1. Apr 22, 2012

ronho1234

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. Apr 22, 2012

jacobrhcp

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.

3. Apr 22, 2012

HallsofIvy

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 $(x^2-2y+6)/(3x^2+y)= 0$ and $(x^2-2y+6)/(3x^2+y)= 1$.

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