Вычислить длину дуги кривой x=2(2cos(t)-cos(2t)) , y=2(2sin(t)-sin(2t)) 0 меньше = t меньше = pi/3

x(t) = 2 * (2 * cos t - cos 2t), y(t) = 2 * (2 * sin t - sin 2t), 0 <= t <= pi/3. L = ? Решение. L = int (0 pi/3) ((x'(t))^2 + (y'(t))^2)^(1/2) dt x'(t) = (2 * (2 * cos t - cos 2t))' = 2 * (2 * cos t - cos 2t)' = = 2 * (2 * (-sin t) - 2 * (-sin 2t)) = -4 * sin t + 4 * sin 2t y'(t) = (2 * (2 * sin t - sin 2t))' = 2 * (2 * sin t - sin 2t)' = = 2 * (2 * cos t - 2 * cos 2t) = 4 * cos t - 4 * cos 2t (x'(t))^2 + (y'(t))^2 = (-4 * sin t + 4 * sin 2t)^2 + (4 * cos t - 4 * cos 2t)^2 = = 16 * sin^2 t - 32 * sin t * sin 2t + 16 * sin^2 2t + + 16 * cos^2 t - 32 * cos t * cos 2t + 16 * cos^2 2t = = 16 + 16 - 32 * (sin t * sin 2t + cos t * cos 2t) = = 32 - 32 * cos (2t - t) = 32 - 32 * cos t = 32 - 32 * (1 - 2 * sin^2 (t/2)) = = 32 - 32 + 64 * sin^2 (t/2) = 64 * sin^2 (t/2) Получаем, что L = int (0 pi/3) (64 * sin^2 (t/2))^(1/2) dt = 8 * int (0 pi/3) |sin (t/2)| dt = = 8 * int (0 pi/3) sin (t/2) dt = 8 * (-2 * cos (t/2))_{0}^{pi/3} = = 8 * (-2 * cos (pi/6) + 2 * cos 0) = 8 * (-2 * 3^(1/2)/2 + 2) = = -8 * 3^(1/2) + 16 = 8 * (2 - 3^(1/2)) Ответ: L = 8 * (2 - 3^(1/2)).