Results 1 to 5 of 5

Math Help - Intersection of surfaces

  1. #1
    Senior Member chella182's Avatar
    Joined
    Jan 2008
    Posts
    267

    Intersection of surfaces

    The surfaces z=\sqrt{x^2+y^2} and x^2+y^2+z^2=1, z\geq intersect along a horizontal circle. Find an equation describing this circle. The angle between their normals at the intersection points never changes. What is this angle?

    And the hint is an intersection line of two surfaces, f(x,y,z)=0 and g(x,y,z)=0, is obtained as a solution of two simultaneous equations f(x,y,z)=0, g(x,y,z)=0, but I have no idea how to solve that
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by chella182 View Post
    The surfaces z=\sqrt{x^2+y^2} and x^2+y^2+z^2=1, z\geq intersect along a horizontal circle. Find an equation describing this circle. The angle between their normals at the intersection points never changes. What is this angle?

    And the hint is an intersection line of two surfaces, f(x,y,z)=0 and g(x,y,z)=0, is obtained as a solution of two simultaneous equations f(x,y,z)=0, g(x,y,z)=0, but I have no idea how to solve that
    1. Your first equation becomes:

    x^2+y^2-z^2=0 and the 2nd equation is
    x^2+y^2+z^2=1

    Subtract columnwise. You'll get 2z^2=1~\implies~z=\frac12 \sqrt{2}

    2. Plug in this value into the first equation. You'll get:

    x^2+y^2=\frac12

    3. This is the equation of a circle. Include the value of z you'll get a circle parallel to the x-y-plane in a distance of z from this plane.
    Attached Thumbnails Attached Thumbnails Intersection of surfaces-kug_an_kegel.png  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2011
    From
    Newcastle
    Posts
    14

    Re: Intersection of surfaces

    Hi! I know that this thread was started years ago but I have exactly the same question, weirdly enough, and thanks earboth because your explanation was really easy to understand BUT I was wondering about the part where you need to find the angle between their normals at intersection point....

    Now I might be completely wrong but I got that the normals would be:
    n_1=2(x,y,-z) and
    n_2=2(x,y,z)

    which are in opposite z directions so I would think that the angle would be pi radians.

    Am I completely wrong, and if so then how is it meant to be calculated?


    (Also sorry, I can't seem to figure out how to do the Pi sign!!)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328

    Re: Intersection of surfaces

    Two vectors with opposite z components but non-zero x and y components are not opposite. for example, the vector, (1, 0, 1) and (1, 0, -1) are perpendicular (their dot product is 0) and so have angle between them \pi/2, not \pi. Opposite vectors, with angle \pi have all three components opposite.

    Remember that the dot product of two vectors, \vec{u} and \vec{v}, can be written as \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta) where \theta is the angle between the two vectors. In particular, 2<x, y, -z> and 2<x, y, z> both have length 2\sqrt{x^2+ y^2+ z^2}= 2 because the points satisfy x^2+ y^2+ z^2= 1. That is, \cos(\theta)= 2(x^2+ y^2- z^2)/2= 0 because points on this curve also satisfy x^2+ y^2- z^2= 0. That is, the vectors do NOT have angle \pi radians, they have angle \pi/2.

    You can get \pi by using \pi in LaTeX: [ tex ]\pi[ /tex ] without the spaces.

    You are also wrong about the normal to the circle, n1. That circle, as said before, is of the form x^2+ y^2= 2 with z= \sqrt{2}/2. Since the z component is constant, the circle lies in a plane parallel to the xy-plane. The center of the circle is at (0, 0, \sqrt{2}/2) and any point on the circle is of the form (x, y, \sqrt{2}/2). A normal to the circle is (x, y, 0), not (x, y, -z). You are correct that a normal to the sphere with center at (0, 0, 0) is parallel to a radius of that sphere and so is (x, y, z). The vector (x, y, 0) has length \sqrt{x^2+ y^2}= \sqrt{1/2}= \sqrt{2}/2 still. The dot product of the vectors (x, y, 0) and (x, y, z) is x^2+ y^2= 1/2.

    That is, the correct angle between the vectors is given by cos(\theta)= \frac{1/2}{(1)(\sqrt{2}/2}= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} so that \theta= \pi/4.
    Last edited by HallsofIvy; November 20th 2011 at 04:03 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2011
    From
    Newcastle
    Posts
    14

    Re: Intersection of surfaces

    Thank you so much!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Intersection of two surfaces
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 2nd 2011, 04:50 AM
  2. intersection between two surfaces
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 11th 2010, 04:02 AM
  3. Trajectory as an intersection of two surfaces
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 28th 2010, 01:21 AM
  4. Intersection of Surfaces
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 15th 2009, 03:25 AM
  5. Intersection of surfaces
    Posted in the Geometry Forum
    Replies: 2
    Last Post: April 16th 2009, 09:57 PM

Search Tags


/mathhelpforum @mathhelpforum