Задача 35812 ...

qwe2

2019-04-13 22:39:04

Найдите длину дуги кривой
[b](x=e^(t)cost
(y=e^(t)sint
0 ≤ t ≤ π/2[/b]

sova

2019-04-13 22:52:18

★

x(t)=e^(t)*cost

y(t)=e^(t)*sint

L= ∫ ^(t_(2))_(t_(1))sqrt((x`(t))^2+(y`(t))^2)dt

x`(t)=e^(t)*cost+e^(t)*(-sint)
y`(t)=e^(t)*sint+e^(t)*cost

(x`(t))^2=(e^(t)*(cost-sint))^2

(y`(t))^2=(e^(t)*(cost+sint))^2

(x`(t))^2+(y`(t))^2=

=(e^(t))^(2)*(cos^2t-2sint*cost+sin^2t+cos^2t+2sintcost+sin^2t)=

=4(e^(t))^2



L= ∫ ^(π/2)_(0)sqrt(4e^(t))^2)dt=

=2 ∫ ^(π/2)_(0)e^(t)dt=

=2(e^(t))^(π/2)_(0)=

=2*(e^(π/2)-e^(0))= [b]2*(e^(π/2)-1)[/b]