# Orthogonal Trajectories

• Jun 22nd 2011, 05:06 PM
aldrincabrera
Orthogonal Trajectories
• Jun 22nd 2011, 05:28 PM
Ackbeet
Re: Orthogonal Trajectories
The basic idea is this. Suppose you have a first-order differential equation

$\displaystyle y'=f(x,y).$

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 $\displaystyle y_{\perp}$ that are orthogonal to the solutions of the DE above are going to satisfy the following DE:

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

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.
• Jun 23rd 2011, 02:49 AM
aldrincabrera
Re: Orthogonal Trajectories
,.,wow,.,.thnx so much mr.Ackbeet
• Jun 23rd 2011, 03:22 AM
Ackbeet
Re: Orthogonal Trajectories
You're welcome!
• Jun 26th 2011, 06:15 PM
HallsofIvy
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: $\displaystyle (1/2)y^2= (1/2)x^2+ C$ which can be written $\displaystyle \frac{y^2}{2C}- \frac{x^2}{2C}= 1$ 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: $\displaystyle x^2+ y^2= a$ is the family of circles, with center at the origin and different radii. Differentiating both sides with respect to x, $\displaystyle 2x+ 2y(dy/dx)= 0$. That is, the solutions to the differential equation $\displaystyle \frac{dy}{dx}= -\frac{x}{y}$ are those circles. Again, inverting and multiplying by -1, $\displaystyle \frac{dy}{dx}= \frac{y}{x}$ describes the family of "orthogonal trajectories" to those circles. It should be no surprise that we can write that as $\displaystyle \frac{dy}{y}= \frac{dx}{x}$ and, integrating, $\displaystyle ln y= ln x+ C$. Taking exponentials of both sides, y= C' x where $\displaystyle C'= e^C$ and we see the 'obvious'- that the lines through the origin are all perpendicular to those circles.