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Thread: plane curves

  1. #1
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    plane curves

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

    So, if you know that the "global curvature" $\displaystyle \int_0^{\ell(\alpha)}\kappa(s)ds$ for a convex curve equals $\displaystyle 2\pi$, 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):

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

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