# Hypocycloid: A Mysterious Curve.

1. Feb 28, 2013

### Tenshou

1. The problem statement, all variables and given/known data
The hypocycloid is the plane curve generated by a point $P$ on the circumference of a circle $C$, as this circle rolls with out sliding on the interior of the fixed circle $C_0$. If $C$ has a fixed radius of $r$ and $C_0$ is at the origin with radius $r_0$ and the initial location of the point $P$ is at $(r_{0}, 0)$, what is the representation of a hypocycloid

2. Relevant equations

3. The attempt at a solution

I didn't know how to start but to draw a picture. then I thought about the Epicycloid... these curves are related? I didn't know what to do so I played around, and I don't know if they are right or not. s.th. $\delta_{0} = ((r+r_{0})/(r_{0}))\theta$ $(r+r_{0})cos(\delta_{0})-r cos(\delta_{0}) + (r+r_{0})sin(\delta_{0}) + r sin(\delta_{0})$ I know this isn't right but just what I was thinking they could be...

2. Mar 1, 2013

3. Mar 1, 2013

### Tenshou

That isn't much help, =/ I thought someone would be able to work out the problem and tell me how they got the answer because it doesn't make sense, I am not looking for the answer I am looking for a solution... I know it sounds redundant, but look... I know the answer already but I do not know how to reach the solution, so if anyone can help me out that would be awesome

4. Mar 1, 2013

### Whovian

Well, first you can try to express it parametrically. Let t be a time parameter (letting the circle roll with constant angular velocity,) and try to find x and y in terms of t. Then it's possible to just convert back to rectangular by solving for a relationship between x and y.

5. Mar 1, 2013

### haruspex

Ok, that was not clear. In the equations at http://en.wikipedia.org/wiki/Hypocycloid#Properties, the centre of the small circle, O', is at location (R-r, θ) in polar coordinates. Suppose P is at (r, ψ) in polar coordinates relative to O'. rψ = (R-r)θ (do you see why?). Converting to Cartesian and adding up the x and y coordinates separately gives the equations.