Вычислить длину дуги кривой
{x=t-sin(t)
{y=1-cos(t)
0<=t<=2п

8

Пошаговое объяснение:

x=t-sint; y=1-cost; 0≤t≤2π

x'(t)=(t-sint)'=1-cost

y'(t)=(1-cost)'=sint

(x'(t))²+(y'(t))²=(1-cost)²+(sint)²=1-2cost+cos²t+sin²t=

=1-2cost+1=2-2cost=2-2(1-2sin²(t/2))=4sin²(t/2)

0≤t≤2π⇒0≤t/2≤π⇒sint/2≥0⇒|sint/2|=sint/2

L=∫√((x'(t))²+(y'(t))²)dt=∫√(4sin²(t/2))dt=

=∫|2sin(t/2)|dt=2∫sin(t/2)dt=2·(-2cos(t/2))=

=-4cos(t/2)=-4[cos(2π/2)-cos(0/2)]=-4(-1-1)=8