = dt[/tex]

    [tex]-\sqrt(1 - x^2) = t + C[/tex]

    [tex]C=-1[/tex]

    Therefore:-

    [tex]t = 1 - \sqrt(1 - x^2)[/tex]

    However i cant get the answer [tex]t = \sqrt(15)[/tex] when [tex]x = 1/4[/tex]. I think i messed up somewhere in my integration. I would really appreciate any help with this.

    Thank you,
    Bob
     
    Last edited: Mar 6, 2006
  2. jcsd
  3. Mar 6, 2006 #2
    Did you add the constants well after integration? If you show your works of integration it would be easy to check.
     
  4. Mar 6, 2006 #3
    [tex]\frac{dv}{dt}= -x^{-3}[/tex]
    [tex]\frac{1}{2}v^2 = \frac{x^-2}{2} + C[/tex]

    As at rest when x = 1, v = 0
    then C = -1

    [tex]v = \sqrt(x^-2 - 1)[/tex]

    Then i continue to integrate as above.
     
    Last edited: Mar 6, 2006
  5. Mar 6, 2006 #4
    You didn't multiply C by 2 in the third step.
     
  6. Mar 6, 2006 #5
    But isn't C just a constant. Which wouldnt matter if i multiplied it by 2 or not.
     
  7. Mar 6, 2006 #6
    Sorry it was a typo. I preesed enter by mistake.
    Check the sign at t = -(1 - x^2)^-2. It is correct till there.
     
  8. Mar 6, 2006 #7
    Thank you very much. My maths has been pretty bad over the last couple of days (you can tell by the number of posts).
     
  9. Mar 6, 2006 #8

    benorin

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    Homework Helper

    By thw way, you should use \sqrt{} rather than \sqrt()
     
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