I have the following parametrization of a flower curve:
$\theta \mapsto [(R+r\cos(6\pi\theta))\cos(2\pi \theta), (R+r\cos(6\pi \theta))\sin(2\pi \theta)]$
I'd like to know a level set $\phi(x,y)$ such that $\phi(x,y)=0$ associated to this curve.
Basically, I know that the map above is a parametrisation of my curve $\gamma$. However, I need a level set description of it, so I'm looking for a $\phi(x,y)$ such that the level set $\{ \phi =0 \}$ is $\gamma$.
On
Another way is to use a computer algebra package and eliminate the system of equations. This gives the following 8th degree polynomial: $$\Phi(x,y)=-R^2x^6 - 3R^2x^4y^2 - 3R^2x^2y^4 - R^2y^6 + r^2x^6 - 6r^2x^4y^2 + 9r^2x^2y^4 - 2rx^7 + 2rx^5y^2 + 10rx^3y^4 + 6rxy^6 + x^8 + 4x^6y^2 + 6x^4y^4 + 4x^2y^6 + y^8$$
You get this from, for example, eliminating $s$ and $c$ from the system of three equations: $[x-(R+r(4c^3-3c))c,y-(R+r(4c^3-3c))s,s^2+c^2-1]$ where it was used that: $$cos(6\pi\theta)=4cos^3(2\pi\theta)-3cos(2\pi\theta)$$ and $$sin^2(2\pi\theta)+cos^2(2\pi\theta)-1=0$$ and where: $$c=cos(2\pi\theta)$$ and $$s=sin(2\pi\theta)$$
Start with
$$ \begin{align*} x &= (R + r \cos(6 \pi \theta)) \cos(2 \pi \theta) \\ y &= (R + r \cos(6 \pi \theta)) \sin(2 \pi \theta) \end{align*} $$
And we want an equation involving $x,y,r,R$ but eliminating $r$ and $\theta$. There's no general approach for that, but let's see what we can get from these.
$$ \begin{align*} \sqrt{x^2+y^2} &= R + r \cos(6\pi\theta) \\ \frac{x}{\sqrt{x^2+y^2}} &= \cos(2\pi\theta) \\ \frac{y}{\sqrt{x^2+y^2}} &= \sin(2\pi\theta) \\ \cos(6\pi \theta) &= \cos^3(2\pi\theta) - 3 \cos(2\pi\theta)\sin^2(2\pi\theta) \\ \cos(6\pi\theta) &= (x^3-3xy^2)(x^2+y^2)^{-3/2} \\ \sqrt{x^2+y^2} &= R + r(x^3-3xy^2)(x^2+y^2)^{-3/2} \end{align*} $$
So one solution is
$$ \phi(x,y) = \sqrt{x^2+y^2} - r(x^3-3xy^2)(x^2+y^2)^{-3/2} - R $$
$\phi$ could also be multiplied by any function which is never zero.