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Curve of intersection of plane and surface

  1. Sep 27, 2006 #1
    this is the problem and i've worked for mostof it, but im not sure i've got the right way to go about it, or if the ansewr i have found it right.
    here, im using the method for finding the vector equation

    given the plane 2x + 6y + z = 6 intersects the paraboloid z = x^2 + y^2, find and name this curve of intersection.
    __________________________________________

    so far, i rearranged the plane equation to a function of z and let it equal te equation of the paraboloid.

    this gave me:
    x^2 + y^2 = 6 -2x - 6y
    6 = x^2 + 2x + y^2 + 6y
    6 = [x^2 + 2x + 1^2] + [y^2 + 6y +3^2] -10
    16 = [x + 1]^2 + [y + 3]^2
    4^2 = [x + 1]^2 + [y + 3]^2

    this then gave me the projection downwards of the function, a circle. but the function should be an ellipse. (which cant be called a function y of x, because it fails the vertical test, not to clear on this)

    to get the function of the curve of intersection, (here is where im not sure about where im going) i wanted x infunction of t and y in function of t (for parametric equations)

    x = f(t)
    y = g(t)

    so i went:

    rule: (cos(t))^2 + (sin(t))^2 = 1

    [(x +1)/4]^2 + [(y + 3)/4]^2 = 1

    (x +1)/4 = cos(t) => x + 1 = 4*cos(t) => x = 4*cos(t) - 1
    (y + 3)/4 = sin(t) => y + 3 = 4*sin(t) => y = 4*sin(t) - 3
    [(respectively): eqn, first eqn, second eqn]
    [where o<= t<= 2*pi]

    from,
    4^2 = [x + 1]^2 + [y + 3]^2
    => 4^2 = [4*cos(t)]^2 + [4*sin(t)]^2
    4^2 = 16*[cos(t)]^2 + 16*[sin(t)]^2
    4^2 = 16*([cos(t)]^2 + [sin(t)]^2)
    4^2 = 16*([cos(t)]^2 + [sin(t)]^2)


    substituting second eqn x and y values into the plane:
    z = 6 - 2x - 6y
    => z = 6 - 2*[4*cos(t) -1] - 6*[4*sin(t) - 3]
    z = 6 - 8*cos(t) +2 - 24*sin(t) + 18
    z = -8*cos(t) - 24*sin(t) + 26

    this is what i have done so far (stop me if im doing something completely wrong please ^^') but i'm a little lost in what the text book is telling me

    therefore the parametric equations for the equation
    here is where everything goes iffy (if not already somewhere above).
    when i use the parametric equation, because the projection is a circle the parametric equations would be:


    x = 4*cos(t) - 1
    y = 4*sin(t) - 3
    z = -8*cos(t) - 24*sin(t) + 26
    [where o<= t<= 2*pi]

    so the corresponding vector equation is:

    r(t) = [4*cos(t) - 1]i + [4*sin(t) - 3]j + [-8*cos(t) - 24*sin(t) + 26]k
    [where o<= t<= 2*pi]

    sorry if my explaination is unclear in some parts. im not sure if this method gets me to the answer i want, but is there possibly another way to solve this problem? this was the only way i could think of.
    the text book didnt exactly teach the topic but there was an example that was similar to this problem im working on so i was working alot of this from that example (which was a really simple example and only had four lines for working).
     
  2. jcsd
  3. Sep 27, 2006 #2
    I didn't check the details of your calculation but the logic seems fine to me.
     
  4. Sep 27, 2006 #3

    HallsofIvy

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    That seems a bit more complicated than what I would do.

    You have determined that x and y on the curve of intersection satisfy
    (x + 1)2 + (y + 3)2= 16, a circle with center at (-1, -3) (in the xy-plane) and radius 4.
    Parametric equations for that circle are [itex]x= -1+ 4cos(\theta), y= -3+ 4sin(\theta)[/itex]
    Now, we also know that x, y, z, satisfy 2x + 6y + z = 6, the equation of the plane, so z= 6- 2x- 6y or [itex]z= 6- 2(-1+ 4cos(\theta))- 6(-3+ 4 sin(\theta))= 6+ 2+ 18- 8cos(\theta)- 24sin(\theta)[/itex]
    [itex]= 25- 8 cos(\theta)- 23 sin(\theta)[/itex].
     
  5. Sep 27, 2006 #4
    @jpr0: thanks for checking

    yeah, i figured there would have been a much easier way to get this done but i couldnt find one. how else could it be done?

    was that last line supposed to be 26 and 24 instead of the 25 and 23 (respectively)

    in that case should i have stopped at
    instead of subbing all the variable values into the vector equation? or is it still right to do the subbing?
     
  6. Sep 28, 2006 #5
    Your algebra is good but you should always think of the meaning of equations.
    You first solved simultaniously the two equations by equating z-s. Which means you got an equation wrt x and y, which is the projection of the intersection, and you recognized by its form to represent a circle.

    You have then expressed it parametrically, and substituted in either plane or paraboloid equation to get the equation of the intersection. You then just name that curve and that's the solution.
    You also shouldn't worry about whether something is function or not in questions like these. You are looking for equations of the lines\surfaces.
     
  7. Oct 1, 2006 #6
    Does this mean that the equation for the intersection of the plane and surface is z= 26-8*cos(t)-24*sin(t)?

    I don't understand this, as this equation does not represent a circle. :confused: Couldn't you just leave it as 4^2 = [x + 1]^2 + [y + 3]^2. Could somebody explain why we need to find the parametric equations?

    Also, if I wanted to use Lagrange multipliers to find points on this curve which are nearest to and furthest from the origin, how would I go about that? I know that if you want to find the distance D from a point (in this case the origin) to a plane, you need to minimise D subject to the constraint of the equation of the plane, however for the intersection above we do not have a plane only a curve where the plane and surface intersect. We also need to find a maximum.

    Sorry if that makes no sense! :smile:
     
  8. Oct 1, 2006 #7
    Shouldn't this be in Homework Help, folks?

    Not quite (and someone please correct me if I'm wrong - I'm doing the same course as Locst and Yura so I'm not a master!)

    The equation for the z-component of the ellipse is z= 26-8*cos(t)-24*sin(t), and you need the similar equations for x and y before you can claim to have an equation (or set of parametric equations) for the intersection.

    4^2 = [x + 1]^2 + [y + 3]^2 is the projection of this ellipse into the x-y plane. It has no z-component, and is most definitely not the equation for the intersection.

    To find the maximum and minimum distances from the origin you need to find maximum and minimum values for d^2=x^2+y^2+z^2 subject to the constraints 2x + 6y + z = 6 and z = x^2 + y^2. (Normally this would be d=(x^2+y^2+z^2)^-1, but you can square both sides to remove the nasty square root operator.)

    Just set up your simultaneous equations with two LaGrange multipliers and solve them to get values for x, y and z. You end up with five equations in five unknowns. There is a good example of this on page 1006 of Stewart 5th ed.

    Finally, since the assignment is due in in a little under two hours, best of luck!:redface:

    Russel
     
  9. Oct 1, 2006 #8
    oh im done. i was just getting some reassurance for my method i used. (thus why i didnt post this into the homework forum)
     
  10. Oct 2, 2006 #9

    HallsofIvy

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    A one-dimensional object such as a curve (or circle in particular) in 3 dimensions, cannot be written in terms of a single equation. Each equation reduces the "degrees of freedom" by one so anything like z= f(x,y) must represent a 3-1= 2 dimensional object- a surface. To represent a curve in 3 dimensions you need either two equations (like representing a line as the intersection of two planes) so that you have 2 equations in 3 unknows- 3- 2= 1 "degree of freedom"- or, using parametric equations, to give you 3 equations if 4 unknowns (x, y, z, and the parameter), so again you have 4- 3= 1 "degree of freedom". You could not "just leave it as 4^2 = [x + 1]^2 + [y + 3]^2" because that tells you nothing about the z-coordinate. In three dimensions that is the equation of a cylinder, just as, in 2 dimensions, y= 1 is the equation of a line.

    [/quote]Also, if I wanted to use Lagrange multipliers to find points on this curve which are nearest to and furthest from the origin, how would I go about that? I know that if you want to find the distance D from a point (in this case the origin) to a plane, you need to minimise D subject to the constraint of the equation of the plane, however for the intersection above we do not have a plane only a curve where the plane and surface intersect. We also need to find a maximum.

    Sorry if that makes no sense! :smile:[/QUOTE]
    Then you would minimize D subject to the point being on that curve.
     
  11. Oct 2, 2006 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    A one-dimensional object such as a curve (or circle in particular) in 3 dimensions, cannot be written in terms of a single equation. Each equation reduces the "degrees of freedom" by one so anything like z= f(x,y) must represent a 3-1= 2 dimensional object- a surface. To represent a curve in 3 dimensions you need either two equations (like representing a line as the intersection of two planes) so that you have 2 equations in 3 unknows- 3- 2= 1 "degree of freedom"- or, using parametric equations, to give you 3 equations if 4 unknowns (x, y, z, and the parameter), so again you have 4- 3= 1 "degree of freedom". You could not "just leave it as 4^2 = [x + 1]^2 + [y + 3]^2" because that tells you nothing about the z-coordinate. In three dimensions that is the equation of a cylinder, just as, in 2 dimensions, y= 1 is the equation of a line.

    [/quote]Also, if I wanted to use Lagrange multipliers to find points on this curve which are nearest to and furthest from the origin, how would I go about that? I know that if you want to find the distance D from a point (in this case the origin) to a plane, you need to minimise D subject to the constraint of the equation of the plane, however for the intersection above we do not have a plane only a curve where the plane and surface intersect. We also need to find a maximum.

    Sorry if that makes no sense! :smile:[/QUOTE]
    Then you would minimize D subject to the point being on that curve.
     
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