Find the length of the cycloid

[eoghan]

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=\int\sqrt{1-cos^2 t -2cost + sin^2 t}dt from 0 to 2p
\sqrt{2}\int\sqrt{1-cos t}dt from 0 to 2p
with the substitution w=cost I get:
length=2\sqrt{2}\int\frac{dw}{2\sqrt{1+w}}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?

 

[gabbagabbahey]

I assume you mean: 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...

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

w=\cos t \quad \Rightarrow dw=-\sin t dt

But(!) \sin t \neq \sqrt{1-w^2} since that would imply \sin t was always positive for t \in [0,2\pi]...which is false...clearly when \sin t is positive you will have \sin t =\sqrt{1-w^2}; but when \sin t is negative, you will have \sin t =-\sqrt{1-w^2}...so when is \sin t pos/neg?

 

[eoghan]

I got it!
Thank you