0 energy points
Studying for a test? Prepare with these 13 lessons on Integrating multivariable functions.
See 13 lessons

Line integrals and vector fields

Video transcript
One of the most fundamental ideas in all of physics is the idea of work. Now when you first learn work, you just say, oh, that's just force times distance. But then later on, when you learn a little bit about vectors, you realize that the force isn't always going in the same direction as your displacement. So you learn that work is really the magnitude, let me write this down, the magnitude of the force, in the direction, or the component of the force in the direction of displacement. Displacement is just distance with some direction. Times the magnitude of the displacement, or you could say, times the distance displaced. And the classic example. Maybe you have an ice cube, or some type of block. I just ice so that there's not a lot of friction. Maybe it's standing on a bigger lake or ice or something. And maybe you're pulling on that ice cube at an angle. Let's say, you're pulling at an angle like that. That is my force, right there. Let's say my force is equal to-- well, that's my force vector. Let's say the magnitude of my force vector, let's say it's 10 newtons. And let's say the direction of my force vector, right, any vector has to have a magnitude and a direction, and the direction, let's say it has a 30 degree angle, let's say a 60 degree angle, above horizontal. So that's the direction I'm pulling in. And let's say I displace it. This is all review, hopefully. If you're displacing it, let's say you displace it 5 newtons. So let's say the displacement, that's the displacement vector right there, and the magnitude of it is equal to 5 meters. So you've learned from the definition of work, you can't just say, oh, I'm pulling with 10 newtons of force and I'm moving it 5 meters. You can't just multiply the 10 newtons times the 5 meters. You have to find the magnitude of the component going in the same direction as my displacement. So what I essentially need to do is, the length, if you imagine the length of this vector being 10, that's the total force, but you need to figure out the length of the vector, that's the component of the force, going in the same direction as my displacement. And a little simple trigonometry, you know that this is 10 times the cosine of 60 degrees, or that's equal to, cosine of 60 degrees is 1/2, so that's just equal to 5. So this magnitude, the magnitude of the force going in the same direction of the displacement in this case, is 5 newtons. And then you can figure out the work. You could say that the work is equal to 5 newtons times, I'll just write a dot for times. I don't want you to think it's cross product. Times 5 meters, which is 25 newton meters, or you could even say 25 Joules of work have been done. And this is all review of somewhat basic physics. But just think about what happened, here. What was the work? If I write in the abstract. The work is equal to the 5 newtons. That was the magnitude of my force vector, so it's the magnitude of my force vector, times the cosine of this angle. So you know, let's call that theta. Let's say it a little generally. So times the cosine of the angle. This is the amount of my force in the direction of the displacement, the cosine of the angle between them, times the magnitude of the displacement. So times the magnitude of the displacement. Or if I wanted to rewrite that, I could just write that as, the magnitude of the displacement times the magnitude of the force times the cosine of theta. And I've done multiple videos of this, in the linear algebra playlist, in the physics playlist, where I talk about the dot product and the cross product and all of that, but this is the dot product of the vectors d and f. So in general, if you're trying to find the work for a constant displacement, and you have a constant force, you just take the dot product of those two vectors. And if the dot product is a completely foreign concept to you, might want to watch, I think I've made multiple, 4 or 5 videos on the dot product, and its intuition, and how it compares. But just to give you a little bit of that intuition right here, the dot product, when I take f dot d, or d dot f, what it's giving me is, I'm multiplying the magnitude, well I could just read this out. But the idea of the dot product is, take how much of this vector is going in the same direction as this vector, in this case, this much. And then multiply the two magnitudes. And that's what we did right here. So the work is going to be the force vector, dot, taking the dot part of the force vector with the displacement vector, and this, of course, is a scalar value. And we'll work out some examples in the future where you'll see that that's true. So this is all review of fairly elementary physics. Now let's take a more complex example, but it's really the same idea. Let's define a vector field. So let's say that I have a vector field f, and we're going to think about what this means in a second. It's a function of x and y, and it's equal to some scalar function of x and y times the i-unit vector, or the horizontal unit vector, plus some other function, scalar function of x and y, times the vertical unit vector. So what would something like this be? This is a vector field. This is a vector field in 2-dimensional space. We're on the x-y plane. Or you could even say, on R2. Either way, I don't want to get too much into the mathiness of it. But what does this do? Well, if I were to draw my x-y plane, so that is my, again, having trouble drawing a straight line. All right, there we go. That's my y-axis, and that's my x-axis. I'm just drawing the first quadrant, and but you could go negative in either direction, if you like. What does this thing do? Well, it's essentially saying, look. You give me any x, any y, you give any x, y in the x-y plane, and these are going to end up with some numbers, right? When you put x, y here, you're going to get some value, when you put x, y here, you're going to get some value. So you're going to get some combination of the i- and j-unit vectors. So you're going to get some vector. So what this does, it defines a vector that's associated with every point on x-y plane. So you could say, if I take this point on the x-y plane, and I would pop it into this, I'll get something times i plus something times j, and when you add those 2, maybe I get a vector that something like that. And you could do that on every point. I'm just taking random samples. Maybe when I go here, the vector looks something like that. Maybe when I go here, the victor looks like this. Maybe when I go here, the vector looks like that. And maybe when I go up here, the vector goes like that. I'm just randomly picking points. It defines a vector on all of the x, y coordinates where these scalar functions are properly defined. And that's why it's called a vector field. It defines what a potential, maybe, force would be, or some other type of force, at any point. At any point, if you happen to have something there. Maybe that's what the function is. And I could keep doing this forever, and filling in all the gaps. But I think you get the idea. It associates a vector with every point on x-y plane. Now, this is called a vector field, so it probably makes a lot of sense that this could be used to describe any type of field. It could be a gravitation field. It could be an electric field, it could be a magnetic field. And this could be essentially telling you how much force there would be on some particle in that field. That's exactly what this would describe. Now, let's say that in this field, I have some particle traveling on x-y plane. Let's say it starts there, and by virtue of all of these crazy forces that are acting on it, and maybe it's on some tracks or something, so it won't always move exactly in the direction that the field is trying to move it at. Let's say it moves in a path that moves something like this. And let's say that this path, or this curve, is defined by a position vector function. So let's say that that's defined by r of t, which is just x of t times i plus y of t times our unit factor j. That's r of t right there. Well, in order for this to be a finite path, this is true before t is greater than or equal to a, and less than or equal to b. This is the path that the particle just happens to take, due to all of these wacky forces. So when the particle is right here, maybe the vector field acting on it, maybe it's putting a force like that. But since the thing is on some type of tracks, it moves in this direction. And then when it's here, maybe the vector field is like that, but it moves in that direction, because it's on some type of tracks. Now, everything I've done in this video is to build up to a fundamental question. What was the work done on the particle by the field? To answer that question, we could zoom in a little bit. I'm going to zoom in on only a little small snippet of our path. And let's try to figure out what the work is done in a very small part of our path, because it's constantly changing. The field is changing direction. my object is changing direction. So let's say when I'm here, and let's say I move a small amount of my path. So let's say I move, this is an infinitesimally small dr. Right? I have a differential, it's a differential vector, infinitely small displacement. and let's say over the course of that, the vector field is acting in this local area, let's say it looks something like that. It's providing a force that looks something like that. So that's the vector field in that area, or the force directed on that particle right when it's at that point. Right? It's an infinitesimally small amount of time in space. You could say, OK, over that little small point, we have this constant force. What was the work done over this small period? You could say, what's the small interval of work? You could say d work, or a differential of work. Well, by the same exact logic that we did with the simple problem, it's the magnitude of the force in the direction of our displacement times the magnitude of our displacement. And we know what that is, just from this example up here. That's the dot product. It's the dot product of the force and our super-small displacement. So that's equal to the dot product of our force and our super-small displacement. Now, just by doing this, we're just figuring out the work over, maybe like a really small, super-small dr. But what we want to do, is we want to sum them all up. We want to sum up all of the drs to figure out the total, all of the f dot drs to figure out the total work done. And that's where the integral comes in. We will do a line integral over-- I mean, you could think of it two ways. You could write just d dot w there, but we could say, we'll do a line integral along this curve c, could call that c or along r, whatever you want to say it, of dw. That'll give us the total work. So let's say, work is equal to that. Or we could also write it over the integral, over the same curve of f of f dot dr. And this might seem like a really, you know, gee, this is really abstract, Sal. How do we actually calculate something like this? Especially because we have everything parameterized in terms of t. How do we get this in terms of t? And if you just think about it, what is f dot r? Or what is f dot dr? Well, actually, to answer that, let's remember what dr looked like. If you remember, dr/dt is equal to x prime of t, I'm writing it like, I could have written dx dt if I wanted to do, times the i-unit vector, plus y prime of t, times the j-unit vector. And if we just wanted to dr, we could multiply both sides, if we're being a little bit more hand-wavy with the differentials, not too rigorous. We'll get dr is equal to x prime of t dt times the unit vector i plus y prime of t times the differential dt times the unit vector j. So this is our dr right here. And remember what our vector field was. It was this thing up here. Let me copy and paste it. And we'll see that the dot product is actually not so crazy. So copy, and let me paste it down here. So what's this integral going to look like? This integral right here, that gives the total work done by the field, on the particle, as it moves along that path. Just super fundamental to pretty much any serious physics that you might eventually find yourself doing. So you could say, well gee. It's going to be the integral, let's just say from t is equal to a, to t is equal to b. Right? a is where we started off on the path, t is equal to a to t is equal to b. You can imagine it as being timed, as a particle moving, as time increases. And then what is f dot dr? Well, if you remember from just what the dot product is, you can essentially just take the product of the corresponding components of your of vector, and add them up. So this is going to be the integral from t equals a to t equals b, of p of p of x, really, instead of writing x, y, it's x of t, right? x as a function of t, y as a function of t. So that's that. Times this thing right here, times this component, right? We're multiplying the i-components. So times x prime of t d t, and then that plus, we're going to do the same thing with the q function. So this is q plus, I'll go to another line. Hopefully you realize I could have just kept writing, but I'm running out of space. Plus q of x of t, y of t, times the component of our dr. Times the y-component, or the j-component. y prime of t dt. And we're done! And we're done. This might still seem a little bit abstract, but we're going to see in the next video, everything is now in terms of t, so this is just a straight-up integration, with respect to dt. If we want, we could take the dt's outside of the equation, and it'll look a little bit more normal for you. But this is essentially all that we have to do. And we're going to see some concrete examples of taking a line integral through a vector field, or using vector functions, in the next video.