Здравствуйте!
Помогите, пожалуйста, вычислить длину дуги кривой
r = 4 * (1 + cos fi)

r = 4 * (1 + cos fi)
Используем формулу:
L = int (a b ) (r^2 + (r')^2)^(1/2) dfi
1 + cos fi >= 0 при любых fi, тогда
L = int (0 2pi) (r^2 + (r')^2)^(1/2) dfi
r^2 = (4 * (1 + cos fi))^2 = 16 * (1 + cos fi)^2 = 16 * (1 + 2 * cos fi + cos^2 fi) =
= 16 + 32 * cos fi + 16 * cos^2 fi
r' = (4 * (1 + cos fi))' = 4 * (1 + cos fi)' = 4 * (-sin fi) = -4 * sin fi
Тогда
(r')^2 = (-4 * sin fi)^2 = 16 * sin^2 fi
r^2 + (r')^2 = 16 + 32 * cos fi + 16 * cos^2 fi + 16 * sin^2 fi = 32 + 32 * cos fi =
= 32 * (1 + cos fi) = 32 * (1 + 2 * sin^2 (fi/2) - 1) = 64 * sin^2 (fi/2)
Отсюда
(r^2 + (r')^2)^(1/2) = (64 * sin^2 (fi/2))^(1/2) = 8 * |sin (fi/2)|
L = int (0 2pi) 8 * |sin (fi/2)| dfi = | fi/2 = t, fi = 2 * t, dfi = 2 dt | =
= int (0 pi) 8 * 2 * |sin t| dt = 16 * int (0 pi) |sin t| dt = 16 * int (0 pi) sin t dt =
= 16 * (-cos t)_{0}^{pi} = 16 * (-cos pi + cos 0) = 16 * (1 + 1) = 32
Ответ: L = 32.

Спасибо