求摆线x＝1−costy＝t−sint一拱（0≤t≤2π）的弧长．

Question

求摆线

x＝1−cost

y＝t−sint

一拱（0≤t≤2π）的弧长．

Answer 1

∵ds＝√(x′2(t)+y2(t))dt=√(sin2t+(1−cost)2)dt＝√(2(1−cost))dt=2sin(t)/(2)   (0≤t≤2π)
∴S＝

∫

2π 0

√(x′2(t)+y2(t))dt＝

∫

2π 0

2sin(t)/(2)dt=−4cos(t)/(2)

|

2π 0

＝8