MHB What is the direction of the velocity vector on the rising side of a cycloid?

[Ackbach]

Here is this week's POTW:

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It is known that if you take a circle and roll it without slipping on a flat surface, a single point on the circle traces out the path of a cycloid. Show that the direction of the velocity vector for any point on the rising side of a cycloid is directed toward the highest point on the generating circle.

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[Ackbach]

Congratulations to greg1313 for his correct answer, which I reproduce below:

Spoiler

The parametric equations for a cycloid are given by

$$x=r(T-\sin T)$$

and

$$y=r(1-\cos T)$$

with $$T$$ in radians.

W.L.O.G. let $$r=1$$.

$$\frac{dy}{dx}=\frac{\sin T}{1-\cos T}$$ -- this is the "slope" of the velocity vector.

The coordinates of the highest point of the generating circle are $$(x,y)=(T,2)$$.

The coordinates of the point of the generating circle where it contacts the cycloid are $$(x,y)=(T-\sin T,1-\cos T)$$

The slope of the segment from the point of contact to the highest point of the generating circle is

$$\frac{2-(1-\cos T)}{T-(T-\sin T)}=\frac{1+\cos T}{\sin T}=\frac{\sin T}{1-\cos T}$$

As this is equivalent to $$\frac{dy}{dx}$$ the velocity vector points to the highest point of the generating circle (on the rising side of the cycloid).