How can I parametrize $|x|+|y|=1$ 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
 A: 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 $\theta$:
$\quad\quad\quad\quad\quad\quad$
A: 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.
A: 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.
A: You could use $x = \cos t |\cos t|$, $y = \sin t |\sin t|$, $0 \le t \le 2\pi$.
A: $x = t, \; y = \pm \big|1 - |t|\big|$ such that $-1 \leq t \leq 1$.
A: 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:
$$
f(t)=\begin{cases}
(1-4t)A+4tB & \text{if $0\leq t\leq\frac14$} \\
(2-4t)B+(4t-1)C & \text{if $\frac14\leq t\leq\frac12$} \\
(3-4t)C+(4t-2)D & \text{if $\frac12\leq t\leq\frac34$} \\
(4-4t)D+(4t-3)A & \text{if $\frac34\leq t\leq1$} 
\end{cases}
$$
A: 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.
A: 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
$$t\in[-2,2]\to \\x=1-|t|,\\y=\text{sign}(t)(1-|1-|t||).$$
