Parametrizing a Self-Intersecting Rectangle

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Homework Statement


Let S be the self-intersecting rectangle in ##\mathbb{R}^3## given by the implicit equation ##x^2−y^2z = 0##. Find a parametrization for S.

Homework Equations


The Attempt at a Solution


This is my first encounter with a surface like this. The first thing that came to my mind was letting ##z = f(x,y)## so that the parametrization can be given as:

##\Phi(u,v) = (u,v,\displaystyle \frac{u^2}{v^2})##

The problem I'm having is finding the limits for ##u## and ##v##. When I plugged the implicit equation into Mathematica, the surface looks like an X shape from above, though I had to expand the range for the axes by a very large amount for it to look like that. I do know that ##v## has to be non-zero, but that's pretty much it.
 
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Karnage1993 said:

Homework Statement


Let S be the self-intersecting rectangle in ##\mathbb{R}^3## given by the implicit equation ##x^2−y^2z = 0##. Find a parametrization for S.


Homework Equations





The Attempt at a Solution


This is my first encounter with a surface like this. The first thing that came to my mind was letting ##z = f(x,y)## so that the parametrization can be given as:

##\Phi(u,v) = (u,v,\displaystyle \frac{u^2}{v^2})##

The problem I'm having is finding the limits for ##u## and ##v##. When I plugged the implicit equation into Mathematica, the surface looks like an X shape from above, though I had to expand the range for the axes by a very large amount for it to look like that. I do know that ##v## has to be non-zero, but that's pretty much it.

If you look at the traces when ##z = c^2## for you get ##x^2-y^2c^2=0## so the level curves of that surface are ##(x+cy)(x-cy)=0##, which is two intersecting straight lines. I would think about letting one of the parameters be ##z=c^2,\ y = y## and doing it as two pieces. That would certainly solve your range problem and also eliminates your problem when ##c=0##.
 
I don't quite get why you set ##z = c^2 \ge 0##. For example, if ##x=0,y=0## and ##z## negative, that also satisfies the equation, right?
 
Karnage1993 said:
I don't quite get why you set ##z = c^2 \ge 0##. For example, if ##x=0,y=0## and ##z## negative, that also satisfies the equation, right?

Yes. It's a bit strange though. The ##z## axis solves the equation alright, but there is no surface if ##z<0##. You just have the bare axis. I wouldn't include it as part of the parameterization, but your mileage may vary.
 
Okay, so suppose one piece of the surface is when ##x = cy##. How would I go about finding the limits for ##c,y##?
 
Karnage1993 said:
Okay, so suppose one piece of the surface is when ##x = cy##. How would I go about finding the limits for ##c,y##?

You know ##c## parameterizes the ##z## axis which is ##z\ge 0## for our purposes, and we are using ##c^2## for convience, so ##c\ge 0## is easy. Then those straight lines go forever so you wouldn't have any limit on ##y##. Have you tried a parametric plot to compare with your original plot?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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