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Find the length of the cycloid

  1. Nov 25, 2008 #1
    Hi all!
    I have to find the length of the cycloid given by:
    x(t)=t-sint
    y(t)=1-cost

    I did so:
    length=[tex]\int\sqrt{1-cos^2 t -2cost + sin^2 t}[/tex]dt from 0 to 2p
    [tex]\sqrt{2}\int\sqrt{1-cos t}[/tex]dt from 0 to 2p
    with the substitution w=cost I get:
    length=[tex]2\sqrt{2}\int\frac{dw}{2\sqrt{1+w}}[/tex]dw from 1 to 1 (cos0=cos2p=1)
    Here's the problem: any integral from 1 to 1 is 0! But the cycloid has length=8
    Where am I wrong?
     
  2. jcsd
  3. Nov 25, 2008 #2

    gabbagabbahey

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    Gold Member

    I assume you mean: [tex]s=\int_0^{2\pi} \sqrt{1+\cos^2 t-2\cos t + \sin^2 t}dt=\sqrt{2} \int_0^{2\pi} \sqrt{1- \cos t}dt[/tex]....

    If so, there is a clear problem with your substitution:

    [tex]w=\cos t \quad \Rightarrow dw=-\sin t dt[/tex]

    But(!) [tex]\sin t \neq \sqrt{1-w^2}[/tex] since that would imply [itex]\sin t[/itex] was always positive for [itex]t \in [0,2\pi][/itex]...which is false....clearly when [itex]\sin t[/itex] is positive you will have [tex]\sin t =\sqrt{1-w^2}[/tex]; but when [itex]\sin t[/itex] is negative, you will have [tex]\sin t =-\sqrt{1-w^2}[/tex]....so when is [itex]\sin t[/itex] pos/neg?
     
  4. Nov 25, 2008 #3
    I got it!
    Thank you :smile:
     
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