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Please help with this level curve

  1. Apr 22, 2012 #1
    sketch the level curve z=(x^2-2y+6)/(3x^2+y) at heights z=0 and z=1

    i have already compute the 2 equations for the 2 z values and drawn it in 2d but when it comes to plotting it with the extra z axis i don't know what to do. please help...
     
  2. jcsd
  3. Apr 22, 2012 #2
    As your title suggests, what you need is a 'level curve', not a level surface.

    Maybe I'm wrong, but if I had to do this exercise, I'd find y(x) for z=1 and z=0 and draw them, as you say have done already. I think you're done already.
     
  4. Apr 22, 2012 #3

    HallsofIvy

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    There is no "extra z-axis". There is no z-axis at all! The level curves you are asked to draw are in the xy-plane. Graph [itex](x^2-2y+6)/(3x^2+y)= 0[/itex] and [itex](x^2-2y+6)/(3x^2+y)= 1[/itex].

    The first is easy- it is the same as the graph of [itex]x^2- 2y+ 6= 0[/itex] which is just the parabola [itex]y= (1/2)x^2+ 3[/itex]. The second is not much harder: [itex]x^2- 2y+ 6= 4x^2+ y[/itex] or [itex]y= 6- 3x^2[/itex], also a parabola.
     
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