Example of Closed Line Integral of Conservative Field Example of taking a closed line integral of a conservative field
Example of Closed Line Integral of Conservative Field
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- Let's see if we can apply some of our new tools to solve
- some line integrals.
- So let's say we have a line integral along a closed curve
- -- I'm going to define the path in a second -- of x squared
- plus y squared times dx plus 2xy times dy.
- And then our curve c is going to be defined by
- the parameterization.
- x is equal to cosine of t, and y is equal to sine of t.
- And this is valid for t between 0 and 2 pi.
- So this is essentially a circle, a unit circle,
- in the xy plane, and we know how to solve these.
- Let's see if we can use some of our discoveries in the last
- couple of videos to maybe simplify this process.
- So the first thing you might say, hey, this looks like a
- line integral, but you have a dx and dy, I don't
- see a dot dr here.
- It's not clear to me that this is some type of even
- a vector line integral.
- I don't see any of vectors.
- What I want to do first, and the reason why I wanted to show
- you this example, is just to show you that this is just
- another form of writing really a vector line integral.
- To show you that you just have to realize if I have some are
- r of t -- this is our curve.
- I don't even write these functions in there.
- I'm just going to write it's x of t times i
- plus y of t times j.
- We've seen several videos now that we can write dr dt as
- being equal to dx dt times i plus dy dt times j.
- We've seen this multiple times.
- And we've seen multiple times we want to get the differential
- dr, we can just multiply everything times dt.
- And normally I just put a dt here and a dt there
- and get rid of this dt.
- But if you multiply everything times dt, if you view the
- differentials as actual numbers, you can multiply and
- normally you can treat them like that.
- Then you just get rid of all of the dt's.
- So dr you could imagine is equal to dx times the unit
- vector i plus dy times the unit vector j.
- So put that aside, and you might already
- see a pattern here.
- So if we define our vector field f, f of xy, as being
- equal to x squared plus y squared, i plus 2xy j,
- what is this thing?
- What is this thing over here?
- Well, what is f dot dr going to be?
- Dot products, you just multiply the corresponding components
- of our vectors and then add them up.
- So it's going to be if you take this f and dot it with that dr
- you're going to get the i component, x squared plus y
- squared times that dx plus -- I'll do it in the pink again --
- plus the y component, the j component 2xy times that dy.
- That's the dot product.
- And notice, this thing right here is identical to
- that thing right there.
- So our line integral, just to put it in a form that we're
- familiar with, this is the same exact thing as the line
- integral over this curve c, this closed curve c, of this f
- -- maybe I'll write it in that magenta color, or actually it's
- more of a purple or pink color -- f dot this dr.
- That's what this line integral is, it's just a different
- way of writing it.
- Now that you see it, in the future if you see in kind of
- this differential form, you'll immediately know OK, there's
- one vector field that this is its x component, this is its y
- component, dotting with the dr. This is the x component of dr
- or the i component, and this is the y component or the j
- component of the dr. So you immediately know what the
- vector field is that we're taking a line integral of.
- This is the x, that's the y.
- Now, let's ask ourselves a question.
- Is f conservative?
- So is f equal to the gradient of some scalar field, we'll
- call it capital F -- is this the case?
- So let's assume it is and see if we can solve for a scalar
- field whose grade it really is f.
- Then we know that f is conservative.
- And then if f is conservative, and this is the whole reason we
- want to do it, that means that any closed loop, any line
- integral over a closed curve of f is going to be equal
- to 0 and we'd be done.
- So if we can show this then the answer to this question or this
- question is going to be 0.
- We don't even have to mess with the cosine of t's and the
- sign of t's and all that.
- Actually, we don't even have to take antiderivatives.
- So let's see if we can find an f whose gradient is
- equal to that right there.
- So in order for f's gradient to be that, that means that the
- partial derivative of our capital F with respect to x has
- got to be equal to that right there.
- It's got to be equal to x squared plus y squared.
- And it also tells us that the partial derivative of capital
- F with respect to y has got to be equal to 2xy.
- And just as a review, if I have the gradient of any function,
- of any scalar field is equal to the partial of f with respect
- to x times i plus the partial of capital F with
- respect to y times j.
- So that's why I'm just pattern matching.
- I'm just saying well, gee, if this is the gradient of that,
- then this must be that, which I wrote down right here, and this
- must be that, which I wrote down here.
- So let's see if I can find an f that satisfies both
- of these constraints.
- So we could just take the antiderivative with respect to
- x on both sides -- remember, you just treat y like a
- constant or y squared like a constant -- it's just a number.
- So then we could say that f is equal to the antiderivative of
- x squared is x to the third over 3.
- And then the antiderivative of y squared -- remember,
- this is with respect to x.
- So you just treat it like a number.
- That could just be the number k, or this
- could be the number 5.
- So this is just going to be that times x.
- So plus x times y squared.
- And then there could be some function of y here.
- So plus some, I don't know, I'll call it g of y.
- Because there could have been some function of y here.
- If it's a pure function of y, when you take the derivative or
- the partial with respect to x, this would have disappeared.
- So it would reappear when we take the antiderivative.
- And just to be clear, let me make it clear that f is going
- to be a function of x and y.
- So we just have the, I guess you could say
- the antiderivative with respect to x.
- Let's see if we take the antiderivative with respect
- to y and then we can reconcile the two.
- So based on this, f of xy, f of xy is going to have to look
- like -- so let's take the antiderivative with
- respect to y here.
- So remember, you just treat x like it's just some number --
- it could be a k, it could be an m, it could be a 5.
- It's just some number.
- So if x is just some -- the antiderivative
- of 2y is y squared.
- And if x is just a number there, the antiderivative of
- this with respect to y is just going to be xy squared.
- Don't believe me?
- Take the partial of this with respect to y.
- Treat x like a constant you'll get 2 times xy
- with no exponent there.
- And, of course, if you took the antiderivative with respect to
- x, there might be some function of x here.
- We were just basing it off of that information.
- Now given that, this information says f of xy
- is going to have to look something like this.
- This information tells us f of xy's going to have to
- look something like that.
- Let's see if there is an f of xy that looks like
- both of them essentially.
- So let's see.
- On this one we have xy squared here, we have
- an xy squared there.
- So good.
- That looks good.
- And it over here we have an f of x -- we have something
- that's a pure function of x.
- And here we have something that is a pure function of x.
- So these two things could be the same thing.
- Then here we have a pure function of y that might be
- there, but it didn't really show up anywhere over here.
- So we could just say hey, that's going to be 0.
- 0 is a pure function of y.
- You could have something called g of y is equal to 0.
- And then we get that capital F of xy is equal to x to the
- third over 3 plus xy squared.
- And the gradient of this is going to be equal to f.
- And we've already established that.
- But just to hit the point home, let's take the gradient of it.
- Just if you don't believe this little stuff that I did right
- there, let's take the gradient.
- The gradient of f is equal to, and sometimes people put a
- little vector there because you're getting
- a vector out of it.
- You could put a little vector on top of that gradient sign.
- The gradient of f is going to be what?
- The partial of this with respect to x times i.
- So the partial of this with respect to x.
- The derivative here is 3 divided by 3 is 1.
- So it's just x squared plus the derivative of this with respect
- to x is y squared times i plus the partial with respect to y.
- Well, the partial with respect to y of this 0, partial with
- respect to y of this is 2xy or 2xy to the first.
- So it's 2xy times j.
- And this is exactly equal to f, our f that we wrote up there.
- So we've established that f can definitely be written -- f is
- definitely the gradient of some potential scalar
- function there.
- So f is conservative, and that tells us that this closed loop
- integral, line integral, of f is going to be equal to 0.
- And we are done.
- We could even ignore the actual parameterization of the path.
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