# Computing the Envelope of a Family of Curves

The curve showing up on the bottom of the coffee cup looks like a curve that has each of the reflected light rays as a tangent line for some point on the curve, as illustrated below.

### Figure 5: The nephroid as an envelope of light rays.

In this situation, we say that this curve is part of the envelope of the family of reflected lines.

In general, suppose we have a family of curves C varying with t, (like our different lines reflected from the light ray at y=t) and Ct is defined by an equation F(t,x,y)=0. A curve E is said to be part of the envelope of the family if it shares a tangent line with each curve Ct. We would like to be able to compute an equation for curves in the envelope for this family, but this may be difficult. It is also possible that no such envelope curves exist.

However, when F(t,x,y) is a polynomial function of all 3 variables t,x,y, there is a way to use elimination of variables to compute the equation of a curve containing an envelope to the family Ct. This is explained in Projection, profiles, envelopes.

The idea goes like this. The equation F(t,x,y)=0 defines a surface in 3-dimensional (x,y,t)-space. In the case of our family of reflected line this surface is built geometrically as a "ruled surface" by stacking up each reflected light ray in sequence. Click here to see an MPEG movie (478 Kbytes) of this process. Of course, the complete surface is not merely the union of the reflected rays that come in from the left (as in Figure 5), but also includes the union of rays that reflect off the left-hand side of the coffee cup.

As explained in Projection, profiles, envelopes, the points of the envelope are the projections onto the xy-plane of those points on this surface F(t,x,y)=0 for which the tangent plane is parallel to the t-axis. This leads us to look for the equation of the envelope curves by eliminating the variable t from the equations

F(t,x,y) = 0
d/dt F(t,x,y) = 0

In order to do this for our family of reflected lines Ct, we must first express their equations in the form F(t,x,y) = 0, i.e. we must get rid of the square root occurring in the equation that we already found, and then bring all the terms to one side.

## Question 6

• Show that you can manipulate the equation we found for the line reflected from y=t so that there is only a single product involving a square root sitting alone on one side of the equation.
• Now square both sides of the equation to get rid of the square root, and bring all the terms to one side to obtain something of the form F(t,x,y)=0 for some polynomial F(t,x,y) in the variables t,x,y. Compare your answer with the one we got:

F(t,x,y) = ((y-t)(2-t2) + t(4-t2))2 - t2x2(4-t2)

Using Maple, we can eliminate the t variable via the following computation:

```> F:=((y-t)*(2-t^2)+t*(4-t^2))^2-t^2*x^2*(4-t^2);

2            2  2    2  2       2
F := ((y - t) (2 - t ) + t (4 - t ))  - t  x  (4 - t )

> GB:=gbasis({F,diff(F,t)},[t,x,y],plex):
```
Only one polynomial (the 9th and last) in the resulting Gröbner basis does not involve the variable t, so this gives the equation for the envelope:

```> factor(GB[9]);

2  2            2       2       4       2  2       4      6       4  2
x  y  (- 4 + 12 x  - 15 y  - 12 x  - 24 x  y  - 12 y  + 4 x  + 12 x  y

4  2      6
+ 12 y  x  + 4 y )
```

## Question 7

How does this compare with the equation for the nephroid we computing from the parametrization? What extra factors show up in the envelope computation?
Next: Conclusions
Previous: The Geometry of Reflecting Light Rays
Up: Introduction
Vic Reiner <reiner@math.umn.edu>
Frederick J. Wicklin <fjw@geom.umn.edu>