Notes 19: Envelopes - UMass

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 in R3 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 . Thus the envelope is the locus of points (x, y, t) so that the normal to the plane is F t F (x, y, t) 0 and F 0. t Here is a second derivation of these equations. (x(t), y(t)) so that The envelope is a curve γ(t) 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 ) at the point (x0 , y0 ). Please check that this is through (x0 , y0 ) in the direction ( F y x orthogonal to F. Thus the next point on the envelope beyond (x0 , y0 ) is (x0 , y0 ) ( F F , ). 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 ((x0 , y0 ) ( F F , ), t0 ) 0. y x For small this is F (x0 , y0 , t0 ) F F F F F ( ) 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. 3. We conclude that F 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 f1 (x t)2 (y t2 )2 4 0, f2 2(x t)( 1) 2(y t2 )( 2t). The GroebnerBasis of the ideal I f1 , f2 is g1 1156 688x2 191x4 16x6 544y 30x2 y 40x4 y 225y 2 96x2 y 2 16x4 y 2 136y 3 32x2 y 3 16y 4 g2 7327t 6929x 2946x3 224x5 1928ty 2922xy 1480x3 y 128x5 y 768ty 2 792xy 2 224x3 y 2 896ty 3 544xy 3 128x3 y 3 256ty 4 384xy 4 g3 952 431tx 159x2 16x4 320y 12txy 214x2 y 32x4 y 366y 2 48txy 2 32x2 y 2 80y 3 64txy 3 32x2 y 3 32y 4 g4 697t 23x 288tx2 174x3 32x5 108ty 322xy 112x3 y 336ty 2 32xy 2 32x3 y 2 64ty 3 96xy 3 g5 128 135t2 26tx 111x2 16x4 64y 40txy 8x2 y 32y 2 32txy 2 16x2 y 2 16y 3 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 g1 g1 , ). V (g1 , x y To verify it is the same points we check that g1 , g1 g1 , A1 , A2 , A3 and x y A21 , A22 , A23 g1 , 3 g1 g1 , . x y

Figure 1: First Example of an Envelope 4 2 Out[50] 0 -2 -4 -4 -2 0 2 Out[55] 4 4 6

Figure 2: Example of Envelope 5 4 3 Out[58] 2 1 0 0 1 2 3 4 5 5

Figure 3: Derivation of Envelope Equation 6

Figure 4: The Envelope 6 4 2 Out[3] 0 -2 -4 -2 0 2 7 4

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 the normal to the plane is @F @t:Thus the envelope is the locus of points (x .