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Particle's position as a function of time - Integration

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle falling under gravity is subjet to a retarding force proportional to its velocity. Find its position as a function of time, if it starts from rest, and show that it will eventually reach a terminal velocity.

    2. The attempt at a solution

    To save myself some latex time, I skipped a couple of the intermediary steps here:

    [tex]F_{tot}=F_{g}-F_{retarding}[/tex]

    [tex]m\ddot{x}=-mg-\lambda\dot{x}[/tex]

    [tex]\ddot{x}=-g-\frac{\lambda}{m}\dot{x}[/tex]

    let

    [tex]\gamma=\frac{\lambda}{m}[/tex]

    then

    [tex]\frac{d}{dt}dx=-gdt-\gamma dx[/tex]

    integrating both sides

    [tex]\frac{d}{dt}\int_{o}^{x}dx=-g\int_{0}^{t}dt-\gamma\int_{0}^{x}dx[/tex]

    [tex]\dot{x}+\gamma x=-gt[/tex]

    This final equation I get is correct (hopefully my procedure is too) and at this point, the question offers a hint:

    [HINT: After the first integration, use an integrating factor, i.e. a function f(t) such when the equation is multiplied by f(t), the left-hand side becomes an exact derivative, in fact, the derivative of xf. The final stage requires an integration by parts]

    This is where I got stuck. How do I determine the integrating factor f(t)? And what exactly is meant by "the left-hand side becomes an exact derivative, in fact, the derivative of xf"?

    Your help is appreciated, thanks!
    phyz
     
  2. jcsd
  3. Mar 10, 2009 #2

    tiny-tim

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    Hi phyz! :smile:

    (have a lambda: λ :wink:)

    Wouldn't it be easier to put x' = v, so mv' = -mg - λv? :wink:

    An exact derivative means something like vcos(t) + v'sin(t) …

    it's exactly the derivative of vsin(t) :smile:
     
  4. Mar 10, 2009 #3
    Ho Mr Tim! :smile:

    I'd love to say that I see where you're going, but that would be a blatant lie! :tongue:

    So please bear with me...

    How would this be easier necessarily? I'd still end up with

    [tex]
    \dot{x}+\gamma x=-gt
    [/tex]

    wouldn't I? :confused:

    Oh, OK...I get this bit...But how? Grrrr!!! I hate feeling so daft :cry:
     
  5. Mar 10, 2009 #4
    So if x is a function of t and we know f is a function of t then the derivative of xf will be

    [tex]\frac{d}{dt}xf(t)=\dot{x}f(t)+xf'(t)[/tex]

    Right?...but that doesn't bring me any closer to determining f(t)...

    *banging head on desk*
     
  6. Mar 10, 2009 #5
    Hold on...methinks the grey matter is kicking in (albeit slowly)...if what I have above is right, then

    [tex]f'(t)=\gamma[/tex]

    so

    [tex]f(t)=\gamma t[/tex]

    ?????

    ...hang in there, I need to get to pen and paper! :biggrin:
     
  7. Mar 10, 2009 #6
    *sigh*

    Nope...screwed up...

    *back to banging head on desk*
     
  8. Mar 10, 2009 #7

    tiny-tim

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    It's easier because your equation had x' x and t, but mine only has v' and v, and can be integrated immediately to give v as a function of t. :wink:
    You need to find a function f(t) to multiply x' + γx,

    so that f(t)x' + γf(t)x is an exact derivative, of the form g(t)x' + g'(t)x …

    so the equation relating f f' g and g' is … ? :smile:
     
  9. Mar 10, 2009 #8
    I think I may actually have it now...if we let

    [tex]f(t)=e^{\gamma t}[/tex]

    and we multiply it with

    [tex]\dot{x}+\gamma x=-gt[/tex]

    throughout, then we get

    [tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-gt[/tex]

    and

    [tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}[/tex]

    is an exact derivative of

    [tex]x e^{\gamma t}[/tex]

    How's that? Or am I still missing the mark?

    Thanks for your help Tiny-Tim! It really is much appreciated! :smile:
     
    Last edited: Mar 11, 2009
  10. Mar 10, 2009 #9

    tiny-tim

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    Yes, that's exactly right! :biggrin:

    (btw, the general method would be g'/g = γ, so ln(g) = ∫γdt + C, or g = Ae∫γdt

    in this case, γ is constant, so it's just Aeγt)
     
  11. Mar 11, 2009 #10
    Can you believe that I actually understand what you're saying! :biggrin:

    OK, so progress has been made...However, it seems as if last night I used up my allowed number of "Eureka" moments for the week and haven't been able to complete the solution :frown:

    I know that the answer to this problem is

    [tex]x=(\frac{g}{\gamma^2})(1-e^{-\gamma t})-\frac{gt}{\gamma}[/tex]

    But can't seem to make the connection required to move there from

    [tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-gt[/tex]

    At first I thought that, since we now have an exact derivative, we could just integrate both sides with respect to t as follows

    [tex]\int(\dot{x}e^{\gamma t}+x\gamma e^{\gamma t})dt=-g\int tdt[/tex]
    [tex]xe^{\gamma t}=-\frac{\gamma}{2}t^2[/tex]

    Which is clearly going the wrong way :grumpy: Then I thought

    [tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-gt[/tex]
    [tex]\frac{dx}{dt}e^{\gamma t}+x\gamma e^{\gamma t}=-gt[/tex]
    [tex]e^{\gamma t}dx + x\gamma e^{\gamma t}dt=-gt dt[/tex]
    [tex]dx=-\gamma x dt -\frac{gt}{e^{\gamma t}}dt[/tex]

    But what then to do with the term

    [tex]-\gamma x dt[/tex] :confused:

    Where am I going wrong?
     
  12. Mar 11, 2009 #11

    tiny-tim

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    Hi phyzmatix! :smile:

    (btw, I'm having difficulty reading your x-dots … I wish you'd use dashes instead of dots and maybe not even use latex :redface:)
    :biggrin: Woohoo! :biggrin:
    You need to recharge by having a bath in between. :wink:
    erm :redface: …[tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-e^{\gamma t} gt[/tex] :wink:
     
  13. Mar 11, 2009 #12
    Done! :smile:

    Ye gods! What a muppet! :redface:
     
  14. Mar 11, 2009 #13
    Mr Tim, you're a legend! :biggrin:

    Thank you so much for your patience! :smile:
     
  15. Mar 11, 2009 #14

    tiny-tim

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    thanks :smile:

    and now try it again, starting with v'/(g + γv) = -1, just to see the difference :wink:
     
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