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Solution to the Rubber Rope Problem

Here is the restatement to the problem posed in my last post along with its solution.


A worm is at one end of a rubber rope that can be stretched indefinitely. For the sake of this problem, the worm can live forever and is, for all intents and purposes, dimensionless. At the beginning of the problem, the rope is one kilometer in length. The worm travels from one end of the rope to the other at the rate of one centimeter per second. In other words, each second that elapses, the worm travels a distance of one centimeter. Thus, after the first second, the worm has traveled a distance of one centimeter and the rope is stretched instantly to a total length of two kilometers. After two seconds, the worm has traveled two centimeters, and the rope is stretched instantly until it is now three kilometers in length, and so on. The stretching of the rope is uniform and only the rope is stretched. The rate of speed of the worm and the unit of time is uniform, as well. The question is this: Will the worm ever reach the other end of the rope? And, if so, how long will it take.


Regardless of the parameters given for this problem (the initial length of rope, the uniformity in stretching the rope, and the speed of the worm), the worm will eventually make it to the end of the rope in a finite time. This is true regardless of whether the process is a continuous one or not, but the solution to the problem is much easier to understand if taken in discrete steps rather than looking at it from a continuous perspective.

The worm’s progress along the rope is viewed as a series of decreasing fractions of the rope’s length. This series could be infinite but converge to a single point far short of the end of the rope. Indeed, this is the case if the rope’s length doubles each second. However, since in this case the rope’s length is not doubling each second, the worm does make it to the end. There are 100,000 centimeters in a kilometer, so after the first second, the worm will have traveled 1/100,000th the length of the rope. During the next second, the worm travels a distance of 1/200,000th of the rope’s length, after the rope is stretched to two kilometers. In the third second, 1/300,000th of the rope’s length when the rope has been stretched to three kilometers, and so on. Thus, the worm’s progress, expressed as the fractional part of the rope’s overall length is:

expression 1

The series within brackets is the familiar harmonic series that diverges. As soon as its value exceeds 100,000, the expression  will exceed 1, and, therefore, the worm will ultimately reach the end of the rope as the value of the expression becomes  1. The number of terms, n, in this partial harmonic series will be the number of seconds required for the value of 1 to be reached and is equivalent to the time needed for the worm to reach the end of the rope. Since the worm progresses one centimeter per second along the rope, the value, n, is also the final length of the rope in kilometers.
The enormous number, n, in seconds, correct to within one minute, is equal to:

expression 2

where y is Euler’s constant. defined as the limiting difference between the harmonic series and the natural logarithm whose value is approximately 0.5772156649….

Thus, the value of, n, which represents both the length of the rope in kilometers and the time in seconds for the worm to reach the end is so large that the length will exceed the diameter of the known Universe and the time required vastly exceeds the age of the Universe itself.