Question

Вычислить длины дуг кривых, заданных параметрическими уравнениями.x=3(t-sint)y=3(1-cost)pi=

Answer 1

$\displaystyle L=\int_{\pi}^{2\pi}\sqrt{(y_t')^2+(x_t')^2}\mathrm{dt}=*\\\\ x_t'=\left(3(t-\sin{t})\right)'_t=3(1-\cos{t})\\y_t'=\left(3(1-\cos{t})\right)'_t=3\sin{t}\\ (y_t')^2+(x_t')^2=9-18\cos{t}+9\cos^2{t}+9\sin^2{t}=18(1-\cos{t}) \\\\ *=\int_{\pi}^{2\pi}\sqrt{18(1-\cos{t})}\mathrm{dt}=3\sqrt{2}\int_{\pi}^{2\pi}\sqrt{1-\cos{t}}\mathrm{dt}=3\sqrt{2}\int_{\pi}^{2\pi}\sqrt{1-\cos{2t\over2}}\mathrm{dt}=3\sqrt{2}\int_{\pi}^{2\pi}\sqrt{1-\cos^2{t\over2}+\sin^2{t\over2}}\mathrm{dt}=3\sqrt{2}\int_{\pi}^{2\pi}\sqrt{\sin^2{t\over2}+\cos^2{t\over2}-\cos^2{t\over2}+\sin^2{t\over2}}\mathrm{dt}=6\int_{\pi}^{2\pi}\sqrt{\sin^2{t\over2}}\mathrm{dt}=12\int_{\pi}^{2\pi}\sqrt{\sin^2{t\over2}}\mathrm{d{t\over2}}=*\\\\t\in[\pi;2\pi]\Rightarrow\;\; \sin^2{t\over2}\geq0\\\\ *=12\int_{\pi}^{2\pi}\sin{t\over2}\mathrm{d{t\over2}}=-12\cos{t\over2}\bigg|_{\pi}^{2\pi}=-12(\cos{\pi}-\cos{\pi\over2})=12$