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## Line integrals in vector fields

# Line integrals and vector fields

## Video transcript

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. It was this thing up here. 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 components of your of
vector, and add them up. 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.