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I never had seen parametric equations, but while trying to learn line integrals through Wikipedia, quickly found these equations are remarkable. Some can represent things for which more normal equations or functions are needed, if at all possible.

However, the page about them in Wikipedia is not very long, and I didn't learn as much about them as I would like. Being so, I didn't find described any limitation for them. They can make many more finite shapes than normal equations.

But can define stuff with clear ends, such as a curve with two separate end-points, or just a line segment? I have no idea how to find these hypothetical parametric equations, as I said, I didn't find limitations, nor examples to this. Is this possible, or does it require "parametric inequations"?

Jim
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JMCF125
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  • You're only limited by the set of functions you know - $+, \times, \sin, \log$, etc. There is a function $f$ such that $f(x) = 0, x < 0$ and $f(x) = x, x < 1$, otherwise $f(x) = 1$. What's stopping you from using it? – Karolis Juodelė Dec 06 '13 at 20:11

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The parametric equations $$x = t^2 \qquad y = 0$$ give a line with one endpoint. The parametric equations $$x = \sin(t) \qquad y = 0$$ give a line with two end points.

Jim
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    Note to the OP: each of the examples here are degenerate ("extreme") cases of more familiar objects. The family of equations $x = t^2, ; y = at$ define parabolas with vertex at the origin that open to the right. As $a$ shrinks they get "skinnier" until at $a = 0$, you have a ray.

    Similarly, $x = \sin t, ; y = a \cos t$ gives a family of ellipses that all have horizontal radius of $1$, and whose vertical radius is $a$. As $a$ vanishes, the ellipses get squashed down to a horizontal line segment.

     – Sammy Black Dec 06 '13 at 20:07
  • @Sammy, thanks, something did sound fishy. It sounds like cheating now. – JMCF125 Dec 06 '13 at 20:09
  • There's nothing wrong with these parametrizations, per se; they're just unusual. (They do fail to be smooth at those endpoints, so all of the usual tools of calculus breakdown there.) – Sammy Black Dec 06 '13 at 20:37
  • @Sam, I'll add more criteria to my question (as referred in the last comment) ASAP. – JMCF125 Dec 06 '13 at 21:35
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$$\begin{align} x &= 0 \\ y &= \sin t \end{align}$$

Karolis Juodelė
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You do this by defining an interval for the parameter.

Take a simple line: $x = y = z = t.$ The (infinite) line implies that the parameter $t$ can take on any value.

The line segment from $(-1, -1, -1)$ to $(2, 2, 2)$ means restricting $t$ to the interval $[-1, 2]$.

John
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    I mean, how can I do it without defining the interval, only with parametric equations, and nothing else. – JMCF125 Dec 06 '13 at 20:08
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    A parametrization is a set of equations together with explicit bounds on the parameter. Each coordinate is a function $x = f(t)$, $y = g(t)$, etc., and every function has a domain on which it is defined. When you don't describe the domain of the parametrization, you are probably assuming that it is all of $\Bbb{R}$ by default. – Sammy Black Dec 06 '13 at 20:11
  • @SammyBlack, but when part of the result is outside the default $\mathbb R$, such as in a function $f:\mathbb R\to\mathbb R,\ f(x)=y=\sqrt{x^2-1}$ I have half a circle. Why? I'm profiting to create imaginary numbers for $-1<x<1$, and may omit that inequation, as imaginary numbers are cut out by the default $\mathbb R$. This is the kind of thing I was searching for, as referred in other answers. You do, nonetheless, make a good case, and I'll consider adding something alike what I'm writing in this comment to the question. – JMCF125 Dec 06 '13 at 20:33