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Parameterize the curve of intersection

  1. May 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Parameterize the curve of intersection of the cylinder x^2 + y^2 = 16 and the plane x + z = 5

    2. Relevant equations



    3. The attempt at a solution

    i think i must first parameterize the plane

    x = 5t, y = 0, z = -5t

    then i think i plug those into the eq. of the cylinder

    25t^2 = 16
    t = .8

    so x = 4, y = o, z = -4, am i on the right track, i feel like i made it up
     
  2. jcsd
  3. May 4, 2009 #2

    Mark44

    Staff: Mentor

    You have x = 4, y = 0, z = -4. That represents a single point. Do you think that the cylinder and the plane intersect in a single point? Doesn't seem likely to me.

    Have you drawn a sketch of the two surfaces? That would at least give you an idea of how the two intersect.
     
  4. May 4, 2009 #3
    heres a rather crude sketch
     
  5. May 4, 2009 #4
    oops
     

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  6. May 5, 2009 #5

    Mark44

    Staff: Mentor

    OK, the sketch doesn't help much, but I have to give you credit for doing it. The circular cylinder looks very nice, but what you show as a straight line (x + z = 5) is actually a plane. If the plane were horizontal, it would intersect the cylinder in circle. Since the plane is canted (it makes an angle of 45 degrees with the x-y plane), the intersection will be an ellipse.

    How do you normally find the intersection of two curves? You need to get that first, and then parametrize it, rather than the other way around, as you tried to do in your OP.
     
  7. May 5, 2009 #6
    hmm ok i solved the second equation for x = 5-z and pluged it into the other one

    (5-z)^2 + y^2 = 16

    y^2 + z^2 - 2z = -9 i think im lost
     
  8. May 5, 2009 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You can certainly parameterize the cylinder by [itex]x= 4 cos(t)[/itex], [itex]y= 4 sin(t)[/itex], [itex]z= z[/itex] using t and z as parameters. You can parameterize the plane x+ z= 5 as, x= 5-z, y= y, z= z using y and z as parameters. The fact that I could use the same z as parameter in both equations reduces the problem slightly. They will intersect when [itex]x= 4 cos(t)= 5- z[/itex], [itex]y= 4 sin(t)= y[/itex], and [itex]z= z[/itex]. Since the intersection of two 2-dimensional surfaces is 1-dimensional, we need to reduce those 3 parameter, t, y, and z, to only one parameter. I would recommend squareing and adding the first two equations.
     
  9. May 6, 2009 #8
    x = 4cost = 5-z Should i solve this for z, giving me a new z, or should i just make it x = 4cost - 5 + z
    y = sint
    z=z

    is this what u meant by squaring and adding the first 2 eq.

    x4 + y4 = 162
    x2 + y2 = 52
    -------------------------------------------
    x4 + x2 + y4 + y2 = 281 How does this help
     
  10. May 6, 2009 #9

    Mark44

    Staff: Mentor

    It doesn't help because you can't do what you did. If you have an equation
    x2 + y2 = 16, and you square both sides, you don't get
    x4 + y4 = 162! And the same with the other equation you showed.

    You have these equations...

    For x: 4 cos(t) = 5 - z
    For y: 4 sin(t) = y
    For z: z = z

    What HallsOfIvy said was to square both sides of the first two equations, and then adding them together. That will get rid of the t variable.

    I've done this work and I'm ending up with a circle, but I'm pretty certain (and said this earlier) that the two surfaces intersect in an ellipse. So I'm still looking into this.
     
  11. May 8, 2009 #10
    wow what a stupid mistake haha, i forgot the basics of math

    ok so i squared my x and y and then added them together

    x^2 = 16cos(t)^2 - 25 + z^2

    y^2 = 16sin(t)^2

    x^2 + y^2 = 16cos(t)^2 - 25 + z^2 + 16sin(t)^2 but i still got my t's, what i mess up on
     
  12. May 9, 2009 #11

    Mark44

    Staff: Mentor

    A way to do this is described here: http://mathforum.org/library/drmath/view/66054.html

    Any point P on the ellipse is given by this vector formula:
    P = C + a*cos(Z)*U + b*sin(Z)*V, 0 <= Z < 2pi
    C is the point at the center of the ellipse (or a vector from the origin to the center).
    U and V are unit vectors in the directions of the major and minor axes.
    Z is the parameter.
    a and b are constants that determine the distances from the center to one or another vertex.
     
  13. May 9, 2009 #12
    is this the easiest way to solve this problem. we never learned anything similar to that in the course
     
  14. May 11, 2009 #13
    Parametrize elipse

    1. The problem statement, all variables and given/known data

    Parameterize the curve of intersection of the cylinder x^2 + y^2 = 16 and the plane x + z = 5

    2. Relevant equations



    3. The attempt at a solution

    I found a lecture from MIT on itunes that solved a generic problem like this
    http://deimos3.apple.com/WebObjects/Core.woa/Browse/mit.edu.1885187900.01885187905.1890174426?i=1832321151 [Broken]

    This is the jist of what he says

    Given a cylinder x^2 + y^2 = a^2, and an intersecting plante z = c(1)x + c(2)y + d,
    the parameterized elipse would follow in the form

    x = a cos(t)
    y = a sin(t)
    z = a*c(1)*cos(t) + a*c(2)*sin(t) + d

    so given my problem where a = 4, c(1) = -1, and c(2) = 0, d = 5, I got

    x = 4 cos(t)
    y = 4 sin(t)
    z = -4cos(t) + 5


    does anybody agree or disagree with this method
     
    Last edited by a moderator: May 4, 2017
  15. May 11, 2009 #14
    Re: Parametrize elipse

    Seems good to me. You can check by substituting the parametrized values for x,y,z in the original 2 equations.
     
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