7. Let (s) : [0,L)-R2 be a smooth (C\") curve with ll , (s) 1-1 and curvature Ke

Question

7. Let (s) : [0,L)-R2 be a smooth (C") curve with ll , (s) 1-1 and curvature Ke(s) > 0 so that Na exists on all of [0,L]. A curve is called parallel to if it is defined by (s) = (s)-rNa(s) where r e R is a chosen constant and Na(s) is the normal of a. Different choices of r give different parallels to . a. Prove that the curvature Kf satisfies that Ka(s)-l i + r(s(s -1 when is regular at (s). b. Give an example of regular curve (s), a point (so) on it, an r value, but is NOT regular at (s). c. Prove that (The length of ) = 2nr + (The length ofa), if is a simple closed strictly convex smooth plane curve and r >-(max Ka)-1 A simple closed smooth curve (s) : [QL] R2 is called strictly convex if its curvature Ka > 0 on all of [O,L], and the derivatives of all orders k match at the initial terminal point: (k)(0) = (A)(L). For example, circles and ellipses are strictly convex. You may assume that the rotation index of a simple closed curve is 1 if oriented counterclockwise, without proof. However, provide an explanantion why the signed curvature and standard curvature are the same for a strictly convex counterclockwise oriented closed curve, if you use such a fact.

Explanation / Answer

Solutions to problem 7

First let us solve the ‘a’ part of the question.

Given is smooth function from [0,L] to R 2 . Also, given norm ’ (s) is 1 and is positive so that the normal N (s) to curve exists for all in [0,L].

(s) = (s) – r N (s) is a family of curves parallel to (s), where r is a constant.

For a specific value of r, we have to prove that the curvature (s) satisfies the given formula : (s) = modulus    (s) /   [ 1 + r (s) ] , when is regular.

Solution: is regular implies that ‘(s) is non zero for all s in [0, L] .

curve is defined as : (s) = (s) – r N (s)                             (1)

Differentiating both sides of equation (1) w.r.to t, we get

   =      -   r

ie.      =         -   r ( -    ;  

                   = + r )
( 1/ ) = 1 + r )( 1/   

Since    have absolute values 1, we get

( 1/ ) = 1 + r )( 1/

In other words,    =    / 1 + r ) , which is the result to be proved.

This proves 7(a)

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