Orthogonal Trajectories of Curves

Use D.E.’s to find curves that run perpendicular to the ones we already know:

"orthogonal trajectories"

Simple example: have a "parallel" on the globe and find the meridians

This is an issue you most likely will run into again and again throughout your career:

You know the function of a curve - or several curves - very well but now you would like to find those curves that run perpendicular to the known ones.

This is quite important in imaging, for example.

Imagine you have an object, e.g. a new car model. In order to describe the surface you are measuring the distance between a reference point and the models surface (using sonar, for example). You are doing this systematically for several heights, let's say in intervals of 1 cm.

The result will be a number equidistant curves, describing a thin "onion ring" like layer of the surface. How can we get from this representation to the more practical and better looking wire-type representation?

Here we go….

As so often, the basic idea is very simple:

Simple idea behind this:

a) Find a D.E. **the given curves** are solution of (the one’s we already
know):

First of all we have to look for a differential equation our known curves (or, better, their function) will satisfy - means, our measured curves are solution of some D.E. - we only have to figure out which one….

b) The curves we are looking for, the orthogonal trajectories then must satisfy the following differential equation:

O.K., let's assume we have found such a D.E. Keep in mind, are the known functions and is that particular D.E. ….

We know that is nothing but the slope of our curve. Then, of course the negative and inverse value of describes the slope of the function perpendicular to y!

c) Solve the D.E. above (that might be tricky) . The solutions will be the trajectories we were looking for.

All that's left to do is solving this new differential equation

This not always is as easy as it is in the following example:

Example:

with c arbitrary. (parabolas)

Orthogonal trajectories?

Still contains a "c" !

In order to find the orthogonal trajectories we first solve the equation
for . Note that
equals to a function of x – we are trying to find it as a function of x
**and** y ! .

Fortunately we can do this by first rewriting our the equation of function so
it equals to "c" , **and then** substituting this expression into the first
derivative:

But

As a result we get as an alternative way to write our differential equation. You might want to go back and follow again step by step what we did so far ….

Now we can use to set up the D.E. for the trajectories (Note the red marked "-"sign!

We are lucky: The newly retrieved differential equation for the trajectories, , can be easily solve by the method of "Separation of Variables" !

And the results are – surprised? – ellipses!!!

- So, now we know: If we have a set of parabolas their orthogonal trajectories will be ellipses. That’s good to know. Let’s assume you have some parabolic graphs and you are trying to translate your "slice"-graphics into a wire-mesh picture - your graphics program should better comes up with some ellipse-like shapes!

- You know even more now: The orthogonal trajectories of ellipses will be parabolas!

- And, besides of having learned the method
**how**to find the orthogonal trajectories of functions you again have increased your ability to do some quality control of your work as an engineer!

By now you should be able to solve most 1^{st} order ODE’s with ease
(If you still have trouble to do so, you better catch up: This issue will come
up again and again!)

However, the way to a solution can be long and tedious, and sometimes (after torturous hours of calculations and fruitless tries it turns out there is no solution at all. Or, there are many possible solutions. Or, …..

Before wasting valuable time, let us discuss if and when a solution of
1^{st} order ODE exists and, if yes, if this is an unique solution or
not:

Try this one yourself!

Copyrights 1999, 2000 by Peter Dragovitsch and Ben A. Fusaro