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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. The Ladder Problem:. How long a ladder can you carry around a corner?. The Traditional Approach. Reverse the question
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Ladders, Envelopes, and p-Norms An old technique gives a new approach to an old problemDan Kalman American UniversityFall 2014 www.dankalman.net
The Ladder Problem: How long a ladder can you carry around a corner?
The Traditional Approach • Reverse the question • Instead of the longest ladder that will go around the corner … • Find the shortest ladder that will not
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 machineinternet site
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
Solution to Ladder Problem • Ladder will fit if (a,b) is outside the region W • Ladder will not fit if (a,b) is inside the region • Longest L occurs when (a,b) is on the curve:
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
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
x = a cosq • y = b sin q
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
Finding the Envelope • Family of curves given by F(x,y,a) = constant • For each a the equation defines a curve • Take the partial derivative with respect to a • Use the equations of F and Fato eliminate the parameter a • Resulting equation in x and y is the envelope
Parameterize Lines • L is the length of ladder • Parameter is angle a • Note x and y intercepts
Double Parameterization • Parameterize line for each a:x(t) = L cos(a)(1-t)y(t) = L sin(a) t • This defines mapping R2→ R2F(a,t) = (L cos(a)(1-t), L sin(a) t) • Fixed a line in family of lines • Fixed t ellipse in family of ellipses • Envelope points are on boundary of image: Jacobian F = 0
Mapping R2→ R2 • Jacobian F vanishes when t = sin2a • Envelope curve parameterized by( x , y ) = F (a , sin2a) = ( L cos3a, L sin3a)
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.)
Alternate Derivation of x2/3 + y2/3 = L2/3 • Begin with the curve x2/3 + y2/3 = L2/3 • 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 x2/3 + y2/3 = L2/3
Details • The equation x2/3 + y2/3 = L2/3 defines a level curve of the function f (x,y) = x2/3 + y2/3 • 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 s1/3(s2/3 + t2/3 ) = s1/3 L2/3and t1/3(s2/3 + t2/3 ) = t1/3 L2/3 • Distance between the intercepts is L2/3(s2/3 + t2/3)1/2 = L
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 = Lany tangent line meets the first quadrant in a segment v of length ||v||2 = L • Generalization: For the curve ||(x,y)||p = Lany tangent line meets the first quadrant in a segment v of length ||v||q = Lwhere 1/p – 1/q = 1 • Call p and q are neoconjugates
Example 1 • If p = 2, q = -2. ||(x,y)||2 = L on circle of radius L • Tangent segment is v = L(sec a,-csc a) • ||v||-2 = L((sec a)-2 +(csc a)-2)-1/2 = L
Example 2 • If p = 1/2, q = 1. • ||(x,y)||1/2 = L(restricted to 1st quadrant) (x1/2 + y1/2)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
Restated Ladder Problem • Slide a segment of 2-length L around acorner. • The corner gap is specified by the vector v = (a,b) • The maximum length given by L = ||v||2/3