generalising cycloid equation

Question

I'm being asked to find 'generalised' parametric equations for a cycloid where P is at a distance that's greater than the radius so d > r, but isn't the distance the circle has rolled the length of the arc of the circle from point P, which is just $r\theta$. But I'm a bit confused about this, how can the distance traveled be greater than $r\theta$?

So, are the equations we know; $x=r(\theta - sin\theta)$ and $y=r(1-cos\theta)$ only true when d=r?

Answer 1

Drawing a diagram is a good start. I don't have the tools for that at the moment though, so we'll have to hope that I can describe this well enough.

$P$ is no longer a point on the circumference of the circle. By generalising we mean that we'll allow $P$ to be somewhere else: either outside or inside the circle. This is where drawing a diagram is a good idea, so you can see how the geometry changes when we move $P$ around. So, let's put $P$ at a distance $\alpha r$ from the centre of the circle, where $\alpha \geq 0$ (you can experiment by letting $\alpha <0$ but it's hard to ascribe a geometric meaning to it then). What changes when we calculate the $x$ and $y$ co-ordinates of $P$?

Well, the distance that the circle moves doesn't change because that's determine by the arc-length of the circumference of the circle. So the term $r\theta$ is unchanged. However, since $P$ is now a distance $\alpha r$ from the centre of the circle, its contribution changes and we have to consider $\sin (\alpha r)$ and $\cos (\alpha r)$ instead.

The generalised parametric form then has co-ordinates $(r\theta - \alpha r \sin \theta, r\theta - \alpha r \cos \theta)$. When you look at the curve that this generates you'll get loops instead of cusps (and, though I'm repeating myself, you should really draw a diagram to appreciate this!)