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Homework Help: Sketch a plane

  1. Jun 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Sketch the indicated traces, and the graph z = f(x,y)

    f(x,y) = x - 2y z = 0; z = 1; x = 0; y = 0

    3. The attempt at a solution

    z = f(x,y) = 0 = x - 2y -> y = -x/2 the trace is a line

    z = f(x,y) = 1 = x - 2y -> y = x-1/2 another line going through x = 1/2

    z = f(0,y) = 0 - 2y -> z = -2y a line with slope -1 in zy plane

    z = f(x,0) = x - 0 -> z = x

    I drew the graph, and i wanted to verify. The one on wolframalpha is too hard to verify...

    If it is wrong, please tell me what went wrong. Thanks...

    When I drew the graph, I noticed that the third and fourth traces were simply x and y axis...

    Attached Files:

  2. jcsd
  3. Jun 19, 2010 #2


    User Avatar
    Science Advisor

    If f(x, y, z)= x- 2yz= 0 then 2yz= x and z= x/2y. That is a "hyperoloid of one sheet".

    Your graph doesn't really show much- it looks to me like a part of a plane.
  4. Jun 19, 2010 #3


    Staff: Mentor

    It's actually a plane, but jwxie didn't include punctuation to separate the function from the traces required.

    I'm pretty sure this is what was intended:
    For this function f(x,y) = x - 2y, sketch these traces: z = 0; z = 1; x = 0; y = 0.
  5. Jun 19, 2010 #4


    Staff: Mentor

    The equation of the line in the x-y plane is y = x/2. Its slope is m = 1/2.
    The trace is a line in the plane z = 1. Its slope is 1/2, as before. The line goes through (0, -1/2, 1).
    No, the slope is -2.
    No, the 3rd and 4th traces are lines in the y-z plane and x-z plane, respectively.
  6. Jun 19, 2010 #5
    First to get sense what figure are you drawing notice:


    That is clearly a plane which passes through (0,0,0) and it is perpendicular of the vector (1,2,1).

    You've found

    well. You did not sketch them right.

  7. Jun 19, 2010 #6


    Staff: Mentor

    Your algebra is off, here. z = x - 2y <==> x - 2y - z = 0. A normal to this plane is the vector <1, -2, -1>.
  8. Jun 19, 2010 #7
    Hi, Mark. Sorry for all the mistakes. I just checked what I had on the paper, apparently I should have looked at my paper while posting the message.

    I had the exact thing as you mentioned.

    I agreed with you that
    when z = 0,, we have y = x/2, and thus slope m = 1/2
    when z = 1,, we have y = x-1/2, and thus slope m = 1/2 because the equation is 1/2 (x-1)
    when z = 0,, we have z = -2y and thus m = -2
    when y = 0,, we have z = x

    i know how the curves each individual trace in R2 (for example, when z = 0, we have xy axis graph, y = x/2) look like.

    how do i combine them???
    thank you.
  9. Jun 19, 2010 #8


    Staff: Mentor

    You already have the y-z trace. Do another one in the plane x = 2, for example. That should give you a good idea of what your function looks like (the graph is a plane).
  10. Jun 19, 2010 #9
    Hi. Thank you. I have the xy traces (two lines, x/2 and x-1/2)
    i also have the xz trace (when y = k), and this gives me the xz plane

    you mean i need to sketch an yz plane with x = k?
  11. Jun 19, 2010 #10


    Staff: Mentor

    There is only one x-y trace: the line in the x-y plane whose equation is y = x/2. The other line is the cross-section (not trace) in the plane z = 1, with equation y = x/2 - 1/2.
    The x-z trace is the plane y = 0, not y = k. This trace is not the x-z plane; it is the line z = x in the plane y = 0.
    What I mean is that you should sketch the cross section in some plane parallel to the y-z plane, such as in the plance x = 5.

    There are only three traces:
    x-y trace, the cross section of the graph of your function in the x-y plane (the plane z = 0).
    x-z trace, the cross section of the graph in the x-z plane (the plane y = 0).
    y-z trace, the cross section of the graph in the y-z plane (the plane x = 0).

    Intersections of the graph of the function with other planes are called cross sections. Cross sections in planes parallel to the x-y plane are called level curves.
  12. Jun 22, 2010 #11
    thank you mark. i got it.
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