Mathematics Homework

So you have your text open to the homework exercise, you have a homework notebook (in which you actually work out the problems -- you write out clean copies to turn in later), pencil, calculator, and ... what next?

First of all, turn off the TV: mathematics and poetry tend to require more attention (and less distraction) than most subjects. Then ...

Next, you should have reviewed your notes and read the text. I do not mean skim: you should have read it through carefully because otherwise, despite a feeling of familiarity you may have of it, you still won't know the material well enough to use it. For more on reading, see the pages on reading textbooks.

Mathematical work involves dealing with rigid objects: you can't bend, squeeze, stretch or twist things the way people do in some other fields. This makes mathematical work both harder (since we are not used to dealing with rigidity) and easier (since when it is done right, the results are more checkable and reliable). For more on this, see the pages on precision.

With all this in mind, first read the problem. Many difficulties arise simply by not knowing what the question is. So work out:

• What is the situation? What are the objects in the problem? How are they related? What do you know? What are the quantities involved? Which quantities are known and which are unknown? How are they related? What do you want to know?

For example, suppose that you were asked the following question. Suppose that you had a car going along the x-axis so that at time t, the car's x-coordinate was 20t. Suppose that another car was going up the y-axis so that at time t, the car's y-coordinate was 20t - 40. How close do the cars get?

First, we want to draw a picture, rather like the one here, labelling everything:

The situation. When we label the coordinates x(t) = 20t and y(t) = 20t - 40, and the distance between the cars, we see how they are related: they form the sides and hypotenuse of a right triangle. What we know. We know, at any time t, the x- and y-coordinates of the two cars, respectively. What we want to know. We want to know the minimum distance between the two cars: notice that we are not given a formula for the distance, so we may should make one ourselves as an intermediate step. The relationship between the known and the unknown. We know that the known and the unknown are related by being the sides and hypotenuse of a right triangle respectively, so we can achieve the intermediate step by using the Pythagorean Theorem.

The situation now is very typical of word problems: from a given situation, obtain a formula. Here the situation is expressed by a picture of a right triangle, and the intermediate step is to use the Pythagorean Theorem to get a formula for the distance between the two cars:

We know come to a common situation: a little insight could spare us a lot of grief. Insight is a tricky thing: the idea is to "see" a trick that can either make a difficult problem easier, or an otherwise impossible problem soluble. In this case, the problem we have is that the distance is the square root of 800t² - 1600t + 1600, and square roots are sometimes unpleasant things. It would be nice to solve this problem without square roots. How to do that?

Insight is somewhat unpredictable. You might get a useful idea at once; you might not. It seems to arise from the operations of the unconscious part of the mind: in his lecture on Mathematical Invention, the mathematician Henri Poincare suggested that mental work begun consciously is continued unconsciously. Poincare imagined that the unconscious somehow combined many components of the problem in many ways, presenting potential solutions to the conscious as they appeared. (See the page on unconscious thinking.) This vision of thinking suggests that an insight occurs after investing a lot of time and energy to the problem.

So here is an insight: the unique t that minimizes the distance also minimizes the square of the distance. So if we can find the t that minimizes distance² = 800t² - 1600t + 1600, that would be the same t that minimizes the distance. So we have reduced the problem to an easier problem: many problems are solved by reducing them to easier problems and then solving the easier problem.

So we now have the easier problem: for what t is distance² = 800t² - 1600t + 1600 minimized? (Now, notice that this is not the entire problem: since we want the minimum distance, we actually need this t to compute the minimum distance.) There are at least two ways to do this: and quite often there are several ways to solve a problem.

Precalculus method: completing the square. Observe that
distance² = 800t² - 1600t + 1600
= 800 (t² - 2t + 2)
= 800 (t² - 2t + 1 + 1)
= 800 [(t² - 2t + 1) + 1]
= 800 [(t - 1)² + 1]
= 800 (t - 1)² + 800
= "nonnegative number equalling 0 at t = 1" + 800,
and distance² is minimized, with minimum 800, at t = 1. It follows that the minimum distance is the square root of 800 = 20 (2)^1/2, or about 28 miles.

Calculus method: finding critical points. Writing distance² as a function of t, we get:

• D'(t) switches from negative to positive at t = 1 (first derivative test), or
• D''(t) = 1600 > 0 (second derivative test).

There are two main phases of doing a homework problem: figuring out how to do it, and then checking if it is correct. The checking part is not as trivial as it seems, and in fact, one kind of homework problem is the verificatio or proof: check that this is right (for more on this kind of exercise, see the page on mathematical proofs). Then you have to write it up.