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?



It looks like you want the acceleration to be to the left of the direction of motion (or in the same direction) at all times as you go counterclockwise around the origin.
That is, taking $(\dot{x},\dot{y})$ as the velocity vector and
$(\ddot{x},\ddot{y})$ as the acceleration vector,
$$\ddot{x}\dot{y} - \dot{x}\ddot{y} \leq 0.$$
