Question

4.175 A projectile launched from the ground with initial velocity components vg and will travel a distance equal to 2t4%/g. The energy that you initially give to the projectile is (I/2)m(+ v) (a) Use the method of Lagrange multipliers to find the maximum possible distance a projectile can go subject to the con- straint that its initial energy is E (b) Find the value of λ from the equations you wrote. (c) Using that value of λ, how much farther n the projectile go if you increase the energy you launch it with by dE?

Answer 1

Let, f(v_x, v_y) = 2v_x v_y/g and constant h(v_x, v_y) = 1/2 m(v^2_x + v^2_y) - E = 0 The constraint h(v_x, v_y) is identically zero on the circle of radius squareroot 2E/m Any multiple of h(v_x, v_y) may be added to f(v_x, v_y) leaving f(v_x, v_y) unchanged in the region on that circle where h(v_x, v_y) is satisfied. Now, Lagrange multiplier method L(v_x, v_yy lambda) = f(v_x, v_y) + lambda h(v_x, v_y) = 2v_x v_y/g + lambda_m/2 (v^2_x + v^2_y) - lambda E Now, gradient nabla_v_x, v_y, lambda L(v_x, v_y, lambda) = (partial differential L/partial differential v_x, partial differential L/partial differential v_y, partial differential L/partial differential lambda) = (2v_y/g + lambda mv_x, 2v_x/g + lambda mv_y, m/2 (v^2_x + v^2_y) - E) and therefore nabla_v_x, v_y, lambda L(v_x, v_y, lambda) = 0 to maximize therefore 2v_y/g + lambda mv_x = 0 rightarrow (i) 2v_x/g + lambda mv_y = 0 rightarrow (ii) m/2 (v^2_x + v^2_y) - E = 0 rightarrow (iii) Here (iii) is original constraint Let lambda = 0 in (i)