Notes 19: Envelopes

For each value of t we are given the circle with center at (x = t, y = 0) and radius 1. Along the lines y = ±1 the curves become infinitely dense. Another way of stating this is to say that as we travel along the curve y = 1 or y = −1, we are tangent to each of the circles we touch. Such a curve is called an envelope. We give a second example. Consider the family of lines y − (1/a) = (−1/a2)(x − a). These are all tangent to the curve xy − 1 = 0. See the figures below. Assume that we are given a family, parametrized by t, of curves, that is, for each value of t we have a curve F (x, y, t) = 0. We show how to find the envelopes of this family of curves. We derive in two ways the equations for doing this. The first way is to consider the family as arising from the surface 3 in R given by F (x, y, t) = 0 and taking slices t = t0, t = t1, ··· at regular intervals and then projecting each slice to the x − y plane. The envelope occurs when the curves get close to each other. The projected curves are closer to each other exactly when the surface is steeper. Indeed, the curves become infinitely close exactly when the surface is vertical. The ”direction” of the surface at a point is given by its tangent plane. The envelope is the projection of the points where the tangent plane is vertical. The tangent plane is vertical when the normal to the tangent has no vertical component. The vertical component of ∂F the normal to the plane is ∂t . Thus the envelope is the locus of points (x, y, t) so that ∂F F (x, y, t) = 0 and = 0. ∂t

Here is a second derivation of these equations. The envelope is a curve γ(t) = (x(t), y(t)) so that

1. γ(t0) = (x0, y0) is a point of F (x, y, t0) = 0, and

2. the tangent line to γ(t) at t = t0 is the same as the tangent line to F (x, y, t0) = 0 at (x0, y0).

We ﬁnd conditions that insure that if (x0, y0) is on the envelope for t = t0), then the point gotten by starting at (x0, y0) and moving along the curve in the direction of the common tangent to γ and F (x, y, t0) is on ’the next’ curve F (x, y, t0 + ) = 0, very small. Please see the ﬁgure below. The tangent to F (x, y, t0) at (x0, y0) is orthogonal to ∇F. Thus the tangent is the line ∂F ∂F through (x0, y0) in the direction ( ∂y , − ∂x ) at the point (x0, y0). Please check that this is orthogonal to ∇F. Thus the next point on the envelope beyond (x0, y0) is ∂F ∂F (x , y ) + ( , − ). 0 0 ∂y ∂x

1 We want this to be on the next curve in our family of curves, that is, we want it to be on the curve F (x, y, t0 + ). Explicity we want ∂F ∂F F ((x , y ) + ( , − ), t + ) = 0. 0 0 ∂y ∂x 0 For small this is ∂F ∂F ∂F ∂F ∂F F (x , y , t ) + + (− ) + ≈ 0. 0 0 0 ∂x ∂y ∂y ∂x ∂t

1. The ﬁrst term F (x0, y0, t0) = 0, since we have stipulated that (x0, y0) is on the level curve for t = t0. 2. The second and third summands sum to zero.

∂F 3. We conclude that ∂t = 0.

An Envelope We consider the family of circles of radius 2 and centers at the points of the parabola parmetrized by (x = t, y = t2). We can write this as the family of curves

F (x, y, t) = (x − t)2 + (y − t2)2 − 4 = 0.

The envelope is given by

2 2 2 f1 = (x − t) + (y − t ) − 4 = 0, 2 f2 = 2(x − t)(−1) + 2(y − t )(−2t).

The GroebnerBasis of the ideal I =< f1, f2 > is

2 4 6 2 4 g1 = −1156 + 688x − 191x + 16x + 544y + 30x y − 40x y+ 225y2 − 96x2y2 + 16x4y2 − 136y3 − 32x2y3 + 16y4 3 5 3 g2 = 7327t + 6929x − 2946x + 224x − 1928ty + 2922xy − 1480x y+ 128x5y − 768ty2 − 792xy2 − 224x3y2 − 896ty3 − 544xy3 + 128x3y3 + 256ty4 − 384xy4 2 4 g3 = −952 − 431tx + 159x + 16x − 320y + 12txy+ 214x2y − 32x4y + 366y2 + 48txy2 + 32x2y2 + 80y3 + 64txy3 − 32x2y3 − 32y4 2 3 5 g4 = −697t − 23x + 288tx + 174x − 32x − 108ty − 322xy+ 112x3y + 336ty2 − 32xy2 − 32x3y2 − 64ty3 + 96xy3 2 2 4 g5 = −128 + 135t + 26tx + 111x − 16x + 64y + 40txy+ 8x2y + 32y2 + 32txy2 − 16x2y2 − 16y3

We see that the envelope is contained in the curve g1 = 0.

2 Which points on the this curve actually arise from points on the envelope? From the Closure Theorem we know that all of the points except those on a subvariety arise from points on the envelope. To see which points lift to points on the envelope we examine the polynomials g2, ··· , g5. We write these as polynomials in the variable t and extract the coeﬃcient of the highest power of t. We see that for g5 the highest degree of t that appears is in the term 135t2. Since 135 6= 0 we can say that every point on the curve g1 lifts to a curve on the envelope. We ae concerned with real points. The extension theorem only works over the the complex numbers. Problems may arise. Given a point (x, y) on the envelope how many circles in our family pass through that point? For each value of t there is a circle in our family with center at (t, t2). Look at equation g4. It is of the form A4(x, y))t + B4(x, y) = 0. Hence if A(x, y) 6= 0, then there is a unique circle in the family passing through that point since we can solve

t = −B4(x, y)/A4(x, y).

This argument also works with g3 and g2.

What happens at the points where A2(x, y) = A3(x, y) = A4(x, y) = 0.

We ﬁnd the points where A2 = A3 = A4 = 0. We do this by using Groebner bases. We ﬁnd three points: (x = 0, y = 4.25), (x = ±.9368, y = 1.63988). These look as though they may be the singular points of the envelope. As an exercies draw the two circles that are ’tangent’ to the envelope at these points. Are these points the singular points of the envelope?

The envelope is given by the equations g1(x, y) = 0. The singular points are the points of the variety ∂g ∂g V (g , 1 , 1 ). 1 ∂x ∂y To verify it is the same points we check that ∂g ∂g g , 1 , 1 ∈< A ,A ,A > and 1 ∂x ∂y 1 2 3 ∂g ∂g A2,A2,A2 ∈< g , 1 , 1 > . 1 2 3 1 ∂x ∂y

3 Figure 1: First Example of an Envelope

0 Out[50]=

-2

-4 -4 -2 0 2 4 6

Out[55]=

4 Figure 2: Example of Envelope

Out[58]=

0 0 1 2 3 4 5

5 Figure 3: Derivation of Envelope Equation

6 Figure 4: The Envelope

Out[3]=

-2

-4 -2 0 2 4