# Avgrensede kurveintegraler av konservative vektorfelt

## Video transcript

In the last video, we saw that
if a vector field can be written as the gradient of a
scalar field-- or another way we could say it: this would be
equal to the partial of our big f with respect to x times i
plus the partial of big f, our scalar field with respect to y
times j; and I'm just writing it in multiple ways just so you
remember what the gradient is --but we saw that if our vector
field is the gradient of a scalar field then we
call it conservative. So that tells us that f is a
conservative vector field. And it also tells us, and this
was the big take away from the last video, that the line
integral of f between two points-- let me draw two points
here; so let me draw my coordinates just so we know
we're on the xy plane. My axes: x-axis, y-axis. Let's say I have the point, I
have that point and that point, and I have two different paths
between those two points. So I have path 1, that goes
something like that, so I'll call that c1 and it
goes in that direction. And then I have, maybe in
a different shades of green, c2 goes like that. They both start here
and go to there. We learned in the last video
that the line integral is path independent
between any two points. So in this case the line
integral along c1 of f dot dr is going to be equal to the
line integral of c2, over the path c2, of f dot dr. The line,
if we have a potential in a region, and we may be
everywhere, then the line integral between any two points
is independent of the path. That's the neat thing about
a conservative field. Now what I want to do in this
video is do a little bit of an extension of the take
away of the last video. It's actually a pretty
important extension; it might already be obvious to you. I've already written this
here; I could rearrange this equation a little bit. So let me do it. So let me a rearrange this. I'll just rewrite
this in orange. So the line integral on path c1
dot dr minus-- I'll just go subtract this from both sides
--minus the line integral c2 of f dot dr is going
to be equal to 0. All I did is I took this take
away from the last video and I subtracted this
from both sides. Now we learned several videos
ago that if we're dealing with a line integral of a vector
field-- not a scalar field --with a vector field,
the direction of the path is important. We learned that the line
integral over, say, c2 of f dot dr, is equal to the negative of
the line integral of minus c2 of f dot dr where we denoted
minus c2 is the same path as c2, but just in the
opposite direction. So for example, minus c2 I
would write like this-- so let me do it in a different color
--so let's say this is minus c2, it'd be a path just like
c2-- I'm going to call this minus c2 --but instead of going
in that direction, I'm now going to go in that direction. So ignore the old c2 arrows. We're now starting from
there and coming back here. So this is minus c2. Or we could write, we could
put, the minus on the other side and we could say that
the negative of the c2 line integral along the path of c2
of f dot dr is equal to the line integral over the reverse
path of f dot dr. All I did is I switched the negative on
the other side; multiplied both sides by negative 1. So let's replace-- in this
equation we have the minus of the c2 path; we have that right
there, and we have that right there --so we could just
replace this with this right there. So let me do that. So I'll write this
first part first. So the integral along the curve
c1 of f dot dr, instead of minus the line integral along
c2, I'm going to say plus the integral along minus c2. This-- let me switch to the
green --this we've established is the same thing as this. The negative of this curve, or
the line integral along this path, is the same thing as the
line integral, the positive of the line integral along
the reverse path. So we'll say plus the line
integral of minus c2 of f dot dr is equal to 0. Now there's something
interesting. Let's look at what the
combination of the path of c1 and minus c2 is. c1 starts over here. Let me get a nice,
vibrant color. c1 starts over here
at this point. It moves from this point
along this curve c1 and ends up at this point. And then we do the minus c2. Minus c2 starts at this point
and just goes and comes back to the original point;
it completes a loop. So this is a closed
line integral. So if you combine this,
we could rewrite this. Remember, this is just a loop. By reversing this, instead of
having two guys starting here and going there, I now can
start here, go all the way there, and then come all
the way back on this reverse path of c2. So this is equivalent to
a closed line integral. So that is the same thing as an
integral along a closed path. I mean, we could call the
closed path, maybe, c1 plus minus c2, if we wanted to be
particular about the closed path. But this could be, I drew c1
and c2 or minus c2 arbitrarily; this could be any closed path
where our vector field f has a potential, or where it is the
gradient of a scalar field, or where it is conservative. And so this can be written as
a closed path of c1 plus the reverse of c2 of f dot dr.
That's just a rewriting of that, and so that's
going to be equal to 0. And this is our take
away for this video. This is, you can view
it as a corollary. It's kind of a low-hanging
conclusion that you can make after this conclusion. So now we know that if we have
a vector field that's the gradient of a scalar field in
some region, or maybe over the entire xy plane-- and this is
called the potential of f; this is a potential function. Oftentimes it will be the
negative of it, but it's easy to mess with negatives --but if
we have a vector field that is the gradient of a scalar field,
we call that vector field conservative. That tells us that at any point
in the region where this is valid, the line integral from
one point to another is independent of the path; that's
what we got from the last video. And because of that, a closed
loop line integral, or a closed line integral, so if we take
some other place, if we take any other closed line integral
or we take the line integral of the vector field on any closed
loop, it will become 0 because it is path independent. So that's the neat take away
here, that if you know that this is conservative, if you
ever see something like this: if you see this f dot dr and
someone asks you to evaluate this given that f is
conservative, or given that f is the gradient of another
function, or given that f is path independent, you can now
immediately say, that is going to be equal to 0, which
simplifies the math a good bit.