Create paramatric shape wihtout 'dents' I am plotting a shape with the following equation
$$\left\{
\begin{array}{c}
x=r_{in} \cos(4 t)+r_{out} \cos(t)\\
y=r_{in} \sin(4 t)+r_{out} \sin( t)
\end{array}
\right.
$$
Given various parameters for $r_{in}$ and $r_{out}$ this will give a 'smooth' shape or a 'dented' shape (see below, I don't know how to scale images down...). In other words, in some cases a line tangential to the shape will intersect the shape, in some cases not. I would like the 'smooth' version, but it seems silly to just use trial-and-error to determine the $r_{in}$ and $r_{out}$ values (or ratio).
I figured that for this to be the case, I would need the acceleration to be always negative (or zero) in the radial direction.
Determining the acceleration is easy,
$$\left\{
\begin{array}{c}
\ddot{x}=-16\ r_{in} \cos(4 t)-r_{out} \cos(t)\\
\ddot{y}=-16\ r_{in} \sin(4 t)-r_{out} \sin( t)
\end{array}
\right.
$$
But now how do I 'extract' the radial component so I can say create and solve the inequality $$\vec{a}_{r}(t)<0\ \forall\ t$$
Or is there a much easier way I'm overlooking?



A little bit of playing shows that $r_{in}$ and $r_{out}$ are not independent.
"Sharp" corners occur at the point when we change from a curve that intersects itself to one that doesn't. Whether you want those solutions is up to you, but it will be a matter of including equality in an inequation (or not).
We get a curve which intersects itself when there exist $s,t$ both in the half-open interval $(0, 2\pi)$ such that $s \ne t$ and $x(s) = x(t)$ and $y(s)= y(t)$. Thus we have two equations, and given a value of $t$ and a value of $r_in$ we can determine values of $s$ and $r_out$ that give solutions.
We want the complement to such a set of solutions.
