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Length of parallel curve

  1. Feb 8, 2010 #1

    I found this problem in Do Carmos "differential geometry of curves and surfaces".
    it asks to show that the length of a parallel curve B to A given by:

    where r is a positive constant, and n is the normal vector, and A is a closed convex plane curve, positively oriented.
    is given by


    The obvious start would be to integrate both sides under a standard parametrization or the curves, but why is it true that integrating rn will give 2*pi*n exactly?

    any insight is appreciated.
  2. jcsd
  3. Feb 9, 2010 #2
    I think I know one way of thinking about it, which can probably be made more rigorous if you want. Just to make it easier to think, consider a uniform grid defined on the closed curve B. Think of B as a deformed circle, so if you follow it with a tangent vector, the tangent vector will have rotated through an angle 2*pi after one going through the whole curve once.

    Assume that curve A is "outside" B at a distance r. So curve B will turn out as the longest. This can be accurately defined by specifying that if the tangent vector mentioned above rotates to the left, then the defined A by choosing n at an angle pi/2 to the right of the tangent vector.

    Transport the grid over to A using the normal vectors.

    Consider one of the sections along curve B, the "inner" curve, e.g. between two division defined by the parametrization. It has some length dl. If the curvature of B is positive (like a circle), the corresponding length element of the "outer" curve A will be a bit longer. By how much? Well, by r*dphi, where dphi is a small angle.

    Since we are going around the closed curve once, we must add all these dphi contributions along the curve. Somce will be positive, and some negative. In the end, the total must be 2*pi since we are going around once. Each contribution as r*dphi, so the total is 2*pi*r. They must add up to this, because the same angles define the rotation of the tangent vector to B, and it returns to its starting position after one travel around.

    If the curve goes m times around (rotation of tangent vector), the result will be m*2*pi*r.

  4. Feb 9, 2010 #3
    I hope this helps.

    Parameterize A by arc length. Then B'(s) = A'(s) -rn'(s) = A'(s)(1 + rk(s)) where k(s) is the curvature.

    |B'(s)| =(1 + rk(s)) |A'(s)| = 1 + rk(s)

    BTW: I think a convex curve will always have non-negative curvature.

    So the length of B is the length of A plus r times the integral of the curvature over the parameter interval.

    The Gauss map of the curve takes the curve into the unit sphere by mapping each point on the curve to its unit normal translated to the origin. The derivative is just the curvature (I may be off by a sign here) times the unit tangent to the sphere. So the integral of the curvature is the integral of the pull back of dtheta to the curve, A. By the change of variables formula this is the integral of dtheta over the circle times the degree of the mapping and so is a multiple of 2 pi. For a convex plane curve I think the degree is plus or minus 1.
  5. Feb 11, 2010 #4
    I get it now,
    thanks everyone!

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