Designing the Farrell Surface with Differential Equations

Peter Farrell
peter@farrellpolymath.com
December, 2010

Here's an exploration you can do into optimum design. I want to make a million dollars by designing the perfectly shaped surface for reflecting light. It'll be used in every flashlight housing, because the light bulb's rays will reflect off the surface and radiate out parallel to each other. How efficient!

The surface will also be used in every satellite dish, because the radio waves from space will hit the surface and reflect right into the receiver in the focus of the dish.

You’ll have to do some modeling with a graphing program (I use Geogebra), and set up some differential equations. If you can access the internet, WolframAlpha.com can solve them for you.

First, fix a point on the y-axis that will act as the focus for the light bulb or satellite receiver. Here I called it A. Then make some vertical lines to represent the parallel rays coming in or out of the dish. I graphed x = 1, x = 2 and x = 3. Feel free to use negative numbers if you like.
Now add a point to each line to act as a point on the surface. I figured the surface would curve up somewhat but otherwise the points are placed pretty randomly. Draw a segment between the surface points and the focus. I colored them red. I drew some red rays over the vertical lines for cosmetic effect. Now you can hide the black lines and the points at the top.
So we’ve made sure the parallel rays hit the surface points and reflect straight into the focus. But what direction is the surface facing to do this perfect job of reflection? It’s the derivative or slope. We can’t directly find the slope line yet, but we can bisect the angle between the incident ray and the reflected ray. That gives the normal line (in blue at right) to the mirror, and the tangent line is its perpendicular. Hide the normal lines and display the slopes of the tangent lines.
Sure, the points do a perfect job of reflecting right where they are, but the surface has to be defined by a single equation. Trying to make the points themselves follow a pattern would take a lot of work and trial and error. But if we work with the derivatives (the slopes), not only will we get an equation for this curve, but for a whole family of curves that work. That means the slopes have to have a pattern to them. Move the points up and down to put the slopes in some proportion k to their x-values.
I got lazy and just made k = 1. That means the slope at any point equals the x-value. Now we have 3 points on the reflective surface doing their job reflecting parallel rays into the focus. Even better, the slopes are all equal to the x-value. That’s a differential equation:
Now we can go to any Computer Algebra System or WolframAlpha.com and ask it to solve our differential equation. Here’s the form of the solution we’re given: The x^2 term means it’s a parabola. Just from knowing a pattern to the slopes we found the equation of the curve! Let’s graph and see how far we need to move it up or down to fit the points (that’s why it has a constant added).
So close! Change the formula to make the parabola go up a bit until it contains all the points. (Look at the point where x = 1. What do you have to add to to make the y-value 1?)
A half a unit up works. So the equation of the curve is If we rotate it around the y-axis we’ll get a 3-D surface that reflects all incoming or outgoing rays the way we want it to. Next time you see a satellite dish, be sure to call it a Farrell Surface! Problem: Can you create a parabola with its focus exactly 1 unit away from the vertex? Challenge problem: If we call the distance from the focus to the vertex f, find the general equation for the surface. ( but what is k in terms of f?)