**2.6: Velocity and Position by Integration**

We haven’t gone over integrals in the “Calculus!” section yet. We will eventually, and I don’t want to muddle anything by attempting a 2-3 sentence explanation. So for now I’m going to treat it like you already understand the integral.

Recall that the derivative of position is velocity and the derivative of velocity is acceleration. I like to think of it this way: Taking the derivative allows you to go “up” in how complicated your concept is. Position is just a location. Velocity is a change in location per second. Acceleration is a change per second in the change in location per second.

By taking the integral, you can go “down” in how complicated the concept is. When you take the integral of acceleration, you’re essentially multiplying time back in. So, you go from distance per time per time, to just distance per time (aka velocity). By multiplying in time again, you can get to position.

This is the way you should think about integrals, in my opinion. You’ll often be told “the integral is finding the area under a curve.” This is technically true, but not terribly useful. As Kalid from BetterExplained notes, saying that be like saying multiplication is finding the area of a rectangle. It’s a way to think about it, but it isn’t what’s actually going on.

Remember, calculus is the study of changing quantities. The integral, for all the complexity it can present, is essentially the multiplying of changing quantities. When you take the integral of acceleration with respect to time, you’re taking a function of acceleration and multiplying by time at each infinitesimal step, resulting in velocity.

Got it? If not, I heartily recommend Kalid’s article. It’s one of my favorite insights of all time.

Overall, this is fairly straightforward stuff if you get integrals. The important thing to realize is that by using calculus, you can go from position to velocity to acceleration to jerk and then back down again fairly simply.

I’m not quite sure I’d agree with assigning notions of “more” or “less” complicated to derivatives and integrals, at least in terms of pedagogy and the complexity of the computation. While taking a derivative yields a rate of change, and integrating “undoes” a rate of change, it is algorithmically less complex to take a derivative than an integral. Thus I wonder if the association of “down in complicatedness” with derivatives might confuse people who are just beginning to learn calculus. I think I mean a few related things by this:

1. Given any elementary function, there’s always a recipe to compute its derivative, and that recipe is quite simple, as you’ve outlined in previous posts. This is not true in general for computing integrals. There’s no one trick to compute them all, and prior to the advent of computer algebra systems like Mathematica, we had to rely on gargantuan tables of integrals just to integrate functions that had closed-form anti-derivatives. This is to say nothing of the functions which don’t have closed-form anti-derivatives, like sin(x^2). (We’ve invented a transcendental function, the Fresnel integral, to be the anti-derivative of sin(x^2), but it’s not the same thing.)

2. Let’s consider the class of simplest non-trivial functions, which I would argue would be polynomials. Given a finite polynomial, if you take enough derivatives, you eventually get 0, and all higher derivatives are also 0. Taking an integral makes the function “more complicated” as it will go up in degree. This process can be repeated indefinitely, perpetually yielding higher-order polynomials.

3. I think derivatives in the context of linear motion may appear more complicated than integrals, as velocity is “more complicated” than position (and acceleration more so than velocity). However, this may be because we conceptually deal with the notion of position as the “basic” physical quantity, and when one’s position begins to change, it becomes relevant to talk about velocity. Instead, let’s consider another physical scenario. When I go to buy a light bulb, what physical parameter do I want to know about it? I want to know its wattage, that is, how much power it puts out. Power is the rate of energy consumption. (Power = d Energy/dt.) Then, after purchasing and installing my 60 W incandescent (I really should switch to CFLs!), I turn it on and wonder how much it will cost me to run the bulb for the month. To compute this “auxilliary” quantity, I remember that 1 W = 1 joule/second, and I integrate my usage of the bulb over the time I have it on. I do this, and I get a quantity in joules (If instead of a lightbulb, I’m considering a power plant, I might instead compute my total energy in terms of kilowatt-hours). I guess the point here is that power is the “fundamental” quantity here, and its integral, not its derivative, is the “derived” quantity of usual interest.

Anyhow, the post overall is very good (as usual!), but I would be careful about saying one kind of computation makes us go “up” or “down” in complicatedness.

(I, of course, could be dead wrong! I’d be very interested to hear what others think in terms of pedagogy; especially those who are learning calculus right now, or did so recently.)

Very good points. I hope the readers all check out the comments to get a different take from me on this stuff. But yeah, you’re right. It’s really a matter of perspective and yours is just as, if not more, reasonable.