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Time derivative of rotating vector

  1. Oct 29, 2011 #1
    Trying to teach myself physics and I've run into a problem I don't quite understand.

    "The magnitude of dA/dt can be found by the following geometrical argument. The change of A in the time interval t to Δt is"

    ΔA = A(t + Δt) - A(t)

    And then somehow it gets to
    A| = 2Asin(Δ[itex]\theta[/itex]/2)

    I just can't see how those are equal. Especially where theta/2 comes in.
    This comes out of Kleppner pages 25 and 26, if that helps. Supposedly there's a sketch, but I can't see one.
  2. jcsd
  3. Oct 29, 2011 #2


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    Science Advisor

    Work it out. Suppose at t = 0, A points along the x-axis, so A = (A,0), If you rotate by an amount theta, the A = (A cos(theta),A sin(theta). So Delta A = (A(1-cos(theta)),A sin(theta)). So:

    [tex] |\Delta A| =A \sqrt{(1-\cos(\theta)^2 + \sin(\theta)^2} =A \sqrt{1-2\cos(\theta) + \cos(\theta)^2 + \sin(\theta)^2} =A \sqrt{2-2\cos(\theta)} = 2A \sin(\frac{\theta}{2})[/tex]

    This last is a half-angle trig identity:

    [tex]\sin(\frac{\theta}{2}) = \sqrt{\frac{1-\cos(\theta)}{2}}[/tex]
  4. Oct 29, 2011 #3
    Thanks, I actually just got it myself by drawing a picture and using the Pythagorean theorem. Essentially equivalent to what you did.
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