Talk:Length of Arc of Cycloid

I'm confused a bit. I'm not sure what happens at $\theta = 0$ and $\theta = 2\pi$. At those points the function isn't differentiable, so the curve isn't continuously differentiable, am I wrong? So is this integral improper? That seems quite bizarre. --GFauxPas 00:51, 2 January 2012 (CST)


 * Doesn't have to be. It's continuous on the closed interval $\theta \in \closedint 0 {2 \pi}$ and differentiable on $\theta \in \openint 0 {2 \pi}$, that is, not including the end points.


 * We had this discussion a few weeks ago on the chat page for derivative or something (I can't remember exactly, you'll remember), where it was pointed out that a differentiable function does not need to be (and frequently isn't) differentiable at the end points.


 * Perhaps this point needs to be specified somehow on this page, or (more to the point) on the source pages for arc length (as this point will crop up repeatedly in this context). --prime mover 01:41, 2 January 2012 (CST)


 * I edited all the pages I made about arc lengths to try to address that point, thanks for your explanation. --GFauxPas 00:38, 3 January 2012 (CST)