1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Orthogonal polynomials

  1. Jul 31, 2015 #1
    Given a quadric equation (F(x,y) = 0), exist other quadric equation (G(x,y) = 0) such that the poinst of intersection between the graphics are ortogonals.

    So, how to find the coefficients of the new quadic equation?

    EDIT: I think that F and G needs to satisfyF • G = 0. So, if F is known, how discover G?
    Last edited: Jul 31, 2015
  2. jcsd
  3. Jul 31, 2015 #2
    How can points be orthogonal?

    The intersections of two surfaces are most commonly curves.

    I am not sure what is being asked.
  4. Aug 1, 2015 #3
    OblateSpheroidalCoordinate_650.gif 65465.46545.png

    The crossings are orthogonals.
  5. Aug 1, 2015 #4


    User Avatar
    Science Advisor

    You mean "the curves are orthogonal at every point of intersection".

  6. Aug 2, 2015 #5
    Yeah! So, exist some way of find the G function?
  7. Aug 7, 2015 #6


    User Avatar
    Science Advisor

    First, in order to have what you are talking about, you have to have, not a single function, F(x, y), but a family of functions.

    To take a simple example, look at [itex]x^2+ y^2= R^2[/itex], the family of all circles with center at the origin. Differentiating both sides of the equation with respect to x, [itex]2x+ 2y dy/dx= 0[/itex] so that [itex]dy/dx= -x/y[/itex] at every point. To be orthogonal to that, a function, y(x), must have derivative equal to the negative of the reciprocal. That is, we want [itex]dy/dx= -1/(-x/y)= y/x[/itex]. That is, the "orthogonal complement" of this family of curves must satisfy [itex]dy/dx= y/x[/itex].
    That is an easily separable equation- it can be written [itex]dy/y= dx/x[/itex]. Integrating both sides, [itex]ln(|y|)= ln(|x|)+ C[/itex] so that [itex]|y|= C'|x|[/itex] where C' is equal to the exponential of C. By allowing C to take on non-positive values also, we can drop the absolute values and have [itex]y= Cx[/itex]. That is the family of all straight lines that go through the origin. They are diameters of the original circles so always perpendicular to them.

    More generally, given a family of curves, functions of x and y that depend a constant, to find the orthogonal complement, differentiate the equation defining the family to eliminate that constant. Find dy/dx= m(x,y) from that equation and then solve the differential equation dy/dx= -1/m(x,y).
  8. Aug 8, 2015 #7
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook