Question

Вычислить длины дуг кривых, заданных параметрическими уравнениями.

x=3(t-sint)

y=3(1-cost)

pi=

Answer 1

$displaystyle L=int_{pi}^{2pi}sqrt{(y_t')^2+(x_t')^2}mathrm{dt}=*\\ x_t'=left(3(t-sin{t})right)'_t=3(1-cos{t})\y_t'=left(3(1-cos{t})right)'_t=3sin{t}\ (y_t')^2+(x_t')^2=9-18cos{t}+9cos^2{t}+9sin^2{t}=18(1-cos{t}) \\ *=int_{pi}^{2pi}sqrt{18(1-cos{t})}mathrm{dt}=3sqrt{2}int_{pi}^{2pi}sqrt{1-cos{t}}mathrm{dt}=3sqrt{2}int_{pi}^{2pi}sqrt{1-cos{2tover2}}mathrm{dt}=3sqrt{2}int_{pi}^{2pi}sqrt{1-cos^2{tover2}+sin^2{tover2}}mathrm{dt}=3sqrt{2}int_{pi}^{2pi}sqrt{sin^2{tover2}+cos^2{tover2}-cos^2{tover2}+sin^2{tover2}}mathrm{dt}=6int_{pi}^{2pi}sqrt{sin^2{tover2}}mathrm{dt}=12int_{pi}^{2pi}sqrt{sin^2{tover2}}mathrm{d{tover2}}=*\\tin[pi;2pi]Rightarrow;; sin^2{tover2}geq0\\ *=12int_{pi}^{2pi}sin{tover2}mathrm{d{tover2}}=-12cos{tover2}bigg|_{pi}^{2pi}=-12(cos{pi}-cos{piover2})=12$