Envelopes and String Art Gregory Quenell 1 Activity: 1 Draw - PDF document

Envelopes and String Art Gregory Quenell 1

Activity: 1 Draw line segments connecting 0 . 8 (0 , x ) with (1 − x, 0) 0 . 6 for x = 0 . 1 , 0 . 2 , . . . , 0 . 9. 0 . 4 Benefits: 0 . 2 • Gives you something to do during calculus class 0 . 2 0 . 4 0 . 6 0 . 8 1 • Makes a pleasing pattern of inter- secting lines 2

Activity: 1 Draw line segments connecting 0 . 8 (0 , x ) with (1 − x, 0) 0 . 6 for x = 0 . 1 , 0 . 2 , . . . , 0 . 9. 0 . 4 Benefits: 0 . 2 • Gives you something to do during calculus class 0 . 2 0 . 4 0 . 6 0 . 8 1 • Makes a pleasing pattern of inter- secting lines • Provides an interesting curve to study 3

Question: 1 What curve is this? 0 . 8 0 . 6 Observation: 0 . 4 The curve’s defining property is that 0 . 2 the sum of the x - and y -intercepts of each of its tangent lines is 1. 0 . 2 0 . 4 0 . 6 0 . 8 1 That gives us the condition y − xdy y dx + x − dy/dx = 1 4

Different approach: For each α ∈ [0 , 1], let ℓ α be the line segment connecting (0 , β ) (0 , α ) with (1 − α, 0) . (0 , α ) If α and β are close together, then the ℓ α � ✠ intersection point of ℓ α and ℓ β is close to a point on the curve. ✲ ℓ β Exercise: (1 − β, 0) (1 − α, 0) For α � = β , the segments ℓ α and ℓ β in- tersect at the point ( αβ, (1 − α )(1 − β )) . 5

Result: As β → α , the point ( αβ, (1 − α )(1 − β )) approaches a point on the curve. Thus, each point on the curve has the ℓ α ✠ � form ✲ ℓ β β → α ( αβ, (1 − α )(1 − β )) lim for some α . This is an easy limit, and we get the parametrization ( α 2 , (1 − α ) 2 ) , 0 ≤ α ≤ 1 for our envelope curve. 6

Remarks: 1 • The coordinates 0 . 8 x = α 2 and y = (1 − α ) 2 0 . 6 satisfy 0 . 4 √ x + √ y = 1 0 . 2 so our curve is (one branch of) a hypocircle with exponent 1 2 . 0 . 2 0 . 4 0 . 6 0 . 8 1 • Stewart, p. 234, problem 8 says “Show that the sum of the x - and y -intercepts of any tangent line to the curve √ x + √ y = √ c is equal to c .” 7

Exercise: The coordinates 2 x = α 2 and y = (1 − α ) 2 satisfy 1 2( x + y ) = ( x − y ) 2 + 1 Result: Our envelope curve lies on a parabola 1 2 in the uv -plane, where u = x + y and v = x − y . 8

Exercise: The coordinates 2 x = α 2 and y = (1 − α ) 2 satisfy 1 2( x + y ) = ( x − y ) 2 + 1 Result: Our envelope curve lies on a parabola 1 2 in the uv -plane, where u = x + y and v = x − y . 9

Activity: String Art Drive nails at equal intervals along two lines, and connect the nails with decorative string. 10

Activity: String Art The envelope curves are the images, under a linear transformation, of parabolas tangent to the coordinate axes. That is, they are parabolas tangent to the nailing lines. 11

Digression: Game Theory Colin Consider a two-person, non-zero-sum IIA IIB game in which each player has two IA (2 , 0) (3 , 6) strategies. Rose IB (4 , 2) (0 , 0) (IA,IIB) 6 Such a game has four possible payoffs. We list them in a payoff matrix . 4 Payoff to Colin We can show the payoffs to Rose and Colin as points in the payoff plane . (IB,IIA) 2 (IB,IIB) (IA,IIA) 2 4 Payoff to Rose 12

Assumptions: Colin We assume each player adopts a IIA IIB mixed strategy : IA (2 , 0) (3 , 6) Rose • Rose plays IA with probability p IB (4 , 2) (0 , 0) and IB with probability 1 − p . • Colin plays IIA with probability q and IIB with probability 1 − q The expected payoff is then pq (2 , 0) + p (1 − q )(3 , 6) + (1 − p ) q (4 , 2) + (1 − p )(1 − q )(0 , 0) or p [ q (2 , 0) + (1 − q )(3 , 6)] + (1 − p ) [ q (4 , 2) + (1 − q )(0 , 0)] or q [ p (2 , 0) + (1 − p )(4 , 2)] + (1 − q ) [ p (3 , 6) + (1 − p )(0 , 0)] 13

(IA,IIB) Possible payoff points: 6 Each value of q determines one point on the line from (2 , 0) to (3 , 6) and one point on the line from (4 , 2) to 4 Payoff to Colin (0 , 0). Then p is the parameter for a line segment between these points. (IB,IIA) 2 p [ q (2 , 0) + (1 − q )(3 , 6)] +(1 − p ) [ q (4 , 2) + (1 − q )(0 , 0)] (IB,IIB) 4 (IA,IIA) Payoff to Rose 14

(IA,IIB) Possible payoff points: 6 Alternatively, each value of p determines one point on the line from (2 , 0) to (4 , 2) and one point on the 4 Payoff to Colin line from (3 , 6) to (0 , 0). Then q is the parameter for a line segment between these points. (IB,IIA) 2 q [ p (2 , 0) + (1 − p )(4 , 2)] +(1 − q ) [ p (3 , 6) + (1 − p )(0 , 0)] (IB,IIB) 4 (IA,IIA) Payoff to Rose 15

(IA,IIB) Possible payoff points: 6 Either way, the expected payoff is contained in a region bounded by four lines and a parabolic envelope 4 Payoff to Colin curve. If the game is played a large number of times and the average payoff converges (IB,IIA) 2 to a point outside this region, then the players’ randomizing devices are not independent. (IB,IIB) 4 (IA,IIA) Payoff to Rose This could be due to collusion, espionage, or maybe just poor random-number generators. 16

Generalization: Unequal Spacing (0 , Y ( α )) Draw line segments ℓ α connecting (0 , Y ( β )) ( X ( α ) , 0) with (0 , Y ( α )) for arbitrary differentiable functions X and Y . These are “spacing functions”. ( X ( α ) , 0) ( X ( β ) , 0) Exercise: Segments ℓ α and ℓ β intersect at the point � X ( α ) X ( β )( Y ( β ) − Y ( α )) � X ( α ) Y ( β ) − Y ( α ) X ( β ) , Y ( α ) Y ( β )( X ( α ) − X ( β )) X ( α ) Y ( β ) − Y ( α ) X ( β ) 17

Generalization: Unequal Spacing (0 , Y ( α )) To find a point on the envelope curve, we need to compute ( X ( α ) , 0) � X ( α ) X ( β )( Y ( β ) − Y ( α )) � X ( α ) Y ( β ) − Y ( α ) X ( β ) , Y ( α ) Y ( β )( X ( α ) − X ( β )) lim X ( α ) Y ( β ) − Y ( α ) X ( β ) β → α 18

Calculation: “Plugging in” α for β gives � X ( α ) X ( α )( Y ( α ) − Y ( α )) � X ( α ) Y ( α ) − Y ( α ) X ( α ) , Y ( α ) Y ( α )( X ( α ) − X ( α )) X ( α ) Y ( α ) − Y ( α ) X ( α ) � 0 � 0 , 0 = 0 So we try something else . . . The x -coordinate of a point on the envelope is X ( α ) X ( β )( Y ( β ) − Y ( α )) lim X ( α ) Y ( β ) − Y ( α ) X ( β ) β → α 19

X ( α ) X ( β )( Y ( β ) − Y ( α )) Calculation: lim X ( α ) Y ( β ) − Y ( α ) X ( β ) β → α X ( α ) X ( β )( Y ( β ) − Y ( α )) = lim X ( α ) Y ( β ) − X ( α ) Y ( α ) + X ( α ) Y ( α ) − Y ( α ) X ( β ) β → α X ( α ) X ( β )( Y ( β ) − Y ( α )) = lim X ( α )( Y ( β ) − Y ( α )) − Y ( α )( X ( β ) − X ( α )) β → α X ( α ) X ( β )( Y ( β ) − Y ( α ) ) β − α = lim X ( α )( Y ( β ) − Y ( α ) ) − Y ( α )( X ( β ) − X ( α ) β → α ) β − α β − α Y ( β ) − Y ( α ) X ( α ) X ( α ) · lim β → α β − α = Y ( β ) − Y ( α ) X ( β ) − X ( α ) X ( α ) · lim − Y ( α ) · lim β − α β − α β → α β → α ( X ( α )) 2 Y ′ ( α ) = X ( α ) Y ′ ( α ) − Y ( α ) X ′ ( α ) 20

Result: Do the same thing for the y -coordinate Y ( α ) Y ( β )( X ( α ) − X ( β )) lim X ( α ) Y ( β ) − Y ( α ) X ( β ) β → α − ( Y ( α )) 2 X ′ ( α ) = X ( α ) Y ′ ( α ) − Y ( α ) X ′ ( α ) We get the parametrization � � ( X ( α )) 2 Y ′ ( α ) − ( Y ( α )) 2 X ′ ( α ) X ( α ) Y ′ ( α ) − Y ( α ) X ′ ( α ) , X ( α ) Y ′ ( α ) − Y ( α ) X ′ ( α ) for the envelope curve. 21

Example: The picture shows lines generated by � � 3 α − 1 + 1 X ( α ) = 4 2 2 along the x -axis and Y ( α ) = 1 − α 2 along the y -axis. The formula from the previous slide gives the parametrization � − 2 α 3 (4 α 2 − 6 α + 3) 4 α 4 − 15 α 2 + 12 α − 3 , − 3(2 α − 1) 2 ( α 2 − 1) 2 � 4 α 4 − 15 α 2 + 12 α − 3 for the envelope curve. 22

Example: A ladder of length L slides down a  wall.         What is the envelope curve?          Solution: Y ( α )   We want ( X ( α )) 2 + ( Y ( α )) 2 = L 2 ,            so we may as well take     X ( α ) = L sin( α ) and Y ( α ) = L cos( α ) . � �� � X ( α ) 23

Example: A ladder of length L slides down a  wall.         What is the envelope curve?          Solution: Y ( α )   We want ( X ( α )) 2 + ( Y ( α )) 2 = L 2 ,            so we may as well take     X ( α ) = L sin( α ) and Y ( α ) = L cos( α ) . � �� � X ( α ) We get � � ( X ( α )) 2 Y ′ ( α ) − ( Y ( α )) 2 X ′ ( α ) X ( α ) Y ′ ( α ) − Y ( α ) X ′ ( α ) , X ( α ) Y ′ ( α ) − Y ( α ) X ′ ( α ) = ( L sin 3 ( α ) , L cos 3 ( α )) 24

Remarks: The envelope curve, parametrized by          x = L sin 3 ( α ) and y = L cos 3 ( α )          has equation Y ( α )     2 2 2  3 + y 3 = L  x  3         (This is called an astroid .)   � �� � X ( α ) 25