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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
    polar-coordinates_discretization.gif

    ?temp_hash=13cdb39cd76b0e50e6bc9a89ab8c68d8.png
    The crossings are orthogonals.
     
  5. Aug 1, 2015 #4

    HallsofIvy

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    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

    HallsofIvy

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    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
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