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Frenet formulas

  1. Mar 25, 2006 #1
    I'm reading Advanced Calculus by Wilfred Kaplan 1952. He is demonstrating how to find the decomposition of the acceleration vector into its normal and tangential components. I'm following along until he replaces the magnitude of the derivative of the angle with respect to the distance traveled by the particle along the curve with 1/p where p is the radius of curvature of the path. Later in the book, he mentions Frenet formulas. Can anyone explain what these Frenet formulas are about and what he means by radius of curvature of the path? Or did I not supply enough information? Thanks.
     
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  3. Mar 26, 2006 #2

    HallsofIvy

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    You should have covered radius of curvature in calculus. It's the reciprocal of the curvature of a path so that your [itex]\frac{1}{\rho}[/itex] is just the curvature itself.

    As for the Frenet formulas, you might try this:
    http://mathworld.wolfram.com/FrenetFormulas.html
     
  4. Mar 27, 2006 #3
    Well, a community college isn't exactly the best way to learn multivariable calculus. That is why I'm reading this book. Are there any online references that can give me an intuitive understanding of what you are talking about. I still do not know what the radius of curvature is. Thanks.
     
  5. Mar 27, 2006 #4

    arildno

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    Here's an intuitive idea:
    Think of some (smooth) curve lying in a plane.
    At any point of the curve, you may ask:
    Which circle will be the best fit of the curve's behaviour at that point?
    (This is somewhat analogous to saying that the "tangent" at a point is the best line fit to the curve's behaviour there.)

    That circle is called the "osculating" (kissing) circle at the point, and its radius is the radius of curvature at that point.
     
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