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Meme Yann
Apr 03, 2020

Converting into parametrics equations

in Math Problems

Hello everyone!


I would like to know if it was possible to convert the implicit equation: x^2 - y^2 = 0 into a parametric equation.



13 comments
0
Eric Steckx
Apr 04, 2020

If you don't put bounds, I think it's not possible. The graphic of your equation is a cross with four infinite branches and with one paramater, you can describe complicated curves but only with two infinite extensions.

If your variables are bounded, maybe it's possible.

Meme Yann
Apr 04, 2020

So basically, I wanted to try to make a parametric equation of this equation to infinity and I did some research to transfer an implicit to parametric equation. Still, the case of x^2 + y^2 = 0 is quite complex as you actually said. Because from what I see, when I switch from implicit to parametric, I find some weird things. I actually think you're right, that's impossible.

When you say "If your variables are bounded, maybe it's possible", what do you mean?

0
Eric Steckx
Apr 15, 2020

Hello ! Sorry to be so late.

If x (and y) are bounded between -n and n (open interval), you may try :

x=n.sin(t) and y=n.tan(t).∣cos(t)∣ for 0 ≤ t ≤ 2𝜋 but excluding 𝜋/4 and 3𝜋/4

I think it's ok : check, please!

Ian Fowler
Apr 15, 2020

Not trying to be a PITA, but:


1) |cos(t)| / cos(t) = +/- 1 for all values of t ( except when t = multiple of pi/2) so y reduces to +/- sin(t) and x = sin(t) except at t = mult pi/2 when you get a jump discontinuity. But, as t--> pi/2, y --> +/- n.


2) Restricting t to 0<= t <= 2pi only gives you a portion of the 2 lines where x and y are trapped in the square between +/- n. And it doesn't change anything if you let t range over all the real numbers. You are still trapped in the square where x and y are stuck between +/- n.


3) So x = n.sin(t) and y = +/- n.sin(t) except when t = mult of pi/2 where y does not have a value. You might as well use x = n and y = +/-n where you get all of the infinite pairs of solutions and no discontinuities.

Ian Fowler
Apr 07, 2020  ·  Edited: Apr 07, 2020

x^2 - y^2 = 0 describes 2 straight lines: y = x and y = -x


So let x = t and then y = t or y = -t where t is any real number.

This is sometimes referred to an a "degenerate hyperbola".


For the rectangular hyperbola, x^2 - y^2 = a^2, the asymptotes are the 2 lines y = x and y = -x. As a tends to 0, then the branches of the hyperbola itself actually approach the the 2 lines y = x and and y = -x and when a actually becomes 0 then the hyperbola turns into the 2 lines - hence the term "degenerate"


As a tends to infinity nothing really special happens. The hyperbola still exists has as the same 2 asymptotes, y = x and y = -x . This prevents the hyperbola approaching the 2 vertical lines x = a and x = -a. It just takes larger and larger of values of x to get arbitrarily close to to the asymptotes.


The case where x^2 + y^2 = 0 is the simplest case of all. It describes a single point (0,0)

For the circle x^2 + y^2 = a^2, the radius is a and as a tends to 0 the circle shrinks and approaches the center (0,0) and when a actually takes on the value of zero we get just the point (0,0). Hence a single point is sometimes called a "degenerate circle". As a tends to infinity we just get a larger and larger circle.


The curvature of the circle x^2 + y^2 = a^2 is 1/a so I suppose we can argue the as a tends to infinity the curvature tends to 0 - which is the curvature of a straight line but there is no single equation which can represent such a line.



Devansh Singh
Apr 17, 2020

y=t, x=t or y=-t, x=-t.

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Ian Fowler
Apr 17, 2020  ·  Edited: Apr 17, 2020

both of these pairs of equations only pick up the line y=x (i.e. each pair plots the same line) and not the line y=-x and therefore you lose half of the solutions.


The only way to pick up the other half is x = t ; y = -t.


So from above: y=t, x=t or y=-t, x=-t :

You have to change drop one of the - signs to get : y=t, x=t or y=t, x=-t.

Otherwise you lose the line y = -x.