Infinite number of monkeys

From Wikipedia:

The infinite monkey theorem states that a monkey hitting keys at random on a typewriter keyboard for an infinite amount of time will almost surely type a given text, such as the complete works of William Shakespeare.

This can be formally proven using the second Borel-Cantelli Lemma, a topic which requires measure-theoretic probability. Thus leading me to one of the driest observations that I’ve ever read in a graduate-level textbook, following the proofs of the Borel-Cantelli Lemmas:

The record of a prolonged coin-tossing game is bound to contain every conceivable book in the Morse code [using heads for dot and tails for dash], from Hamlet to eight-place logarithmic tables. It has been suggested that an army of monkeys might be trained to pound typewriters at random in the hope that ultimately great works of literature would be produced. Using a coin for the same purpose may save feeding and training expenses and free the monkeys for other monkey business.

W. Feller, An Introduction to Probability Theory and Its Applications, Volume 1 (Chapter 8.3), page 202.

Acceleration

The following two questions came from a middle-school math textbook. The first is reasonable, while the second is a classic example of an author being overly cute when writing a homework problem.

  1.  A car slams on its brakes, coming to a complete stop in 4 seconds.  The car was traveling north at 60mph. Calculate the acceleration.
  2. A rocket blasts off. At 10 seconds after blast off, it is at 10,000 feet, traveling at 3600mph.  Assuming the direction is up, calculate the acceleration.

For the first question, we’ll assume constant deceleration (after all, this comes from a middle-school textbook). First, let’s convert from miles per hour to feet per second:

60 ~ \frac{\hbox{mile}}{\hbox{hour}} = 60 ~ \frac{\hbox{mile}}{\hbox{hour}} \times \frac{1 ~ \hbox{hour}}{3600 ~ \hbox{second}} \times \frac{5280 ~ \hbox{feet}}{1 ~ \hbox{mile}} = 88~ \frac{\hbox{feet}}{\hbox{second}}

The deceleration is therefore equal to the change in velocity over time, or

\frac{-88 ~ \hbox{feet/second}}{4 ~ \hbox{second}} = -22 ~\hbox{ft/s}^2

Now notice the word north in the statement of the first question. This bit of information is irrelevant to the problem. I presume that the writer of the problem wants students to practice picking out the important information of a problem from the unimportant… again, a good skill for students to acquire.

green line

Let’s now turn to the second question. At first blush, this also has irrelevant information…  it is at 10,000 feet. So I presume that the author wants students to solve this in exactly the same way:

3600 ~ \frac{\hbox{mile}}{\hbox{hour}} = 3600 ~ \frac{\hbox{mile}}{\hbox{hour}} \times \frac{1 ~ \hbox{hour}}{3600 ~ \hbox{second}} \times \frac{5280 ~ \hbox{feet}}{1 ~ \hbox{mile}} = 5280 ~ \frac{\hbox{feet}}{\hbox{second}}

for an acceleration of

\frac{5280 ~ \hbox{feet/second}}{10 ~ \hbox{second}} = 528 ~\hbox{ft/s}^2

The major flaw with this question is that the acceleration of the rocket completely determines the distance that the rocket travels. While middle-school students would not be expected to know this, we can use calculus to determine the distance. Since the initial position and velocity are zero, we obtain

x''(t) = 528

x'(t) = \int 528 \, dt = 528t + C

x'(0) = 528(0) + C

0 = C

\therefore x'(t) = 528t + 0 = 528t

x(t) = \int 528t \, dt = 264t^2 + C

x(0) = 264(0)^2 + C

\therefore x(t) = 264t^2 + 0 = 264t^2

Therefore, the rocket travels a distance of 264 ~ \hbox{feet/second}^2 \times (10 ~ \hbox{second})^2 = 26400 ~ \hbox{feet}. In other words, not 10,000 feet.

As a mathematician, this is the kind of error that drives me crazy, as I would presume that the author of this textbook should know that he/she just can’t make up a distance in the effort of making a problem more interesting to students.