Finding Coefficients of Orthogonal Quadric Equations

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    Orthogonal Polynomials
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Discussion Overview

The discussion revolves around finding coefficients of a quadric equation (G(x,y) = 0) that intersects orthogonally with another given quadric equation (F(x,y) = 0). The focus is on the mathematical conditions required for orthogonality at the points of intersection, particularly the gradient condition ∇F • ∇G = 0.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that for two quadric equations to intersect orthogonally, their gradients must satisfy the condition ∇F • ∇G = 0.
  • Another participant questions the concept of orthogonality in the context of points and intersections, noting that intersections of surfaces are typically curves.
  • A later reply clarifies that the curves are orthogonal at every point of intersection.
  • One participant proposes that to find the function G, a family of functions rather than a single function F(x, y) is necessary.
  • Another participant provides an example using circles to illustrate how to derive the orthogonal complement of a family of curves, detailing the process of differentiation and solving the resulting differential equation.

Areas of Agreement / Disagreement

Participants express differing views on the nature of orthogonality and the requirements for finding the function G. There is no consensus on the method for deriving G or the interpretation of orthogonality in this context.

Contextual Notes

The discussion includes assumptions about the nature of the functions involved and the conditions under which orthogonality is defined. The mathematical steps and definitions may vary among participants, leading to potential ambiguities.

Bruno Tolentino
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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?
 
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How can points be orthogonal?

The intersections of two surfaces are most commonly curves.

I am not sure what is being asked.
 
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The crossings are orthogonals.
 
Bruno Tolentino said:
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.
You mean "the curves are orthogonal at every point of intersection".

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?
 
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Yeah! So, exist some way of find the G function?
 
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).
 
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