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Curve of Intersection in two Three-Dimensional EQs

  1. Mar 5, 2008 #1
    Determine the curve of intersection of the surfaces x^2 + y^2 + z^2 = 4 and x + y + z = 1. The curve should be in parametric form.

    With this problem, I'm really not sure what direction to go in. I had thought about using the quadratic formula in some manner, but really unclear. Any advice would be great! Thanks!
     
  2. jcsd
  3. Mar 5, 2008 #2
    First, think about it. Do you know what kind of surfaces each of these are? If so, do you see how they can intersect?
     
  4. Mar 5, 2008 #3
    I'm actually not too sure; I want to graph them in Mathematica, but I don't know the command.

    edit: Ok, I graphed them in Mathematica as follows

    first = ContourPlot3D[{x^2 + y^2 + z^2 - 4}, {x, -10, 10}, {y, -10,
    10}, {z, -10, 10}];
    second = ContourPlot3D[{(x + y + z - 1)}, {x, -10, 10}, {y, -10,
    10}, {z, -10, 10}];
    Show[first, second]

    The first function is simply a sphere while the second is just a plane intersecting it.
    I'm still unsure how to find the curve of intersection, although I can now visually see it.
     
    Last edited: Mar 5, 2008
  5. Mar 5, 2008 #4
    The next step you should work on is parametrizing the plane. Can you do this?
     
  6. Mar 5, 2008 #5
    By parametrizing the plane, are you referring to the second equation? If so, can't you just do something like:

    x = u
    y = v
    z = 1 - u - v
     
    Last edited: Mar 5, 2008
  7. Mar 5, 2008 #6
    Yep. Now how about the sphere?
     
  8. Mar 5, 2008 #7
    Since its a sphere of radius 2 it should be:
    x = 2sinΦcosΘ
    y = 2sinΦsinΘ
    z = 2cosΦ
     
  9. Mar 5, 2008 #8
    Now the tricky part. You need to eliminate a parameter so that your curve of intersection expresses the set of points of the intersection with one parameter (this way it's a curve rather than a surface).
     
  10. Mar 5, 2008 #9
    I'm really not sure how to attempt doing this: I tried messing around in Mathematica by trying to somehow equate the two parameterizations together such as...

    x = 2 Sin Cos[v],
    y = 2 Sin Sin[v]
    z = 1 - 2 SinCos[v] - 2 Sin Sin[v]

    Graphing this in Mathematica gives a slanted circular plane and not the curve that we're looking for.
     
    Last edited: Mar 5, 2008
  11. Mar 6, 2008 #10
    Above you aren't using the spherical definition for z.

    Edit: Hmmm, this is harder than I thought, hope I didn't bring you to a dead end. I'll keep working on it.

    Edit 2: Somehow you need to combine the two equations together to get something like U^2 + V^2 = constant^2, at which point you then declare U = (constant) cos theta and V = (constant) sin theta, but how to get U and V here is not something I am able to discover yet. Anybody else have a clue here?
     
    Last edited: Mar 6, 2008
  12. Mar 6, 2008 #11
    Ok, thanks for your help!
     
  13. Mar 6, 2008 #12
  14. Mar 6, 2008 #13
    Ai. I'm also trying to solve this problem. The examples they show only have 2 variables for the Planes instead of three variables, so their methods work easily.
     
  15. Mar 6, 2008 #14
    Reflected, are you by any chance in my class? lol
     
  16. Mar 6, 2008 #15

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Frankly, I am concerned that Reflected seems more concerned with learning to use Mathematica than with learning mathetmatics! If you are expected to be able to do problems like this one, you certainly should not need Mathematica to realize that x2+ y2+ z2= 4 is a sphere with center at the origin and radius 2, or that x+ y+ z= 1 is a plane through (0, 0, 1), (0, 1, 0), and (1, 0, 0). Tools are good, crutches are not.

    The intersection of the two is a circle with center at (1/3, 1/3, 1/3) (that is the point on x+ y+ z= 1 closest to (0,0,0), the center of the sphere). The distance from (0, 0, 0) to that point is [itex]\sqrt{(1/3)^2+ (1/3)^2+ (1/3)^2}= \sqrt{3}/3[/itex]. From the Pythagorean theorem then, the radius of the circle of intersection is [itex]\sqrt{4- 1/3}= \sqrt{33}/3[/itex].

    It's not really necessary to know all that to find the parameterization of the circle. Here's how I would do it.

    From the plane equation, z= 1- x- y. Putting that into the equation of the sphere, [itex]x^2+ y^2+ (1- x- y)^2= x^2+ y^2+ 1- 2x- 2y+ x^2+ y^2+ 2xy= 2x^2+ 2y^2+ 2xy- 2x- 2y= 3[/itex]. Dividing through by 2, [itex]x^2+ y^2+ xy- x-y= 3/2[/itex]. We can eliminate the "xy" term by making the substitutions u= x+ y, v= x- y so that x= (1/2)u+ (1/2)v, y= (1/2)u- (1/2)v and the equation becomes [itex](3/4)u^2- u+ (1/4)v^2= 3[/itex], eliminating the "mixed" term. completing the square in "u" part, we have [itex](u- 2/3)^2+ (1/3)v^2= 40/9[itex] or [itex](9/40)(u- 2/3)^2+ (3/40)v^2= 1[/itex]. An obvious parameterization for that is [itex]u= (3/2\sqrt{10})cos(\theta)+ 2/3[/itex], v= (\sqrt{3}/2\sqrt{10})sin(\theta)[/itex]. Now, work backwards using u= x+ y, v= x- y to get x and y in terms of [itex]\theta[/itex] and then use z= 1- x- y to get z in terms of [itex]\theta[/itex].

    Modulo any arithmetic errors that should do it.
     
    Last edited: Mar 7, 2008
  17. Mar 6, 2008 #16
    I don't understand how you parametrizted the final equation after all of the substitution and completing the square.
     
    Last edited: Mar 6, 2008
  18. Mar 7, 2008 #17
    Not sure if you need still this or not but what I did was set z = 1 - x - y
    plug it in x^2+y^2+z^2=4 and get this long equation, simplified it and got a xy term. then I tried to put equation so that it's quadratic ax^2+bx+c = 0 then used quadratic formula and go x = f(y); set y = t so x = f(t) , plug everything back in z = 1- x -y and got the parametric function for z. Boy, this problem was driving me nuts and now it's over. LOL
     
    Last edited: Mar 7, 2008
  19. Mar 7, 2008 #18
    Reflected, this should work. I got my equation already and ParametricPlot3D'ed in Mathematica and showed me this beautiful (T_T) circle around the sphere and on the plane. Try it. =)
     
  20. Mar 7, 2008 #19
    yes, i tried doing that as well. How did you set that expression into the quadratic formula?
    You have 2x^2 + 2xy - 2x + 2y^2 -2y = 3
    so how does that go into the quad form?
     
    Last edited: Mar 7, 2008
  21. Mar 7, 2008 #20
    erm i set it so that
    2y^2 +(2x-2)y+(2x^2-2x-3) .. do you see the a b and c now?
    OH.. one thing.. you in my class tho? Lol i got the exact same problem. Ohlone. Bradshaw?
    Hope this helps you a lot.
    a = 2
    b = 2x-2
    c = 2x^2-2x-3
     
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