# Vector field line integrals dependent on path direction | Multivariable Calculus | Khan Academy

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
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.