Burn Math Class: And Reinvent Mathematics for Yourself (2016)

6. Two in One

6.2. Field-Testing the Fundamental Hammer

6.2.1Fundamentally Hammering Constants

We know how to find the area of a rectangle, so let’s test the idea on the graph of the constant machine m(x) ≡ #. This machine is just a horizontal line whose height is #. The graph of this machine between two points x = a and x = b is just a rectangle with height # and width b − a, so the area should be # · (b − a). Now, if all our ideas about anti-derivatives are really on the right track, then it also has to be the case that

where m(x) ≡ #, and M(x) is an “anti-derivative” of m(x). That is, M(x) is just any machine whose derivative is m(x), which is to say M′(x) = m(x) = #. We don’t have any methods of finding anti-derivatives, so we basically just have to use our familiarity with derivatives, together with some hopeful optimism, and stare blankly at the machine m(x) ≡ # until we can think of something whose derivative is that. Fortunately, in this case, it’s not too difficult, since we know that (#x)′ = #. This tells us that M(x) = #x, and we can use this to check and see whether our idea seems to be on the right track.

Figure 6.2: We can compute the area of a rectangle without any calculus. This fact provides a simple test of our fundamental hammer idea. In this case, the fundamental hammer reproduces just what we’d expected, namely, .

Between any two points x = a and x = b, the area under the graph of a horizontal line m(x) ≡ # has to be # · (b − a), because it’s just a rectangle. So in this particular case, our strange new symbol refers to the number # · (b − a). We don’t yet know for sure whether this new symbolreally is just the difference of the anti-derivatives, but now we can check! Since we found M(x) = #x, we also know that M(b) − M(a) = #b − #a, but this is just # · (b − a).

Hooray! We got the same answer both ways. Notice that we never used equation 6.5. That’s good. We don’t want to assume it’s true yet. Instead, we computed the two sides of it in different ways and then showed that equation 6.5 was true in this simple case. Let’s keep going.

6.2.2Fundamentally Hammering Lines

Figure 6.3: Picturing the reason why . This allows us to check that the fundamental hammer works by examining another simple case, in which we already know how to compute the area without using the fundamental hammer.

What about lines? Back in Chapter 1, we discovered that the graphs of machines like m(x) ≡ cx + b are straight lines. As a simple example, let’s look at . This just refers to the area under m(x) ≡ cx between x = 0 and x = b. Fortunately, this is just a triangle with a width of b and a height ofcb. Two such triangles let us build a rectangle with area (b)(cb) = cb², so the area of the triangle is just . Without any calculus, we know that

We didn’t use any calculus here. The ∫ symbol is just an abbreviation for the area, and we already know how to compute that. Now we can test our fundamental hammer again by seeing if it reproduces what we already know. Can we think of a machine whose derivative is cx? Let’s guess cx² and see if it works.

If we differentiate cx², we’ll get 2cx. There’s an extra 2 out front that we didn’t want, so let’s throw a in front of our original guess and try again. If we differentiate , then we’ll get cx, and that’s what we wanted. Now we can use this to check our idea again. Starting with m(x) ≡ cx, we just found that this machine’s anti-derivative is , and therefore,

Perfect. First, we found that by thinking about the area of a triangle. Then, we thought of an anti-derivative of cx, and found that . Both sides of the fundamental hammer match up, just like we hoped. To reiterate, we aren’t using the fundamental hammer at this point. Rather, we’re testing it by computing both sides of the hammer (equation 6.5) in different ways, and verifying that they’re equal. This gives us more confidence that the line of reasoning we used to derive the fundamental hammer was indeed on the right track.

6.2.3A Worry

As we already know, derivatives kill constants, so if a machine m has one anti-derivative, it must necessarily have an infinite number of them. Why? Well, if , then as well, for any fixed number #. So each possible number # gives us an equally valid “anti-derivative” of m, namely M(x) + #. This is worrisome at first. Our fundamental hammer requires us to find an anti-derivative in order to compute an area. The area is something concrete, and intuitively there should be only one correct answer for it. However, if there are an infinite number of different anti-derivatives of a machine m, they had better all give the same answer for the area, or else our fundamental hammer is broken. While we’re not yet sure if our hammer will always work, we can at least convince ourselves that the worrisome example above is no cause for worry. The fundamental hammer tells us to compute an area using a difference: M(b) − M(a). As such, any anti-derivative of the form W(x) ≡ M(x) + # will give us the same answer for the area, because W(b) − W(a) = [M(b) + #] − [M(a) + #] = M(b) − M(a). Having put that worry to rest, let’s keep moving.

6.2.4Full Speed Ahead! Using the Fundamental Hammer

What have we done so far? First we made an argument in which we discovered the fundamental hammer of calculus:

But we weren’t entirely sure whether we should trust our derivation of it. So we tested it on several simple cases where we already knew what to expect. So far it has worked in every case, so we’re a bit more confident. Let’s see what it gives us when we’re not sure what to expect.

6.2.5An Unfamiliar Case

As always, the machine m(x) ≡ x² provides a simple test case. Let’s see what the fundamental hammer gives us when we ask it about the area between (say) x = 0 and x = 3:

where M(x) is some machine whose derivative is x². Can we think of such a machine? Well, since derivatives knock the power down by one, it had better look like #x³. Then, when we take the derivative and bring the power down, we’ve got to end up with x², so we need it to be true that (#x³)′ = 3#x² = x², which tells us that the number # has to be . So is an anti-derivative of x², which lets us extend the above equation a few steps further, writing

Interesting. . . The area of this particular curvy thing is just a whole number. That’s eerily simple. What else could we do?