# Andre eksempel på kurveintegralet av konservative vektorfelt

## Video transcript

Let's do another problem. Very similar to the last one,
but with a subtle difference. And that subtle difference
will make a big difference. Let's say we take the line
integral over some curve c-- I'll define the curve in a
second-- of x squared plus y squared dx plus 2xy dy-- and
this might look very familiar. This was very similar to what
we saw last time, except last time we had a closed
line integral. This is not a closed
line integral. And our curve, c, the
parameterization is x is equal to cosine of t, y
is equal to sine of t. So far-- it looks like sit. Let me write sine of t-- so
far, it looks very similar to the closed line integral
example we did in the last video, but instead of t going
from 0 to 2 pi, we're going to have t go from 0 to pi. t is greater than or equal to
0, is less than or equal to pi. So now we're essentially, our
path-- if I were to draw it on the x-y plane-- so that is my
y-axis, that is my x-axis. So now our path isn't all the
way around the unit circle. Our path-- our curve c now--
just starts at t is equal to 0. You can imagine t is
almost the angle. t is equal to 0, and we're
going to go all the way to pi. So that's what our path is
right now, in this example. So it's not a curved path. It's not a closed path. So we can't just show that f,
in this example-- and we're going to re-look at what f
looks like-- hey, if that's a conservative vector field, if
it's a closed loop that equals 0, this isn't a closed loop. So we can't apply that. But let's see if we can apply
some of our other tools. So like we saw in the last
video, this might look a little bit foreign to you. But if you say that f is equal
to that times i, plus that times j, then it might look
a little bit more familiar. If we say that f of xy-- the
vector field f is equal to x squared plus y squared times i
plus 2xy times j and dr-- I don't even have to look at this
right now. dr, you can always write it as dx times
i plus dy times j. You'll immediately see, if you
take the dot product of these 2 things, if you take f dot dr--
they're both vector valued, vector valued differential,
vector valued field, or vector valued function-- if you take f
dot dr, you'll get this right here. You'll get what we have
inside of the interval. You'll get that
right there, right? That times that-- you take
the product of the i terms-- that times that is equal to
that, and add it to the product of the j terms. 2xy times dy. Write like that. So our integral, we can
rewrite it as this. Along this curve of f dot
dr, where this is our f. Now, we still might want
to ask ourselves, is this a conservative field? Or does it have a potential? Is f equal to the gradient of
some function, capital F? I guess I could write the
gradient like that, because it creates a vector. This is a vector, too. Is this true? And we saw in the
last video, it is. I'll redo it a little
bit fast this time. Because if this is true, we
can't say this is a closed loop and say, oh, it's just
going to be equal to 0. But if this is true, then we
know that this-- that the integral is path independent. And we'll know that this is
going to be equal to capital F, if we say that t is going
from-- well, in this case t is going from 0 to pi-- we could
say that this is going to be equal to capital F of pi
minus capital F of 0. Or if we want to write it in
terms of x and y-- because f is going to be a function of x and
y-- we could write-- and this right here, these are t's. We could also write that this
is equal to f of x of pi, y of pi, minus f of x of 0, y of 0. That's what I mean
when I say f of pi. If we were to write f
purely as a function of t. But we know that this capital
F is going to be a function. It's a scalar function
defined on xy. So we could say f of
x of pi, y of pi. These are the t's now. These are all
equivalent things. So if it is path dependent,
we can find our f. We can just evaluate this
thing by just taking our f, evaluating it at
these two points. At this point, and at
that point right there. Because it would be
path independent. If this is a conservative, if
this has a potential function, if this is the gradient of
another scalar field, then this is a conservative vector field,
and its line integral is path independent. It's only dependent on that
point and that point. So let's see if we
can find our f. So I'm going to do exactly what
we did in the last video. If you watch that last
video, it might be a little bit monotonous. But I'll do it a little
bit faster here. So we know that the partial of
f with respect to x is going to have to be equal to
this right here. So that's x squared
plus y squared. Which tells us, if we take the
antiderivative, with respect to x, then f of xy is going to
have to be equal to x to the third over 3 plus xy squared--
right? y squared is just a constant in terms of
x-- plus f of y. There might be some function
of y that, when you take the partial with respect
to x, it just disappears. And then we know that the
partial of f with respect to y has got to be equal to
that thing or that thing. We're saying that this
is the gradient of f. So this has to be the
partial with respect to y. 2xy. And you might want to
watch the other video. I go through this just a little
bit slower in that one. So the antiderivative of this
with respect to y-- so we get f of xy-- would be equal to xy
squared plus some function of x. Now we did this in
the last video. These 2 things have to be the
same thing, in order for the gradient of capital F
to be lowercase f. And we have xy
squared, xy squared. We have a function of x, we
have a function purely of x. And then we don't have a
function purely of y here, so this thing right
here must be 0. So we've solved. Our capital F of xy must
be equal to x to the 3 over 3 plus xy squared. So we know that lowercase f
is definitely conservative. It is path independent. It has its potential. It is the gradient of
this thing right here. And so to solve our integral--
this was a 0-- to solve our integral, we just have to
figure out x of pi, y of pi, x of 0, y of 0. Evaluate the bullet points,
and then subtract the 2. So let's do that. So x was cosine of
t, y is sine of t. Let me rewrite it down here. So x is equal to cosine of t. y is equal to sine of t. So x of 0 is equal to cosine
of 0, which is equal to 1. x of pi is equal to cosine of
pi, which is equal to minus 1. y of 0 is sine of
0, which is 0. y of pi, which is equal to sine
of pi, which is equal to 0. So f of x of pi, y of pi--
this is the same thing, so let me rewrite this. Our integral is simplified to--
our integral along that path of f dot dr-- is going to be equal
to capital F of x of pi. x of pi is minus 1. y of pi is equal to 0. Minus capital F of x of 0
is 1, comma y of 0 is 0. And so what is this equal to? Just remember, this right
here is the same thing is that right there. That is x of pi. That is y of pi. That term right there. You can imagine this whole f of
minus 1, 0-- that's the same thing as f of pi, if you
think in terms of just t. That could be a little
confusing, so I want to make that clear. So this is just
straightforward to evaluate. What is f of minus 1, 0? x is minus 1. y is 0. So it's going to be minus 1
to the third power-- right, that's our x-- over 3. So it's minus 1/3. It's going to be minus 1/3
plus minus 1 times 0 squared. So that's just going to be a 0. In both cases, the y is 0. So this term is
going to disappear. So we can ignore that. And then we have minus
f of 1, comma 0. We put a 1 here. 1 to the third over 3. That is 1/3 plus 1
times 0 squared. That's just 0. So this is going to be
equal to minus 1/3. Minus 1/3 is equal
to minus 2/3. And we're done. And once again, because this is
a conservative vector field, and it's path independent, we
really didn't have to mess with the cosine of t's and sines of
t's when we actually took our antiderivative. We just have to find the
potential function and evaluate it at the 2 end points to get
the answer of our integral, of our line integral, minus 2/3.