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Question about graphing a shape on the cartesian coordinate system

  1. Aug 4, 2012 #1
    How would you graph a square of any dimension on the cartesian coordinate system?
     
  2. jcsd
  3. Aug 4, 2012 #2

    chiro

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    Hey mesa.

    What exactly do you mean? Do you mean instead of representing x you represent x2 for a particular vector (where x is the 1st element of the vector)?
     
  4. Aug 4, 2012 #3
    Any way you like, I just want to see other solutions than mine.
     
  5. Aug 4, 2012 #4

    chiro

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    Well you can do it in a few different ways.

    If you represent your system as <x2,y,z> then your co-ordinate system will look like a normal Cartesian system with the negative region for x < 0 removed entirely.

    However if you want to represent <x,y,z>, then you will get a contracted transformed version of the x-axis, but if you are going from the x2 to x, then you have the branch problem since you lose information of the sign by squaring the term.

    What happens is that for x as opposed to its square, the rate of change along the axis is not uniform or linear, but non-linear: in the linear case if you double the magnitude on the x-axis you double the x value but in a non-linear if you double the magnitude the value is not simple double: it is less or greater than.
     
  6. Aug 4, 2012 #5

    Mentallic

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    Maybe mesa means an actual square (quadrilateral)?
     
  7. Aug 5, 2012 #6
    Yes, an actual square.
    Sorry about that Chiro, I should have read your post more carefully and been more specific.
     
  8. Aug 5, 2012 #7

    chiro

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    Ohh I see what you're saying now.

    It depends on the number of dimensions and the particular attribute you are trying to visualize.

    Typically what happens in high-dimensional spaces, is that a specific kind of projection is used to create a visual of that attribute in a two or three dimensional space.

    The short answer is that it depends on the attribute you are trying to visualize, but the kinds of things would most likely be similar if not equivalent to the ones used in a normal cartesian geometry. (It's hard to give specifics without specifics).
     
  9. Aug 5, 2012 #8
    Basically I am looking for (an)other function(s) that will result in a graph of a plain old square on a two dimensional cartesian coordinate system.

    If there is more than one solution then pick your favorite, this is new territory for me although it was a great deal of fun working up a solution.

    Now my function can graph a square about the origin although I am having difficulty limiting the range/domain outside of it lol. I have also played with trying to make rectangles too off the original function but I changed x to being a dependent variable along with y so transformations and deformations seem to work very differently than what I am used to... Yikes!,
    should be fun :)
     
  10. Aug 5, 2012 #9

    LCKurtz

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    ##|x|+|y|=1##

    ##max \{ |x|,|y|\} = 1##
     
  11. Aug 5, 2012 #10
    That's a nice solution, and takes care of the domain/range problems I was having.
    Got anything else?
     
  12. Aug 5, 2012 #11

    LCKurtz

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    You can rotate and translate it to anywhere and resize it by changing the 1:$$

    |(x-a)\cos(\alpha)-(y-b)\sin(\alpha)|+|(x-a)\sin(\alpha)+(y-b)\cos(\alpha)| = 1$$
     
  13. Aug 5, 2012 #12
    Very interesting, okay from your answers I think I see how I might fix my domain/range issues.

    As it stands I have:

    [(y^2)^(1/2)-n]^2 = x^2

    with n being the number that fixes the size of the square. I used a trig based function that produced a square root of -1 to keep the domain/range in check outside the 'square' but it's just tacked on and even though it sort of works it's kind of crappy so I removed it in search of a proper solution.
     
  14. Aug 5, 2012 #13
    [tex]|x+y|-|x-y| = 1[/tex]
     
  15. Aug 5, 2012 #14
    Isn't there a problem at x=0 or am I missing something?
     
  16. Aug 5, 2012 #15

    LCKurtz

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    That's just a typo, he means ##|x+y|+|x-y|=1##. Same as my first one but squared up with the axes.
     
  17. Aug 6, 2012 #16
    You're right, thanks.
     
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