How to teach integrals motivated by the work done in moving an object?

Question

I am now teaching Calculus of several variables this semester. In apllications of integrals, the problem of finding the work done in moving an object under a force $F$ is one of the most common problems.

Let's assume that an object is moving from a point A to a point B (on a horizontal line) under a constant force $F$ pointing an angle $\alpha$ (w.r.t. the horizontal line). Then in most textbooks (Calculus), the work done in this case is defined by $W=|F||AB|\cos(\alpha)$. By accepting this definition, then one can meaningfully find a way to evaluate the work done in moving an object along a curve (in 2D or 3D) and under a non-constant force (this leads to line integrals).

When teaching this, the most uncomfortable fact is "why do we have such a definition of the work" in the first place. Since the word "work" should say something, it might give confusion.

My question: how do we give intuition to the definition of work above? For example, how do we give an explanation for the fact that the work is negative if $\alpha$ is greater than 90 degrees?

Thanks so much for any hints.

Answer 1

The best way to illustrate the sense of $W=|F||AB|\cos(\alpha)$ is to show that it makes sense that each of the factors is proportional to the work done.

• If you double the amount of force to an object to move it a certain distance, it doubles the amount of work needed.

• If you double the distance to move an object by applying a certain force, it doubles the amount of work needed.

• If you apply the force in the same direction as the direction the object moves, you are being completely efficient in your work. The more your work is applied downward, the more effort you are wasting pushing the object into the ground instead of in the direction you want. At the end, if you are pointing straight down, you are doing no work no matter how much force you expend. And you if your force is in the opposite direction, you are doing "negative work" in the sense of moving the object further from your goal. These intuitive facts turn out to be modeled precisely by the cosine of the angle between the direction of your work and the direction the object is to move.

Answer 2

I mainly teach physics, only a little math on the side. If I was teaching this math topic and wanted to spend 3 minutes giving the appropriate physical motivation, I would do something like the following.

There is something called energy. It comes in various forms. Food has energy, which is what we're talking about when we talk about calories. Heat, sound, and light all have energy. When a material object like this bowling ball is moving, then it has energy because of its motion.

Now I have this bowling ball, which I borrowed from the physics department's stockroom, and I'm going to roll it down the aisle of the classroom. Please pick up your legs and backpacks in case I don't do it perfectly straight. The ball started from rest, and then I made a force on it, which transferred energy into it. The amount of this energy transfer is called work. When the force is constant, we have $W=Fd$.

Now, Sally, the ball is near your desk. Could you please roll it back up the aisle toward me at a speed similar to the one I used? Now I slow the ball to a stop with my foot. This time the work I did was negative: I took energy out of the ball. This makes sense because if we pick some direction to call positive, then the force and the distance will have opposite signs, and their product will be negative.

Now when the force is not constant, we can't just multiply the distance by "the" force. This is a pattern that you'll see over and over again in applications of calculus. When we need to generalize a relationship involving multiplication to the case where one of the factors is varying, we do that using an integral.

Answer 3

I would start with the scalar. Work = force times distance. Starting with the scalar makes things easier, more intuitive. (Not a math logic point, just a psychological one.) You can then easily generalize, perhaps without showing a derivation and say "well this is what work is in these trickier situations (vectors, non constant force, more dimensions, etc.) At least they see the similarity of the look of the equation to what they know from high school basic physics.

As far as why W=F*d, you can devolve to "this is the definition". Or probably make some argument of how work is just another name for energy. (The simplest and most intuitive example of work done is moving an object up vertically in a gravity field. The work done is equal to the potential energy created.)

All that said, I would really try to avoid getting too tied up in the physics of the thing, especially when you are a little uncomfortable or unfamiliar with it yourself. I suspect a lot of what bothers you, here, is your own questions, not those of the students. But you are going to transfer that to them and bog the class down with a bunch of physics talk when they really ought to be practicing calculations. (This is frequent the commenters/questioners here want to explain to the class something that really bugs them from not being explained versus something that bugs the class from the omission.)

Remember, they will have other classes (most will) that go over this from more of a physics background and less of a math one. For a student taking both (at or near same time) it can be useful to see the same concept from two slants. One more physical and one more calculational. The physics class makes sense when you say "I've seen that math before, can concentrate on physical meaning of what is happening" (and conversely in math class). But I would try to keep math class as more calcuational versus deep digressions to physics. This supports the physics class better.