# Homework Help: Help with level curves

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

### Mentallic

Well if you have your 3-d axes, then at z=0 you simply draw the equation you found (always stick on the plane z=0 and draw it as you would have drawn the equation in 2-d on the x-y axis) then at z=1 draw the second equation.

3. Apr 22, 2012

### ehild

You do not need to draw the z axis. Plot the level curves in 2D on the x,y plane and mark the curves with z=0 and z=1. Think of a map, the level lines are marked with the height.

ehild

4. Apr 22, 2012

### ronho1234

umm okay so if i draw my two equations on a xy plane i get a concave up with intersect 3 for z=0 and for z=1 i get a concave down intersect 2 and roots -root3 and root3... so are you telling just to leave it at that?...

and for a 3d axis because i have situated my x at the bottom the y then z anticlockwise this will mean that my concave up parabola would be flat on the y axis and my concave down will be like a saddle-like... i just can't see what the final level curve is suppose to look like when i have two separate parabolas

5. Apr 22, 2012

### Mentallic

Yes sorry, as ehild said, your level curves should be drawn in 2d on the x-y plane with a label of which z value it's following.

If you're struggling to figure out what the shape of curve in 3d space is supposed to look like, try solve the equation for y and see if that sheds any light:

$$z=\frac{x^2-2y+6}{3x^2+y}$$

$$3zx^2+3zy=x^2-2y+6$$

$$(3z-2)y=(1-3z)x^2+6$$

$$y=\frac{(1-3z)x^2+6}{3z-2}$$

Now what we can gather from this is that for each level curve of z, we will get some parabola of the form $ax^2+b$ for some constants a,b so the parabola will vary from being concave up and concave down, with differing intersections on the y-axis. This would suggest it has the shape of a saddle, yes?

6. Apr 22, 2012

### ronho1234

yes i kinda get what you're saying its just that because i have two separate parabolas on my xyz axis i don't know where the critical point on the saddle would be and where it would connect and how exactly its suppose to look like....

7. Apr 22, 2012

### ronho1234

since because the question asks only for z=0 and z=1 should i just leave 2 separate parabolas on my xyz axes completely separate of should i have them opposite each other on a xy plane labelled z=0 and z=1???

8. Apr 22, 2012

### Mentallic

Well can you figure out what value of z the parabola goes from a positive to a negative concavity? Take a look at the function y=f(x,z).

On your xyz axes you should have the parabola drawn at z=0, and the other parabola drawn at z=1. This is to be used as a template to help you draw the curve for all z.

For example, if I wanted to draw $z=x^2+y^2$ then if I take a few level curves, say for z=0,1,2 then I can use those to get a rough idea of what it should look like in 3-d, and it'll turn out as so: