Помогите, пожалуйста, найти длину дуги, заданной уравнением
x(t) = 2 * (2 * cos t - cos 2t), y(t) = 2 * (2 * sin t - sin 2t), 0 <= t <= 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)).