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Math Help - plane curves

  1. #1
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    plane curves

    Let alpha(s), s in [0, l] be a closed convex plane curve positively oriented. The curve beta(s) = alpha(s) - rn(s), where r is a positive constant and n is the normal vector, is called a parallel curve to alpha.

    Show that length of beta = length of alpha + 2(pi)r

    2(pi)r is the circumference of a circle. l is the length of alpha

    Can someone help me with this?
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  2. #2
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    I have a partial answer:

    You can always assume that \alpha is an arc length parametrization ( \|\vec{\alpha'}(s)\|=1 for all s)
    You have \beta(s)=\alpha(s)-r \vec{n}(s), so that \vec{\beta'}(s)=\vec{\alpha'}(s)-r \vec{n'}(s)=\vec{\tau}(s) + r \kappa(s) \vec{\tau}(s), where \vec{\tau}, \vec{n} are the unit tangent, normal vectors at time s (the normal \vec{n} is assumed to be the image of \vec{\tau} by the rotation of angle \frac{\pi}{2}) and \kappa(s) is the (algebraic) curvature at time s. As a consequence, the length of the curve is: \ell(\beta)=\int_0^{\ell(\alpha)} \|\vec{\beta'}(s)\|ds = \int_0^{\ell(\alpha)}|1+r\kappa(s)|ds =<br />
 \ell(\alpha)+r\int_0^{\ell(\alpha)}\kappa(s)ds since the convexity of the curve gives \kappa(s)\geq 0 (I assume in fact that the curve is oriented positively, so that the result you give is right; otherwise we could not check whether 1+r\kappa(s)\geq 0).

    So, if you know that the "global curvature" \int_0^{\ell(\alpha)}\kappa(s)ds for a convex curve equals 2\pi, you are done. This is a classical result, and I have a two page long proof in a book I don't feel like rewriting here. So I hope you know that...
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  3. #3
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    Thank you very much for helping. I also figured that out this morning. But do you know how to show:
    Area of beta = Area of alpha + lr + (pi)r^2?
    I asked my classmates, but no one could get it...
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  4. #4
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    Before any justification, here is my computation (which I did before justifying):

    \mathcal{A}(\beta)-\mathcal{A}(\alpha)=\int_0^r \ell(\alpha-\rho \vec{n}) d\rho = \int_0^r (\ell(\alpha)+2\pi \rho) dr=r\ell(\alpha)+\pi r^2.

    Because of the convexity, the space between \alpha and \beta is the image of [0,\ell(\alpha)]\times[0,r] by the (injective) application s,\rho)\mapsto \alpha(s)+\rho \vec{n}(s)" alt="\phis,\rho)\mapsto \alpha(s)+\rho \vec{n}(s)" />. So, the area you want is: \int_0^{\ell(\alpha)}\int_0^r |({\rm Jac }\,\phi)(\rho,s)|d\rho\,ds where I integrate the jacobian determinant of \phi at (\rho,s). Remember the Jacobian is the determinant of the differential of \phi written in the canonical bases. However, the determinant is not altered if you write d\phi in another orthonormal basis, even if this basis depends on s... A not so stupid choice would be to compute the Jacobian of \phi at the point (\rho,s) in the (orthonormal) basis (\vec{\tau}(s),\vec{n}(s)). We have \partial_\rho \phi=\vec{n}(s) and \partial_s \phi=(1-\kappa(s)\rho)\vec{\tau(s)} (like for the arc length), so the Jacobian is the determinant of a diagonal matrix with diagonal coefficients 1 and 1-\kappa(s)\rho. This gives that the area equals: \int_0^r\int_0^{\ell(\alpha)}|1-\kappa(s)\rho|ds\,d\rho=\int_0^r\ell(\alpha-\rho\vec{n})d\rho, using Fubini and the arclength computation. This is the desired justification. qed.
    Last edited by Laurent; September 25th 2008 at 01:09 PM.
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