Вычислить длины дуг кривых, заданных параметрическими уравнениями.
{ x = 3 ( cos t + t sin t ) y = 3 ( sin t − t cos t ) {\displaystyle {\begin{cases}x=3\left(\cos {t}+t\sin {t}\right)\\y=3\left(\sin {t}-t\cos {t}\right)\end{cases}}}
x ′ ( t ) = 3 ( − sin t + sin t + t cos t ) = 3 t cos t {\displaystyle \quad x'(t)=3(-\sin {t}+\sin {t}+t\cos {t})=3t\cos {t}}
y ′ ( t ) = 3 ( cos t − cos t + t sin t ) = 3 t sin t {\displaystyle \quad y'(t)=3(\cos {t}-\cos {t}+t\sin {t})=3t\sin {t}}
L = ∫ 0 π 3 9 t 2 cos 2 t + 9 t 2 s i n 2 t d t = 3 ∫ 0 π 3 t d t = 3 2 t 2 | 0 π 3 = π 2 6 {\displaystyle L=\quad \int \limits _{0}^{\frac {\pi }{3}}{\sqrt {9t^{2}\cos ^{2}{t}+9t^{2}sin^{2}{t}}}\,dt=\quad 3\int \limits _{0}^{\frac {\pi }{3}}t\,dt={\Bigl .}{\frac {3}{2}}t^{2}{\Bigr |}_{0}^{\frac {\pi }{3}}={\frac {\pi ^{2}}{6}}}