# generalising cycloid equation

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?

$$P$$ 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!)