The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

The Traveling Bee

Date: 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 

But this problem is driving me nuts:


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 

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 

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   
Associated Topics:
High School Puzzles
High School Sequences, Series

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.