# Line integrals and vector fields | Multivariable Calculus | Khan Academy

One of the most fundamental ideas in all of physics
is the idea of work.
Now when you first learn work, you just say, oh, that's
just force times distance.
But then later on, when you learn a little bit about
vectors, you realize that the force isn't always going in
the same direction as your displacement.
So you learn that work is really the magnitude, let me
write this down, the magnitude of the force, in the direction,
or the component of the force in the direction
of displacement.
Displacement is just distance with some direction.
Times the magnitude of the displacement, or you could say,
times the distance displaced.
And the classic example.
Maybe you have an ice cube, or some type of block.
I just ice so that there's not a lot of friction.
Maybe it's standing on a bigger lake or ice or something.
And maybe you're pulling on that ice cube at an angle.
Let's say, you're pulling at an angle like that.
That is my force, right there.
Let's say my force is equal to-- well, that's
my force vector.
Let's say the magnitude of my force vector, let's
say it's 10 newtons.
And let's say the direction of my force vector, right, any
vector has to have a magnitude and a direction, and the
direction, let's say it has a 30 degree angle, let's say a 60
degree angle, above horizontal.
So that's the direction I'm pulling in.
And let's say I displace it.
This is all review, hopefully.
If you're displacing it, let's say you displace it 5 newtons.
So let's say the displacement, that's the displacement vector
right there, and the magnitude of it is equal to 5 meters.
So you've learned from the definition of work, you can't
just say, oh, I'm pulling with 10 newtons of force and
I'm moving it 5 meters.
You can't just multiply the 10 newtons times the 5 meters.
You have to find the magnitude of the component going in the
same direction as my displacement.
So what I essentially need to do is, the length, if you
imagine the length of this vector being 10, that's the
total force, but you need to figure out the length of the
vector, that's the component of the force, going in the same
direction as my displacement.
And a little simple trigonometry, you know that
this is 10 times the cosine of 60 degrees, or that's equal to,
cosine of 60 degrees is 1/2, so that's just equal to 5.
So this magnitude, the magnitude of the force going
in the same direction of the displacement in this
case, is 5 newtons.
And then you can figure out the work.
You could say that the work is equal to 5 newtons times, I'll
just write a dot for times.
I don't want you to think it's cross product.
Times 5 meters, which is 25 newton meters, or you could
even say 25 Joules of work have been done.
And this is all review of somewhat basic physics.
But just think about what happened, here.
What was the work?
If I write in the abstract.
The work is equal to the 5 newtons.
That was the magnitude of my force vector, so it's the
magnitude of my force vector, times the cosine of this angle.
So you know, let's call that theta.
Let's say it a little generally.
So times the cosine of the angle.
This is the amount of my force in the direction of the
displacement, the cosine of the angle between them, times the
magnitude of the displacement.
So times the magnitude of the displacement.
Or if I wanted to rewrite that, I could just write that as, the
magnitude of the displacement times the magnitude of
the force times the cosine of theta.
And I've done multiple videos of this, in the linear algebra
playlist, in the physics playlist, where I talk about
the dot product and the cross product and all of that, but
this is the dot product of the vectors d and f.
So in general, if you're trying to find the work for a constant
displacement, and you have a constant force, you just take
the dot product of those two vectors.
And if the dot product is a completely foreign concept to
you, might want to watch, I think I've made multiple, 4
or 5 videos on the dot product, and its intuition,
and how it compares.
But just to give you a little bit of that intuition right
here, the dot product, when I take f dot d, or d dot f,
what it's giving me is, I'm multiplying the magnitude, well
I could just read this out.
But the idea of the dot product is, take how much of this
vector is going in the same direction as this vector,
in this case, this much.
And then multiply the two magnitudes.
And that's what we did right here.
So the work is going to be the force vector, dot, taking the
dot part of the force vector with the displacement vector,
and this, of course, is a scalar value.
And we'll work out some examples in the future where
you'll see that that's true.
So this is all review of fairly elementary physics.
Now let's take a more complex example, but it's
really the same idea.
Let's define a vector field.
So let's say that I have a vector field f, and we're
going to think about what this means in a second.
It's a function of x and y, and it's equal to some scalar
function of x and y times the i-unit vector, or the
horizontal unit vector, plus some other function, scalar
function of x and y, times the vertical unit vector.
So what would something like this be?
This is a vector field.
This is a vector field in 2-dimensional space.
We're on the x-y plane.
Or you could even say, on R2.
Either way, I don't want to get too much into
the mathiness of it.
But what does this do?
Well, if I were to draw my x-y plane, so that is my, again,
having trouble drawing a straight line.
All right, there we go.
That's my y-axis, and that's my x-axis.
I'm just drawing the first quadrant, and but you could
go negative in either direction, if you like.
What does this thing do?
Well, it's essentially saying, look.
You give me any x, any y, you give any x, y in the x-y plane,
and these are going to end up with some numbers, right?
When you put x, y here, you're going to get some value, when
you put x, y here, you're going to get some value.
So you're going to get some combination of the i-
and j-unit vectors.
So you're going to get some vector.
So what this does, it defines a vector that's associated with
every point on x-y plane.
So you could say, if I take this point on the x-y plane,
and I would pop it into this, I'll get something times i plus
something times j, and when you add those 2, maybe I get a
vector that something like that.
And you could do that on every point.
I'm just taking random samples.
Maybe when I go here, the vector looks
something like that.
Maybe when I go here, the victor looks like this.
Maybe when I go here, the vector looks like that.
And maybe when I go up here, the vector goes like that.
I'm just randomly picking points.
It defines a vector on all of the x, y coordinates where
these scalar functions are properly defined.
And that's why it's called a vector field.
It defines what a potential, maybe, force would be,
or some other type of force, at any point.
At any point, if you happen to have something there.
Maybe that's what the function is.
And I could keep doing this forever, and
filling in all the gaps.
But I think you get the idea.
It associates a vector with every point on x-y plane.
Now, this is called a vector field, so it probably makes a
lot of sense that this could be used to describe
any type of field.
It could be a gravitation field.
It could be an electric field, it could be a magnetic field.
And this could be essentially telling you how much force
there would be on some particle in that field.
That's exactly what this would describe.
Now, let's say that in this field, I have some particle
traveling on x-y plane.
Let's say it starts there, and by virtue of all of these crazy
forces that are acting on it, and maybe it's on some tracks
or something, so it won't always move exactly in the
direction that the field is trying to move it at.
Let's say it moves in a path that moves something like this.
And let's say that this path, or this curve, is defined by
a position vector function.
So let's say that that's defined by r of t, which is
just x of t times i plus y of t times our unit factor j.
That's r of t right there.
Well, in order for this to be a finite path, this is true
before t is greater than or equal to a, and less
than or equal to b.
This is the path that the particle just happens to
take, due to all of these wacky forces.
So when the particle is right here, maybe the vector field
acting on it, maybe it's putting a force like that.
But since the thing is on some type of tracks, it moves
in this direction.
And then when it's here, maybe the vector field is like that,
but it moves in that direction, because it's on some
type of tracks.
Now, everything I've done in this video is to build up
to a fundamental question.
What was the work done on the particle by the field?
To answer that question, we could zoom in a little bit.
I'm going to zoom in on only a little small
snippet of our path.
And let's try to figure out what the work is done in a very
small part of our path, because it's constantly changing.
The field is changing direction.
my object is changing direction.
So let's say when I'm here, and let's say I move a
small amount of my path.
So let's say I move, this is an infinitesimally
small dr. Right?
I have a differential, it's a differential vector, infinitely
small displacement.
and let's say over the course of that, the vector field is
acting in this local area, let's say it looks
something like that.
It's providing a force that looks something like that.
So that's the vector field in that area, or the force
directed on that particle right when it's at that point.
Right?
It's an infinitesimally small amount of time in space.
You could say, OK, over that little small point, we
have this constant force.
What was the work done over this small period?
You could say, what's the small interval of work?
You could say d work, or a differential of work.
Well, by the same exact logic that we did with the simple
problem, it's the magnitude of the force in the direction of
our displacement times the magnitude of our displacement.
And we know what that is, just from this example up here.
That's the dot product.
It's the dot product of the force and our super-small
displacement.
So that's equal to the dot product of our force and our
super-small displacement.
Now, just by doing this, we're just figuring out the work
over, maybe like a really small, super-small dr. But
what we want to do, is we want to sum them all up.
We want to sum up all of the drs to figure out the total,
all of the f dot drs to figure out the total work done.
And that's where the integral comes in.
We will do a line integral over-- I mean, you could
think of it two ways.
You could write just d dot w there, but we could say, we'll
do a line integral along this curve c, could call that c
or along r, whatever you want to say it, of dw.
That'll give us the total work.
So let's say, work is equal to that.
Or we could also write it over the integral, over the same
curve of f of f dot dr.
And this might seem like a really, you know, gee, this
is really abstract, Sal.
How do we actually calculate something like this?
Especially because we have everything parameterized
in terms of t.
How do we get this in terms of t?
And if you just think about it, what is f dot r?
Or what is f dot dr?
Well, actually, to answer that, let's remember
what dr looked like.
If you remember, dr/dt is equal to x prime of t, I'm writing it
like, I could have written dx dt if I wanted to do, times the
i-unit vector, plus y prime of t, times the j-unit vector.
And if we just wanted to dr, we could multiply both sides, if
we're being a little bit more hand-wavy with the
differentials, not too rigorous.
We'll get dr is equal to x prime of t dt times the unit
vector i plus y prime of t times the differential dt
times the unit vector j.
So this is our dr right here.
And remember what our vector field was.
Let me copy and paste it.
And we'll see that the dot product is
actually not so crazy.
So copy, and let me paste it down here.
So what's this integral going to look like?
This integral right here, that gives the total work done by
the field, on the particle, as it moves along that path.
Just super fundamental to pretty much any serious physics
that you might eventually find yourself doing.
So you could say, well gee.
It's going to be the integral, let's just say from t is equal
to a, to t is equal to b.
Right? a is where we started off on the path, t is equal
to a to t is equal to b.
You can imagine it as being timed, as a particle
moving, as time increases.
And then what is f dot dr?
Well, if you remember from just what the dot product is, you
can essentially just take the product of the corresponding
So this is going to be the integral from t equals a to t
equals b, of p of p of x, really, instead of writing x,
y, it's x of t, right? x as a function of t, y as
a function of t.
So that's that.
Times this thing right here, times this component, right?
We're multiplying the i-components.
So times x prime of t d t, and then that plus, we're going
to do the same thing with the q function.
So this is q plus, I'll go to another line.
Hopefully you realize I could have just kept writing, but
I'm running out of space.
Plus q of x of t, y of t, times the component of our dr. Times
the y-component, or the j-component.
y prime of t dt.
And we're done!
And we're done.
This might still seem a little bit abstract, but we're going
to see in the next video, everything is now in terms of
t, so this is just a straight-up integration,
with respect to dt.
If we want, we could take the dt's outside of the equation,
and it'll look a little bit more normal for you.
But this is essentially all that we have to do.
And we're going to see some concrete examples of taking a
line integral through a vector field, or using vector
functions, in the next video.