|Apr22-12, 12:54 AM||#1|
please help with this level curve
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...
|Apr22-12, 12:09 PM||#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.
|Apr22-12, 02:38 PM||#3|
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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