Introduction to Modern Economic Growth Exercise 7.3 The key equation of the calculus of variations is the Euler-Legrange equation, which characterizes the solution to the following problem (under similar regularity conditions to those of Theorem 7.2): Z t1 F (t, x (t) , x˙ (t)) dx max x(t) subject to x (t) = Suppose that F is differentiable in all of its arguments and an interior continuously differentiable solution exists The so-called Euler-Legrange equation, which provides the necessary conditions for an optimal solution, is ∂F (t, x (t) , x˙ (t)) ∂ F (t, x (t) , x˙ (t)) − = ∂x (t) ∂ x˙ (t) ∂t Derive this equation from Theorem 7.2 [Hint: define y (t) ≡ x˙ (t)] 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 + [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 q + [x0 (t)]2 dt − Prove that the solution to this problem requires Ê Ă ÂÔ d x0 (t) + (x0 (t))2 = dt 364