Kindly assist me. It is calculus of variations

Question 4 - 22 marks This is a past examination question. Consider the functional Stu] = dx F(1,B.V/), v(a) = A, where the right-hand end point v is determined so that the path lies on the curve r(x, y) = 0, Le. r(v, (v)) =0, and a and A are constants. Consider a perturbed path to = y + ch, where A is an admissible perturbation, i.e. satisfying h(a) = 0 and r(u, m(r.)) =0, where u. = u te[ + O(e ) is the perturbed value of v. (a) Show that to first order in e, "() = me) +e('()E the)) + o(.), and hence that at (x, y) = (v, g(v)). {(Tz + /(v) T, ) + h(u), =0. (b) Using the formula for the Gateaux differential show, by integrating by parts or otherwise, that a stationary path must satisfy the transversality condition TI Fy + Ty ( #' Fy - F) =0 at (r, y) = (0, y(v)). (e) Consider the case when y(0) = 6, where y(v) > 0. 6 2 0 and the right-hand end point (v, y()) lies on the line ary + Ar + 7 =0, where a, A, y are constants with # # 0. Show that the first-integral may be written as for some constant e >0, and that the solutions of the first-integral are circles centred at (-ca, 0) with radius e, Le. where 6 =e- 8 *Using the transversality condition and by differentiating implicitly the equation y' + (r to) = e, show that y' = 0/8 and es =7/8. Finally, show that in the limit as # +0, the stationary path becomes By' + ( 8x + 7)2 =73. 191