I'm interested in constructing a unit normal vector on the boundary of a rectangle in $\;\mathbb R^2\;$ and so I found these steps:
However I'm having a really hard time completing step 0! How can I parametrize the rectangle in $\;\mathbb R^2\;$?
For example, I know that if I had a circle in $\;\mathbb R^2\;$ then my equation on the coordinate plane would be : $\;(x-x_0)^2+(y-y_0)^2=r^2\;$ and hence in order to parametrize this equation, I would set $\;y-y_0=rcost\;,\;x-x_0=rsint\;$.
I'm searching to find something similar for the case of rectangle but it must be so obvious that nobody mentions about it! What is the parametric equation of a rectangle in 2 dimensions?
Any help would be valuable... Thanks in advance!


Not all "curves" in $\mathbb R^2$ admits a normal vector, and not all "curves" in $\mathbb R^2$ admits a "parametrization". For example, pictorially it is not possible to find a normal vector at the four corners of your rectangle $R$. Similarly:
To prove this, for simplicity assume that the rectangle $R$ has the four corners $(0,0), (1,0),(0,1),(1,1)$. I claim that if $\gamma : (-\epsilon, \epsilon) \to R$ is differentiable, one to one and $\gamma (0) = (0,0)$, then $\gamma'(0) = 0$.
Write $\gamma (t) = (\gamma_1(t), \gamma_2(t))$. We need to show $$\gamma_1'(0) = \gamma_2'(0) = 0.$$
Since $\gamma$ is one to one, we may assume that $\gamma(s)$ lies in the horizontal sides for $s>0$ and $\gamma(u)$ lies in the vertical sides when $u<0$. So $\gamma_2(s) = 0$ for $s>0$ and $\gamma_1(u) = 0$ for $u<0$. This implies
$$ \gamma_2(0) = \lim_{s\to 0^+} \frac{\gamma_2(s) - \gamma_2(0)}{s-0} = 0$$
and similarly $\gamma_1'(0) = 0$.