Let alpha(s), s in be a closed convex plane curve positively oriented. The curve , 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. is the length of alpha
Can someone help me with this?
I have a partial answer:
You can always assume that is an arc length parametrization ( for all )
You have , so that , where are the unit tangent, normal vectors at time (the normal is assumed to be the image of by the rotation of angle ) and is the (algebraic) curvature at time . As a consequence, the length of the curve is: since the convexity of the curve gives (I assume in fact that the curve is oriented positively, so that the result you give is right; otherwise we could not check whether ).
So, if you know that the "global curvature" for a convex curve equals , 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...
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...
Before any justification, here is my computation (which I did before justifying):
Because of the convexity, the space between and is the image of by the (injective) application . So, the area you want is: where I integrate the jacobian determinant of at . Remember the Jacobian is the determinant of the differential of written in the canonical bases. However, the determinant is not altered if you write in another orthonormal basis, even if this basis depends on ... A not so stupid choice would be to compute the Jacobian of at the point in the (orthonormal) basis . We have and (like for the arc length), so the Jacobian is the determinant of a diagonal matrix with diagonal coefficients and . This gives that the area equals: , using Fubini and the arclength computation. This is the desired justification. qed.