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How do I integrate this?

  1. Mar 31, 2016 #1
    1. The problem statement, all variables and given/known data
    dv/dt = -βv
    Integrate to find velocity as a function of time, assume the particle's initial velocity is v0.
    β is constant
    v = velocity (not constant)
    t = time

    2. Relevant equations


    3. The attempt at a solution
    dv = -βvdt
    ∫dv = -β∫vdt --------> limits of integration for the right side are from 0 to t
    v(t) = -βt + v0
    I know i'm probably supposed to separate the variables so that v is with dv and β is with t, but this way seems to work too. Both ways seems correct, but there can only be one right answer... which one is it? Why does my attempt work out mathematically but is still wrong?
     
  2. jcsd
  3. Mar 31, 2016 #2
    Using the supposedly correct separation of variables I get:
    v(t) = e ^ (ln(v0) - βt)
    Is this okay?
     
  4. Mar 31, 2016 #3

    SteamKing

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    Why don't you show your work to obtain this solution.
     
  5. Mar 31, 2016 #4
    dv/dt = -βv

    dv/v = -βdt
    ∫dv/v = -β∫dt-----> left integral from v0 to v, right integral from 0 to t
    ln(v) from v0 to v = -βt
    ln(v) - ln(v0) = -βt
    ln(v) = ln(v0) - βt
    v(t) = e ^ (ln(v0) - βt)----------> "e" both sides
     
  6. Mar 31, 2016 #5
    The better way to do it is to use the ln(A)-ln(B)=ln(A/B) rule, so in the step where you have ln(v)-ln(v0), make it ln(v/v0). Then you can raise both sides to e and then get the following:

    e^(ln(v/v0))= e^(-βt)

    v/v0 = e^-βt

    v(t)=(v0) e^-βt

    This is a much easier form to have it in, and it really helps to visually model the relationship between v and time. I hope this helps=)
     
  7. Apr 1, 2016 #6
    Thanks, it looks a lot like one of those exponential growth/decay relationships - which says a lot about how the object behaves. Do you have any thoughts on my original attempt at the solution though? The math works out, but it ends up being the wrong answer. Here's what I did:
    dv = -βvdt
    ∫dv = -β∫vdt --------> limits of integration for the right side are from 0 to t
    v(t) = -βt + v0
     
  8. Apr 1, 2016 #7
    In any differential equation, you have to separate the variables, and putting v on the side of dt instead of with the dv shouldn't come up with the right answer. Besides, if you do the integration out, it would actually be:
    -β∫vdt from 0 to t---> -β(vt). Adding the right side, you get--> -βvt + v0.
    So that method would not get you the answer of -βt + v0 anyways. Is -βt + v0 the right answer? I hope this helps=)
     
  9. Apr 1, 2016 #8

    LCKurtz

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    The reason you can't do the problem this way is that since ##v(t)## is unknown, you can't calculate ##\int v(t)~dt##
     
  10. Apr 1, 2016 #9

    vela

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    This isn't correct either. As the OP noted, you shouldn't get inconsistent answers, and your result clearly isn't the same as the exponential solution.
     
  11. Apr 1, 2016 #10

    Ray Vickson

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    Your attempt replaces a differential equation by an integral equation; it still does not solve the problem. Written out properly (with all integration limits included), your method yields
    [tex] v(t) - v_0 = - \beta \int_0^t v(\tau) \, d\tau [/tex]
    You cannot easily figure out ##\{v(t)\}## from this!
     
  12. Apr 1, 2016 #11
    I'm thinking v(t) as a velocity function, and you're integrating that over dt, so

    ∫dv = -β∫vdt right side from 0 ---> t
    v - v0 = -β∫(dx/dt)dt right side from 0 ---> t
    v - v0 = -β∫dx right side from 0 ---> t
    v(t) = -βt + v0 ?
     
  13. Apr 1, 2016 #12

    pasmith

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    If you integrate by substitution, you have to change the limits. Setting [itex]v = dx/dt[/itex] means that [tex]
    \int_0^t v\,dt = \int_{x(0)}^{x(t)} \,dx.[/tex] You end up with [tex]
    v(t) - v_0 = - \beta(x(t) - x(0))[/tex] which of course doesn't help because you don't know what [itex]x(t)[/itex] is.
     
  14. Apr 1, 2016 #13
    Oh I see, integrating the velocity function over time gives us the position function - which in this case doesn't matter.

    On a separate note,
    v(t)−v0=−β(x(t)−x(0))
    is also equal
    Δv = -βΔx
    right?
     
  15. Apr 1, 2016 #14

    SammyS

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    Yes, OP can use ln(v) - ln(v0) = ln(v/v0) as you suggested.

    However, it's possible to simply take OP's result and use rules of exponents.

    ##\ v(t) = e^{\,\ln(v_0) - \beta t}\ ##
    ##=e^{\,\ln(v_0)}\cdot e^{-\beta t}\ ##
    ##=v_0\cdot e^{-\beta t}\ ##​
    .
     
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