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Cycloid question

  1. Sep 3, 2007 #1
    The parametric equations of a cycloid are [tex] x=a(\theta - \sin\theta) \mbox{ and } y = a(1-\cos\theta) \\[/tex] Where a is a constant. Show that s^2=8ay, where s is the arc length measured from the point theta =0

    [tex] \frac{dx}{d\theta} =a(1-\cos\theta) \mbox{ and } \frac{dy}{d\theta}= a\sin\theta\\ [/tex]
    [tex] \int\sqrt{a^2(1-\cos\theta)^2 +a^2\sin^2 \theta}d\theta \\ = \sqrt{2}a\int\sqrt{1 - \cos\theta}d\theta\\[/tex]. I don't know how to do the integration, a hint would be welcome. Thanks.
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  3. Sep 3, 2007 #2

    Dick

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    There's a half angle formula that can help you a lot.
     
  4. Sep 3, 2007 #3
    IS it [tex] \cos2\theta = \frac{1-\tan^2 \theta}{1+tan^2\theta} [/tex]
     
  5. Sep 3, 2007 #4
    What is [tex]\sin{\frac{\theta}{2}}[/tex] ?
     
  6. Sep 4, 2007 #5
    I got this formula [tex] \sin^2 \theta = \frac{1}{2}(1-\cos2\theta) \\ \mbox{ therefore we have the integral =} \sqrt{2}a\int_0^{2\pi}\sqrt{2\sin^2 \frac{\theta}{2}}d\theta \\[/tex] . When I evaluate this integral I get s=8a not s^2=8ay. I have to look up what the [tex] \sin\frac{\theta}{2} \[/tex] equals.
     
  7. Sep 4, 2007 #6
    The half-angle formula is: [tex] \sin\frac{\theta}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}} \mbox{ where } \ s=\frac{a+b+c}{2} \\[/tex]. How do I use this? Thanks.
     
  8. Sep 4, 2007 #7

    Dick

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    That looks like a triangle formula - why would you want to use that? I agree with you that the integral from 0 to 2pi is 8a. y(2pi)=0. So s^2=8ay doesn't work. It does work for pi. But it doesn't work for pi/2. I don't think its a very good formula.
     
  9. Sep 5, 2007 #8
    A cycloid is the path traced out by a point on a car wheel as the car is moving at constant speed, I think. Therefore the limits of integration would be 0 to pi.Thanks for the help.
     
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