Question

Вычислить длину дуги, заданной параметрически.

 

$left { {{x=2(t-sint)} atop {y=2(1-cost)}} right.$        $0leq t leq pi$

 

L=   $intlimits^beta_alpha {sqrt{[phi'(t)]^2+[Psi'(t)]^2}} , dt$

Answer 1

$intlimits^pi_0 {sqrt((x'(t))^2+(y'(t))^2)} , dt =$

=$intlimits^pi_0 {sqrt((2(t-sint))')^2+(2(1-cos t))')^2)} , dt =$

=$intlimits^pi_0 {sqrt((2-2cost)^2+(2sin t)^2)} , dt =$

=$intlimits^pi_0 {sqrt(4-8cos t+4cos^2 t+4sin^2 t)} , dt =$

=$intlimits^pi_0 {sqrt(4-8cos t+4)} , dt =$

=$intlimits^pi_0 {sqrt(8-8cost)} , dt =$

=$sqrt(8) intlimits^pi_0 {sqrt(1-cos t)} , dt =$

=$sqrt(8)intlimits^pi_0 {sqrt(2sin^2 (t/2))} , dt =$

=$sqrt(8)*2*sqrt(2) intlimits^pi_0 {sin (t/2)} , dt/2 =$

=$8 *(-cos (t/2)) |limits^pi_0 =$

=$8 (-cos (pi/2)+cos (0/2))=8*(-0+1)=8$