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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:

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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!

kaithkolesidou
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2 Answers2

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If the rectangle is $a\leq x\leq b$, $c\leq y\leq d$, its patametrization on the boundary: on the bottom side $y=c,\, x=t, a\leq t\leq b$; on the top side $y=d,\, x=t, a\leq t\leq b$; on the left side $x=a,\, y=t, c\leq t\leq d$ and on the right side $x=b,\, y=t, c\leq t\leq d$, where $t$ is parameter.

daulomb
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  • This is the parametrization I 've already had done before asking but I was hoping to find a more "general" parametric equation (which now I learned that it doesn't exist). However thanks a lot! – kaithkolesidou Jan 02 '18 at 15:19
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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:

There isn't a parametrization $\gamma : \mathbb S^1 \to R$ so that $\gamma'(t)$ is nonzero for all $t$.

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 $\gamma_1'(0) = 0$.

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    I see what you 're saying here and if you can I would also like to see the link of the proof. So in the case of a rectangle I can only find unit normal vectors on its sides since only its sides can be parametrized, am I right? – kaithkolesidou Jan 02 '18 at 15:17
  • Yes, away from the corners you can use the parametrization given by another answer. I will find/write a proof @kaithkolesidou –  Jan 02 '18 at 15:19
  • @Jonh Ma Ok. thanks a lot for your time! I'll wait – kaithkolesidou Jan 02 '18 at 15:20
  • @John Ma: I agree thanks for clarification. Does this mean that the divergence theorem is not applicable on the cube? – daulomb Jan 02 '18 at 15:28
  • @kaithkolesidou please see the edit. –  Jan 02 '18 at 15:31
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    @daulomb Actually divergence theorem is applicable on the cube, since the boundary is not too bad: the place where normal vector is not well-defined is very small (One can make that precise). –  Jan 02 '18 at 15:32
  • @JohnMa I don't get why $;γ_2(s)=0;$ for $;s \gt 0;$. If $;γ(s);$ lies in the horizontal sides then $;γ_2(s)=const;$...Why is the constant $;0;$? – kaithkolesidou Jan 02 '18 at 15:44
  • @kaithkolesidou I set the rectangle to be the standard one (but it suffices to know that it is constant, zero or not) –  Jan 02 '18 at 15:47
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    @JohnMa Ok, I think I got it now. Thanks again for your time! :) – kaithkolesidou Jan 02 '18 at 15:51