Arc Length of Cycloid: Show s^2=8ay

[John O' Meara]

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

\frac{dx}{d\theta} =a(1-\cos\theta) \mbox{ and } \frac{dy}{d\theta}= a\sin\theta\\
\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\\. I don't know how to do the integration, a hint would be welcome. Thanks.

 

[Dick]

There's a half angle formula that can help you a lot.

 

[John O' Meara]

IS it \cos2\theta = \frac{1-\tan^2 \theta}{1+tan^2\theta}

 

[nicktacik]

What is \sin{\frac{\theta}{2}} ?

 

[John O' Meara]

I got this formula \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 \\ . When I evaluate this integral I get s=8a not s^2=8ay. I have to look up what the \sin\frac{\theta}{2} \ equals.

 

[John O' Meara]

The half-angle formula is: \sin\frac{\theta}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}} \mbox{ where } \ s=\frac{a+b+c}{2} \\. How do I use this? Thanks.

 

[Dick]

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.

 

[John O' Meara]

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.