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Sketching level curves

  1. Jan 29, 2009 #1
    [tex]x^{2}-y^{2}-2x+4y+5;[/tex]

    let [tex]x^{2}-y^{2}-2x+4y+5 \ = \ c; [/tex]

    To sketch this as a level curve, I'm not sure how to proceed. I can't seem to rearrange the function into anything familiar.

    For the sake of trying to find a reference point, I let x=0 and found

    [tex] y \ = \ 2 \ ^{+}_{-}\sqrt{9-c};[/tex]

    then y=0 =>

    [tex] x \ = \ 1 \ ^{+}_{-}\sqrt{-4+c};[/tex]

    If I let c = 5, I get

    [tex]x^{2}-y^{2}-2x+4y = 0; [/tex]

    which gives x = 0, x = 2, y = 0, y = 4.

    What should I do next?
     
  2. jcsd
  3. Jan 29, 2009 #2
    Level curves are usually for functions f(x,y) = blah i.e. when you have 2 ind. variables. Here you only have one ind.... so you can solve for y by completing the square
     
  4. Jan 29, 2009 #3
    The function that I am trying to draw level curves for is

    [tex] V(x,y) =x^{2}-y^{2}-2x+4y+5;[/tex]
     
  5. Jan 29, 2009 #4
    Well you should be able to rewrite it by completing the square, I'll work this one as example

    [tex]V(x,y) = x^2 - y^2 - 2x + 4y + 5 = (x-1)^2 - (y-2)^2 + 8 [/tex]

    Assuming my algebra is right, does that help you see it?
     
  6. Jan 30, 2009 #5
    Ok, yeah I see that. It's a translated hyperbola?

    I keep getting

    [tex]V(x,y) = (x-1)^{2} - (y-2)^{2} +5; [/tex]

    Which is a hyperbola centered at (1,2) and translated somewhere 5 units?
     
  7. Jan 30, 2009 #6
    How are you getting that?

    [tex] (x-1)^2 - (y-2)^2 + 5 = x^2 - 2x + 1 - (y^2 - 4y + 4) + 5 = x^2 - y^2 - 2x + 4y + 2 [/tex]

    That's not what you started with?
     
  8. Jan 30, 2009 #7
    I should have been paying better attention.

    How would I graph this? My cal books just have all these guys equal to 1.

    [tex] V(x,y) = (x-1)^2 \ - \ (y-2)^2 \ + \ 8 \ = \ c; [/tex]
    [tex] V(x,y) = (x-1)^2 \ - \ (y-2)^2 \ = \ c \ - 8; [/tex]
     
  9. Feb 1, 2009 #8
    Well, if [itex](x - 1)^2 - (y - 2)^2 = c - 8[/itex], then

    [tex]\frac{(x - 1)^2}{c - 8} - \frac{(y - 2)^2}{c - 8} = 1.[/tex]

    That should be in a slightly more familiar form.
     
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