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Is there a more rigorous way to prove this?

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that d(v^2)/dt = 2 . (d^2r/dt^2) . (dr/dt)

    HINT: v^2 = ||dr/dt||^2 = dr/dt . dr/dt

    2. Relevant equations



    3. The attempt at a solution

    I did it another method:

    d(v^2)/dt = d/dv(v^2) . dv/dt --------------chain rule
    = 2v . dv/dt
    since v=dr/dt and dv/dt = d^2r/dt^2
    therefore = 2 . (d^2r/dt^2) . (dr/dt)

    Is there a more rigorous method to solving this, using the 'HINT' it gave me. I can't quite figure it out.
     
    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2

    tiny-tim

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    Hi Keshroom! :smile:

    (try using the X2 button just above the Reply box :wink:)

    Sorry, but this doesn't work at all. :redface:
    (your "." is ordinary multiplication of two scalars, unlike the question, where it's the dot-product of two vectors; but anyway:)

    no, dv/dt is not d2r/dt2 :redface:

    (except in one-dimensional motion)
    it's easier if you rewrite the hint as:

    v2 = r' . r' :wink:
     
  4. Mar 10, 2012 #3
    Sorry i'm still learning with the notation and everything. Awww ok. You mind telling me how to prove this please?
    thanks

    btw there were errors in the question which i have edited now
     
    Last edited: Mar 10, 2012
  5. Mar 10, 2012 #4

    tiny-tim

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  6. Mar 10, 2012 #5
    lol what....so simple. So it's just
    r'r'' + r'r'' = 2r'r'' ?
     
    Last edited: Mar 10, 2012
  7. Mar 10, 2012 #6

    tiny-tim

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    no, it's r'.r'' + r'.r'' = 2r'.r'' :wink:
     
  8. Mar 10, 2012 #7
    --_--

    haha thanks!
     
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