DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

§4. Calculus of Variations and Surfaces of Constant Mean Curvature 105the locus of one focus as we roll the ellipse along the x-axis. By definition of an ellipse, we have‖ −−→ F 1 Q‖ + ‖ −−→ F 2 Q‖ =2a, and by Exercise 7, we have yy 2 = b 2 (see Figure 4.3). On the other hand, wededuce from Exercise 8 that −−→ F 1 Q is normal to the curve, and that, therefore, y = ‖ −−→ F 1 Q‖ cos φ. Sincethe “reflectivity” property of the ellipse tells us that ∠F 1 QP 1∼ = ∠F2 QP 2 ,wehave y 2 = ‖ −−→ F 2 Q‖ cos φ.Since cos φ = dx/ds and ds/dx = √ 1+(dy/dx) 2 ,wehaveF 1φxyP 1F 2y 2Q P 2Figure 4.3y + b2y = y + y 2 =2a cos φ =2a dxdsand so0=y 2 − 2ay dxds + b2 = y 2 − √ 2ay + 1+y ′2 b2 =0.Setting H 0 = −1/2a, wesee that this matches the equation (†) above.▽EXERCISES 3.4♯ 1. Suppose g :[0, 1] × (−1, 1) → R is continuous and let G(ε) = g(t, ε)dt. Prove that if ∂g0∂ε is∫ 1∫continuous, then G ′ ∂gε ∫ 1(0) =∂ε (t, 0)dt. (Hint: Consider h(ε) = ∂g(t, u)dtdu.)∂ε♯ 2. *a. Suppose f is a continuous function on [0, 1] and0∫ 1functions ξ on [0, 1]. Prove that f =0. (Hint: Take ξ = f.)b. Suppose f is a continuous function on [0, 1] and0∫ 10∫ 100f(t)ξ(t)dt = 0 for all continuousf(t)ξ(t)dt = 0 for all continuousfunctions ξ on [0, 1] with ξ(0) = ξ(1) = 0. Prove that f =0. (Hint: Take ξ = ψf for anappropriate continuous function ψ.)c. Deduce the same result for C 1 functions ξ.d. Deduce the same result for vector-valued functions f and ξ.3. Use the Euler-Lagrange equations to show that the shortest path joining two points in theEuclidean plane is a line segment.