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How to identify the curve at intersection of level surfaces

  1. Feb 14, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Sketch a picture of the cone x = sqrt(y^2+z^2) , and elliptic paraboloid x = 2−y^2−z^2 on the same grid.
    Although the picture does not have to be perfect, indicate clearly the orientation of both figures relative to coordinate axes. Identify the curve at the intersection of the surfaces.

    2. Relevant equations


    3. The attempt at a solution
    What does it mean to identify the curve at the intersection?

    My thinking is that this means that I have to set these two equations equal to each other, and solve, resulting in some curve?

    Am I on the right track?
     
  2. jcsd
  3. Feb 14, 2015 #2

    Mark44

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    Yes.
     
  4. Feb 14, 2015 #3

    RJLiberator

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    Perfect.
    So I set
    2-y^2-z^2=sqrt(y^2+z^2)
    I then add y^2 and z^2 to both sides.
    2 = sqrt(y^2+z^2)+y^2+z^2.

    Something tells me this needs to be simplified and is not a correct answer. Correct?
     
  5. Feb 14, 2015 #4

    RJLiberator

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    Ok, it seems I was being lazy, perhaps:

    So, I set the equal to each other in my first step here:
    sqrt(y^2+z^2)=2-y^2-z^2

    Second step: I square both sides, the right hand side gives me a long equation that I can deal with
    y^2+z^2 = 4-4y^2+y^4-4 z^2+2y^2z^2+z^4

    Third step: Algebraic manipulation to set the equation with respects to y or z
    5y^2+5z^2=(y^2+z^2)^2+4
    5(y^2+z^2)=(y^2+z^2)^2+4

    Divide both sides by (y^2+z^2) I'm not sure if I can do this step... on the right hand side I have the +4 in the numerator, my math intuition is telling me this anot be simplified like I am doing, please let me know if I made an error

    5=(y^2+z^2)+4
    1-y^2=z^2
    z=sqrt(1-y^2)

    This should be the correct answer, according to wolfram simplification.
     
  6. Feb 15, 2015 #5

    Mark44

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    It's legitimate to divide by y2 + z2, provided that y and z are not both zero. However, your work below is incorrect, because you didn't divide the constant term. When you divide one side of an equation, you have to divide all terms.


    I found a much simpler way to do this.
    Squaring both sides of ##x = \sqrt{y^2 + z^2}## yields ##x^2 = y^2 + z^2##. In the new equation, we must have ##x \ge 0## because the square root produces only nonnegative numbers.
    The other equation is x = 2 - y2 - z2 = 2 - (y2 + z2.
    Replace y2 + z2 in the equation immediately above with x2.
    Solve, keeping only the positive value for x.
     
  7. Feb 15, 2015 #6

    RJLiberator

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    I see what you did there, however:

    When we get to x=2-(y^2+z^2) and substitute our x^2=y^2+z^2
    It becomes x=2-x^2
    which results in
    sqrt(2-x)=x

    Isn't this an incomplete result?
     
  8. Feb 15, 2015 #7

    RJLiberator

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    Anyone have any more insight in this problem?
     
  9. Feb 16, 2015 #8

    HallsofIvy

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    This is a quadratic equation. Surely you know how to solve x^2+ x- 2= 0?

     
  10. Feb 16, 2015 #9

    RJLiberator

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    Ah, I see the light now. The quadratic equation represents the curve.

    Thank you kindly for the words.
     
  11. Feb 17, 2015 #10

    LCKurtz

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    What quadratic equation? I have yet to see in this thread where anyone has specifically said that the curve of intersection is a circle nor have I seen an equation or parametric representation of it, which is the usual way to express a curve in 3d. Can you do that?
     
  12. Feb 17, 2015 #11

    RJLiberator

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    You are absolutely right. I jumped the gun and didn't reason:

    x^2+x-2=0.

    The quadratic equation:
    [-1+/- sqrt(1-4*1*-2)]2
    we get the solutions of 1 and -2
    So at x=1 and x=-2 we have points of intersection.

    Am I correct up until this point?
     
  13. Feb 17, 2015 #12

    LCKurtz

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    So you have a couple of values of ##x##. What do they represent? Do they help you solve the problem? What are you going to do next?

    What I am trying to get at is whether you have a plan for solving this problem. Are you making progress? Do you know whether you are making progress? Is the problem almost solved? Or are you just shuffling symbols around?
     
  14. Feb 17, 2015 #13

    RJLiberator

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    To be honest, this is the last part of the assignment that I am unsure of. Most likely due to the way it is worded.

    So we have the values x=1 and x=-2. By the looks of the graph
    It appears that these values represent the y-intercepts and z-intercepts for equations x=2-y^2-z^2.
    I don't really have much else. I'm not quite sure how they help me solve the problem. I was hoping that they would be the intersection points, but it doesn't appear to be so.
    I have to believe that the problem is very closed to being solved, but I do feel like I am merely shuffling symbols around :/.

    I imagine the curve must represent a circle of intersection points based on the graph.
     
  15. Feb 17, 2015 #14

    LCKurtz

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    You do have the graph of the two surfaces? Do both ##x=1## and ##x=-2## (remember, those are both equations of planes in 3d) look equally relevant to the intersection? Can you describe in words what the intersection curve must look like and where it is?
     
  16. Feb 17, 2015 #15

    RJLiberator

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    Assuming my graph is correct here is my interpretation of the information:

    x=1 and x=-2 are both planes.
    x=1 plane seems to be a major point of intersection.

    x=-2 seems to be of no relevance.

    based on this, the curve must be a part of x=1.
    I cant imagine that the answer must be the plane of x=1, that doesn't make sense to me.
    So it must be something deeper. when x=1 the curves intersect. But y and z values can vary in proportion so they must form some sort of circle of intersection.
    Does this mean that the radius must be 1 and the circle must be y^2+z^2=1 ?
     
  17. Feb 17, 2015 #16

    LCKurtz

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    Now you are getting somewhere. Instead of guessing for that last question, check what happens when you put ##x=1## in both of your original equations.
     
  18. Feb 17, 2015 #17

    RJLiberator

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    1=sqrt(y^2+z^2)
    square both sides
    1=y^2+z^2

    1=2-y^2-z^2
    -1=-y^2-z^2
    1=y^2+z^2

    Aha. That's a thing of beauty.
    So what threw me off was the x=-2. The -2 from the quadratic equation means nothing in this case as it is not relevant due to the domain, however, using the value of 1 i can then plug it into both equations and get a curve of intersection.

    I appreciate you for taking the extra time to teach me this.
     
  19. Feb 17, 2015 #18

    LCKurtz

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    A couple more things. The reason you got the extraneous ##x=-2## was you squared one of the equations during the process of solving, which generated the extra root. So now you know the intersection is the circle ##y^2+z^2=1## in the plane ##x=1##. That is an acceptable form for the answer. For extra credit, now lets see a parametric equation for it$$
    \vec R(t) = \langle x(t),y(t),z(t)> = \langle ?,?,?\rangle$$Think "polar" coordinates.
     
  20. Feb 17, 2015 #19

    RJLiberator

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    The erroneous solution from the square makes sense.

    Polar coordinates... hm, let's try:

    we know the radius is 1.
    rcos(theta)
    rsin(theta)

    with r being 1 and theta being an angle from 0 to 2pi.

    This doesn't make complete sense since we have three dimensions and I'm not sure how to proceed beyond this at this point.

    Perhaps this is the answer:
    <0, cos(theta), sin(theta) >
     
  21. Feb 17, 2015 #20

    LCKurtz

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    Call the variable theta instead of "t" if it makes you feel more comfortable and you like to type. Only two questions left:
    1. Where is the left side and the = sign for the parametric equation?
    2. Why did you put the x coordinate zero? Especially after all that work...
     
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