求摆线x=1?costy=t?sint一拱(0≤t≤2π)的弧长

求摆线x=1?costy=t?sint一拱(0≤t≤2π)的弧长.
2025-06-28 18:11:51
推荐回答(1个)
回答1:

ds=

x2(t)+y2(t)
dt=
sin2t+(1?cost)2
dt=
2(1?cost)
dt
=2sin
t
2
   (0≤t≤2π)

S=
x2(t)+y2(t)
dt=
2sin
t
2
dt
=?4cos
t
2
|
=8