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OK.
Today we have a new topic,

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00:00:26 --> 00:00:33
and we are going to start to
learn about vector fields and

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line integrals.
Last week we had been doing

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double integrals.
For today we just forget all of

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that, but don't actually forget
it.

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00:00:43 --> 00:00:45
Put it away in a corner of your
mind.

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00:00:45 --> 00:00:48
It is going to come back next
week, but what we do today will

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include line integrals.
And these are completely

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different things,
so it helps,

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actually, if you don't think of
double integrals at all while

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doing line integrals.
Anyway, let's start with vector

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fields.
What is a vector field?

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Well, a vector field is
something that is of a form,

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while it is a vector, 
but while M and N, 

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the components, 
actually depend on x and y,on

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the point where you are.
So, they are functions of x and

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y.
What that means,

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00:01:32 --> 00:01:36
concretely, is that every point
in the plane you have a vector.

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00:01:36 --> 00:01:39
In a corn field,
every where you have corn.

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00:01:39 --> 00:01:43
In a vector field,
everywhere you have a vector.

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That is how it works.
A good example of a vector

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00:01:49 --> 00:01:55
field, I don't know if you have
seen these maps that show the

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00:01:55 --> 00:01:59
wind, but here are some cool
images done by NASA.

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00:01:59 --> 00:02:04
Actually, that is a picture of
wind patterns off the coast of

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California with Santa Ana winds,
in case you are wondering what

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00:02:09 --> 00:02:13
has been going on recently.
You have all of these vectors

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00:02:13 --> 00:02:17
that show you the velocity of
the air basically at every

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00:02:17 --> 00:02:18
point.
I mean, of course you don't

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00:02:18 --> 00:02:21
draw it every point,
because if you drew a vector at

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00:02:21 --> 00:02:24
absolutely all the points of a
plane then you would just fill

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00:02:24 --> 00:02:27
up everything and you wouldn't
see anything.

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00:02:27 --> 00:02:33
So, choose points and draw the
vectors at those points.

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00:02:33 --> 00:02:38
Here is another cool image,
which is upside down.

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00:02:38 --> 00:02:44
That is a hurricane off the
coast of Mexico with the winds

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00:02:44 --> 00:02:49
spiraling around the hurricane.
Anyway, it is kind of hard to

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00:02:49 --> 00:02:52
see.
You don't really see all the

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00:02:52 --> 00:02:57
vectors, actually,
because the autofocus is having

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00:02:57 --> 00:03:03
trouble with it.
It cannot really do it,

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00:03:03 --> 00:03:11
so I guess I will go back to
the previous one.

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00:03:11 --> 00:03:30
Anyway, a vector field is
something where at each point --

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00:03:30 --> 00:03:50
-- in the plane we have vector F
that depends on x and y.

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00:03:50 --> 00:03:54
This occurs in real life when
you look at velocity fields in a

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fluid.
For example, the wind.

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00:03:56 --> 00:03:58
That is what these pictures
show.

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00:03:58 --> 00:04:01
At every point you have a
velocity of a fluid that is

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00:04:01 --> 00:04:04
moving.
Another example is force fields.

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00:04:04 --> 00:04:08
Now, force fields are not
something out of Star Wars.

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00:04:08 --> 00:04:10
If you look at gravitational
attraction,

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you know that if you have a
mass somewhere,

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00:04:13 --> 00:04:16
well, it will be attracted to
fall down because of the gravity

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field of the earth,
which means that at every point

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you have a vector that is
pointing down.

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And, the same thing in space,
you have the gravitational

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field of planets,
stars and so on.

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That is also an example of a
vector field because,

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wherever you go,
you would have that vector.

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And what it is depends on where
you are.

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The examples from the real
world are things like velocity

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in a fluid or force field where
you have a force that depends on

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the point where you are.
We are going to try to study

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vector fields mathematically.
We won't really care what they

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are most of the time,
but, as we will explore with

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them defined quantities and so
on,

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we will very often use these
motivations to justify why we

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would care about certain
quantities.

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The first thing we have to
figure out is how do we draw a

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vector field,
you know, how do you generate a

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plot like that?
Let's practice drawing a few

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vector fields.
Well, let's say our very first

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vector field will be just 2i j.
It is kind of a silly vector

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field because it doesn't
actually depend on x and y.

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That means it is the same
vector everywhere.

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I take a plane and take vector
.

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I guess it points in that
direction.

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It is two units to the right
and one up.

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And I just put that vector
everywhere.

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You just put it at a few points
all over the place.

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00:06:13 --> 00:06:16
And when you think you have
enough so that you understand

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what is going on then you stop.
Here probably we don't need

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that many.
I mean here I think we get the

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picture.
Everywhere we have a vector

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.
Now, let's try to look at

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slightly more interesting
examples.

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00:06:37 --> 00:06:42
Let's say I give you a vector
field x times i hat.

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There is no j component.
How would you draw that?

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00:06:55 --> 00:07:00
Well, first of all,
we know that this guy is only

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in the i direction so it is
always horizontal.

94
00:07:05 --> 00:07:08
It doesn't have a j component.
Everywhere it would be a

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horizontal vector.
Now, the question is how long

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00:07:10 --> 00:07:15
is it?
Well, how long it is depends on

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x.
For example,

98
00:07:16 --> 00:07:20
if x is zero then this will
actually be the zero vector.

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00:07:20 --> 00:07:28
x is zero here on the y-axis.
I will take a different color.

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00:07:28 --> 00:07:33
If I am on the y-axis,
I actually have the zero

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00:07:33 --> 00:07:36
vector.
Now, if x becomes positive

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00:07:36 --> 00:07:41
small then I will have actually
a small positive multiple of i

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00:07:41 --> 00:07:45
so I will be going a little bit
to the right.

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00:07:45 --> 00:07:51
And then, if I increase x,
this guy becomes larger so I

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00:07:51 --> 00:07:54
get a longer vector to the
right.

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00:07:54 --> 00:08:03
If x is negative then my vector
field points to the left

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00:08:03 --> 00:08:10
instead.
It looks something like that.

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00:08:10 --> 00:08:16
Any questions about that
picture?

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00:08:16 --> 00:08:17
No.
OK.

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00:08:17 --> 00:08:20
Usually, we are not going to
try to have very accurate,

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00:08:20 --> 00:08:23
you know, we won't actually
take time to plot a vector field

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00:08:23 --> 00:08:26
very carefully.
I mean, if we need to,

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00:08:26 --> 00:08:30
computers can do it for us.
It is useful to have an idea of

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00:08:30 --> 00:08:32
what a vector field does
roughly.

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00:08:32 --> 00:08:36
Whether it is getting larger
and larger, in what direction it

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00:08:36 --> 00:08:39
is pointing, what are the
general features?

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00:08:39 --> 00:08:45
Just to do a couple of more,
actually, you will see very

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00:08:45 --> 00:08:50
quickly that the examples I use
in lecture are pretty much

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00:08:50 --> 00:08:55
always the same ones.
We will be playing a lot with

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00:08:55 --> 00:08:59
these particular vector fields
just because they are good

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00:08:59 --> 00:09:04
examples.
Let's say I give you xi yj.

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00:09:04 --> 00:09:08
That one has an interesting
geometric significance.

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00:09:08 --> 00:09:16
If I take a point (x,
y), there I want to take a

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00:09:16 --> 00:09:20
vector x, y.
How do I do that?

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00:09:20 --> 00:09:23
Well, it is the same as a
vector from the origin to this

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point.
I take this vector and I copy

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00:09:27 --> 00:09:30
it so that it starts at one
point.

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00:09:30 --> 00:09:40
It looks like that.
And the same thing at every

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00:09:40 --> 00:09:45
point.
It is a vector field that is

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00:09:45 --> 00:09:53
pointing radially away from the
origin, and its magnitude

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00:09:53 --> 00:09:59
increases with distance from the
origin.

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You don't have to draw as many
as me, but the idea is this

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00:10:03 --> 00:10:08
vector field everywhere points
away from the origin.

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00:10:08 --> 00:10:14
And its magnitude is equal to
the distance from the origin.

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00:10:14 --> 00:10:15
If these were,
for example,

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00:10:15 --> 00:10:18
velocity fields,
well, you would see visually

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00:10:18 --> 00:10:23
what is happening to your fluid.
Like here maybe you have a

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00:10:23 --> 00:10:28
source at the origin that is
pouring fluid out and it is

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00:10:28 --> 00:10:32
flowing all the way away from
that.

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00:10:32 --> 00:10:45
Let's do just a last one.
Let's say I give you minus y, x.

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00:10:45 --> 00:10:55
What does that look like?
That is an interesting one,

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00:10:55 --> 00:11:01
actually.
Let's say that I have a point

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00:11:01 --> 00:11:06
(x, y) here.
This vector here is <x,

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00:11:06 --> 00:11:10
y>.
But the vector I want is <-

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00:11:10 --> 00:11:14
y, x>.
What does that look like?

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00:11:14 --> 00:11:18
It is perpendicular to the
position to this vector.

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00:11:18 --> 00:11:24
If I rotate this vector,
let me maybe draw a picture on

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00:11:24 --> 00:11:28
the side, and take vector x,
y.

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00:11:28 --> 00:11:36
A vector with components
negative y and x is going to be

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00:11:36 --> 00:11:43
like this.
It is the vector that I get by

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00:11:43 --> 00:11:50
rotating by 90 degrees
counterclockwise.

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00:11:50 --> 00:11:51
And, of course,
I do not want to put that

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00:11:51 --> 00:11:54
vector at the origin.
I want to put it at the point

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00:11:54 --> 00:11:57
x, y.
In fact, what I will draw is

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00:11:57 --> 00:12:06
something like this.
And similarly here like that,

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00:12:06 --> 00:12:13
like that, etc.
And if I am closer to the

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origin then it looks a bit the
same, but it is shorter.

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00:12:19 --> 00:12:23
And at the origin it is zero.
And when I am further away it

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00:12:23 --> 00:12:27
becomes even larger.
See, this vector field, 

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00:12:27 --> 00:12:29
if it was the motion of a
fluid,

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00:12:29 --> 00:12:36
it would correspond to a fluid
that is just going around the

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origin in circles rotating at
uniform speed.

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00:12:41 --> 00:12:54
This is actually the velocity
field for uniform rotation.

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00:12:54 --> 00:13:01
And, if you figure out how long
it takes for a particle of fluid

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00:13:01 --> 00:13:06
to go all the way around,
that would be actually 2(pi)

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00:13:06 --> 00:13:10
because the length of a circle
is 2(pi) times the radius.

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00:13:10 --> 00:13:16
That is actually at unit
angular velocity,

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00:13:16 --> 00:13:21
one radiant per second or per
unit time.

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00:13:21 --> 00:13:29
That is why this guy comes up
quite a lot in real life.

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00:13:29 --> 00:13:33
And you can imagine lots of
variations on these.

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00:13:33 --> 00:13:36
Of course, you can also imagine
vector fields given by much more

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00:13:36 --> 00:13:38
complicated formulas,
and then you would have a hard

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00:13:38 --> 00:13:40
time drawing them.
Maybe you will use a computer

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00:13:40 --> 00:13:44
or maybe you will just give up
and just do whatever calculation

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00:13:44 --> 00:13:47
you have to do without trying to
visualize the vector field.

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00:13:47 --> 00:13:51
But if you have a nice simple
one then it is worth doing it

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00:13:51 --> 00:13:56
because sometimes it will give
you insight about what you are

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00:13:56 --> 00:14:02
going to compute next.
Any questions first about these

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00:14:02 --> 00:14:07
pictures?
No.

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00:14:07 --> 00:14:17
OK.
Oh, yes?

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00:14:17 --> 00:14:22
You are asking if it should be
y, negative x.

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00:14:22 --> 00:14:26
I think it would be the other
way around.

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00:14:26 --> 00:14:30
See, for example,
if I am at this point then y is

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00:14:30 --> 00:14:34
positive and x is zero.
If I take y,

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00:14:34 --> 00:14:39
negative x, I get a positive
first component and zero for the

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00:14:39 --> 00:14:42
second one.
So, y, negative x would be a

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00:14:42 --> 00:14:46
rotation at unit speed in the
opposite direction.

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00:14:46 --> 00:14:48
And there are a lot of tweaks
you can do to it.

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00:14:48 --> 00:14:54
If you flip the sides you will
get rotation in the other

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00:14:54 --> 00:14:59
direction.
Yes?

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00:14:59 --> 00:15:04
How do know that it is at unit
angular velocity?

192
00:15:04 --> 00:15:09
Well, that is because if my
angular velocity is one then

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00:15:09 --> 00:15:14
that means the actually speed is
equal to the distance from the

194
00:15:14 --> 00:15:17
origin.
Because the arch length on a

195
00:15:17 --> 00:15:23
circle of a certain radius is
equal to the radius times the

196
00:15:23 --> 00:15:25
angle.
If the angle varies at rate one

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00:15:25 --> 00:15:28
then I travel at speed equal to
the radius.

198
00:15:28 --> 00:15:32
That is what I do here.
The length of this vector is

199
00:15:32 --> 00:15:34
equal to the distance of the
origin.

200
00:15:34 --> 00:15:37
I mean, it is not obvious on
the picture.

201
00:15:37 --> 00:15:39
But, really,
the vector that I put here is

202
00:15:39 --> 00:15:43
the same as this vector rotated
so it has the same length.

203
00:15:43 --> 00:15:46
That is why the angular
velocity is one.

204
00:15:46 --> 00:15:56
It doesn't really matter much
anyway.

205
00:15:56 --> 00:15:59
What are we going to do with
vector fields?

206
00:15:59 --> 00:16:08
Well, we are going to do a lot
of things but let's start

207
00:16:08 --> 00:16:13
somewhere.
One thing you might want to do

208
00:16:13 --> 00:16:19
with vector fields is I am going
to think of now the situation

209
00:16:19 --> 00:16:24
where we have a force.
If you have a force exerted on

210
00:16:24 --> 00:16:28
a particle and that particles
moves on some trajectory then

211
00:16:28 --> 00:16:33
probably you have seen in
physics that the work done by

212
00:16:33 --> 00:16:37
the force corresponds to the
force dot product with the

213
00:16:37 --> 00:16:41
displacement vector,
how much you have moved your

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00:16:41 --> 00:16:43
particle.
And, of course,

215
00:16:43 --> 00:16:47
if you do just a straight line
trajectory or if the force is

216
00:16:47 --> 00:16:51
constant that works well.
But if you are moving on a

217
00:16:51 --> 00:16:56
complicated trajectory and the
force keeps changing then,

218
00:16:56 --> 00:17:01
actually, you want to integrate
that over time.

219
00:17:01 --> 00:17:09
The first thing we will do is
learn how to compute the work

220
00:17:09 --> 00:17:16
done by a vector field,
and mathematically that is

221
00:17:16 --> 00:17:23
called a line integral.
Physically, remember the work

222
00:17:23 --> 00:17:28
done by a force is the force
times the distance.

223
00:17:28 --> 00:17:33
And, more precisely,
it is actually the dot product

224
00:17:33 --> 00:17:39
between the force as a vector
and the displacement vector for

225
00:17:39 --> 00:17:44
a small motion.
Say that your point is moving

226
00:17:44 --> 00:17:48
from here to here,
you have the displacement delta

227
00:17:48 --> 00:17:50
r.
It is just the change in the

228
00:17:50 --> 00:17:53
position vector.
It is the vector from the old

229
00:17:53 --> 00:17:57
position to the new position.
And then you have your force

230
00:17:57 --> 00:18:00
that is being exerted.
And you do the dot product

231
00:18:00 --> 00:18:03
between them.
That will give you the work of

232
00:18:03 --> 00:18:08
a force during this motion.
And the physical significance

233
00:18:08 --> 00:18:11
of this, well,
the work tells you basically

234
00:18:11 --> 00:18:15
how much energy you have to
provide to actually perform this

235
00:18:15 --> 00:18:17
motion.
Just in case you haven't seen

236
00:18:17 --> 00:18:20
this in 8.01 yet.
I am hoping all of you have

237
00:18:20 --> 00:18:25
heard about work somewhere,
but in case it is completely

238
00:18:25 --> 00:18:30
mysterious that is the amount of
energy provided by the force.

239
00:18:30 --> 00:18:33
If a force goes along the
motion, it actually pushes the

240
00:18:33 --> 00:18:36
particle.
It provides an energy to do it

241
00:18:36 --> 00:18:38
to do that motion.
And, conversely,

242
00:18:38 --> 00:18:42
if you are trying to go against
the force then you have to

243
00:18:42 --> 00:18:46
provide energy to the particle
to be able to do that.

244
00:18:46 --> 00:18:49
In particular,
if this is the only force that

245
00:18:49 --> 00:18:54
is taking place then the work
would be the variation in

246
00:18:54 --> 00:18:58
kinetic energy of a particle
along the motion.

247
00:18:58 --> 00:19:00
That is a good description for
a small motion.

248
00:19:00 --> 00:19:04
But let's say that my particle
is not just doing that but it's

249
00:19:04 --> 00:19:09
doing something complicated and
my force keeps changing.

250
00:19:09 --> 00:19:15
Somehow maybe I have a
different force at every point.

251
00:19:15 --> 00:19:19
Then I want to find the total
work done along the motion.

252
00:19:19 --> 00:19:24
Well, what I have to do is cut
my trajectory into these little

253
00:19:24 --> 00:19:27
pieces.
And, for each of them,

254
00:19:27 --> 00:19:30
I have a vector along the
trajectory.

255
00:19:30 --> 00:19:34
I have a force,
I do the dot product and I sum

256
00:19:34 --> 00:19:37
them together.
And, of course,

257
00:19:37 --> 00:19:41
to get the actual answer,
I should actually cut into

258
00:19:41 --> 00:19:44
smaller and smaller pieces and
sum all of the small

259
00:19:44 --> 00:19:49
contributions to work.
So, in fact,

260
00:19:49 --> 00:19:59
it is going to be an integral.
Along some trajectory,

261
00:19:59 --> 00:20:04
let's call C the trajectory for
curve.

262
00:20:04 --> 00:20:26
It is some curve.
The work adds up to an integral.

263
00:20:26 --> 00:20:34
We write this using the
notation integral along C of F

264
00:20:34 --> 00:20:39
dot dr.
We have to decode this

265
00:20:39 --> 00:20:45
notation.
One way to decode this is to

266
00:20:45 --> 00:20:54
say it is a limit as we cut into
smaller and smaller pieces of

267
00:20:54 --> 00:21:03
the sum over each piece of a
trajectory of the force of a

268
00:21:03 --> 00:21:11
given point dot product with
that small vector along the

269
00:21:11 --> 00:21:15
trajectory.
Well, that is not how we will

270
00:21:15 --> 00:21:16
compute it.
To compute it,

271
00:21:16 --> 00:21:20
we do things differently.
How can we actually compute it?

272
00:21:20 --> 00:21:27
Well, what we can do is say
that actually we are cutting

273
00:21:27 --> 00:21:32
things into small time
intervals.

274
00:21:32 --> 00:21:36
The way that we split the
trajectory is we just take a

275
00:21:36 --> 00:21:38
picture every,
say, millisecond.

276
00:21:38 --> 00:21:41
Every millisecond we have a new
position.

277
00:21:41 --> 00:21:46
And the motion,
the amount by which you have

278
00:21:46 --> 00:21:51
moved during each small time
interval is basically the

279
00:21:51 --> 00:21:56
velocity vector times the amount
of time.

280
00:21:56 --> 00:22:01
In fact, let me just rewrite
this.

281
00:22:01 --> 00:22:08
You do the dot product between
the force and how much you have

282
00:22:08 --> 00:22:11
moved,
well, if I just rewrite it this

283
00:22:11 --> 00:22:13
way,
nothing has happened, 

284
00:22:13 --> 00:22:16
but what this thing is,
actually,

285
00:22:16 --> 00:22:28
is the velocity vector dr over
dt.

286
00:22:28 --> 00:22:36
What I am trying to say is that
I can actually compute by

287
00:22:36 --> 00:22:44
integral by integrating F dot
product with dr / dt over time.

288
00:22:44 --> 00:22:51
Whatever the initial time to
whatever the final time is,

289
00:22:51 --> 00:22:56
I integrate F dot product
velocity dt.

290
00:22:56 --> 00:22:59
And, of course,
here this F,

291
00:22:59 --> 00:23:04
I mean F at the point on the
trajectory at time t.

292
00:23:04 --> 00:23:11
This guy depends on x and y
before it depends on t.

293
00:23:11 --> 00:23:17
I see a lot of confused faces,
so let's do an example.

294
00:23:17 --> 00:23:23
Yes?
Yes.

295
00:23:23 --> 00:23:30
Here I need to put a limit as
delta t to zero.

296
00:23:30 --> 00:23:32
I cut my trajectory into
smaller and smaller time

297
00:23:32 --> 00:23:35
intervals.
For each time interval,

298
00:23:35 --> 00:23:39
I have a small motion which is,
essentially,

299
00:23:39 --> 00:23:43
velocity times delta t,
and then I dot that with a

300
00:23:43 --> 00:23:50
force and I sum them.
Let's do an example.

301
00:23:50 --> 00:23:58
Let's say that we want to find
the work of this force.

302
00:23:58 --> 00:24:00
I guess that was the first
example we had.

303
00:24:00 --> 00:24:04
It is a force field that tries
to make everything rotate

304
00:24:04 --> 00:24:08
somehow.
Your first points along these

305
00:24:08 --> 00:24:11
circles.
And let's say that our

306
00:24:11 --> 00:24:16
trajectory, our particle is
moving along the parametric

307
00:24:16 --> 00:24:24
curve.
x = t, y = t^2 for t going from

308
00:24:24 --> 00:24:28
zero to one.
What that looks like -- Well,

309
00:24:28 --> 00:24:31
maybe I should draw you a
picture.

310
00:24:31 --> 00:24:43
Our vector field.
Our trajectory.

311
00:24:43 --> 00:24:48
If you try to plot this,
when you see y is actually x

312
00:24:48 --> 00:24:53
squared, so it a piece of
parabola that goes from the

313
00:24:53 --> 00:24:58
origin to (1,1).
That is what our curve looks

314
00:24:58 --> 00:25:01
like.
We are trying to get the work

315
00:25:01 --> 00:25:04
done by our force along this
trajectory.

316
00:25:04 --> 00:25:07
I should point out;
I mean if you are asking me how

317
00:25:07 --> 00:25:10
did I get this?
That is actually the wrong

318
00:25:10 --> 00:25:13
question.
This is all part of the data.

319
00:25:13 --> 00:25:15
I have a force and I have a
trajectory, and I want to find

320
00:25:15 --> 00:25:17
what the work done is along that
trajectory.

321
00:25:17 --> 00:25:24
These two guys I can choose
completely independently of each

322
00:25:24 --> 00:25:29
other.
The integral along C of F dot

323
00:25:29 --> 00:25:34
dr will be -- Well,
it is the integral from time

324
00:25:34 --> 00:25:41
zero to time one of F dot the
velocity vector dr over dt times

325
00:25:41 --> 00:25:46
dt.
That would be the integral from

326
00:25:46 --> 00:25:50
zero to one.
Let's try to figure it out.

327
00:25:50 --> 00:25:58
What is F?
F, at a point (x,

328
00:25:58 --> 00:26:01
y), is <- y,
x>.

329
00:26:01 --> 00:26:06
But if I take the point where I
am at time t then x is t and y

330
00:26:06 --> 00:26:09
is t squared.
Here I plug x equals t,

331
00:26:09 --> 00:26:15
y equals t squared,
and that will give me negative

332
00:26:15 --> 00:26:19
t squared, t.
Here I will put negative t

333
00:26:19 --> 00:26:24
squared, t dot product.
What is the velocity vector?

334
00:26:24 --> 00:26:33
Well, dx over dt is just one,
dy over dt is 2t.

335
00:26:33 --> 00:26:43
So, the velocity vector is 1,2t
dt.

336
00:26:43 --> 00:26:47
Now we have to continue the
calculation.

337
00:26:47 --> 00:26:51
We get integral from zero to
one of, what is this dot

338
00:26:51 --> 00:26:54
product?
Well, it is negative t squared

339
00:26:54 --> 00:26:57
plus 2t squared.
I get t squared.

340
00:26:57 --> 00:27:01
Well, maybe I will write it.
Negative t squared plus 2t

341
00:27:01 --> 00:27:06
squared dt.
That ends up being integral

342
00:27:06 --> 00:27:12
from zero to one of t squared
dt, which you all know how to

343
00:27:12 --> 00:27:18
integrate and get one-third.
That is the work done by the

344
00:27:18 --> 00:27:23
force along this curve.
Yes?

345
00:27:23 --> 00:27:29
Well, I got it by just taking
the dot product between the

346
00:27:29 --> 00:27:36
force and the velocity.
That is in case you are

347
00:27:36 --> 00:27:47
wondering, things go like this.
Any questions on how we did

348
00:27:47 --> 00:27:51
this calculation?
No.

349
00:27:51 --> 00:28:01
Yes?
Why can't you just do F dot dr?

350
00:28:01 --> 00:28:05
Well, soon we will be able to.
We don't know yet what dr means

351
00:28:05 --> 00:28:10
or how to use it as a symbol
because we haven't said yet,

352
00:28:10 --> 00:28:14
I mean, see,
this is a d vector r.

353
00:28:14 --> 00:28:17
That is kind of strange thing
to have.

354
00:28:17 --> 00:28:21
And certainly r is not a usual
variable.

355
00:28:21 --> 00:28:24
We have to be careful about
what are the rules,

356
00:28:24 --> 00:28:28
what does this symbol mean?
We are going to see that right

357
00:28:28 --> 00:28:30
now.
And then we can do it,

358
00:28:30 --> 00:28:33
actually, in a slightly more
efficient way.

359
00:28:33 --> 00:28:35
I mean r is not a scalar
quantity.

360
00:28:35 --> 00:28:39
R is a position vector.
You cannot integrate F with

361
00:28:39 --> 00:28:47
respect to r.
We don't know how to do that.

362
00:28:47 --> 00:28:52
OK.
Yes?

363
00:28:52 --> 00:29:10
 
 

364
00:29:10 --> 00:29:12
The question is if I took a
different trajectory from the

365
00:29:12 --> 00:29:15
origin to that point (1,1),
what will happen?

366
00:29:15 --> 00:29:19
Well, the answer is I would get
something different.

367
00:29:19 --> 00:29:21
For example,
let me try to convince you of

368
00:29:21 --> 00:29:23
that.
For example,

369
00:29:23 --> 00:29:31
say I chose to instead go like
this and then around like that,

370
00:29:31 --> 00:29:35
first I wouldn't do any work
because here the force is

371
00:29:35 --> 00:29:38
perpendicular to my motion.
And then I would be going

372
00:29:38 --> 00:29:40
against the force all the way
around.

373
00:29:40 --> 00:29:42
I should get something that is
negative.

374
00:29:42 --> 00:29:46
Even if you don't see that,
just accept it at face value

375
00:29:46 --> 00:29:50
that I say now.
The value of a line integral,

376
00:29:50 --> 00:29:54
in general, depends on how we
got from point a to point b.

377
00:29:54 --> 00:29:57
That is why we have to compute
it by using the parametric

378
00:29:57 --> 00:30:00
equation for the curve.
It really depends on what curve

379
00:30:00 --> 00:30:01
you choose.

380
00:30:01 --> 00:30:14

381
00:30:14 --> 00:30:22
Any other questions.
Yes?

382
00:30:22 --> 00:30:25
What happens when the force
inflects the trajectory?

383
00:30:25 --> 00:30:28
Well, then, actually,
you would have to solve a

384
00:30:28 --> 00:30:31
differential equation telling
you how a particle moves to find

385
00:30:31 --> 00:30:35
what the trajectory is.
That is something that would be

386
00:30:35 --> 00:30:39
a very useful topic.
And that is probably more like

387
00:30:39 --> 00:30:42
what you will do in 18.03,
or maybe you actually know how

388
00:30:42 --> 00:30:45
to do it in this case.
What we are trying to develop

389
00:30:45 --> 00:30:49
here is a method to figure out
if we know what the trajectory

390
00:30:49 --> 00:30:52
is what the work will be.
It doesn't tell us what the

391
00:30:52 --> 00:30:55
trajectory will be.
But, of course,

392
00:30:55 --> 00:30:57
we could also find that.
But here, see,

393
00:30:57 --> 00:30:59
I am not assuming,
for example,

394
00:30:59 --> 00:31:02
that the particle is moving
just based on that force.

395
00:31:02 --> 00:31:04
Maybe, actually,
I am here to hold it in my hand

396
00:31:04 --> 00:31:06
and force it to go where it is
going,

397
00:31:06 --> 00:31:10
or maybe there is some rail
that is taking it in that

398
00:31:10 --> 00:31:14
trajectory or whatever.
I can really do it along any

399
00:31:14 --> 00:31:16
trajectory.
And, if I wanted to,

400
00:31:16 --> 00:31:18
if I knew that was the case,
I could try to find the

401
00:31:18 --> 00:31:20
trajectory based on what the
force is.

402
00:31:20 --> 00:31:36
But that is not what we are
doing here.

403
00:31:36 --> 00:31:41
Let's try to make sense of what
you asked just a few minutes

404
00:31:41 --> 00:31:45
ago, what can we do directly
with dr?

405
00:31:45 --> 00:31:50
dr becomes somehow a vector.
I mean, when I replace it by dr

406
00:31:50 --> 00:31:55
over dt times dt,
it becomes something that is a

407
00:31:55 --> 00:32:01
vector with a dt next to it.
In fact -- Well,

408
00:32:01 --> 00:32:07
it is not really new.
Let's see.

409
00:32:07 --> 00:32:15
Another way to do it,
let's say that our force has

410
00:32:15 --> 00:32:21
components M and N.
I claim that we can write

411
00:32:21 --> 00:32:27
symbolically vector dr stands
for its vector whose components

412
00:32:27 --> 00:32:31
are dx, dy.
It is a strange kind of vector.

413
00:32:31 --> 00:32:38
I mean it is not a real vector,
of course, but as a notion,

414
00:32:38 --> 00:32:45
it is a pretty good notation
because it tells us that F of dr

415
00:32:45 --> 00:32:51
is M dx plus N dy.
In fact, we will very often

416
00:32:51 --> 00:32:58
write, instead of F dot dr line
integral along c will be line

417
00:32:58 --> 00:33:03
integral along c of M dx plus N
dy.

418
00:33:03 --> 00:33:06
And so, in this language,
of course, what we are

419
00:33:06 --> 00:33:08
integrating now,
rather than a vector field,

420
00:33:08 --> 00:33:11
becomes a differential.
But you should think of it,

421
00:33:11 --> 00:33:12
too, as being pretty much the
same thing.

422
00:33:12 --> 00:33:17
It is like when you compare the
gradient of a function and its

423
00:33:17 --> 00:33:20
differential,
they are different notations

424
00:33:20 --> 00:33:24
but have the same content.
Now, there still remains the

425
00:33:24 --> 00:33:27
question of how do we compute
this kind of integral?

426
00:33:27 --> 00:33:30
Because it is more subtle than
the notation suggests.

427
00:33:30 --> 00:33:35
Because M and N both depend on
x and y.

428
00:33:35 --> 00:33:38
And, if you just integrate it
with respect to x,

429
00:33:38 --> 00:33:40
you would be left with y's in
there.

430
00:33:40 --> 00:33:41
And you don't want to be left
with y's.

431
00:33:41 --> 00:33:46
You want a number at the end.
See, the catch is along the

432
00:33:46 --> 00:33:51
curve x and y are actually
related to each other.

433
00:33:51 --> 00:33:54
Whenever we write this,
we have two variables x and y,

434
00:33:54 --> 00:33:58
but, in fact,
along the curve C we have only

435
00:33:58 --> 00:34:00
one parameter.
It could be x.

436
00:34:00 --> 00:34:02
It could be y.
It could be time.

437
00:34:02 --> 00:34:04
Whatever you want.
But we have to express

438
00:34:04 --> 00:34:06
everything in terms of that one
parameter.

439
00:34:06 --> 00:34:11
And then we get a usual single
variable integral.

440
00:34:11 --> 00:34:17
How do we evaluate things in
this language?

441
00:34:17 --> 00:34:21
Well, we do it by substituting
the parameter into everything.

442
00:34:21 --> 00:34:50
 
 

443
00:34:50 --> 00:35:06
The method to evaluate is to
express x and y in terms of a

444
00:35:06 --> 00:35:17
single variable.
And then substitute that

445
00:35:17 --> 00:35:20
variable.
Let's, for example,

446
00:35:20 --> 00:35:25
redo the one we had up there
just using these new notations.

447
00:35:25 --> 00:35:31
You will see that it is the
same calculation but with

448
00:35:31 --> 00:35:40
different notations.
In that example that we had,

449
00:35:40 --> 00:35:52
our vector field F was negative
.

450
00:35:52 --> 00:35:55
What we are integrating is
negative y dx plus x dy.

451
00:35:55 --> 00:35:58
And, see, if we have just this,
we don't know how to integrate

452
00:35:58 --> 00:35:59
that.
I mean, well,

453
00:35:59 --> 00:36:01
you could try to come up with
negative x, y or something like

454
00:36:01 --> 00:36:03
that.
But that actually doesn't make

455
00:36:03 --> 00:36:08
sense.
It doesn't work.

456
00:36:08 --> 00:36:13
What we will do is we will
actually have to express

457
00:36:13 --> 00:36:18
everything in terms of a same
variable,

458
00:36:18 --> 00:36:21
because it is a single integral
and we should only have on

459
00:36:21 --> 00:36:25
variable.
And what that variable will be,

460
00:36:25 --> 00:36:30
well, if we just do it the same
way that would just be t.

461
00:36:30 --> 00:36:34
How do we express everything in
terms of t?

462
00:36:34 --> 00:36:37
Well, we use the parametric
equation.

463
00:36:37 --> 00:36:45
We know that x is t and y is t
squared.

464
00:36:45 --> 00:36:46
We know what to do with these
two guys.

465
00:36:46 --> 00:36:49
What about dx and dy?
Well, it is easy.

466
00:36:49 --> 00:37:01
We just differentiate.
dx becomes dt, dy becomes 2t dt.

467
00:37:01 --> 00:37:03
I am just saying,
in a different language,

468
00:37:03 --> 00:37:05
what I said over here with dx
over dt equals one,

469
00:37:05 --> 00:37:09
dy over dt equals 2t.
It is the same thing but

470
00:37:09 --> 00:37:17
written slightly differently.
Now, I am going to do it again.

471
00:37:17 --> 00:37:20
I am going to switch from one
board to the next one.

472
00:37:20 --> 00:37:30
My integral becomes the
integral over C of negative y is

473
00:37:30 --> 00:37:39
minus t squared dt plus x is t
times dy is 2t dt.

474
00:37:39 --> 00:37:44
And now that I have only t
left, it is fine to say I have a

475
00:37:44 --> 00:37:49
usual single variable integral
over a variable t that goes from

476
00:37:49 --> 00:37:51
zero to one.
Now I can say,

477
00:37:51 --> 00:37:55
yes, this is the integral from
zero to one of that stuff.

478
00:37:55 --> 00:38:00
I can simply it a bit and it
becomes t squared dt,

479
00:38:00 --> 00:38:04
and I can compute it,
equals one-third.

480
00:38:04 --> 00:38:08
I have negative t squared and
then I have plus 2t squared,

481
00:38:08 --> 00:38:11
so end up with positive t
squared.

482
00:38:11 --> 00:38:20
It is the same as up there.
Any questions?

483
00:38:20 --> 00:38:32
Yes?
dy is the differential of y,

484
00:38:32 --> 00:38:38
y is t squared,
so I get 2t dt.

485
00:38:38 --> 00:38:46
I plug dt for dx,
I plug 2t dt for dy and so on.

486
00:38:46 --> 00:38:51
And that is the general method.
If you are given a curve then

487
00:38:51 --> 00:38:56
you first have to figure out how
do you express x and y in terms

488
00:38:56 --> 00:38:58
of the same thing?
And you get to choose,

489
00:38:58 --> 00:39:00
in general, what parameter we
use.

490
00:39:00 --> 00:39:14
You choose to parameterize your
curve in whatever way you want.

491
00:39:14 --> 00:39:28
The note that I want to make is
that this line integral depends

492
00:39:28 --> 00:39:40
on the trajectory C but not on
the parameterization.

493
00:39:40 --> 00:39:43
You can choose whichever
variable you want.

494
00:39:43 --> 00:39:49
For example, 
what you could do is when you

495
00:39:49 --> 00:39:52
know that you have that
trajectory,

496
00:39:52 --> 00:39:56
you could also choose to
parameterize it as x equals,

497
00:39:56 --> 00:40:03
I don't know, sine theta, 
y equals sine square theta, 

498
00:40:03 --> 00:40:09
because y is x squared where
theta goes from zero to pi over

499
00:40:09 --> 00:40:11
two.
And then you could get dx and

500
00:40:11 --> 00:40:15
dy in terms of d theta.
And you would be able to do it

501
00:40:15 --> 00:40:19
with a lot of trig and you would
get the same answer.

502
00:40:19 --> 00:40:22
That would be a harder way to
get the same thing.

503
00:40:22 --> 00:40:27
What you should do in practice
is use the most reasonable way

504
00:40:27 --> 00:40:31
to parameterize your curve.
If you know that you have a

505
00:40:31 --> 00:40:35
piece of parabola y equals x
squared, there is no way you

506
00:40:35 --> 00:40:38
would put sine and sine squared.
You could set x equals,

507
00:40:38 --> 00:40:40
y equals t squared,
which is very reasonable.

508
00:40:40 --> 00:40:43
You could even take a small
shortcut and say that your

509
00:40:43 --> 00:40:47
variable will be just x.
That means x you just keep as

510
00:40:47 --> 00:40:50
it is.
And then, when you have y,

511
00:40:50 --> 00:40:53
you set y equals x squared,
dy equals 2x dx,

512
00:40:53 --> 00:40:56
and then you will have an
integral over x.

513
00:40:56 --> 00:41:09
That works.
So, this one is not practical.

514
00:41:09 --> 00:41:11
But you get to choose.

515
00:41:11 --> 00:41:45

516
00:41:45 --> 00:41:52
Now let me tell you a bit more
about the geometry.

517
00:41:52 --> 00:41:55
We have said here is how we
compute it in general,

518
00:41:55 --> 00:41:59
and that is the general method
for computing a line integral

519
00:41:59 --> 00:42:00
for work.
You can always do this, 

520
00:42:00 --> 00:42:04
try to find a parameter, 
the simplest one, 

521
00:42:04 --> 00:42:07
express everything in terms of
its variable and then you have

522
00:42:07 --> 00:42:11
an integral to compute.
But sometimes you can actually

523
00:42:11 --> 00:42:14
save a lot of work by just
thinking geometrically about

524
00:42:14 --> 00:42:21
what this all does.
Let me tell you about the

525
00:42:21 --> 00:42:29
geometric approach.
One thing I want to remind you

526
00:42:29 --> 00:42:34
of first is what is this vector
dr?

527
00:42:34 --> 00:42:41
Well, what is vector delta r?
If I take a very small piece of

528
00:42:41 --> 00:42:47
the trajectory then my vector
delta r will be tangent to the

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00:42:47 --> 00:42:50
trajectory.
It will be going in the same

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00:42:50 --> 00:42:53
direction as the unit tangent
vector t.

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00:42:53 --> 00:42:58
And what is its length?
Well, its length is the arc

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00:42:58 --> 00:43:02
length along the trajectory,
which we called delta s.

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00:43:02 --> 00:43:09
Remember, s was the distance
along the trajectory.

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00:43:09 --> 00:43:21
We can write vector dr equals
dx, dy, but that is also T times

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00:43:21 --> 00:43:24
ds.
It is a vector whose direction

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00:43:24 --> 00:43:29
is tangent to the curve and
whose length element is actually

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00:43:29 --> 00:43:35
the arc length element.
I mean, if you don't like this

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00:43:35 --> 00:43:41
notation, think about dividing
everything by dt.

539
00:43:41 --> 00:43:45
Then what we are saying is dr
over dt, which is the velocity

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00:43:45 --> 00:43:47
vector.
Well, in coordinates,

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00:43:47 --> 00:43:51
the velocity vector is dx over
dt, dy over dt.

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00:43:51 --> 00:43:57
But, more geometrically,
the direction of a velocity

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00:43:57 --> 00:44:03
vector is tangent to the
trajectory and its magnitude is

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00:44:03 --> 00:44:10
speed ds over dt.
So, that is really the same

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00:44:10 --> 00:44:14
thing.
If I say this,

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00:44:14 --> 00:44:20
that means that my line
integral F to dr,

547
00:44:20 --> 00:44:28
well, I say I can write it as
integral of M dx plus N dy.

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00:44:28 --> 00:44:32
That is what I will do if I
want to compute it by computing

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00:44:32 --> 00:44:36
the integral.
But, if instead I want to think

550
00:44:36 --> 00:44:42
about it geometrically,
I could rewrite it as F dot T

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00:44:42 --> 00:44:45
ds.
Now you can think of this,

552
00:44:45 --> 00:44:50
F dot T is a scalar quantity.
It is the tangent component of

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00:44:50 --> 00:44:53
my force.
I take my force and project it

554
00:44:53 --> 00:44:58
to the tangent direction to a
trajectory and the I integrate

555
00:44:58 --> 00:45:04
that along the curve.
They are the same thing.

556
00:45:04 --> 00:45:07
And sometimes it is easier to
do it this way.

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00:45:07 --> 00:45:15
Here is an example.
This is bound to be easier only

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00:45:15 --> 00:45:18
when the field and the curve are
relatively simple and have a

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00:45:18 --> 00:45:20
geometric relation to each
other.

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00:45:20 --> 00:45:24
If I give you an evil formula
with x cubed plus y to the fifth

561
00:45:24 --> 00:45:28
or whatever there is very little
chance that you will be able to

562
00:45:28 --> 00:45:32
simplify it that way.
But let's say that my

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00:45:32 --> 00:45:42
trajectory is just a circle of
radius a centered at the origin.

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00:45:42 --> 00:45:52
Let's say I am doing that
counterclockwise and let's say

565
00:45:52 --> 00:46:00
that my vector field is xi yj.
What does that look like?

566
00:46:00 --> 00:46:07
Well, my trajectory is just
this circle.

567
00:46:07 --> 00:46:11
My vector field,
remember, xi plus yj,

568
00:46:11 --> 00:46:15
that is the one that is
pointing radially from the

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00:46:15 --> 00:46:18
origin.
Hopefully, if you have good

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00:46:18 --> 00:46:20
physics intuition here,
you will already know what the

571
00:46:20 --> 00:46:26
work is going to be.
It is going to be zero because

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00:46:26 --> 00:46:34
the force is perpendicular to
the motion.

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00:46:34 --> 00:46:41
Now we can say it directly by
saying if you have any point of

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00:46:41 --> 00:46:49
a circle then the tangent vector
to the circle will be,

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00:46:49 --> 00:46:51
well, it's tangent to the
circle,

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00:46:51 --> 00:46:54
so that means it is
perpendicular to the radial

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00:46:54 --> 00:47:00
direction,
while the force is pointing in

578
00:47:00 --> 00:47:09
the radial direction so you have
a right angle between them.

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00:47:09 --> 00:47:16
F is perpendicular to T.
F dot T is zero.

580
00:47:16 --> 00:47:22
The line integral of F dot T ds
is just zero.

581
00:47:22 --> 00:47:26
That is much easier than
writing this is integral of x

582
00:47:26 --> 00:47:29
over dx plus y over dy.
What do we do?

583
00:47:29 --> 00:47:33
Well, we set x equals a cosine
theta, y equals a sine theta.

584
00:47:33 --> 00:47:37
We get a bunch of trig things.
It cancels out to zero.

585
00:47:37 --> 00:47:42
It is not much harder but we
saved time by not even thinking

586
00:47:42 --> 00:47:45
about how to parameterize
things.

587
00:47:45 --> 00:47:50
Let's just do a last one.
That was the first one.

588
00:47:50 --> 00:47:55
Let's say now that I take the
same curve C,

589
00:47:55 --> 00:48:03
but now my vector field is the
one that rotates negative yi

590
00:48:03 --> 00:48:11
plus xj.
That means along my circle the

591
00:48:11 --> 00:48:23
tangent vector goes like this
and my vector field is also

592
00:48:23 --> 00:48:28
going around.
So, in fact,

593
00:48:28 --> 00:48:34
at this point the vector field
will always be going in the same

594
00:48:34 --> 00:48:41
direction.
Now F is actually parallel to

595
00:48:41 --> 00:48:47
the tangent direction.
That means that the dot product

596
00:48:47 --> 00:48:52
of F dot T, remember,
if it is the component of F in

597
00:48:52 --> 00:48:57
this direction that will be the
same of the length of F.

598
00:48:57 --> 00:49:02
But what is the length of F on
this circle if this length is a?

599
00:49:02 --> 00:49:05
It is just going to be a.
That is what we said earlier

600
00:49:05 --> 00:49:08
about this vector field.
At every point,

601
00:49:08 --> 00:49:11
this dot product is a.
Now we know how to integrate

602
00:49:11 --> 00:49:13
that quite quickly.

603
00:49:13 --> 00:49:25

604
00:49:25 --> 00:49:30
Because it becomes the integral
of a ds, but a is a constant so

605
00:49:30 --> 00:49:34
we can take it out.
And now what do we get when we

606
00:49:34 --> 00:49:38
integrate ds along C?
Well, we should get the total

607
00:49:38 --> 00:49:43
length of the curve if we sum
all the little pieces of arc

608
00:49:43 --> 00:49:47
length.
But we know that the length of

609
00:49:47 --> 00:49:52
a circle of radius a is 2pi a,
so we get 2(pi)a squared.

610
00:49:52 --> 00:50:01
If we were to compute that by
hand, well, what would we do?

611
00:50:01 --> 00:50:08
We would be computing integral
of minus y dx plus x dy.

612
00:50:08 --> 00:50:13
Since we are on a circle,
we will probably set x equals a

613
00:50:13 --> 00:50:16
times cosine theta,
y equals a times sine theta for

614
00:50:16 --> 00:50:23
theta between zero and 2pi.
Then we would get dx and dy out

615
00:50:23 --> 00:50:27
of these.
So, y is a sine theta,

616
00:50:27 --> 00:50:33
dx is negative a sine theta d
theta, if you differentiate a

617
00:50:33 --> 00:50:40
cosine, plus a cosine theta
times a cosine theta d theta.

618
00:50:40 --> 00:50:45
Well, you will just end up with
integral from zero to 2pi of a

619
00:50:45 --> 00:50:51
squared time sine squared theta
plus cosine square theta times d

620
00:50:51 --> 00:50:53
theta.
That becomes just one.

621
00:50:53 --> 00:50:56
And you get the same answer.
It took about the same amount

622
00:50:56 --> 00:50:59
of time because I did this one
rushing very quickly,

623
00:50:59 --> 00:51:03
but normally it takes about at
least twice the amount of time

624
00:51:03 --> 00:51:06
to do it with a calculation.
That tells you sometimes it is

625
00:51:06 --> 00:51:08
worth thinking geometrically.
 

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00:51:08 --> 00:51:13