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The curve formed by the intersection of paraboloid and ellipsoid

  1. Jan 31, 2014 #1
    I will state the specifics to this problem if necessary.

    I need to find the parametric equations for the the tan line at point, P(x1,y1,z1) on the curve formed from paraboloid intersection with ellipsoid.

    The parametric equations for the level surfaces that make up paraboloid and ellipsoid are NOT given.

    The level functions for paraboloid and the level function for ellipsoid are given.

    This is what I've done so far.

    I found the equation of the curve that forms from the intersection.

    [c(x,y)] = curve of paraboloid and ellipsoid intersection.

    The tangent vector at p(x1,y1,z1) on curve should be the same as the tangent vector at same point on paraboloid and ellipsoid.

    I have taken the gradient of curve dot tangent vector = 0

    I have taken the gradient of level fxn of paraboloid dot tangent vector = 0

    I have taken the gradient of level fxn of ellipsoid dot tangent vector = 0


    Solving the three linear equations that all equal to zero gets me nowhere.


    I am stuck. Help me.
     
  2. jcsd
  3. Jan 31, 2014 #2

    HallsofIvy

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    The problem is that it "should be the same as" a tangent vector to the paraboloid and ellipsoid. A smooth surface, such as a paraboloid or ellipsoid, has a tangent surface at each point, so an infinite number of tangent vectors.

    I don't know what you mean by the gradient of a curve. And which tangent vector do you mean here?

    Again, I don't know which tangent vector you mean.


     
  4. Jan 31, 2014 #3
    The tangent vector at point p(x1,y1,z1) on paraboloid is the same as the tangent vector at point p(x1,y1,z1) on ellipsoid and same as the tangent vector at Point P(x1,y1,x1) on curve, [c(x,y)]

    This tangent vector is NOT defined and I have to find the parametric equations for this tangent line.

    Gradient of [c(x,y)]

    Remember... [c(x,y)] = curve that forms from intersection between paraboloid and ellipsoid

    The tangent vector is what I'm supposed to find. The tangent vector at point P(x1,y1,z1)


    Again, it is the tangent vector that is undefined in the problem.
     
    Last edited: Jan 31, 2014
  5. Jan 31, 2014 #4
    Am I correct with all my statements above?
     
  6. Jan 31, 2014 #5

    haruspex

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    Unless you are given the parameter, you must mean you need to find some parametric equations. There could be many ways of parameterising it.
    I think Halls' problem with that statement is that it reads as though there is a well-defined tangent vector at a given point on a paraboloid etc. You probably mean
    The tangent vector at p(x1,y1,z1) on curve should be a tangent vector at same point on paraboloid and ellipsoid. ​
    Halls and I are puzzled by that statement. How are you defining the gradient of a curve in 3-space?
    These are the normals to the surfaces, so that sounds right. (In fact, I would have approached it by finding the tangent planes then taking the intersection.)
    Please post your working as far as you got.
     
  7. Jan 31, 2014 #6
    Is the intersection between paraboloid and ellipsoid a surface or a curve?

    I have a 2nd degree polynomial in xy form for this intersection between paraboloid and ellipsoid which indicates I have a surface. However, I can't seem to accept that the intersection between paraboloid and ellipsoid is a surface.
     
  8. Jan 31, 2014 #7

    LCKurtz

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    I second Haruspex's request. Please post your original equations and your work. A polynomial in "xy form" doesn't represent anything but a polynomial. You need an equation to represent a surface. And, yes, you would expect a paraboloid and ellipsoid to intersect in a curve or curves in the typical case if they intersect at all. Please post your work.
     
  9. Jan 31, 2014 #8

    haruspex

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    If you have an equation involving x and y but not z, that can be thought of as the projection of the curve (not a surface) onto the XY plane. It will constitute a (presumably correct) statement about the relationship of y to x along the 3D curve (line). You could go back to your surface equations and eliminate y instead, giving you an equation in x and z. That will be the projection of the curve onto the XZ plane.
     
  10. Feb 1, 2014 #9
    Here is the exact problem.

    Paraboloid [z(x,y)] = x^2 + y^2

    Ellipsoid 4x^2 + y^2 + z^2 =9

    Find parametric equations for tangent line to the curve of intersection of paraboloid and ellipsoid at point P(-1,1,2)


    My work done so far.

    The curve of intersection is
    [C(x,y)] = 4x^2 + y^2 + x^4 + 2x^2 y^2 + y^4 = 9

    Gradient of [C(x,y)] dot tangent vector = 0

    Yes or no?

    Gradient of ellipsoid, [E(x,y,z)] dot tangent vector = 0

    Yes or no?
     
    Last edited: Feb 1, 2014
  11. Feb 1, 2014 #10

    LCKurtz

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    That is not the equation of the curve of intersection. That is the equation of a cylindrical surface in 3D which passes through the curve of intersection of the surfaces. It is also the 2D equation of the projection of that curve onto the xy plane.

    But you don't need the equation of the curve in the first place. You are after the equation of its tangent line at the given point. Think about the cross product of the normals to the given surfaces at the given point. What direction would it point?
     
  12. Feb 1, 2014 #11
    Didn't even consider the cross product.

    That helps a lot!

    Brb
     
  13. Feb 3, 2014 #12
    Okay... I found that the cross product between gradient of paraboloid function and gradient of ellipsoid function at point P(-1,1,2) produces

    Tangent vector T = <9/2, 6, 6>
    And unit tangent vector = <3/41 * rad41 , 4/41 * rad41, 4/41 * rad41>


    Thus,

    Parametric equation for tangent line is

    -1 + 3/41 rad41 t = [x(t)]

    1 + 4/41 rad41 t = [y(t)]

    2 + 4/41 rad41 t = [z(t)]


    This does NOT match answer in back of book.

    HELP ME
     
  14. Feb 3, 2014 #13

    haruspex

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    Not what I get. Please post your working.
     
  15. Feb 4, 2014 #14
    Paraboloid
    D(x,y,z) = (z^-1) (x^2 + y^2) = 1

    ELLIPSOID
    L(x,y,z) = 4x^2 + y^2 + z^2 = 9


    Gradient of paraboloid at P(-1,1,2)

    <-1,1,-1/4>

    Gradient of ellipsoid at P(-1,1,2)

    <-8,2,4>

    Cross these two to get

    <9/2,6,6>
     
  16. Feb 4, 2014 #15
    Help. I am stuck
     
  17. Feb 4, 2014 #16
    Okay!

    I am close to the answer, but I need a fresh outlook from some new brains.

    Curve of intersection is

    C(x,y) = x^4 + 4x^2 + 2x^2 y^2 + y^4 + y^2 = 9

    Gradient of C(x,y) is

    <-16,10,0>

    Cross gradient of ELLIPSOID to gradient of curve and I get

    8 * <-5,-8,-6>


    I'm not sure why I can't simple cross gradient of paraboloid and ellipsoid to get same as above.
     
  18. Feb 4, 2014 #17
    The answer says

    X = -1 - 10t

    Y = 1 - 16t

    Z = 2 - 12t


    My answer says

    X=-1-5t
    Y=1-8t
    Z=2-6t

    I'm quite sure my answer is the same as the answer in textbook for this parametric equation of tangent line.

    Yes or no?
     
  19. Feb 4, 2014 #18
    2 questions.

    1. Why can't I cross gradient of paraboloid to gradient of curve of intersection?


    2. Why can't I simply cross gradient of paraboloid to gradient of ellipsoid to get the right answer?



    IT IS NECESSARY TO CROSS ELLIPSOID GRADIENT TO CURVE GRADIENT.

    WEIRD!!!!!
     
  20. Feb 4, 2014 #19

    LCKurtz

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    Your paraboloid is ##z = x^2+y^2## or ##x^2+y^2 - z = 0##. Easier than what you wrote.
     
    Last edited: Feb 4, 2014
  21. Feb 4, 2014 #20

    LCKurtz

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    Yes. Your direction vector is a constant multiple of theirs. Just different parameterizations of the same line.
     
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