A very basic definition of a mathematical envelope is a bunch of lines put together to form a curve (see Webster 1913's definition). However, it doesn’t have to be just lines. Envelopes can use circles to form cardioids, planes to form 3-D figures, or any number of things. The simplest envelopes aren’t very hard to draw/create. In fact, I made them in grade school as a young child without even realizing it. In first grade, my teacher had us make string art. She gave us a piece of wood with nails hammered into it in a square formation. Everyone in class was given one piece of string and told to loop the string around the nails in straight lines to see if we could form curves. Ahhh, the joys of discovery. I stumbled onto the creation of my first envelope. Now that I am a little more mature, I tend to use graph paper, a pencil, and a ruler instead of wood, nails, and string.

Here is a simple envelope and the method used to find its equation:

On a Cartesian coordinate system, connect the points (0,8) and (1,0), (0,6) and (2,0), and (0,4) and (3,0). Continue the pattern until the envelope can be seen.

For a line to be a part of this curve, it must be tangent to the curve. In other words, it must touch the curve and have the same slope as the curve at that point. This takes two equations; one for the point on the curve (the purpose of the point-slope equation below) and one to be sure it has the same slope (the derivative of the point-slope equation).

The first thing that must be done in the process of finding the equation to any envelope is to choose a parameter that will result in only one line of the envelope. For this particular problem, I chose the y-intercept, using the generic point (0,β) to write an equation for all the lines. The corresponding x-intercept for the point (0, β) is (5-β/2,0). Using the point-slope equation for a line, y_{1}-y_{2}=m(x_{1}-x_{2}), I plugged in a point and my slope to get this equation.

y – β = ((-β)/(5 – β/2)) x *or* y = ((-β)/(5 – β/2)) x + β

From here I simplified to get the equation 10y – yβ = -2βx + 10β – β^{2}

I then took the derivative with β as my variable and obtained –y = -2x + 10 – 2β. Set these two equations up as a system of equations. Solve the second equation for β, and plug into the first equation of the system. After quite a bit of simplifying, I arrived at the equation

0 = ¼ y^{2} - xy + x^{2} -5y –10x + 25

This fits the standard formula for a conic 0 = Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F. This particular envelope happens to be a parabola.

All it takes to solve envelopes is the right parameter and the willingness to plow through some nasty, and occasionally impossible, algebra.