plane curves

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September 24th 2008, 12:53 PM
dori1123

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?
September 25th 2008, 10:06 AM
Laurent

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...
September 25th 2008, 11:00 AM
dori1123

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...
September 25th 2008, 12:25 PM
Laurent

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 $\phi:(s,\rho)\mapsto \alpha(s)+\rho \vec{n}(s)$ . 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.