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Describe the level surface

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Describe the level surface

    2. Relevant equations

    f(x,y,z) = 3x - y + 2z

    3. The attempt at a solution

    f(x,y,z) = 3x - y + 2z

    3x - y + 2z = w


    f(-1,1,2) for w <= 0 <= is a plane

    regards
    Brendan
     
  2. jcsd
  3. Mar 17, 2009 #2
    What do you mean by

    ?

    The solution may be easier then you think.
     
  4. Mar 17, 2009 #3
    The only example I've been given is:

    Describe the level surface of
    w = x^2+(y-1)^2+(z-2)^2

    we use
    x^2+(y-1)^2+(z-2)^2= k where K is a constant

    Which describes a surface which is a sphere of radius sqrt(k) and center (0,1,2)
     
  5. Mar 18, 2009 #4

    lanedance

    User Avatar
    Homework Helper

    so as as you said a level surface is given by
    F(x,y,z) = w for a constant w

    so for a given k, what is the surface described by
    f(x,y,z) = 3x - y + 2z = w

    if you're unsure where to start, pick a w, say w = 0 and figure out what surface it is. Then look at what happens for different w. I find it also helps to look at the coordinate axis, say so what happens to z for different w values with x=y=0

    as for your post
    has no real meaning, f(-1,1,2) represents a scalar value and i'm not sure what you mean by your comaprison sybols for w

    however as a hint
    is close to the money
     
  6. Mar 18, 2009 #5
    I know that the w is a constant so if w = 0

    3x - y + 2z = w

    This represents all the values of x y z which make the function = 0

    thats where I got the f(-1,1,2) from because the function = 0 at those values.
    But I know there are many vaules that would make the function = 0.

    I know that there are no squares so its probabably not an ellipsoid or sphere.
    I'm having a lot of trouble knowing how to graph it.

    The example I've been given has a center of (0,1,2). I know that those figures make the function = 0 but why did the figures get chosen?

    Brendan
     
  7. Mar 18, 2009 #6

    Mark44

    Staff: Mentor

    For each value of w, you can graph 3x - y + 2z = w.

    For w = 0, the equation is
    3x - y + 2z = 0
    What does this look like. With one equation in three unknowns, you can pick any two and find the third.

    Pick about 10 points and graph this surface.

    If w = 5, the equation is
    3x - y + 2z = 5
    Do the same as above.

    If w = -10, the equation is
    3x - y + 2z = -10
    Do the same as above.

    Repeat until it clicks what the level surfaces look like.
     
  8. Mar 18, 2009 #7
    Hi ,
    I've been graphing the surface 3x - y + 2z = w and it looks like flat surfaces with center
    x=0, y= 0, z = w/2 There doesn't seem to be any resitrictions on the value that z can take.

    Now if w = 0 , 5 or -10 the su`rface is just centered on 0 , 2.5 or -5

    To describe the function is it enough to say that it is a surface with no curvature centered on x=0, y= 0, z = w/2 ?
     
  9. Mar 18, 2009 #8

    Mark44

    Staff: Mentor

    These flat surfaces you finding are called "planes." They don't have centers. You can describe a plane completely by specifying three points that are on the plane, or by specifying a normal to the plane and one point on it.

    To describe the function f(x, y, z), it's probably sufficient to say that the level surfaces are all planes.
     
  10. Mar 18, 2009 #9
    Thanks about the method of graphing the function it helped alot!

    Would you say that when

    "describing the function f(x, y, z), it's probably sufficient to say that the level surfaces are all planes."

    would you mention the magnitude of the plan eg

    The level surfaces of the function f(x, y, z) = w are all planes at w/2

    regards
    Brendan
     
  11. Mar 18, 2009 #10

    Mark44

    Staff: Mentor

    It doesn't make any sense to talk about the magnitude of a plane. And I don't know what "The level surfaces of the function f(x, y, z) = w are all planes at w/2" means.
     
  12. Mar 18, 2009 #11

    lanedance

    User Avatar
    Homework Helper

    ummmm, I'm not sure what your last sentance means...

    I would mention all the level surface represent parallel planes and try and specify the unique plane defined by a w value

    As Mark said a plane can be described by either
    - 3 points, or
    - a normal and one point

    so you could find the 3 points for each plane in terms of w (the axis corssings spring to mind)

    or I think its easier to notice all the planes have the same noraml vector (can you find it?) and give a single axis crossing in terms of w
     
  13. Mar 19, 2009 #12
    So the level surface represent parallel planes.

    For f(x,y,z) 3x - y + 2z = 0

    We give three points

    P1(0,3,1.5)
    P2(0,1,1/2)
    P3(0,2,1)

    All are point of a unique plane.

    With Normal vector

    P1->P2(0,-2,-1)
    P1->P3(0,-1,-1/2)

    P1->P2 X P1->P3

    normal vector (0,0,0)
     
  14. Mar 19, 2009 #13

    Mark44

    Staff: Mentor

    Each level surface is a plane.
    For f(x, y, z) = 3x -y + 2z, the level surface f(x, y, z) = 0 is a plane that contains the three points below.
    Your vectors are calculated correctly, but you have unfortunately picked points that are on a line, so your cross product is zero. The normal vector to this level surface is not (0, 0, 0).
     
  15. Mar 19, 2009 #14
    Okay I've picked three new points (I'm assuming they can be any points on the plane)

    For f(x,y,z) 3x - y + 2z = 0

    P1(1,2,-1/2)
    P2(0,1,1/2)
    P3(3,1,-4)



    P1->P2(-1,-1,1)
    P1->P3(2,-1,-3.5)

    P1->P2 X P1->P3

    normal vector (4.5,-1.5,3)

    Is this right ?

    Or am I using the wrong furmula for this problem ?
     
  16. Mar 19, 2009 #15

    Mark44

    Staff: Mentor

    This is correct. The most obvious normal to this plane is (3, -1, 2), which is a scalar multiple of the one you found (and vice versa). It is no coincidence that (3, -1, 2) is a normal to the family of planes 3x -y + 2z = k, where k is any real number.
     
  17. Mar 19, 2009 #16
    Thanks alot for your help I really appreciate it.
     
  18. Mar 20, 2009 #17

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm not sure why you are focusing so much on w= 0. It should be immediately clear, from what you have already learned about planes, that f(x,y,z)= 3x - y + 2z = Constant is a plane with normal vector (3, -1, 2) for any Constant.
     
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