# Find the length of the cycloid

1. Nov 25, 2008

### 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?

2. Nov 25, 2008

### 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?

3. Nov 25, 2008

I got it!
Thank you