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Justification for differential manipulation in work-KE theorem proof

  1. Jul 2, 2011 #1
    Hi, this question is about the mathematical justification for a physics topic so hopefully this is the right forum.
    All the proofs of the work-KE theorem I have found go something like this:
    W= int(F)dx from x1 to x2
    = m(int(dv/dt))dx from x1 to x2
    = m(int((dv/dt)v)dt from t1 to t2
    = (m/2)(vf^2-vi^2) by the chain rule/substitution
    My question is regarding the substitution "dx=vdt," and the associated change of variables from x to t, which is obviously true but how do you justify it from the substitution rule or something else not involving differentials? (I'm sure that expression makes perfect sense when it comes to differential forms but I have not learned those yet)
     
  2. jcsd
  3. Jul 2, 2011 #2

    tiny-tim

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    hi shooba! :smile:

    (have an integral: ∫ and try using the X2 and X2 icons just above the Reply box :wink:)
    no, that's wrong, there's no need for a change in variable from dx to dt

    it should only use the https://www.physicsforums.com/library.php?do=view_item&itemid=353"

    m(∫(dv/dt))dx from x1 to x2

    = m(∫(dv/dx)(dx/dt))dx from x1 to x2 (from the chain rule)

    = m(∫(dv/dx)v)dx from x1 to x2

    = m(∫(dv/dx)v)dx from x1 to x2

    = m(∫(d(v2/2)/dx))dx from x1 to x2

    = (m/2)(v22-v12)
     
    Last edited by a moderator: Apr 26, 2017
  4. Jul 2, 2011 #3
    Wow thanks, that's a lot clearer than the other derivations!
    Just one question; when we write dv/dx we are treating velocity as a function of x; does this somehow conflict with the fact that velocity is the time derivative of x and thus a function of t? Is "v" somehow a function of both? (this might be more of a physics question)
     
  5. Jul 3, 2011 #4

    tiny-tim

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    hi shooba! :smile:

    (just got up :zzz: …)
    no, it's a maths question …

    suppose you're in a car looking at the speedometer, and you want to draw a graph of the speed,v …

    you can look at your watch, and draw a graph of v against t, or you can look at the milometer, (or mileposts), and draw a graph of v against x …

    the speedometer doesn't know how you're graphing it …

    all it produces is v :wink:

    v is a function of whatever we want it to be
     
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