The Traveling BeeDate: 09/18/98 at 01:00:53 From: David Emerling Subject: Taking limits with boundaries I am brushing up on math since I am getting ready to do some substitute teaching. I've been cruising the Internet and have found many fantastic problems that brought back to mind many things that I have long since forgotten. But this problem is driving me nuts: BUSY BEE: Two trains leave two towns that are fifty miles apart. They travel toward each other at rates of 30 mph and 20 mph respectively. A bumblebee flying at the rate of 50 mph starts out just as the faster train departs the train station, and flies to the slower train. The bee then turns around and goes back to meet the faster train. Then it turns around again, etc., and it and keeps flying back and forth between the trains until the trains meet. How far does the tired bumblebee fly? My approach has been as follows: I put the faster train at the origin and the slower train 50 miles away, so the faster train begins at x = 0 and the slower train at x = 50. The position of the faster train can be expressed as: x = 30t. The position of the slower train can be expressed as: x = 50 - 20t. I plotted these two lines on a Cartesian system with the position of the trains (x) represented by the vertical axis and time (t) represented by the horizontal axis. It can be easily calculated and it is easily seen that the two lines cross at t = 1. This is when the two trains meet. The bee begins its flight with x = 0 as its position, which as a function of time can be expressed as x = 50t. I plotted this line, which obviously is between the lines representing the two trains. But the line representing the bee stops when it intercepts the upper boundary represented by the x = 50 - 20t and then begins a downward direction, as a new line with a slope of -50t. Looking at the plots, it is quite apparent what is happening, but I can't express it mathematically! I need to take the sum of the distances the bee travels as expressed by x = 50t yet bounded at the top by x = 50 - 20t and at the bottom by x = 30t. This strikes me as a calculus problem where I have to come up with an expression that calculates this distance the bee travels (with the appropriate boundaries) as t = 0 to t = 1. Any help to get me started? Date: 09/18/98 at 08:01:52 From: Doctor Jerry and Doctor Ian Subject: Re: Taking limits with boundaries Hi David, This is a well known problem. It is often told as part of a story about John von Neumann (1903-1957), an extremely quick and bright mathematician. The quick solution is to note that the trains will meet in 1 hour and the bee is flying at the rate of 50 mph. Hence, at the end of the hour the bee will have flown a distance of 50 miles. It is said that this problem was given to von Neumann and that after a few moments he said that the bee had traveled 50 miles. In subsequent conversation it appeared that he had solved the problem, not by the short method mentioned above, but by quickly summing a geometric series in his head. It's not clear from the story which series he summed, but here is one way he might have worked it out: Let's forget for a moment about the whole journey, and just think about what happens during one 'round trip' of the bee (from the faster train to the slower train and back to the faster train). Suppose the bee travels some fraction a/b of the original distance. The combined speed of the trains is the same as the speed of the bee, so the distance between the trains is reduced by this same amount. If the original distance is D, then on the first round trip, the bee travels (a/b)D miles, and we're back in the same situation, except that the new 'original' distance is D - (a/b)D miles, or (1 - a/b)D miles. The same thing happens on each successive trip. So on the second round trip, the bee goes (a/b)(1 - a/b)D miles, and the new 'original' distance is (1 - a/b)^2D miles. On the third round trip, the bee goes (a/b)(1 - a/b)^2 miles, and the new 'original' distance is (1 - a/b)^3D miles. So the total distance that the bee travels must be D[(a/b) + (a/b)(1 - a/b) + (a/b)(1 - a/b)^2 + (a/b)(1 -a/b)^3 + ...] = (a/b)D[1 + (1 - a/b) + (1 - a/b)^2 + (1 - a/b)^3 + ...] The bracketed expression is a geometric series, whose sum is 1/(1 - (1 - a/b)) which simplifies to b/a; so the total distance traveled by the bee is (a/b) D [b/a] or just D, i.e., 50 miles. Isn't it neat the way it works out? - Doctor Jerry and Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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