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# Eksempel på lukket kurveintegral i et konservativt felt

## Video udskrift

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.