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Simple trig derivative

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data
    r(t)=(cos t +tsint)i + (sin t -tcost)j + 3k

    2. Relevant equations

    3. The attempt at a solution
    when I take the derivative of r(t) I get;
    v(t)=(-sin t + tcost)i +(cos t +tsint)j

    the book says;
    v(t)=(t cos t)i +(tsint)j
    could some one tell me why? Where did the -sint and the cos t go?
    I guess it is an identity but I don't see it.
  2. jcsd
  3. Sep 26, 2009 #2
    Recall the chain rule to differentiate t sin t and - t cost t.

    Example: to differentiate (t-1)t^2

    = 1(t^2) + (t-1)(2)t
    = t^2 + 2t(t-1)
  4. Sep 26, 2009 #3
    Note that the rule in question is the product rule, not the chain rule.
  5. Sep 26, 2009 #4
    Are you familiar with the product rule for derivatives? [tex]\frac{d}{dt}(uv)=uv^{'}+u^{'}v[/tex]

    beat me to it :(
  6. Sep 26, 2009 #5
    My bad, *product rule.
  7. Sep 26, 2009 #6
    thank you.
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