Superellipse parametric equation

Question

$$\left|\frac{x}{a}\right|^n+\left|\frac{y}{b}\right|^n=1,\qquad\quad n\gt2$$ $$\text{can be rewritten as}$$ $$x(t)=a\,\cdot\,\left|\cos(t)\right|^{2/n}\quad\vee\quad x(t)=-a\,\cdot\,\left|\cos(t)\right|^{2/n}$$ $$y(t)=b\,\cdot\,\left|\sin(t)\right|^{2/n}\quad\vee\quad y(t)=-b\,\cdot\,\left|\sin(t)\right|^{2/n}$$

How is it that the equation of a superellipse can be rewritten as above. I do understand that they use $\cos^2 t+sin^2 t=1.$ However, I don't understand why $a$ and $b$ can be negative or positive?

Answer 1

Both the parametic expressions satisfy the superellipse equation, the ones with the positive sign represent the superellipse in the first quadrant, the ones x-negative and -positive in the second, and so on.

Indeed:

$$\left|\frac{x}{a}\right|^n+\left|\frac{y}{b}\right|^n=\left|\frac{\pm a\,\cdot\,\left|\cos(t)\right|^{2/n}}{a}\right|^n+\left|\frac{\pm b\,\cdot\,\left|\sin(t)\right|^{2/n}}{b}\right|^n= \\=\left|\pm 1\,\cdot\,\left|\cos(t)\right|^{2/n}\right|^n+\left|\pm 1\,\cdot\,\left|\sin(t)\right|^{2/n}\right|^n=|\pm 1|\,\cdot\,\left|\cos(t)^{2/n}\right|^n+|\pm 1|\,\cdot\,\left|\sin(t)^{2/n}\right|^n=\\=\cos^2(t)+\sin^2(t)=1$$

Suppose wlog $a,b >0$ otherwise just switch the sign, thus we have

Firt quadrant $t\in[0,\frac{\pi}{2})$: $$\begin{cases} x(t)=a\,\cdot\,\left|\cos(t)\right|^{2/n} \\ y(t)=b\,\cdot\,\left|\sin(t)\right|^{2/n} \end{cases}$$

Second quadrant $t\in[\frac{\pi}{2},\pi)$: $$\begin{cases} x(t)=-a\,\cdot\,\left|\cos(t)\right|^{2/n} \\ y(t)=b\,\cdot\,\left|\sin(t)\right|^{2/n} \end{cases}$$

Third quadrant $t\in[\pi,\frac{3\pi}{2})$:

$$\begin{cases} x(t)=-a\,\cdot\,\left|\cos(t)\right|^{2/n} \\ y(t)=-b\,\cdot\,\left|\sin(t)\right|^{2/n} \end{cases}$$

Forth quadrant $t\in[\frac{3\pi}{2},2\pi)$: $$\begin{cases} x(t)=a\,\cdot\,\left|\cos(t)\right|^{2/n} \\ y(t)=-b\,\cdot\,\left|\sin(t)\right|^{2/n} \end{cases}$$

Answer 2

It's required for the symmetry, so that the curve can exist in all four quadrants.Perhaps a better way to write this is

$$ x=a|\cos t|^{2/n}\text{sgn}(\cos t)\\ y=b|\sin t|^{2/n}\text{sgn}(\sin t)\\ t\in[0,2\pi] $$

where $\text{sgn}$ is the sign function.

Answer 3

You don't need to use quadrants for the superellipse equation, it is also possible write it as one single equation:

The polar form of the superellipse is

$\left (\begin{array}{l}r=a\cdot b\cdot \dfrac{1}{\left (\left |a\cdot sin(t)\right |^{n}+\left |b\cdot cos(t)\right |^{n}\right )^{\dfrac{1}{n}}}\\\end{array}\right )$

and to convert the polar equation into a parametric equation:

$\left (\begin{array}{l}x=r\cdot cos(t)\\y=r\cdot sin(t)\end{array}\right )$