Вычислить длину дуги кривой

x=2 (2cos (t) - cos (2t)),

y=2 (2sin (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)).