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Graphs of functions of several variables

  1. Oct 24, 2007 #1
    Consider the graph of z=xy.

    Choose the section y=x, i.e. intersect the plane y=x with the graph z=xy, I get z=(x)(x)=x^2, with y=x. Now is this a parabola? (note that the curve actually lies on the plane y=x)
    If so, why?
    Also, is this parabola going to have exactly the same shape as the one parallel to the x-axis?(i.e. z=x^2, y=0)
    How is the graph of z=xy going to look like?

    I spent an hour thinking of these, but still can't figure it out...

    Can someone please help me? I seem to be skewing up geometrically. I can't understand geometry like this...this is driving me crazy
     
    Last edited: Oct 24, 2007
  2. jcsd
  3. Oct 24, 2007 #2

    Dick

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    z=0 along the x and y axes. As you've noticed along the line y=x, the cross section is parabolic with z positive. Along x=-y it's parabolic with z negative. To picture it just think of z as height above (or below) the xy plane. I'm not sure exactly what the question is beyond that.
     
    Last edited: Oct 24, 2007
  4. Oct 24, 2007 #3

    dynamicsolo

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    Also think about symmetries. What happens to the product in each quadrant of the xy-plane? What happens if you change the sign of either x or y?

    How does the product behave as either variable or both approach zero? approach infinity?
     
  5. Oct 25, 2007 #4
    Say if we consider the xz-plane only, then z=x^2 is surely a parabola (*).

    But on the plane y=x, is z=x^2 going to be the projection of the above parabola (*) on the plane y=x? If this is the case, then this curve and the above parabola (*) won't be the same, right? Would it even look like a parabola at all?

    or

    On the plane y=x, is z=x^2, imagine revolving the parabola (*) until we hit the plane y=x, and then take the intersection of this curve (same shape as parabola (*) ) with the plane y=x, is this the curve of intersection of the plane y=x with the graph z=xy? If this is the case, then certainly this curve and the above parabola (*) have the same shape.
     
    Last edited: Oct 25, 2007
  6. Oct 25, 2007 #5

    Dick

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    Nooooo. It's not a surface of revolution. In the xz (that's y=0) plane it's zero. Look at your equation.
     
    Last edited: Oct 25, 2007
  7. Oct 25, 2007 #6

    dynamicsolo

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    To be on the xz-plane, wouldn't y have to be zero?

    It will be z = x^2 = y^2. Since both advance linearly along that plane, z would follow a parabolic curve.

    What would happen on the yz-plane? on the plane y = -x ?

    While we're about it, what would the equation look like on the plane y = kx , k real?
     
  8. Oct 25, 2007 #7
    z=xy=[(x+y)^2-(x-y)^2]/4

    change coordinates x'=x+y, y'=x-y

    z=(x'^2-y'^2)/4
     
  9. Oct 25, 2007 #8
    Sorry, but I didn't mean the surface...

    I actually mean:
    On the plane y=x, is z=x^2, imagine revolving the parabola (*) until we hit the plane y=x, and then take the intersection of this curve (same shape as parabola (*) ) with the plane y=x, is this the curve of intersection of the plane y=x with the graph z=xy?
     
  10. Oct 25, 2007 #9
    Sure.


    Why?
    Sorry, but I don't understand what you mean by advancing linearly along plane...
     
  11. Oct 25, 2007 #10
    What does this imply? I can't see what this is heading into...sorry
     
  12. Oct 25, 2007 #11
    KW-

    Since z=(x'^2-y'^2)/4

    where x'=x+y, y'=x-y

    It shows that z can be written as a parabolic along directions x' and y', which are at right angles. Note that the x' axis is the diagonal between the x and y axes and y' is between the x and -y directions.

    It's a lot easier to sketch the function doing it this way. Try sketching it in the x' y' axes using z=(x'^2-y'^2)/4 and then draw in the x y axes, where x=(x'+y') /2, y=(x'-y')/2
     
  13. Oct 25, 2007 #12
    OK, I see what you are saying now, it's graph is a hyperbolic paraboloid.

    But, I am still not understanding how the intersection of the plane y=x with the graph z=xy would look like...
     
  14. Oct 25, 2007 #13

    dynamicsolo

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    Quote:
    It will be z = x^2 = y^2. Since both advance linearly along that plane, z would follow a parabolic curve.
    Why?
    Sorry, but I don't understand what you mean by advancing linearly along plane...

    When I was saying here is what christianjb is referring to by rotating the coordinate system 45º: on the line y = x, x and y advance at the same constant rate (what I meant by 'advancing linearly' -- I'm sorry if that wasn't the best choice of words). Since y = x,
    whichever coordinate you look at, the value of z increases quadratically. So the cross-section of the surface z = xy on that plane is an upward-facing parabola starting from z = 0 at the origin. That parabola is z = x^2 or z = y^2 .

    If you choose a different cross-section using y = kx , you get z = x·(kx) = k·x^2 , which is just a parabola starting at z = 0 at the origin and growing at a different rate. All the cross-sections made by vertical planes rotated at various angles to the x-axis are parabolas, including the "degenerate" parabola z = 0 on the yz-plane.

    The exception is that there is a "break" on the xz-plane, where the vertical plane is now perpendicular to the second and fourth quadrants of the xy-plane, corresponding to k < 0. The product abruptly switches from being positive to negative on either side of the xz-plane, with the cross-section again being "degenerate" for y = 0.
     
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