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Show that |x(t)|d/dt*|x(t)| = x(t) . x`(t)

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex] Show that |x(t)| d/dt |x(t)| = x(t) \cdot x`(t) [/itex]

    3. The attempt at a solution

    I don't know what they want to see here but my go at this is to take x(t) = 2t. Differentiating this gives x`(t) = 2.

    Taking the normal of both |x(t)| and |x`(t)| gives 2t and 2, respectively.

    So 2t * 2 = 4t.

    Then just do the same with x(t) and x`(t) to get 4t again?

    I don't know if this will suffice as a proper answer
     
  2. jcsd
  3. Apr 16, 2013 #2

    Bacle2

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    Science Advisor

    I assume they expect you to show the result for any path x(t), not just x(t)=2t .

    I think working with the cases (for the absolute value) ; when is x(t)>0 , when is it <0 ,=0 ,

    etc. will be helpful.
     
  4. Apr 16, 2013 #3
    Hint: Use the fact that [itex] \frac{d}{dt} |t|=\frac{t}{|t|} [/itex] with the chain rule to compute your derivative in the two cases [itex] x(t) > 0 [/itex] and [itex] x(t) < 0 [/itex] . You have to consider the case [itex] x(t)=0 [/itex] carefully.
     
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