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Integration of a velocity function by partial fractions

  • Thread starter dustbin
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  • #1
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Homework Statement



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]

Homework Equations



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

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?
 

Answers and Replies

  • #2
SammyS
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Homework Statement



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]

Homework Equations



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

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.

...
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:
  • #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.
 
  • #4
SammyS
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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.
I agree.

This will make life easier.
 
  • #5
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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.
 
  • #6
SammyS
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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
Good !

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

Thank you for your help. Greatly appreciated!
 

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