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Homework Help: Integration question

  1. Sep 30, 2005 #1
    I'm missing something simple here:

    [tex]\frac{d\mbox{v}}{d\mbox{t}} = g - k\mbox{v}^2[/tex]

    I'm integrating from v=0 to v=v_t (where v_t is a known constant independant of velocity and time) and from t=0 to t=t_f, where t_f is the variable I wish to solve for in the end.

    I'd rather not embarass myself by giving out the answer I produced. The actual answer has a hyperbolic tangent and I honestly am not sure how to get that. Could someone nudge me in the right direction?
     
  2. jcsd
  3. Sep 30, 2005 #2

    arildno

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    The inverse of the hyperbolic tangent, the Artanh function is given by:
    [tex]Artanh(x)=log(\frac{1+x}{1-x})[/tex]
     
  4. Sep 30, 2005 #3

    Galileo

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    You can seperate the variables and integrate:

    [tex]\int_{0}^{v(t
    )}\frac{dv}{g-kv^2}=\int_0^t t'dt'[/tex]
     
  5. Sep 30, 2005 #4
    Aha! And the derivative of that is:

    [tex]\frac{1}{1-x^2}[/tex] Which should take care of the problem. If only it had seemed more obvious before I had to ask! Thank you for your time arildno.
     
  6. Sep 30, 2005 #5
    Thank you too for helping Galileo. I'm afraid this seems rather obvious in retrospect, which I thought it might be.

    Have a good friday.
     
  7. Oct 1, 2005 #6

    arildno

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    I'm sorry, I made a mistake:
    We should have:
    [tex]Artanh(x)=\frac{1}{2}log(\frac{1+x}{1-x})[/tex]
    This can be seen by solving for "y" from the following equation:
    [tex]x=Tanh(y)\equiv\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}[/tex]
     
    Last edited: Oct 1, 2005
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