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Quick Calc Correction

  1. Apr 14, 2014 #1
    I'm brushing up on calculus. I don't see how this derivative works.

    x(t) = A cos(ωt - [itex]\varphi[/itex])

    v(t) = dx / dt = - Aωsin(ωt + [itex]\varphi[/itex])



    I get that the derivative of cosx is -sinx. I get that the omega is brought outside the cos function because of U substitution.

    Why does the the minus sign turn into a plus sign?

    Btw I saw this on wikipedia: http://en.wikipedia.org/wiki/Simple_harmonic_motion
     
  2. jcsd
  3. Apr 14, 2014 #2

    jbunniii

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    Looks like a typo on the Wikipedia page. The correct derivative of ##A \cos(\omega t - \varphi)## is ##-A\omega\sin(\omega t - \varphi)##.

    Since ##-\sin(x) = \sin(-x)##, an equivalent expression is ##A\omega\sin(-\omega t + \varphi)##.

    But ##-A\omega\sin(\omega t + \varphi)## is not equal to either of these, as can easily be seen by evaluating at ##\omega = 0##, ##\varphi = \pi/2##.
     
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