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Let's see if we can apply some of our new tools to solve

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some line integrals.

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So let's say we have a line integral along a closed curve

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-- I'm going to define the path in a second -- of x squared

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plus y squared times dx plus 2xy times dy.

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And then our curve c is going to be defined by

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the parameterization.

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x is equal to cosine of t, and y is equal to sine of t.

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And this is valid for t between 0 and 2 pi.

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So this is essentially a circle, a unit circle,

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in the xy plane, and we know how to solve these.

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Let's see if we can use some of our discoveries in the last

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couple of videos to maybe simplify this process.

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So the first thing you might say, hey, this looks like a

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line integral, but you have a dx and dy, I don't

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see a dot dr here.

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It's not clear to me that this is some type of even

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a vector line integral.

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I don't see any of vectors.

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What I want to do first, and the reason why I wanted to show

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you this example, is just to show you that this is just

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another form of writing really a vector line integral.

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To show you that you just have to realize if I have some are

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r of t -- this is our curve.

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I don't even write these functions in there.

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I'm just going to write it's x of t times i

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plus y of t times j.

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We've seen several videos now that we can write dr dt as

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being equal to dx dt times i plus dy dt times j.

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We've seen this multiple times.

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And we've seen multiple times we want to get the differential

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dr, we can just multiply everything times dt.

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And normally I just put a dt here and a dt there

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and get rid of this dt.

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But if you multiply everything times dt, if you view the

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differentials as actual numbers, you can multiply and

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normally you can treat them like that.

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Then you just get rid of all of the dt's.

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So dr you could imagine is equal to dx times the unit

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vector i plus dy times the unit vector j.

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So put that aside, and you might already

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see a pattern here.

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So if we define our vector field f, f of xy, as being

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equal to x squared plus y squared, i plus 2xy j,

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what is this thing?

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What is this thing over here?

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Well, what is f dot dr going to be?

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Dot products, you just multiply the corresponding components

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of our vectors and then add them up.

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So it's going to be if you take this f and dot it with that dr

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you're going to get the i component, x squared plus y

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squared times that dx plus -- I'll do it in the pink again --

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plus the y component, the j component 2xy times that dy.

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That's the dot product.

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And notice, this thing right here is identical to

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that thing right there.

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So our line integral, just to put it in a form that we're

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familiar with, this is the same exact thing as the line

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integral over this curve c, this closed curve c, of this f

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-- maybe I'll write it in that magenta color, or actually it's

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more of a purple or pink color -- f dot this dr.

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That's what this line integral is, it's just a different

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way of writing it.

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Now that you see it, in the future if you see in kind of

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this differential form, you'll immediately know OK, there's

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one vector field that this is its x component, this is its y

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component, dotting with the dr. This is the x component of dr

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or the i component, and this is the y component or the j

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component of the dr. So you immediately know what the

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vector field is that we're taking a line integral of.

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This is the x, that's the y.

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Now, let's ask ourselves a question.

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Is f conservative?

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So is f equal to the gradient of some scalar field, we'll

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call it capital F -- is this the case?

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So let's assume it is and see if we can solve for a scalar

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field whose grade it really is f.

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Then we know that f is conservative.

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And then if f is conservative, and this is the whole reason we

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want to do it, that means that any closed loop, any line

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integral over a closed curve of f is going to be equal

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to 0 and we'd be done.

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So if we can show this then the answer to this question or this

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question is going to be 0.

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We don't even have to mess with the cosine of t's and the

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sign of t's and all that.

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Actually, we don't even have to take antiderivatives.

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So let's see if we can find an f whose gradient is

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equal to that right there.

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So in order for f's gradient to be that, that means that the

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partial derivative of our capital F with respect to x has

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got to be equal to that right there.

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It's got to be equal to x squared plus y squared.

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And it also tells us that the partial derivative of capital

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F with respect to y has got to be equal to 2xy.

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And just as a review, if I have the gradient of any function,

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of any scalar field is equal to the partial of f with respect

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to x times i plus the partial of capital F with

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respect to y times j.

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So that's why I'm just pattern matching.

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I'm just saying well, gee, if this is the gradient of that,

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then this must be that, which I wrote down right here, and this

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must be that, which I wrote down here.

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So let's see if I can find an f that satisfies both

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of these constraints.

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So we could just take the antiderivative with respect to

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x on both sides -- remember, you just treat y like a

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constant or y squared like a constant -- it's just a number.

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So then we could say that f is equal to the antiderivative of

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x squared is x to the third over 3.

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And then the antiderivative of y squared -- remember,

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this is with respect to x.

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So you just treat it like a number.

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That could just be the number k, or this

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could be the number 5.

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So this is just going to be that times x.

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So plus x times y squared.

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And then there could be some function of y here.

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So plus some, I don't know, I'll call it g of y.

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Because there could have been some function of y here.

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If it's a pure function of y, when you take the derivative or

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the partial with respect to x, this would have disappeared.

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So it would reappear when we take the antiderivative.

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And just to be clear, let me make it clear that f is going

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to be a function of x and y.

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So we just have the, I guess you could say

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the antiderivative with respect to x.

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Let's see if we take the antiderivative with respect

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to y and then we can reconcile the two.

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So based on this, f of xy, f of xy is going to have to look

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like -- so let's take the antiderivative with

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respect to y here.

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So remember, you just treat x like it's just some number --

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it could be a k, it could be an m, it could be a 5.

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It's just some number.

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So if x is just some -- the antiderivative

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of 2y is y squared.

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And if x is just a number there, the antiderivative of

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this with respect to y is just going to be xy squared.

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Don't believe me?

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Take the partial of this with respect to y.

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Treat x like a constant you'll get 2 times xy

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with no exponent there.

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And, of course, if you took the antiderivative with respect to

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x, there might be some function of x here.

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We were just basing it off of that information.

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Now given that, this information says f of xy

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is going to have to look something like this.

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This information tells us f of xy's going to have to

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look something like that.

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Let's see if there is an f of xy that looks like

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both of them essentially.

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So let's see.

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On this one we have xy squared here, we have

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an xy squared there.

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So good.

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That looks good.

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And it over here we have an f of x -- we have something

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that's a pure function of x.

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And here we have something that is a pure function of x.

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So these two things could be the same thing.

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Then here we have a pure function of y that might be

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there, but it didn't really show up anywhere over here.

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So we could just say hey, that's going to be 0.

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0 is a pure function of y.

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You could have something called g of y is equal to 0.

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And then we get that capital F of xy is equal to x to the

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third over 3 plus xy squared.

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And the gradient of this is going to be equal to f.

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And we've already established that.

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But just to hit the point home, let's take the gradient of it.

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Just if you don't believe this little stuff that I did right

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there, let's take the gradient.

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The gradient of f is equal to, and sometimes people put a

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little vector there because you're getting

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a vector out of it.

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You could put a little vector on top of that gradient sign.

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The gradient of f is going to be what?

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The partial of this with respect to x times i.

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So the partial of this with respect to x.

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The derivative here is 3 divided by 3 is 1.

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So it's just x squared plus the derivative of this with respect

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to x is y squared times i plus the partial with respect to y.

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Well, the partial with respect to y of this 0, partial with

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respect to y of this is 2xy or 2xy to the first.

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So it's 2xy times j.

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And this is exactly equal to f, our f that we wrote up there.

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So we've established that f can definitely be written -- f is

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definitely the gradient of some potential scalar

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function there.

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So f is conservative, and that tells us that this closed loop

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integral, line integral, of f is going to be equal to 0.

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And we are done.

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We could even ignore the actual parameterization of the path.

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