> The "well behaved, smooth" function f(x) has endpoints f(0) = f(h) = 0. The > curve of the function has length s1. > > An arc of a circle passing through (0, 0) and (0, h) has fixed curvature k > and its arc length is also s1. > > It is required to show that a point must exist on f(x) where curvature is > also k. > As that's was quickly written, I'll state what I think the problem you're asking is. It this correct?
Let p:[0.h] -> R^2 be a planar loop with p(0) = p(h) = (0,0) Let s0 be the length of p and k(x) the curvature of p at x. Shaw that there's some x in [0,k] with k(x) = k where k is the curvature of an arc from (0,0) to (0,h) with length s0.
> I have set up a CAS program to simulate the situation and the proposition > held up in every case. > > I have been working with the idea that, at the required point, the normal to > f(x) is normal to the circle. > > I am not making much headway. Any ideas appreciated.