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Antiderivative of this?

  1. Jul 23, 2006 #1
    Antiderivative of this? - kinematics

    a kinematics question

    v = 1 + e^x

    find x in terms of t given that x = 0 when t = 0

    what i did:

    dx/dt = 1 + e^x
    dt/dx = 1/ (1+ e^x)

    so t = the antiderivative of 1/(1+ e^x)

    i tried and i kinda did this....

    loge(1 + e^x)/e^x + c


    but it doesnt seem right
     
    Last edited: Jul 23, 2006
  2. jcsd
  3. Jul 23, 2006 #2

    siddharth

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    Notice that if you had an e^x in the numerator of your integral, it will be easy.

    Try
    [tex] \int \frac{(1+e^x-e^x)}{1+e^x} dx [/tex]

    Can you take it from here?
     
  4. Jul 23, 2006 #3
    yeah that helped thanks... i got to t= x + loge(1+e^x)
    but how do i get x in terms of t?
     
  5. Jul 23, 2006 #4

    Hootenanny

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    HINT:

    [tex]\int \frac{f'(x)}{f(x)} dx = \ln\left| f(x) \right| + C[/tex]
     
    Last edited: Jul 23, 2006
  6. Jul 23, 2006 #5

    siddharth

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    That's wrong Hoot, I think you missed a f'(x).

    To get x in terms of t is painful. Write x as log(e^x), you'll get a quadratic in e^x, which you can solve.
     
  7. Jul 23, 2006 #6

    Hootenanny

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    Indeed I have, don't know what I was thinking to be honest. Thanks again for you contribution to the pH tutorial.
     
  8. Jul 23, 2006 #7
    thanks gusys
     
  9. Jul 23, 2006 #8

    siddharth

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    No problem, Hoot.

    meee, I think I found a couple of errors in your work. You missed a minus sign, and your constant of integration. It should be

    t= x - loge(1+e^x) + C

    Put the initial conditions (ie, x=0 when t=0) to find C.

    To get x in terms of t, write x as log(e^x) and play around for a while.
     
    Last edited: Jul 23, 2006
  10. Jul 24, 2006 #9
    oh right... thanks!
     
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