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Constant curvature

  1. Mar 21, 2010 #1
    If we parameterize the arc length of a vector valued
    function, say, r(s) and r(s) has constant curvature
    (not equal to zero), then r(s) is a circle.

    Thus, |T'(s)| = K but to prove it we would need
    to show |T'(s)| = K => <-Kcos(s), -Ksin(s)>
    and integrate component-wise two times,
    right?
     
  2. jcsd
  3. Mar 21, 2010 #2
    If your function is into R^2...else it could be way more complicated. I'd try to find a point x such that |r(s)-x| is constant.
     
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