# Kurveintegral i vektorfelt afhængig af vejens retning

## Video udskrift

Let's say I have a position
vector function that looks like this. r of t is equal to x of t
times the unit vector i plus y of t times the unit vector j. And let me actually graph this. So let's say, r of t, I want
to draw it a little bit straighter than that. So that's my y-axis, that is my
x-axis, and let's say r of t, and this is for t is less
than, let me write this. So this is for t is
between a and b. So when t is equal to a, we're
at this vector right here. So if you actually substitute t
is equal to a here, you'd get a position vector that would
point to that point over there. And then, as t increases, it
traces out a curve, or the endpoints of our position
vectors trace a curve that looks something like that. So when t is equal to b, we get
a position vector that points to that point right there. So this defines a path. And the path is going in
this upward direction, just like that. Now let's say that we have
any another position vector function. Let me call it r of t. It's a different one. It's the green r. r of t. Instead of being x of t times
i, it's going to be x of a plus b minus t times i, and instead
of y of t, it's going to be y of a plus b minus t times i. And we've seen this in
the last two videos. This, the path defined by this
position vector function is going look more like this. Let me draw my axes. This is my y-axis,
that is my x-axis. Maybe I should label
them, y and x. This path is going to
look just like this. But instead of starting here
and going there, when t is equal to 1, let me make it
clear, this is also true for a is less than or equal to
t, which is less than b. So t is going to
go from a to b. But here, when t is equal
to a, you substitute it over here, you're going
to get this vector. You're going to start over
there, and as you increment t, as you make it larger and
larger and larger, you're going to trace out that same path,
but in the opposite direction. And so when t is equal to b,
you put that in here, you're actually going to get x of a
and y of a there, right, the b's cancel out, and so you're
going to point right like that. So these are the same, you
could imagine the shape of these paths are the same, but
we're going in the exact opposite direction. So what we're going to do in
this video is to see what happens, how, I guess you could
say, if I have some vector field f of xy equals p
of xy i plus q of xy j. Right? This is just a vector
field over the x-y plane. How the line integral of this
vector field, of this vector field over this path, compares
to the line integral the same vector field over that path. How that compares to this. We'll call this
the minus curve. So this is the positive
curve, we're going to call this the minus curve. So how does it, going over
the positive curve, compare to going over the minus
curve? f of f dot dr. So before I break into the
math, let's just think about a little bit. Let me draw this
vector field f. So maybe it looks, I'm just
going to draw random stuff. So you know, on every point in
the x-y plane, it has a vector, it defines or maps a vector, to
every point on the x-y plane. But we really care about the
points that are on the curve. So maybe on the curve, you
know, this is the vector field at the points on the curve. And let me draw it
over here, too. So all the points on the curve
where we care about, this is our vector field, that
is our vector field. And let's just get an intuition
of what's going to be going. We're summing over the dot,
we're taking each point along the line, and we're summing,
let me start over here. We're taking each point along
the line, let me do it in a different color. And we're summing the dot
product of the value of the vector field at that point, the
dot product of that, with dr, or the differential of our
position vector function. And dr, you can kind of
imagine, as an infinitesimally small vector going in the
direction of our movement. And we take this dot product
here, it's essentially, it's going to be a scalar value,
but the dot product, if you remember, it's the magnitude
of f in the direction of dr, times the magnitude of dr. So it's this, you can imagine
it's the shadow of f onto dr. Let me zoom into that,
because I think it's useful. So this little thing that I'm
drawing right here, let's say that this is my path. This is f at that point. f at that point looks
something like that. And then dr at this point
looks something like that. Let me do it in different
color. dr looks something like that. So that is f. And so the dot product of these
two says, ok, how much of f is going in the same
direction as dr? And you can kind of
imagine, there's a shadow. If you take the f that's going
in the same direction as dr, the magnitude of that times the
magnitude of dr, that is the dot product. In this case, we're going
to get a positive number. Because this length is
positive, this length is positive, that's going to
be a positive number. Now what if our dr was going
in the opposite direction, as it is in this case? So let me draw maybe that
same part of the curve. We have our f, our f will
look something like that. I'm drawing this exact
same part of the curve. But now our dr isn't
going in that direction. Our dr that at this point
is going to be going in the other direction. We're tracing the curve in
the opposite direction. Our dr is now going to be
going in that direction. So if you do f dot dr, you're
taking the shadow, or how much of f is going in the direction
of dr, you take the shadow down here, it's going in the
opposite direction of dr. So you can imagine that when you
multiply the magnitudes, you should get a negative number. Our direction is now opposite,
they're not going in the-- the shadow of f onto the same
direction is dr is going in the opposite direction as dr.
In this case, it's going in the same direction as dr. So the intuition is that maybe
these two things are the negative of each other. And now we can do some
math and try to see if that is definitely,
definitely the case. So let us first figure out,
let's write an expression for the differential dr. So in this
case, dr, dr dt is going to be equal to x prime of t times i
plus y prime of t times j. In this other example, in the
reverse case, our dr, dr dt, is going to be, what's
it going to be equal to? It's the derivative of
x with respect to t. The derivative of this term
with respect to t, that's the derivative of the inside, which
is minus 1, or minus, times the derivative of the outside
with respect to the inside. So that's going to be,
derivative of the inside is minus 1, times the derivative
of the outside with respect to the inside. x prime of a
plus b minus t times i. And then same thing
for the second term. Derivative of y of this term
with respect to the inside, which is minus 1, times the
derivative of the outside with respect to the inside, which is
y prime of a plus b minus t. So this is going to be the
derivative of the inside, times y prime of a
plus b minus is t j. So this is dr dt in this case,
this is dr dt in that case. And if we wanted to write the
differential dr in the forward curve example, it's going to be
equal to x prime of t times i plus y prime of t times
j times the scalar dt. I could multiply it down into
each of these terms, but it keeps it simple, just
leaving it on the outside. Same logic over here. dr is equal to minus x. I changed my shade of green,
but at least it's still green. a plus b minus t i minus y
prime a plus b minus t j, and I'm multiplying
both sides by dt. Now we're ready to express
this as a function of t. So this curve right here, I'll
do it in pink, the pink one is going to be equal to the
integral from t is equal to a to t is equal to b of f of f of
x of f of x of t y of t dot this thing over here, which is,
I'll just write out here, I could simplify it later. x prime of t i plus
y prime of t j. And then all of that
times the scalar dt. This'll be a scalar value, and
then we'll have another scalar value of dt over there. Now, what is this going
to be equal to if I take this reverse integral? The reverse integral is going
to be the integral from, I'm going to need a little more
space, from a to b, of f of not x of t, but x of a plus b minus
t y of a plus b minus t. I'm writing it small
so I have some space. Dot, this is a vector, so dot
this guy right here, dot dr. Dot minus x prime of a plus b
minus t i, minus y prime of, I'm using up too much space. Let me scroll, go
back a little bit. Actually, let me take it
make it even simpler. Let me take this minus
sign out of it. Let me put a plus, and
then I'll put the minus sign out front. So the minus sign is just a
scalar value, so we could put that minus sign out, you know,
when you take a dot product, and if you multiply a scalar
times a dot product, you could just take the scalar out,
that's all I'm saying. So we take that minus sign
out to this part right here. And then you have x prime of a
plus b minus t i plus y prime of a plus b minus t, scroll
over a little bit, t j dt. So the this is the forward,
this is when we're following it along the forward curve, this
is when we're following it along the reverse curve. Now like we did with the
scalar example, let's make a substitution. I want to make it very
clear what I did. All I did here, is I just
took the dot product, but this negative sign,
I just took it out. I just said, this is the same
thing as negative 1 times this thing, or negative 1 times this
thing is the same thing as that. So let's make a substitution on
this side, because I really just want to show you that this
is the negative of that, right there, because that's what
our intuition was going for. So let me just focus
on that side. So let me make a substitution. u is equal to a plus b minus t. Then we get du is equal
to minus dt, right? Just take the derivative
of both sides. Or you get dt is
equal to minus du. And then you get, when t
is equal to a, u is equal to a plus b minus a. So then, u is equal to b. And then when t is equal to
b, u is equal to a, right? Which is equal to b, a
plus b minus b is a. u is equal to a. So this thing, using that
substitution simplifies to, and this is the whole point, that
simplifies to minus integral from u is, when t is a, u is b. From b, when t is b, u is a. The integral from u is equal
to b to u is equal to a of f of x of u y of u, right? That is u, that is u. Dot x prime of u times i,
that's u right there, plus y prime of u times j. And then, instead of a
dt, I need to put a du. dt is equal to minus du. So I could write minus du
here, or just to not make it confusing, I'll put the du
here, and take the minus out front. I already have a minus out
there, so they cancel out. They will cancel out,
just like that. And so you might say, hey Sal,
these two things look pretty similar to each other. They don't like they're
negative of each other. And I'd say, well, you're
almost right, except this guy's limits of integration are
reversed from this guy. So this thing right here, if
we reverse the limits of integration, we have to
then make it negative. So this is equal to minus the
integral from a to b of the vector f of x of u y of u dot
x prime of u i plus y prime of u j du. And now this is identical. This integral, this definite
integral, is identical to that definite integral. We just have a
different variable. We're doing dt here, we have du
here, but we're going to get the same exact number for any a
or b, and given this vector f and the position
vector path r of t. So just to summarize everything
up, when you're dealing with line integrals over vector
fields, the direction matters. If you go in the reverse
direction, you're going to get the negative version of that. And that's because at any point
we take the dot product, you're not going in the, necessarily,
you're going in the opposite direction, so it'll be the
negative of each other. But when you're dealing with
the scalar field, we saw on the last video, we saw that it
doesn't matter which direction that you traverse the path in. That the positive path
has the same value as the negative path. And that's just because we're
just trying to find the area of that curtain. Hopefully you found
that mildly amusing.