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Feynman Lectures on Physics - Rotation in 2 dimensions

  1. Feb 12, 2012 #1
    This isn't really a 'problem', I'm just trying to follow Feynman's reasoning in section 18-2 of Volume 1 The Feynman Lectures on Physics. I've attached a png of the paragraph in question.

    I have 2 issues with this:

    1. If the length OQ = OP, how can there be a right angle at P(x,y)?

    2. I don't understand why QP = r [itex]\Delta\theta[/itex], rather than r * tan ([itex]\Delta\theta[/itex])

    Am I missing something? I've checked the latest errata and this isn't mentioned.

    Thanks for your help.
     

    Attached Files:

  2. jcsd
  3. Feb 12, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi ZirconiumAce! Welcome to PF! :wink:
    Because that's the way these ∆s work …

    we ignore anything that's small compared with a ∆ (in the same way that we ignore dh2, compared with dh, when we do calculus proofs)

    So ∆θ = sin∆θ = tan∆θ. :wink:

    (sin∆θ = ∆θ - (∆θ)3/3! + …)
     
  4. Feb 12, 2012 #3
    Ah yes a Taylor series. Mathematics back in the days of the slide rule I suppose. Now we don't bother with simplifications as much. Cheers! How about the right angle?
     
  5. Feb 12, 2012 #4
    I've got it.
     
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