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Ladders, Envelopes, and p-Norms An old technique gives a new approach to an old problem Dan Kalman American University Fall 2014 www.dankalman.net

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The Ladder Problem: How long a ladder can you carry around a corner?

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The Traditional Approach Reverse the question Instead of the longest ladder that will go around the corner … Find the shortest ladder that will not

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A Direct Approach Why is this reversal necessary? Look for a direct approach: find the longest ladder that fits Conservative approach: slide the ladder along the walls as far as possible Let’s look at some animations: my machine internet sitemy machineinternet site

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About the Boundary Curve Called the envelope of the family of lines Nice calculus technique to find its equation Technique used to be standard topic Well known curve (astroid, hypocycloid) Gives an immediate solution to the ladder problem

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Solution to Ladder Problem Ladder will fit if (a,b) is outside the region Ladder will not fit if (a,b) is inside the region Longest L occurs when (a,b) is on the curve:

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A famous curve Hypocycloid: point on a circle rolling within a larger circle Astroid: larger radius four times larger than smaller radius Animated graphic from Mathworld.com

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Trammel of Archimedes

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Alternate View Ellipse Model: slide a line with its ends on the axes, let a fixed point on the line trace a curve The length of the line is the sum of the semi major and minor axes Animation on next slide

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Cool Java Applet

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x = a cos y = b sin

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Family of Ellipses Paint an ellipse with every point of the ladder Family of ellipses with sum of major and minor axes equal to length L of ladder These ellipses sweep out the same region as the moving line Same envelope

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Animated graphic from Mathworld.com

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Finding the Envelope Family of curves given by F(x,y, ) = constant For each the equation defines a curve Take the partial derivative with respect to Use the equations of F and F to eliminate the parameter Resulting equation in x and y is the envelope

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Parameterize Lines L is the length of ladder Parameter is angle Note x and y intercepts

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Find Envelope

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Double Parameterization Parameterize line for each : x(t) = L cos( )(1-t) y(t) = L sin( ) t This defines mapping R 2 → R 2 F( ,t) = (L cos( )(1-t), L sin( ) t) Fixed line in family of lines Fixed t ellipse in family of ellipses Envelope points are on boundary of image: Jacobian F = 0

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Mapping R 2 → R 2 Jacobian F vanishes when t = sin 2 Envelope curve parameterized by ( x, y ) = F ( , sin 2 ) = ( L cos 3 L sin 3 )

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Another sample family of curves and its envelope

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Find parametric equations for the envelope:

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Plot those parametric equations:

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Doug Ensley’s Envelope Applet

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Definition of Envelope Curve tangent to each member of a family of curves Under suitable conditions the boundary of the region swept out meets this definition Observation: Any smooth curve is the envelope for its own family of tangent lines This leads to a nice generalization of the ladder problem. (Joint work with Alan Krinik and Chaitanya Rao.)

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Alternate Derivation of x 2/3 + y 2/3 = L 2/3 Begin with the curve x 2/3 + y 2/3 = L 2/3 Consider its family of tangent lines Show that each tangent line intersects the first quadrant in a segment of length L Conclusion: the family of tangent lines is the same as the family of positions of the moving ladder of length L This shows that the envelope of the family of lines has equation x 2/3 + y 2/3 = L 2/3

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Details The equation x 2/3 + y 2/3 = L 2/3 defines a level curve of the function f (x,y) = x 2/3 + y 2/3 At any point (s,t) of the curve, a normal vector is given by f (s,t) = (2/3)(s -1/3, t -1/3 ) For (x,y) on tangent line: (x-s,y-t) (s -1/3, t -1/3 ) s -1/3 (x-s) + t -1/3 (y-t) = 0 Intercepts at s 1/3 (s 2/3 + t 2/3 ) = s 1/3 L 2/3 and t 1/3 (s 2/3 + t 2/3 ) = t 1/3 L 2/3 Distance between the intercepts is L 2/3 (s 2/3 + t 2/3 ) 1/2 = L

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Restatement with p norms ||(x,y)|| p = ( |x| p + |y| p ) 1/p Usual distance is ||(x,y)|| 2 We just saw: For the curve ||(x,y)|| 2/3 = L any tangent line meets the first quadrant in a segment v of length ||v|| 2 = L Generalization: For the curve ||(x,y)|| p = L any tangent line meets the first quadrant in a segment v of length ||v|| q = L where 1/p – 1/q = 1 Call p and q are neoconjugates

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Example 1 If p = 2, q = -2. ||(x,y)|| 2 = L on circle of radius L Tangent segment is v = L(sec ,-csc ) ||v|| -2 = L((sec ) -2 + (csc ) -2 ) -1/2 = L

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Example 2 If p = 1/2, q = 1. ||(x,y)|| 1/2 = L (restricted to 1 st quadrant) (x 1/2 + y 1/2 ) 2 = L 4xy = (L – x – y) 2 Parabola with axis y = x. ||(x,y)|| 1 = |x| + |y| (taxicab metric) Tangent segments make a string art design with uniformly spaced pins

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Restated Ladder Problem Slide a segment of 2-length L around a corner. The corner gap is specified by the vector v = (a,b) The maximum length given by L = ||v|| 2/3

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Generalized Ladder Problem Slide a segment of q-length L around a corner. The corner gap is specified by the vector v = (a,b) The maximum length given by L = ||v|| p where 1/p – 1/q = 1 (Need a ladder with a GPS app …)

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History of Envelopes In 1940’s and 1950’s, some authors claimed envelopes were standard topic in calculus Nice treatment in Courant’s 1949 Calculus text Some later appearances in advanced calculus and theory of equations books No instance in current calculus books I checked Not included in Thomas (1 st ed.) Still mentioned in context of differential eqns What happened to envelopes?

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