Parametric Curve Representation of a Square from a Circle

Question

Given the parametric equation of a unit circle $$ \vec r(\theta) = \begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix}, \quad 0 \leq \theta \leq 2\pi $$

It seems that there is some function $$ f : \mathbb{R} \rightarrow \mathbb{R} $$

such that $$ \vec s(\theta) = f(\theta)\vec r(\theta), \quad 0 \leq \theta \leq 2\pi $$ where $\vec s(\theta)$ is the parametric equation of a square with side length $2$.

Can this function $f$ be found, and if so, what is it?

Answer 1

Such function is just:

$$ f(\theta) = \frac{1}{\max(|\sin\theta|,|\cos\theta|)}.$$

Answer 2

For the parameterization of the square, We can define such a function piecewise. For the first(and last) octant, consider that we have a right triangle, with one leg 1, the adjacent angle $\theta$. Therefore $x=1$ and $y = \tan(\theta)$.

This gives $f = \frac 1 {\cos \theta}$ on this region.

You can construct similar parameterizations for the other 4 pieces with rotations about the origin; yielding $f$ as described by Jack, $\frac{1}{\max(|\sin(\theta)|,|\cos(\theta)|)}$

Answer 3

Found this post and thought I would share an equation I derived for "converting" a circle to a square using a ratio method that is also a parametric equation, I am sure that someone else has probably done something similar.

Start by plotting a circle of radius R:

I have just given the equation, but this is represented as:

You rotate the square just by offsetting theta by 45°, what also comes out of this equation is some nice plots, eg a heart: others: