Exercise 7.4. This exercise asks you to use the Euler-Legrange equation derived in Exercise 7.3 to solve the canonical problem that motivated Euler and Legrange, that of finding the shortest distance between two points in a plane. In particular, consider a two dimensional plane and two points on this plane with coordinates (z0, u0) and (z1, u1). We would like to find the curve that has the shortest length that connects these two points. Such a curve can be represented by a function x : R → R such that u = x (z), together with initial and terminal conditions u0 = x (z0) and u1 = x (z1). It is also natural to impose that this curve u = x (z) be smooth, which corresponds to requiring that the solution be continuously differentiable so that x0 (z) exists. To solve this problem, observe that the (arc) length along the curve x can be represented as A [x (z)] ≡ Z z2 z1 q 1+[x0 (z)]2 dz. The problem is to minimize this object by choosing x (z).

Now, without loss of any generality let us take (z0, u0) = (0, 0) and let t = z to transform the problem into a more familiar form, which becomes that of maximizing − Z t1 0 q 1+[x0 (t)]2 dt. Prove that the solution to this problem requires d h x0 (t) ³ 1+(x0 (t))2 ´i dt = 0. Show that this is only possible if x00 (t)=0, so that the shortest path between two points is a straight-line.