# Closed curve line integrals of conservative vector fields

## Video transcript

In the last video, we saw that if a vector field can be written as the gradient of a scalar field-- or another way we could say it: this would be equal to the partial of our big f with respect to x times i plus the partial of big f, our scalar field with respect to y times j; and I'm just writing it in multiple ways just so you remember what the gradient is --but we saw that if our vector field is the gradient of a scalar field then we call it conservative. So that tells us that f is a conservative vector field. And it also tells us, and this was the big take away from the last video, that the line integral of f between two points-- let me draw two points here; so let me draw my coordinates just so we know we're on the xy plane. My axes: x-axis, y-axis. Let's say I have the point, I have that point and that point, and I have two different paths between those two points. So I have path 1, that goes something like that, so I'll call that c1 and it goes in that direction. And then I have, maybe in a different shades of green, c2 goes like that. They both start here and go to there. We learned in the last video that the line integral is path independent between any two points. So in this case the line integral along c1 of f dot dr is going to be equal to the line integral of c2, over the path c2, of f dot dr. The line, if we have a potential in a region, and we may be everywhere, then the line integral between any two points is independent of the path. That's the neat thing about a conservative field. Now what I want to do in this video is do a little bit of an extension of the take away of the last video. It's actually a pretty important extension; it might already be obvious to you. I've already written this here; I could rearrange this equation a little bit. So let me do it. So let me a rearrange this. I'll just rewrite this in orange. So the line integral on path c1 dot dr minus-- I'll just go subtract this from both sides --minus the line integral c2 of f dot dr is going to be equal to 0. All I did is I took this take away from the last video and I subtracted this from both sides. Now we learned several videos ago that if we're dealing with a line integral of a vector field-- not a scalar field --with a vector field, the direction of the path is important. We learned that the line integral over, say, c2 of f dot dr, is equal to the negative of the line integral of minus c2 of f dot dr where we denoted minus c2 is the same path as c2, but just in the opposite direction. So for example, minus c2 I would write like this-- so let me do it in a different color --so let's say this is minus c2, it'd be a path just like c2-- I'm going to call this minus c2 --but instead of going in that direction, I'm now going to go in that direction. So ignore the old c2 arrows. We're now starting from there and coming back here. So this is minus c2. Or we could write, we could put, the minus on the other side and we could say that the negative of the c2 line integral along the path of c2 of f dot dr is equal to the line integral over the reverse path of f dot dr. All I did is I switched the negative on the other side; multiplied both sides by negative 1. So let's replace-- in this equation we have the minus of the c2 path; we have that right there, and we have that right there --so we could just replace this with this right there. So let me do that. So I'll write this first part first. So the integral along the curve c1 of f dot dr, instead of minus the line integral along c2, I'm going to say plus the integral along minus c2. This-- let me switch to the green --this we've established is the same thing as this. The negative of this curve, or the line integral along this path, is the same thing as the line integral, the positive of the line integral along the reverse path. So we'll say plus the line integral of minus c2 of f dot dr is equal to 0. Now there's something interesting. Let's look at what the combination of the path of c1 and minus c2 is. c1 starts over here. Let me get a nice, vibrant color. c1 starts over here at this point. It moves from this point along this curve c1 and ends up at this point. And then we do the minus c2. Minus c2 starts at this point and just goes and comes back to the original point; it completes a loop. So this is a closed line integral. So if you combine this, we could rewrite this. Remember, this is just a loop. By reversing this, instead of having two guys starting here and going there, I now can start here, go all the way there, and then come all the way back on this reverse path of c2. So this is equivalent to a closed line integral. So that is the same thing as an integral along a closed path. I mean, we could call the closed path, maybe, c1 plus minus c2, if we wanted to be particular about the closed path. But this could be, I drew c1 and c2 or minus c2 arbitrarily; this could be any closed path where our vector field f has a potential, or where it is the gradient of a scalar field, or where it is conservative. And so this can be written as a closed path of c1 plus the reverse of c2 of f dot dr. That's just a rewriting of that, and so that's going to be equal to 0. And this is our take away for this video. This is, you can view it as a corollary. It's kind of a low-hanging conclusion that you can make after this conclusion. So now we know that if we have a vector field that's the gradient of a scalar field in some region, or maybe over the entire xy plane-- and this is called the potential of f; this is a potential function. Oftentimes it will be the negative of it, but it's easy to mess with negatives --but if we have a vector field that is the gradient of a scalar field, we call that vector field conservative. That tells us that at any point in the region where this is valid, the line integral from one point to another is independent of the path; that's what we got from the last video. And because of that, a closed loop line integral, or a closed line integral, so if we take some other place, if we take any other closed line integral or we take the line integral of the vector field on any closed loop, it will become 0 because it is path independent. So that's the neat take away here, that if you know that this is conservative, if you ever see something like this: if you see this f dot dr and someone asks you to evaluate this given that f is conservative, or given that f is the gradient of another function, or given that f is path independent, you can now immediately say, that is going to be equal to 0, which simplifies the math a good bit.