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Scalar Product of a diffrential.

  1. Oct 8, 2008 #1
    Hey, in my text book they keep doing this and I can't follow
    for example r.[tex]\ddot{}r[/tex] = 1/2 [tex]\ddot{}r^{}^2{}[/tex]
    and r.[tex]\dot{}r[/tex] = 1/2 [tex]\dot{}r^2{}[/tex].

    Can anyone explain this to me? I know I should probably know it.

    P.S Can't quite get the dot product to look right apologies.
     
  2. jcsd
  3. Oct 8, 2008 #2

    AKG

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    (r²)' = 2rr' by chain rule. Or if the r's are supposed to be vector functions:

    (r.r)'
    = r'.r + r.r' by product rule for dot product
    = r.r' + r.r' by commutativity of dot product
    = 2r.r'
     
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