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Unformatted text preview: Billiards inside a cusp Jonathan L. King University of Florida, Gainesville 32611-2082, [email protected] Introduction Here is an anecdote about how unexciting homework problems led a student studying calculus to a Good Question, and to the mathematics it engendered. The standard calculus curriculum spends quite a bit of time on logarithms. Yet it is no hyperbole that the curve \y = 1=x" {and the standard drill questions concerning its properties{ leave many students comatose. One such was David Feldman {then an undergraduate at Berkeley{ who, having complained to his dad (the mathematician Jacob Feldman) that all the homework was dull, dull, dull, was challenged in return to invent an interesting problem. David came up with this: Can a mathematical cueball (a point), red into the symmetric funnel 1 1 between y = + x and y = x , escape in any direction other than at out? As shown in gure 1 below, the cueball ricochets o the two \cushions" so that the angle of incidence always equals the angle of re ection. \Escape" means that the x-coordinate of the cueball increases monotonically to +1. Evidently a cueball shot along the x-axis escapes. But can any cueball which actually hits the cushions avoid being eventually turned around? y = 1/ x ? x -axis y = -1/ x Figure 1 For what initial position and direction will a cueball escape to in nity? Not long after it was posed, this problem was solved by Benjamin Weiss. Unaware of its origin, nor that it had been solved, nor that it would eventually connect with what became my eld of study, I was intrigued by this problem when Paul Shields posed it to me during my graduate studies. Since the problem will lead to a harder question and to the tool which is the theme of this Partially supported by NSF grant DMS-9112595. Math. Intelligencer, vol.17 no.1, (1995), 8{16. 2 J.L. King article, I'm going to forthwith present the bare hands solution I found at the time|so if you want to think about the problem further, read no farther : : : A first shot at Billiards in a Cusp Since the problem arose in a calculus class, to get the ball rolling let's see what information can be gleaned by methods taught in an introductory calculus course. The argument below has become a nice capstone to the section on \Convergence tests for in nite series" in my own classes. Since the funnel is symmetric, we can without loss of generality re ect the trajectory over the x-axis and thus consider the cueball to be bouncing o \cushions" y = f (x) = 1=x and the x-axis. So as to focus attention on the horizontal cusp at x = +1, let us do away with the vertical cusp at x = 0 by altering the upper-cushion f near the origin so that it is bounded. This does not a ect whether a cueball can escape, since we keep that f (x) = 1=x for all large x. Suppose, for the sake of contradiction, that in gure 2 the cueball v bounces so that its xcoordinate never decreases: y = f(x) P n (x1 , y1 ) = P 1 αn P2 v θ1 x1 P 3 θ2 x2 θ3 x3 Figure 2 Let Pn = (xn ; yn ) be the coordinates of the nth re ection of the orbit of cueball v o the graph of f . Let n be the angle that the (tangent to the) curve at Pn makes with the horizontal ie. n = arctan jf 0 (xn )j . After Pn the trajectory hits the \ oor", from which the cueball rises at angle n . Thus 0 < n < =2 for all n, where n is the angle after the nth re ection o the oor. Elementary geometry shows that n = n 1 + 2 n, and so the situation in gure2 implies this summation condition: 1 X (3a) n < 1...
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This note was uploaded on 01/26/2012 for the course MTG 6401 taught by Professor Staff during the Fall '09 term at University of Florida.

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