Students Stuff Envelopes

by
Nicholas Migliozzi
Princeton Day School
Princeton NJ 08542

Mathematica in Education and Research
Volume 4 Number 2
Copyright 1995 TELOS/Springer-Verlag

Editor's Preface

Anyone who teaches mathematics wants their students to get excited about the subject.  Nicholas Migliozzi has succeeded in accomplishing this at least once, and Mathematica  played a central role.  Both the symbol manipulation and the graphics capabilities of Mathematica  were used to good effect to demonstrate the idea of the envelope of a family of curves. The students' work shows that they "got the picture".

Students Stuff Envelopes

It was a few days before our Christmas break at Princeton Day School, and
students were busy signing cards and stuffing envelopes to give to their
friends.  I happened to mention to my eight students in one of our advanced
classes that there are envelopes in calculus.

They said, "Go on!"

I said, "Really, " and I proceeded to explain:

The definition of an envelope in mathematics is:

Given a family of equations, f(x, y, a) = 0, find a curve that is tangent
to each member of the family.  This curve, when it exists, is called the
envelope of the family.

I mentioned that, since the envelope is tangent to each member of the
family, a reasonable procedure would be to solve simultaneously f(x, y, a)
= 0  and  fa(x, y, a) = 0 (the derivative with respect to a, with all other
variables held constant).

The next day we went to the computer lab, started Mathematica, used
Schaum's Advanced Calculus as a reference, and tried the following:

Example 1:  Given the family f(x, y, a) = ax - a2 -y = 0, find its envelope.

fam = Table[Plot[a*x - a^2, {x, -3, 3}, 
     PlotRange->{-3, 3}, 
     DisplayFunction->Identity], 
     {a, -4, 4, .2}];

I explained:  We are making a table that contains the graphs of several
family members, and we are also delaying the display.

Show[fam, DisplayFunction->$DisplayFunction];

The students were awestruck!

We proceeded to find the envelope for this family.

da = D[a*x - a^2 - y, a]

This is the derivative of f with respect to the variable a.  (x and y are
held constant.)  Next we solve  f(x, y, a) = 0  and  fa(x, y, a) = 0
simultaneously.

Eliminate[{a*x - a^2 - y == 0, da == 0}, a]

We had our curve, but was it the correct answer?  We used Mathematica to do
a graphical check.

env = Plot[x^2/4, {x, -3, 3}, PlotRange->{-3, 3}, 
     PlotStyle->{{Thickness[.01], GrayLevel[.7]}}];

This sure looked good, but did it "fit" the family?

Show[fam, env, DisplayFunction->$DisplayFunction];

A perfect fit!  Everyone clapped.

We tried a second example:

Example 2:  Given the family x sin(a) + y cos(a) = 1, find its envelope.

Clear[fam, env, x, y, a, da, f];

f = Solve[x*Sin[a] + y*Cos[a] == 1, y]

fam = Table[Plot[y /.f, {x, -1.5, 1.5}, 
     PlotRange->{-1.5, 1.5}, 
     DisplayFunction->Identity], 
     {a, 0, 2Pi, Pi/7}];

Show[fam, DisplayFunction->$DisplayFunction, 
     AspectRatio->1];

The envelope appeared to be a circle.  We proceeded algebraically:

da = D[x*Sin[a] + y*Cos[a] - 1, a]

Since the equations involved Sin and Cos, we gave Mathematica a little help.

Solve[{x*Sin[a] + y*Cos[a] == 1, da == 0, 
     Sin[a]^2 + Cos[a]^2 == 1}, {x, y}]

Since Sin2[a] + Cos2[a] = 1, our equation is clearly x2 + y2 = 1 and this
is the envelope for the family.  We checked it:

env = Plot[{Sqrt[1 - x^2], -Sqrt[1 - x^2]}, 
     {x, -1, 1}, PlotRange->{-1, 1}, 
     DisplayFunction->Identity, 
     PlotStyle->{{Thickness[.01], GrayLevel[.7]}, 
     {Thickness[.01], GrayLevel[.8]}}];

Show[fam, env, DisplayFunction->$DisplayFunction, 
     AspectRatio->1];

Another perfect fit!

At this point, we were bold enough to try a 3-dimensional family of surfaces.

Example 3:  Given the family of surfaces  z = 2ax - a2y, find the envelope.

Clear[fam, env, x, y, a, da, f];

f = z - 2*a*x + a^2*y;

da = D[f, a]

Eliminate[{f == 0, da == 0}, a]

Although we could try to graph the family of surfaces, it would take a very
long time and they would be hard to distinguish in 3-space.  We can easily
plot the envelope, however.

Plot3D[(x^2)/y, {x, -3, 3}, {y, -3, 3}, 
     ViewPoint->{3, 1, 1}];

We used our imagination to see the family!

[Editor's note:  For this surface, Mathematica's Plot3D command gives a
slightly misleading graph.  The trouble is that f(x, y) = x2/y is not
continuous at (0, 0), as can be seen by checking limits along the parabolas
y = kx2.  Mathematica tries to connect pieces of the graph which should not
be joined.  A way around this is to parameterize the surface using polar
coordinates:  x = r Cos[t],  y = r Sin[t],  and  z = r Cos2[t]/Sin[t].  Now
the surface can be plotted in two parts, avoiding t = 0, Pi, and 2Pi by,
for instance, letting Pi/100 <= t <= Pi - Pi/100 in one plot,  and Pi -
Pi/100 <= t <= 2Pi - Pi/100 in the other plot.  Here is the end result, two
surfaces which "kiss without touching" at the origin:

Imagination is still required to see the family!]

As an assignment for the next day, I gave the following eight problems. 
Each student was to choose one problem, one which was not already chosen. 
The last problem was open to everyone.

Exercises:

1)  Find the envelope for the family of curves:  ay = a2x + 1.

2)  Find the envelope for the family of curves:  x2/a + y2/(1 - a) = 1.

3)  Find the envelope for the family of curves:  (x - a)2 + y2 = a2/2.

4)  Find the envelope for the family of curves:  (y - a)2 = x2(1 - x2).

5)  Find the envelope for the family of circles whose centers are on the
parabola  y = x2 and pass through the vertex of the parabola.

6)  Find the envelope for the normals to the parabola  y = .5 x2.  This
referred to in mathematics as the evolute.

7)  Find the envelope for the family of lines whose intercepts with the x
and y axes is always a constant segment of length k, say k = 5.

8)  Find the envelope for the family of surfaces:  a2z = a(x - y) - 1.

Extra:  Start with the circle  x2 + y2 = 9  and a point on the circle, say 
(3, 0).  Draw a chord from (3, 0) to any other point on the circle, and use
this chord for the diameter of a new circle.  Repeat the process
indefinitely.  Find the envelope for the family of all these circles.

On the last day of classes the students said they wanted to postpone their
Christmas break by a few days so that they could study more envelopes.

I said, "Go on!"

Solutions

The students' work follows.  [Only the graphs of the families and the
envelopes are given here so that the reader is not deprived of the pleasure
of discovering the solutions.  Ed.]

1)  Solution by John Ackerman.

2)  Solution by Alex Manka.

3)  Solution by Michael de Senna.

4)  Solution by Yegnaswamy Sermadevi.

5)  Solution by Lindsey Sternberg.

6)  Solution by Maxime Ko.

7)  Solution by Julian Wong.

8)  Solution by Grant Gould.  Only the envelope is shown.

Extra:  All eight students tried the extra problem.