How can I parametrize $|x|+|y|=1$

Question

I need parametrize $|x|+|y|=1$ but I don't know how to parametrize. I know that it is a rotated square, I would like understand so if you can explain to me like if I was still, thanks

Answer 1

Starting at $(1, 0)$ and going counter-clockwise around the origin, something like $$ \cases{x(t) = |t-2| - 1\\y(t) = |t-3| - |t-1| + t -2},\quad t\in[0,4] $$ should work.

Answer 2

$x = t, \; y = \pm \big|1 - |t|\big|$ such that $-1 \leq t \leq 1$.

Answer 3

Assume we want to find an equation of the form $r(\theta)>0$ that will describe the square, where: $$x= r(\theta)\cos(\theta),\quad y=r(\theta)\sin(\theta)$$We have that: $$|r(\theta)\sin \theta|=1-|r(\theta)\cos \theta|$$ Or factoring out $r(\theta)$ (since we know $r>0$): $$r(\theta)= \frac{1}{|\cos\theta|+|\sin\theta|}$$

To visualize $x(\theta)$ and $y(\theta)$, see below as a function of $\theta$:

$\quad\quad\quad\quad\quad\quad$

Answer 4

You observed that all points $(x, y)$ satisfying $|x| + |y| = 1$ is a path forming a diamond. Since paths are 1-dimensional lines, you can describe the path as a function of one variable, as $(x(t), y(t))$.

One option for defining $(x(t), y(t))$ is to start at $(0, 1)$ at $t=0$, and move counter-clockwise around the origin, with $t=1$ being at $(1, 0)$, and when $t=2$ the path is at $(-1, 0)$, at $t=3$ the path is at $(0, -1)$, and at $t=4$ the path has returned to $(1, 0)$. That can be written out as:

$$\begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \begin{cases} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} -1 \\ 1 \end{bmatrix}t \quad &\text{ for } 0 \le t \le 1 \\ % \begin{bmatrix} 0 \\ 1 \end{bmatrix} + \begin{bmatrix} -1 \\ -1 \end{bmatrix}(t - 1) \quad &\text{ for } 1 \le t \le 2 \\ % \begin{bmatrix} -1 \\ 0 \end{bmatrix} + \begin{bmatrix} 1 \\ -1 \end{bmatrix}(t - 2) \quad &\text{ for } 2 \le t \le 3 \\ % \begin{bmatrix} 0 \\ -1 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix}(t - 3) \quad &\text{ for } 3 \le t \le 4 \\ \end{cases}$$

If the problem had been a 2 dimensional shape, such as the face of a cube, we would have had to use 2 parameter variables, such as $(x(s, t), y(s, t))$. But for one dimensional shapes you only need 1 variable.

Answer 5

Let $A=(0,1)$, $B=(1,0)$, $C=(0,-1)$, $D=(-1,0)$. The locus $|x|+|y|=1$ is the square with vertices $ABCD$.

Define $f:[0,1]\rightarrow\Bbb R^2$ as follows: $$ f(t)=\begin{cases} (1-4t)A+4tB & \text{if $0\leq t\leq\frac14$} \\ (2-4t)B+(4t-1)C & \text{if $\frac14\leq t\leq\frac12$} \\ (3-4t)C+(4t-2)D & \text{if $\frac12\leq t\leq\frac34$} \\ (4-4t)D+(4t-3)A & \text{if $\frac34\leq t\leq1$} \end{cases} $$

Answer 6

The most natural parametrization is probably $\gamma:[0,2\pi)\to\mathbb{R}^2$ given by: $$ \gamma(t)=\left(\frac{\cos\theta}{|\sin\theta|+|\cos\theta|},\frac{\sin\theta}{|\sin\theta|+|\cos\theta|}\right) $$ i.e. the re-scaled circle.

Answer 7

From the implicit equation,

$$y=\pm(1-|x|)$$with $x\in[-1,1]$.

Let $x$ run from $-1$ to $1$, following the lower branch and using a parameter $t$:

$$t\in[-2,0]\to x=1+t,y=|x|-1=|1+t|-1.$$

Then let $x$ run from $1$ to $-1$, following the upper branch:

$$t\in[0,2]\to x=1-t,y=1-|x|=1-|1-t|.$$

We can condense in a single expression

$$t\in[-2,2]\to \\x=1-|t|,\\y=\text{sign}(t)(1-|1-|t||).$$

Answer 8

You could use $x = \cos t |\cos t|$, $y = \sin t |\sin t|$, $0 \le t \le 2\pi$.