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Homework Help: Intersection point

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Does the surface [itex] z = x^2 + y^2 - 4 [/itex] * intersects the [itex] yz [/itex]-plane? If so, find the equation of the curve and write down the points of intersection.

    3. The attempt at a solution
    [itex]yz-[/itex]plane, so [itex] x=0 [/itex]

    1) * becomes [itex] z = y^2 - 4 [/itex] and this is the equation of the curve that intersects the [itex] yz[/itex]-plane.
    2) So the intersection points are [itex] (0,-2,0) [/itex] and [itex] (0,2,0) [/itex]

    Are 1) and 2) correct? Thanks in advance!
     
  2. jcsd
  3. Apr 26, 2010 #2
    Shouldn't there be a third intersection point?

    You found the intersection points of the curve on the y-axis, what about the z-axis?
     
  4. Apr 26, 2010 #3
    Is it [itex] (0,0,-4) [/itex] ? But I am slightly confused, doesn't the whole curve intersect yz-plane and so there are infinitely many points of intersection?

    Thanks.
     
  5. Apr 26, 2010 #4

    lanedance

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    the graph of [itex] z = x^2 + y^2 - 4 [/itex], is a parabaloid, rotationally symmetric about the z axis

    the intersection of the surface with the yz plane (x=0) is the parabola [itex] z = y^2 - 4 [/itex] which has, as you say, infinitely many points.

    I'm not too sure what the last part is asking for, maybe it is where the curve intersects the y & z axis....
     
  6. Apr 26, 2010 #5
    I wanted to do the question myself so I only show a part of the question.
    The original question is:
    If the surface intersects the [itex]xy, xz [/itex], and [itex] yz [/itex] coordinate planes, find the equation of each of the parabolic boundary curves. Where do the curves intersect with the three coordinate axes?

    I thought if the surface intersects the three coordinate planes, then the points of intersection must be [itex] (0,0,0) [/itex]?

    Thanks.
     
    Last edited: Apr 26, 2010
  7. Apr 26, 2010 #6

    lanedance

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    nope, that is not a point on the surface, and so none of teh curves pass through that point

    the easiest way to work out what i going on is to try & draw a picture, so i hope you'e attempted that. This problem is essentially showing you agood way to draw simple 3D surfaces, by drawing the intersection on each coordinate plane.

    you should get 2 parabolas & and a circle

    if you think about the shape of the surface and curves, the curve constrained by x = 0, will never intersect the x axis. So each point of intersection with axis will have 2 curves passing through it

    due to the symmetry (which can save you some time) there will be one point of intersection on the z axis and 2 each for the x & y...
     
  8. Apr 26, 2010 #7
    I have drawn the graph, it's a paraboloid. But I don't understand about the 2 parabolas and a circle.

    I understand that the level curves are parabola and circle, but why the intersecting ones are only the specific 2 parabolas and 1 circle that you mentioned?

    Thanks.
     
  9. Apr 26, 2010 #8

    lanedance

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    ok, so what is the equation for each curve...
     
  10. Apr 26, 2010 #9
    [itex] z = x^2 - 4 [/itex]
    [itex] z = y^2 - 4 [/itex]
    [itex] x^2 + y^2 = 4 [/itex]

    Are these correct?
     
  11. Apr 26, 2010 #10

    lanedance

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    yep so what do the forms of those equations look like?
     
  12. Apr 26, 2010 #11
    The first two are parabolas. The last one is circle.
    So are the intersection points
    1) [itex] (-2,0,0),(2,0,0) [/itex] for the first curve.
    2) [itex] (0,-2,0),(0,2,0) [/itex] for the second curve.
    3) [itex](0,0,-4) [/itex] for the third curve mentioned above?

    Thanks.
     
  13. Apr 26, 2010 #12

    lanedance

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    teh points look ok, but they correspond to an axis, not a specific curve, each will have 2 curves passing through it
     
  14. Apr 26, 2010 #13
    I am confused. Could you show me the equation for one the curves so I can try to figure what you mean? Didn't I already show the equation of the 3 curves? Are they different from the ones that you just mentioned? Thanks.
     
    Last edited: Apr 26, 2010
  15. Apr 26, 2010 #14

    lanedance

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    you've got it all, so pulling it together from your posts (though check it as i did pretty quick):
    curve 1 - yz plane (x=0)
    [itex] z = y^2 - 4 [/itex]
    curve 2 - xz plane (y=0)
    [itex] z = x^2 - 4 [/itex]
    curve 3 - xy plane (z=0)
    [itex] x^2 + y^2 = 4 [/itex]

    axes intersection points
    x axis - (-2, 0, 0) and (2,0,0) by curves 2&3,
    y axis - (0 ,-2, 0) and (0,2,0) by curves 3&1,
    z axis - (0 , 0, 4) by curves 1&2
     
  16. Apr 26, 2010 #15
    I see, thanks!
     
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