Hovedinnhold

## Line integrals in vector fields (videos)

Gjeldende klokkeslett:0:00Total varighet:7:26

# Parameterfremstilling av en omvendt bane

## Video transkripsjon

What I want to do in the next
few videos is try to see what happens to a line integral,
either a line integral over a scalar field or a vector field,
but what happens that line integral when we change the
direction of our path? So let's say, when I say change
direction, let's say that I have some curve C that
looks something like this. We draw the x- and y- axis. So that's my y-axis, that is
my x-axis, and let's say my parameterization starts there,
and then as t increases, ends up over there just like that. So it's moving in
that direction. And when I say I reverse
the path, we could define another curve. Let's call it minus C, that
looks something like this. That is my y-axis,
that is my x-axis. And it looks exactly the same,
but it starts up here, and then as t increases, it goes down
to the starting point of the other curve. So it's the exact same shape
of a curve, but it goes in the opposite direction. So what I'm going to do in this
video is just understand how we can construct a
parameterization like this, and hopefully understand
it pretty well. And then next two videos after
this, we'll try to see what this actually does to the line
integral, one for a scalar field, and then one
for a vector field. So let's just say, this
parameterization right here, let's just define it in the
basic way that we've always defined them. Let's say that this is x is
equal to x of t, y is equal to y of t, and let's say this is
from t is equal, or t, let me write this way. t starts at a, so t is
greater than or equal to a, and it goes up to b. So in this example, this was
when t is equal to a, and the point right here is the
coordinate x of a, y of a. And then when t is equal to b
up here, this is really just a review of what we've seen
before, really just a review of parameterization, when t is
equal to b up here, this is the point x of b, y of b. Nothing new there. Now given these functions, how
can we construct another parameterization here that has
the same shape, but that starts here? So I want this to be,
t is equal to a. Let me switch colors. Let me switch to,
maybe, magenta. So I want this to be t is equal
to a, and as t increases, I want this to be t equals b. So I want to move in the
opposite direction. So when t is equal to a,
I want my coordinate to still be x of b, y of b. When t is equal to a, I want a
b in each of these functions, and when t is equal to b, I
want the coordinate to be x of a, y of a. Right? Notice, they're opposites now. Here t is equal to a, x
of a, y of a, here t is equal to b, our endpoint. Now I'm at this coordinate,
x of a, y of a. So how do I construct that? Well, if you think about it,
when t is equal to a, we want both of these functions
to evaluate it at b. So what if we define our x, in
this case, for our minus C curve, what if we say x is
equal to x of, and when I say x of I'm talking about the
same exact function. Actually, maybe I should write
it in that same exact color. x of-- but instead of putting t
in there, instead of putting a straight-up t in there, what if
I put an a plus b minus t in there? What happens? Well, let me do it
for the y as well. So then our y, y, is equal
to y of a plus b minus t. a plus b minus is t. I'm using slightly different
shades of yellow, might be a little disconcerting. Anyway, what happens
when we define this? When t is equal to a, when t is
equal to a, let's say that this parameterization is also
for t starts at a and then goes up to b. So let's just experiment and
confirm that this parameterization really is the
same thing as this thing, but it goes in an
opposite direction. Or at least, confirm in
our minds intuitively. So when t is equal to a, when t
is equal to a, x will be equal to x of a plus b
minus a, right? This is when t is equal to
a, so minus t, or minus a, which is equal to what? Well, a minus a, cancel out,
that's equal to x of b. Similarly, when t is equal
to a, y will be equal to y of a plus b minus a. The a's cancel out, so
it's equal to y of b. So that worked. When t is equal to a, my
parameterization evaluates to the coordinate x of b, y of b. When t is equal to
a, x of b, y of b. Then we can do the exact same
thing when t is equal to b. I'll do it over here, because
I don't want to lose this. Let me just draw a line here. I'm still dealing with this
parameterization over here. Actually, let me scroll over
to the right, just so that I don't get confused. When t is equal to b, when t
is equal to b, what does x equal? x is equal to x of
a plus b minus b, right? a plus b minus b when
t is equal to b. So that's equal to x of a. and then when she's able to be
why is equal to lie of a plus b minus b, and of course, that's
going to be equal to y of a. So the endpoints work, and if
you think about it intuitively, as t increases, so when t is at
a, this thing is going to be x of b, y of b. We saw that down here. Now as t increases, this
value is going to decrease. We started x of b, y of b, and
as t increases, this value is going to decrease to a, right? It starts from b,
and it goes to a. This one obviously starts
at a, and it goes to b. So hopefully, that should give
you the intuition why this is the exact same curve as that. It just goes in a completely
opposite direction. Now, with that out of the way,
if you accept what I've told you, that these are really
the same parameterizations, just opposite directions. I shouldn't say same
parameterizations. Same curve going in an opposite
direction, or same path going in the opposite direction. In the next video, I'm going to
see what happens when we evaluate this line integral, f
of x ds, versus this line integral. So this is a scalar field, a
line integral of a scalar field, using this curve or this
path, but what happens if we take a line integral over the
same scalar field, but we do it over this reverse path? That's what we're going
to do in the next video. And the video after that, we'll
do it for vector fields.