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Help with level curves

  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

    Mentallic

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    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.
     
  4. Apr 22, 2012 #3

    ehild

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    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
     
  5. Apr 22, 2012 #4
    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
     
  6. Apr 22, 2012 #5

    Mentallic

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    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:

    [tex]z=\frac{x^2-2y+6}{3x^2+y}[/tex]

    [tex]3zx^2+3zy=x^2-2y+6[/tex]

    [tex](3z-2)y=(1-3z)x^2+6[/tex]

    [tex]y=\frac{(1-3z)x^2+6}{3z-2}[/tex]

    Now what we can gather from this is that for each level curve of z, we will get some parabola of the form [itex]ax^2+b[/itex] 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?
     
  7. Apr 22, 2012 #6
    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....
     
  8. Apr 22, 2012 #7
    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???
     
  9. Apr 22, 2012 #8

    Mentallic

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    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 [itex]z=x^2+y^2[/itex] 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:

    http://www.google.com.au/search?sourceid=chrome&ie=UTF-8&q=z%3Dx%5E2%2By%5E2
     
  10. Apr 22, 2012 #9
    umm so i did the double derivative of y=f(x,z) and i found z= 1/3 so is it at this point that the concavity changes?
    i did try to draw some level curves on the xy plane for z=2 and z=3 i found out that they were both concave down, so i would have z=o level curve concave up and z=1 z=2 z=3 concave curve concave down , so if these values were to be plotted on an xyz graph would this mean that my final level curve by a saddle that is rising/steep around the xz plane and going flatter around z=0 where the y axis is???

    sorry if this is going on for a long while i just can't seem to get it...
     
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