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In the last video, we saw that if a vector field can be

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written as the gradient of a scalar field-- or another way

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we could say it: this would be equal to the partial of our big

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f with respect to x times i plus the partial of big f, our

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scalar field with respect to y times j; and I'm just writing

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it in multiple ways just so you remember what the gradient is

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--but we saw that if our vector field is the gradient of

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a scalar field then we call it conservative.

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So that tells us that f is a conservative vector field.

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And it also tells us, and this was the big take away from the

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last video, that the line integral of f between two

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points-- let me draw two points here; so let me draw my

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coordinates just so we know we're on the xy plane.

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My axes: x-axis, y-axis.

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Let's say I have the point, I have that point and that point,

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and I have two different paths between those two points.

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So I have path 1, that goes something like that, so

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I'll call that c1 and it goes in that direction.

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And then I have, maybe in a different shades of

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green, c2 goes like that.

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They both start here and go to there.

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We learned in the last video that the line integral

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is path independent between any two points.

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So in this case the line integral along c1 of f dot dr

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is going to be equal to the line integral of c2, over the

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path c2, of f dot dr. The line, if we have a potential in

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a region, and we may be everywhere, then the line

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integral between any two points is independent of the path.

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That's the neat thing about a conservative field.

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Now what I want to do in this video is do a little bit of

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an extension of the take away of the last video.

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It's actually a pretty important extension; it might

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already be obvious to you.

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I've already written this here; I could rearrange

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this equation a little bit.

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So let me do it.

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So let me a rearrange this.

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I'll just rewrite this in orange.

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So the line integral on path c1 dot dr minus-- I'll just go

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subtract this from both sides --minus the line integral c2 of

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f dot dr is going to be equal to 0.

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All I did is I took this take away from the last video and

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I subtracted this from both sides.

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Now we learned several videos ago that if we're dealing with

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a line integral of a vector field-- not a scalar field

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--with a vector field, the direction of the

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path is important.

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We learned that the line integral over, say, c2 of f dot

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dr, is equal to the negative of the line integral of minus c2

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of f dot dr where we denoted minus c2 is the same path as

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c2, but just in the opposite direction.

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So for example, minus c2 I would write like this-- so let

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me do it in a different color --so let's say this is minus

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c2, it'd be a path just like c2-- I'm going to call this

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minus c2 --but instead of going in that direction, I'm now

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going to go in that direction.

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So ignore the old c2 arrows.

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We're now starting from there and coming back here.

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So this is minus c2.

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Or we could write, we could put, the minus on the other

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side and we could say that the negative of the c2 line

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integral along the path of c2 of f dot dr is equal to the

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line integral over the reverse path of f dot dr. All I did is

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I switched the negative on the other side; multiplied

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both sides by negative 1.

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So let's replace-- in this equation we have the minus of

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the c2 path; we have that right there, and we have that right

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there --so we could just replace this with

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this right there.

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So let me do that.

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So I'll write this first part first.

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So the integral along the curve c1 of f dot dr, instead of

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minus the line integral along c2, I'm going to say plus the

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integral along minus c2.

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This-- let me switch to the green --this we've established

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is the same thing as this.

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The negative of this curve, or the line integral along this

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path, is the same thing as the line integral, the positive of

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the line integral along the reverse path.

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So we'll say plus the line integral of minus c2 of

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f dot dr is equal to 0.

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Now there's something interesting.

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Let's look at what the combination of the path

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of c1 and minus c2 is.

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c1 starts over here.

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Let me get a nice, vibrant color.

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c1 starts over here at this point.

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It moves from this point along this curve c1 and

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ends up at this point.

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And then we do the minus c2.

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Minus c2 starts at this point and just goes and comes back

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to the original point; it completes a loop.

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So this is a closed line integral.

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So if you combine this, we could rewrite this.

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Remember, this is just a loop.

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By reversing this, instead of having two guys starting here

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and going there, I now can start here, go all the way

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there, and then come all the way back on this

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reverse path of c2.

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So this is equivalent to a closed line integral.

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So that is the same thing as an integral along a closed path.

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I mean, we could call the closed path, maybe, c1 plus

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minus c2, if we wanted to be particular about

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the closed path.

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But this could be, I drew c1 and c2 or minus c2 arbitrarily;

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this could be any closed path where our vector field f has a

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potential, or where it is the gradient of a scalar field,

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or where it is conservative.

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And so this can be written as a closed path of c1 plus the

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reverse of c2 of f dot dr. That's just a rewriting

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of that, and so that's going to be equal to 0.

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And this is our take away for this video.

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This is, you can view it as a corollary.

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It's kind of a low-hanging conclusion that you can make

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after this conclusion.

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So now we know that if we have a vector field that's the

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gradient of a scalar field in some region, or maybe over the

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entire xy plane-- and this is called the potential of f;

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this is a potential function.

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Oftentimes it will be the negative of it, but it's easy

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to mess with negatives --but if we have a vector field that is

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the gradient of a scalar field, we call that vector

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field conservative.

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That tells us that at any point in the region where this is

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valid, the line integral from one point to another is

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independent of the path; that's what we got from

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the last video.

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And because of that, a closed loop line integral, or a closed

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line integral, so if we take some other place, if we take

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any other closed line integral or we take the line integral of

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the vector field on any closed loop, it will become 0 because

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it is path independent.

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So that's the neat take away here, that if you know that

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this is conservative, if you ever see something like this:

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if you see this f dot dr and someone asks you to evaluate

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this given that f is conservative, or given that f

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is the gradient of another function, or given that f is

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path independent, you can now immediately say, that is going

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to be equal to 0, which simplifies the math a good bit.

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