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The parameterization of the circle in rectangular coordinates is given by the the following functions

$$ y = g_1(x) = \sqrt{(1-x^2)} \\ y = g_2(x) = -\sqrt{(1-x^2)} \\ x = g_3(y) = \sqrt{(1-y^2)} \\ x = g_4(y) = -\sqrt{(1-y^2)} $$

for $-1<x<1$ and $-1<y<1$. However the final two (or the first two) are only required because the interval where $x$ (or $y$) is defined is open. Why does that interval need to be open?

user1790813
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  • This question is a little confusing. By the interval being open, do you mean interval for each equation alone or by both combined such that $(x=...,y=....)$. – Arbuja May 29 '16 at 12:16
  • Sorry for the confusion. I was referring to the intervals $-1< x < 1$ and $-1<y<1$. – user1790813 May 29 '16 at 12:22
  • I found that the intervals are closed as long as the output is defined. Yet this question is out of my understanding so it would great if someonelse can answer this. – Arbuja May 29 '16 at 12:37
  • You don´t need the intervals be open if you require just the corresponding values for, say, $y$ in the two first equations. Another thing is if you asked about other properties such as differentiability because you can see geometrically that the slope at $(0,1)$ shows trouble but clearly $y=0$ – Piquito May 29 '16 at 13:57

1 Answers1

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Parametrizations are usually taken to be regular, i.e., continuously-differentiable with non-zero derivative.

The notion of differentiability at endpoints is a dicey matter in general, but for the functions $x \mapsto \pm\sqrt{1 - x^{2}}$ the problem is blunt: There's no differentiable extension to $[-1, 1]$, since the graphs have vertical tangents at the endpoints.

Andrew D. Hwang
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