Unformatted text preview: Billiards inside a cusp
Jonathan L. King
University of Florida, Gainesville 326112082, [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 xcoordinate of the cueball increases
monotonically to +1. Evidently a cueball shot along the xaxis 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 DMS9112595.
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 timeso 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
xaxis and thus consider the cueball to be bouncing o \cushions" y = f (x) = 1=x and the xaxis.
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 uppercushion 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...
View
Full
Document
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.
 Fall '09
 Staff

Click to edit the document details