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Integration of a velocity function by partial fractions

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to integrate

    [tex] v(t) = V( \frac{1- e^{-2gt/V}}{1+ e^{-2gt/V}}) [/tex]

    to show that the position function is given by

    [tex] s(t) = Vt + \frac{V^2}{g}ln(\frac{1 + e^{-2gt/V}}{2}) [/tex]

    2. Relevant equations

    g is the acceleration due to gravity
    V is the terminal velocity

    3. The attempt at a solution

    I've tried a few different approaches. I've tried letting u = e^(2g/V) but I never get anywhere by doing this after finding du and substituting that in. My closest answer has been:

    [tex]

    -V(\frac{e^{-2gt/V} - 1}{e^{-2gt/V} + 1})

    [/tex]

    and then performing long division to get

    [tex]

    -V(1 - \frac{2}{1 + e^{-2gt/V}})

    [/tex]

    When I put that back into the integral I get

    [tex]

    -V\int dt + 2V\int\frac{dt}{1 + e^{-2gt/V}}

    [/tex]

    but I don't get the position function and have that negative at the start of the expression... can I get some hints on how to approach this?
     
  2. jcsd
  3. Sep 8, 2012 #2

    SammyS

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    It looks as if the substitution [itex]\displaystyle u=e^{-2g\,t/V}[/itex] should work just fine.

    Show how you are trying to implement that.
     
    Last edited: Sep 8, 2012
  4. Sep 9, 2012 #3

    HallsofIvy

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    I think I would be inclined to use the substitution [itex]u= 1+ e^{-2gt/V}[/itex] but it can be done either way.
     
  5. Sep 9, 2012 #4

    SammyS

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    I agree.

    This will make life easier.
     
  6. Sep 9, 2012 #5
    Okay, so I got the answer using Sammy's suggestion (though I did it again with HoI's and got it as well). Here is what I have:

    let [tex] u = e^{-2gt/V}. [/tex] Then [tex] du = -\frac{Vdu}{2g}\frac{1}{u} [/tex] and [tex] lnu = -\frac{2gt}{V}. [/tex]

    Substituting gives [tex] -\frac{V^2}{2g}\int\frac{1-u}{u(1+u)}du. [/tex]

    By partial fractions (on the integrand) I got

    [tex] -\frac{V^2}{2g}\int\frac{du}{u} + \frac{V^2}{2g}\int\frac{2du}{1+u} [/tex]

    which gives

    [tex] = -\frac{V^2}{2g}lnu + \frac{V^2}{g}ln(1+u) + C' [/tex]

    where C' is C - ln2. Subsituting for lnu and u gives

    [tex] = Vt + \frac{V^2}{g}ln(1+e^{-2gt/V}) - ln2 + C [/tex]

    which gives the position function

    [tex] = Vt - \frac{V^2}{g}ln( \frac{1 + e^{-2gt/V}}{2} ) + C. [/tex]

    Since s(0) = 0, we have C = 0 and the position function is obtained.
    Sorry if there are some errors with the Tex stuff.... still learning.
     
  7. Sep 9, 2012 #6

    SammyS

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    Good !

    Of course that should be [itex]\displaystyle dt = -\frac{Vdu}{2g}\frac{1}{u}\ . [/itex]
     
  8. Sep 9, 2012 #7
    Yes, sorry :-p. I get lost in all the tex stuff :-)

    Thank you for your help. Greatly appreciated!
     
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