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Antiderivative of this?

  • Thread starter meee
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  • #1
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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:

Answers and Replies

  • #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?
 
  • #3
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yeah that helped thanks... i got to t= x + loge(1+e^x)
but how do i get x in terms of t?
 
  • #4
Hootenanny
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HINT:

[tex]\int \frac{f'(x)}{f(x)} dx = \ln\left| f(x) \right| + C[/tex]
 
Last edited:
  • #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.
 
  • #6
Hootenanny
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siddharth said:
That's wrong Hoot, I think you missed a f'(x).
Indeed I have, don't know what I was thinking to be honest. Thanks again for you contribution to the pH tutorial.
 
  • #7
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thanks gusys
 
  • #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:
  • #9
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oh right... thanks!
 

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