Ladders, Envelopes, and p -Norms

1. Ladders, Envelopes, and p-Norms An old technique gives a new approach to an old problemDan Kalman American UniversityFall 2014 www.dankalman.net

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

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

5. 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

10. 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

11. 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:

12. 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

13. Trammel of Archimedes

14. 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

18. Cool Java Applet

22. x = a cosq • y = b sin q

23. 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

24. Animated graphic from Mathworld.com

25. Animated graphic from Mathworld.com

26. Animated graphic from Mathworld.com

27. Animated graphic from Mathworld.com

28. Animated graphic from Mathworld.com

29. 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

30. Parameterize Lines • L is the length of ladder • Parameter is angle a • Note x and y intercepts

31. Find Envelope

32. Find Envelope

33. 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

34. Mapping R2→ R2 • Jacobian F vanishes when t = sin2a • Envelope curve parameterized by( x , y ) = F (a , sin2a) = ( L cos3a, L sin3a)

35. Another sample family of curves and its envelope

38. Find parametric equations for the envelope:

40. Plot those parametric equations:

42. Doug Ensley’s Envelope Applet

43. 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.)

44. 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

45. 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

46. 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

48. 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

49. 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

50. 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