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Math Help - Orthogonal Trajectories

  1. #1
    Member aldrincabrera's Avatar
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    Smile Orthogonal Trajectories

    ,.hello there,.,amm,.,can anyone please help me understand about " orthogonal trajectories"???any ideas will do ,.thnx,.,.
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  2. #2
    A Plied Mathematician
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    Re: Orthogonal Trajectories

    The basic idea is this. Suppose you have a first-order differential equation

    y'=f(x,y).

    This DE defines the slope of the curve y(x). It is known that, given a straight line with slope m, the slope of the perpendicular line is -1/m. So, the family of curves y_{\perp} that are orthogonal to the solutions of the DE above are going to satisfy the following DE:

    y_{\perp}'=-\frac{1}{f(x,y_{\perp})}.

    You can see Zill's A First Course in Differential Equations with Modeling Applications for a problem or two involving orthogonal trajectories. Just look it up in the index. Also check out pages 117-118 of Tenenbaum and Pollard's Ordinary Differential Equations.
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  3. #3
    Member aldrincabrera's Avatar
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    Re: Orthogonal Trajectories

    ,.,wow,.,.thnx so much mr.Ackbeet
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    Re: Orthogonal Trajectories

    You're welcome!
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  5. #5
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    Re: Orthogonal Trajectories

    For example, the formula y= a/x defines a family of hyperbolas, all asymptotic to the x and y axes. dy/dx= -a/x^2 so that (dy/dx)/y= (-a/x^2)/(a/x)= -1/x where we have eliminated the parameter a: dy/dx= -y/x is a differential equation satisfied by those functions. Since the slopes of perpendicular lines are "negative reciprocal", any curve satisfying dy/dx= x/y will be perpendicular to all of those hyperbola. Of course, we can write that as ydy= xdx and integrate: (1/2)y^2= (1/2)x^2+ C which can be written \frac{y^2}{2C}- \frac{x^2}{2C}= 1 a family of hyperbolas, all asymptotic to the lines y= x and y= -x and all of which are perpendicular to all of the original hyperbolas.

    Another example: x^2+ y^2= a is the family of circles, with center at the origin and different radii. Differentiating both sides with respect to x, 2x+ 2y(dy/dx)= 0. That is, the solutions to the differential equation \frac{dy}{dx}= -\frac{x}{y} are those circles. Again, inverting and multiplying by -1, \frac{dy}{dx}= \frac{y}{x} describes the family of "orthogonal trajectories" to those circles. It should be no surprise that we can write that as \frac{dy}{y}= \frac{dx}{x} and, integrating, ln y= ln x+ C. Taking exponentials of both sides, y= C' x where C'= e^C and we see the 'obvious'- that the lines through the origin are all perpendicular to those circles.
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