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OK.
Today we have a new topic,
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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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Put it away in a corner of your
mind.
00:00:45.000 --> 00:00:48.000
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
00:00:51.000 --> 00:00:52.000
different things,
so it helps,
00:00:52.000 --> 00:00:56.000
actually, if you don't think of
double integrals at all while
00:00:56.000 --> 00:00:59.000
doing line integrals.
Anyway, let's start with vector
00:00:59.000 --> 00:01:02.000
fields.
What is a vector field?
00:01:02.000 --> 00:01:09.000
Well, a vector field is
something that is of a form,
00:01:09.000 --> 00:01:14.000
while it is a vector,
but while M and N,
00:01:14.000 --> 00:01:21.000
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
00:01:30.000 --> 00:01:32.000
y.
What that means,
00:01:32.000 --> 00:01:36.000
concretely, is that every point
in the plane you have a vector.
00:01:36.000 --> 00:01:39.000
In a corn field,
every where you have corn.
00:01:39.000 --> 00:01:43.000
In a vector field,
everywhere you have a vector.
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That is how it works.
A good example of a vector
00:01:49.000 --> 00:01:55.000
field, I don't know if you have
seen these maps that show the
00:01:55.000 --> 00:01:59.000
wind, but here are some cool
images done by NASA.
00:01:59.000 --> 00:02:04.000
Actually, that is a picture of
wind patterns off the coast of
00:02:04.000 --> 00:02:09.000
California with Santa Ana winds,
in case you are wondering what
00:02:09.000 --> 00:02:13.000
has been going on recently.
You have all of these vectors
00:02:13.000 --> 00:02:17.000
that show you the velocity of
the air basically at every
00:02:17.000 --> 00:02:18.000
point.
I mean, of course you don't
00:02:18.000 --> 00:02:21.000
draw it every point,
because if you drew a vector at
00:02:21.000 --> 00:02:24.000
absolutely all the points of a
plane then you would just fill
00:02:24.000 --> 00:02:27.000
up everything and you wouldn't
see anything.
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So, choose points and draw the
vectors at those points.
00:02:33.000 --> 00:02:38.000
Here is another cool image,
which is upside down.
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That is a hurricane off the
coast of Mexico with the winds
00:02:44.000 --> 00:02:49.000
spiraling around the hurricane.
Anyway, it is kind of hard to
00:02:49.000 --> 00:02:52.000
see.
You don't really see all the
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vectors, actually,
because the autofocus is having
00:02:57.000 --> 00:03:03.000
trouble with it.
It cannot really do it,
00:03:03.000 --> 00:03:11.000
so I guess I will go back to
the previous one.
00:03:11.000 --> 00:03:30.000
Anyway, a vector field is
something where at each point --
00:03:30.000 --> 00:03:50.000
-- in the plane we have vector F
that depends on x and y.
00:03:50.000 --> 00:03:54.000
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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That is what these pictures
show.
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At every point you have a
velocity of a fluid that is
00:04:01.000 --> 00:04:04.000
moving.
Another example is force fields.
00:04:04.000 --> 00:04:08.000
Now, force fields are not
something out of Star Wars.
00:04:08.000 --> 00:04:10.000
If you look at gravitational
attraction,
00:04:10.000 --> 00:04:13.000
you know that if you have a
mass somewhere,
00:04:13.000 --> 00:04:16.000
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.
00:04:21.000 --> 00:04:24.000
And, the same thing in space,
you have the gravitational
00:04:24.000 --> 00:04:27.000
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
00:04:54.000 --> 00:05:01.000
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
00:05:05.000 --> 00:05:08.000
are most of the time,
but, as we will explore with
00:05:08.000 --> 00:05:10.000
them defined quantities and so
on,
00:05:10.000 --> 00:05:15.000
we will very often use these
motivations to justify why we
00:05:15.000 --> 00:05:18.000
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
00:05:28.000 --> 00:05:33.000
plot like that?
Let's practice drawing a few
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vector fields.
Well, let's say our very first
00:05:38.000 --> 00:05:43.000
vector field will be just 2i j.
It is kind of a silly vector
00:05:43.000 --> 00:05:45.000
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.
00:06:01.000 --> 00:06:09.000
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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And when you think you have
enough so that you understand
00:06:16.000 --> 00:06:19.000
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.
00:06:37.000 --> 00:06:42.000
Let's say I give you a vector
field x times i hat.
00:06:42.000 --> 00:06:55.000
There is no j component.
How would you draw that?
00:06:55.000 --> 00:07:00.000
Well, first of all,
we know that this guy is only
00:07:00.000 --> 00:07:05.000
in the i direction so it is
always horizontal.
00:07:05.000 --> 00:07:08.000
It doesn't have a j component.
Everywhere it would be a
00:07:08.000 --> 00:07:10.000
horizontal vector.
Now, the question is how long
00:07:10.000 --> 00:07:15.000
is it?
Well, how long it is depends on
00:07:15.000 --> 00:07:16.000
x.
For example,
00:07:16.000 --> 00:07:20.000
if x is zero then this will
actually be the zero vector.
00:07:20.000 --> 00:07:28.000
x is zero here on the y-axis.
I will take a different color.
00:07:28.000 --> 00:07:33.000
If I am on the y-axis,
I actually have the zero
00:07:33.000 --> 00:07:36.000
vector.
Now, if x becomes positive
00:07:36.000 --> 00:07:41.000
small then I will have actually
a small positive multiple of i
00:07:41.000 --> 00:07:45.000
so I will be going a little bit
to the right.
00:07:45.000 --> 00:07:51.000
And then, if I increase x,
this guy becomes larger so I
00:07:51.000 --> 00:07:54.000
get a longer vector to the
right.
00:07:54.000 --> 00:08:03.000
If x is negative then my vector
field points to the left
00:08:03.000 --> 00:08:10.000
instead.
It looks something like that.
00:08:10.000 --> 00:08:16.000
Any questions about that
picture?
00:08:16.000 --> 00:08:17.000
No.
OK.
00:08:17.000 --> 00:08:20.000
Usually, we are not going to
try to have very accurate,
00:08:20.000 --> 00:08:23.000
you know, we won't actually
take time to plot a vector field
00:08:23.000 --> 00:08:26.000
very carefully.
I mean, if we need to,
00:08:26.000 --> 00:08:30.000
computers can do it for us.
It is useful to have an idea of
00:08:30.000 --> 00:08:32.000
what a vector field does
roughly.
00:08:32.000 --> 00:08:36.000
Whether it is getting larger
and larger, in what direction it
00:08:36.000 --> 00:08:39.000
is pointing, what are the
general features?
00:08:39.000 --> 00:08:45.000
Just to do a couple of more,
actually, you will see very
00:08:45.000 --> 00:08:50.000
quickly that the examples I use
in lecture are pretty much
00:08:50.000 --> 00:08:55.000
always the same ones.
We will be playing a lot with
00:08:55.000 --> 00:08:59.000
these particular vector fields
just because they are good
00:08:59.000 --> 00:09:04.000
examples.
Let's say I give you xi yj.
00:09:04.000 --> 00:09:08.000
That one has an interesting
geometric significance.
00:09:08.000 --> 00:09:16.000
If I take a point (x,
y), there I want to take a
00:09:16.000 --> 00:09:20.000
vector x, y.
How do I do that?
00:09:20.000 --> 00:09:23.000
Well, it is the same as a
vector from the origin to this
00:09:23.000 --> 00:09:27.000
point.
I take this vector and I copy
00:09:27.000 --> 00:09:30.000
it so that it starts at one
point.
00:09:30.000 --> 00:09:40.000
It looks like that.
And the same thing at every
00:09:40.000 --> 00:09:45.000
point.
It is a vector field that is
00:09:45.000 --> 00:09:53.000
pointing radially away from the
origin, and its magnitude
00:09:53.000 --> 00:09:59.000
increases with distance from the
origin.
00:09:59.000 --> 00:10:03.000
You don't have to draw as many
as me, but the idea is this
00:10:03.000 --> 00:10:08.000
vector field everywhere points
away from the origin.
00:10:08.000 --> 00:10:14.000
And its magnitude is equal to
the distance from the origin.
00:10:14.000 --> 00:10:15.000
If these were,
for example,
00:10:15.000 --> 00:10:18.000
velocity fields,
well, you would see visually
00:10:18.000 --> 00:10:23.000
what is happening to your fluid.
Like here maybe you have a
00:10:23.000 --> 00:10:28.000
source at the origin that is
pouring fluid out and it is
00:10:28.000 --> 00:10:32.000
flowing all the way away from
that.
00:10:32.000 --> 00:10:45.000
Let's do just a last one.
Let's say I give you minus y, x.
00:10:45.000 --> 00:10:55.000
What does that look like?
That is an interesting one,
00:10:55.000 --> 00:11:01.000
actually.
Let's say that I have a point
00:11:01.000 --> 00:11:06.000
(x, y) here.
This vector here is 00:11:10.000
y>.
But the vector I want is <-
00:11:10.000 --> 00:11:14.000
y, x>.
What does that look like?
00:11:14.000 --> 00:11:18.000
It is perpendicular to the
position to this vector.
00:11:18.000 --> 00:11:24.000
If I rotate this vector,
let me maybe draw a picture on
00:11:24.000 --> 00:11:28.000
the side, and take vector x,
y.
00:11:28.000 --> 00:11:36.000
A vector with components
negative y and x is going to be
00:11:36.000 --> 00:11:43.000
like this.
It is the vector that I get by
00:11:43.000 --> 00:11:50.000
rotating by 90 degrees
counterclockwise.
00:11:50.000 --> 00:11:51.000
And, of course,
I do not want to put that
00:11:51.000 --> 00:11:54.000
vector at the origin.
I want to put it at the point
00:11:54.000 --> 00:11:57.000
x, y.
In fact, what I will draw is
00:11:57.000 --> 00:12:06.000
something like this.
And similarly here like that,
00:12:06.000 --> 00:12:13.000
like that, etc.
And if I am closer to the
00:12:13.000 --> 00:12:19.000
origin then it looks a bit the
same, but it is shorter.
00:12:19.000 --> 00:12:23.000
And at the origin it is zero.
And when I am further away it
00:12:23.000 --> 00:12:27.000
becomes even larger.
See, this vector field,
00:12:27.000 --> 00:12:29.000
if it was the motion of a
fluid,
00:12:29.000 --> 00:12:36.000
it would correspond to a fluid
that is just going around the
00:12:36.000 --> 00:12:41.000
origin in circles rotating at
uniform speed.
00:12:41.000 --> 00:12:54.000
This is actually the velocity
field for uniform rotation.
00:12:54.000 --> 00:13:01.000
And, if you figure out how long
it takes for a particle of fluid
00:13:01.000 --> 00:13:06.000
to go all the way around,
that would be actually 2(pi)
00:13:06.000 --> 00:13:10.000
because the length of a circle
is 2(pi) times the radius.
00:13:10.000 --> 00:13:16.000
That is actually at unit
angular velocity,
00:13:16.000 --> 00:13:21.000
one radiant per second or per
unit time.
00:13:21.000 --> 00:13:29.000
That is why this guy comes up
quite a lot in real life.
00:13:29.000 --> 00:13:33.000
And you can imagine lots of
variations on these.
00:13:33.000 --> 00:13:36.000
Of course, you can also imagine
vector fields given by much more
00:13:36.000 --> 00:13:38.000
complicated formulas,
and then you would have a hard
00:13:38.000 --> 00:13:40.000
time drawing them.
Maybe you will use a computer
00:13:40.000 --> 00:13:44.000
or maybe you will just give up
and just do whatever calculation
00:13:44.000 --> 00:13:47.000
you have to do without trying to
visualize the vector field.
00:13:47.000 --> 00:13:51.000
But if you have a nice simple
one then it is worth doing it
00:13:51.000 --> 00:13:56.000
because sometimes it will give
you insight about what you are
00:13:56.000 --> 00:14:02.000
going to compute next.
Any questions first about these
00:14:02.000 --> 00:14:07.000
pictures?
No.
00:14:07.000 --> 00:14:17.000
OK.
Oh, yes?
00:14:17.000 --> 00:14:22.000
You are asking if it should be
y, negative x.
00:14:22.000 --> 00:14:26.000
I think it would be the other
way around.
00:14:26.000 --> 00:14:30.000
See, for example,
if I am at this point then y is
00:14:30.000 --> 00:14:34.000
positive and x is zero.
If I take y,
00:14:34.000 --> 00:14:39.000
negative x, I get a positive
first component and zero for the
00:14:39.000 --> 00:14:42.000
second one.
So, y, negative x would be a
00:14:42.000 --> 00:14:46.000
rotation at unit speed in the
opposite direction.
00:14:46.000 --> 00:14:48.000
And there are a lot of tweaks
you can do to it.
00:14:48.000 --> 00:14:54.000
If you flip the sides you will
get rotation in the other
00:14:54.000 --> 00:14:59.000
direction.
Yes?
00:14:59.000 --> 00:15:04.000
How do know that it is at unit
angular velocity?
00:15:04.000 --> 00:15:09.000
Well, that is because if my
angular velocity is one then
00:15:09.000 --> 00:15:14.000
that means the actually speed is
equal to the distance from the
00:15:14.000 --> 00:15:17.000
origin.
Because the arch length on a
00:15:17.000 --> 00:15:23.000
circle of a certain radius is
equal to the radius times the
00:15:23.000 --> 00:15:25.000
angle.
If the angle varies at rate one
00:15:25.000 --> 00:15:28.000
then I travel at speed equal to
the radius.
00:15:28.000 --> 00:15:32.000
That is what I do here.
The length of this vector is
00:15:32.000 --> 00:15:34.000
equal to the distance of the
origin.
00:15:34.000 --> 00:15:37.000
I mean, it is not obvious on
the picture.
00:15:37.000 --> 00:15:39.000
But, really,
the vector that I put here is
00:15:39.000 --> 00:15:43.000
the same as this vector rotated
so it has the same length.
00:15:43.000 --> 00:15:46.000
That is why the angular
velocity is one.
00:15:46.000 --> 00:15:56.000
It doesn't really matter much
anyway.
00:15:56.000 --> 00:15:59.000
What are we going to do with
vector fields?
00:15:59.000 --> 00:16:08.000
Well, we are going to do a lot
of things but let's start
00:16:08.000 --> 00:16:13.000
somewhere.
One thing you might want to do
00:16:13.000 --> 00:16:19.000
with vector fields is I am going
to think of now the situation
00:16:19.000 --> 00:16:24.000
where we have a force.
If you have a force exerted on
00:16:24.000 --> 00:16:28.000
a particle and that particles
moves on some trajectory then
00:16:28.000 --> 00:16:33.000
probably you have seen in
physics that the work done by
00:16:33.000 --> 00:16:37.000
the force corresponds to the
force dot product with the
00:16:37.000 --> 00:16:41.000
displacement vector,
how much you have moved your
00:16:41.000 --> 00:16:43.000
particle.
And, of course,
00:16:43.000 --> 00:16:47.000
if you do just a straight line
trajectory or if the force is
00:16:47.000 --> 00:16:51.000
constant that works well.
But if you are moving on a
00:16:51.000 --> 00:16:56.000
complicated trajectory and the
force keeps changing then,
00:16:56.000 --> 00:17:01.000
actually, you want to integrate
that over time.
00:17:01.000 --> 00:17:09.000
The first thing we will do is
learn how to compute the work
00:17:09.000 --> 00:17:16.000
done by a vector field,
and mathematically that is
00:17:16.000 --> 00:17:23.000
called a line integral.
Physically, remember the work
00:17:23.000 --> 00:17:28.000
done by a force is the force
times the distance.
00:17:28.000 --> 00:17:33.000
And, more precisely,
it is actually the dot product
00:17:33.000 --> 00:17:39.000
between the force as a vector
and the displacement vector for
00:17:39.000 --> 00:17:44.000
a small motion.
Say that your point is moving
00:17:44.000 --> 00:17:48.000
from here to here,
you have the displacement delta
00:17:48.000 --> 00:17:50.000
r.
It is just the change in the
00:17:50.000 --> 00:17:53.000
position vector.
It is the vector from the old
00:17:53.000 --> 00:17:57.000
position to the new position.
And then you have your force
00:17:57.000 --> 00:18:00.000
that is being exerted.
And you do the dot product
00:18:00.000 --> 00:18:03.000
between them.
That will give you the work of
00:18:03.000 --> 00:18:08.000
a force during this motion.
And the physical significance
00:18:08.000 --> 00:18:11.000
of this, well,
the work tells you basically
00:18:11.000 --> 00:18:15.000
how much energy you have to
provide to actually perform this
00:18:15.000 --> 00:18:17.000
motion.
Just in case you haven't seen
00:18:17.000 --> 00:18:20.000
this in 8.01 yet.
I am hoping all of you have
00:18:20.000 --> 00:18:25.000
heard about work somewhere,
but in case it is completely
00:18:25.000 --> 00:18:30.000
mysterious that is the amount of
energy provided by the force.
00:18:30.000 --> 00:18:33.000
If a force goes along the
motion, it actually pushes the
00:18:33.000 --> 00:18:36.000
particle.
It provides an energy to do it
00:18:36.000 --> 00:18:38.000
to do that motion.
And, conversely,
00:18:38.000 --> 00:18:42.000
if you are trying to go against
the force then you have to
00:18:42.000 --> 00:18:46.000
provide energy to the particle
to be able to do that.
00:18:46.000 --> 00:18:49.000
In particular,
if this is the only force that
00:18:49.000 --> 00:18:54.000
is taking place then the work
would be the variation in
00:18:54.000 --> 00:18:58.000
kinetic energy of a particle
along the motion.
00:18:58.000 --> 00:19:00.000
That is a good description for
a small motion.
00:19:00.000 --> 00:19:04.000
But let's say that my particle
is not just doing that but it's
00:19:04.000 --> 00:19:09.000
doing something complicated and
my force keeps changing.
00:19:09.000 --> 00:19:15.000
Somehow maybe I have a
different force at every point.
00:19:15.000 --> 00:19:19.000
Then I want to find the total
work done along the motion.
00:19:19.000 --> 00:19:24.000
Well, what I have to do is cut
my trajectory into these little
00:19:24.000 --> 00:19:27.000
pieces.
And, for each of them,
00:19:27.000 --> 00:19:30.000
I have a vector along the
trajectory.
00:19:30.000 --> 00:19:34.000
I have a force,
I do the dot product and I sum
00:19:34.000 --> 00:19:37.000
them together.
And, of course,
00:19:37.000 --> 00:19:41.000
to get the actual answer,
I should actually cut into
00:19:41.000 --> 00:19:44.000
smaller and smaller pieces and
sum all of the small
00:19:44.000 --> 00:19:49.000
contributions to work.
So, in fact,
00:19:49.000 --> 00:19:59.000
it is going to be an integral.
Along some trajectory,
00:19:59.000 --> 00:20:04.000
let's call C the trajectory for
curve.
00:20:04.000 --> 00:20:26.000
It is some curve.
The work adds up to an integral.
00:20:26.000 --> 00:20:34.000
We write this using the
notation integral along C of F
00:20:34.000 --> 00:20:39.000
dot dr.
We have to decode this
00:20:39.000 --> 00:20:45.000
notation.
One way to decode this is to
00:20:45.000 --> 00:20:54.000
say it is a limit as we cut into
smaller and smaller pieces of
00:20:54.000 --> 00:21:03.000
the sum over each piece of a
trajectory of the force of a
00:21:03.000 --> 00:21:11.000
given point dot product with
that small vector along the
00:21:11.000 --> 00:21:15.000
trajectory.
Well, that is not how we will
00:21:15.000 --> 00:21:16.000
compute it.
To compute it,
00:21:16.000 --> 00:21:20.000
we do things differently.
How can we actually compute it?
00:21:20.000 --> 00:21:27.000
Well, what we can do is say
that actually we are cutting
00:21:27.000 --> 00:21:32.000
things into small time
intervals.
00:21:32.000 --> 00:21:36.000
The way that we split the
trajectory is we just take a
00:21:36.000 --> 00:21:38.000
picture every,
say, millisecond.
00:21:38.000 --> 00:21:41.000
Every millisecond we have a new
position.
00:21:41.000 --> 00:21:46.000
And the motion,
the amount by which you have
00:21:46.000 --> 00:21:51.000
moved during each small time
interval is basically the
00:21:51.000 --> 00:21:56.000
velocity vector times the amount
of time.
00:21:56.000 --> 00:22:01.000
In fact, let me just rewrite
this.
00:22:01.000 --> 00:22:08.000
You do the dot product between
the force and how much you have
00:22:08.000 --> 00:22:11.000
moved,
well, if I just rewrite it this
00:22:11.000 --> 00:22:13.000
way,
nothing has happened,
00:22:13.000 --> 00:22:16.000
but what this thing is,
actually,
00:22:16.000 --> 00:22:28.000
is the velocity vector dr over
dt.
00:22:28.000 --> 00:22:36.000
What I am trying to say is that
I can actually compute by
00:22:36.000 --> 00:22:44.000
integral by integrating F dot
product with dr / dt over time.
00:22:44.000 --> 00:22:51.000
Whatever the initial time to
whatever the final time is,
00:22:51.000 --> 00:22:56.000
I integrate F dot product
velocity dt.
00:22:56.000 --> 00:22:59.000
And, of course,
here this F,
00:22:59.000 --> 00:23:04.000
I mean F at the point on the
trajectory at time t.
00:23:04.000 --> 00:23:11.000
This guy depends on x and y
before it depends on t.
00:23:11.000 --> 00:23:17.000
I see a lot of confused faces,
so let's do an example.
00:23:17.000 --> 00:23:23.000
Yes?
Yes.
00:23:23.000 --> 00:23:30.000
Here I need to put a limit as
delta t to zero.
00:23:30.000 --> 00:23:32.000
I cut my trajectory into
smaller and smaller time
00:23:32.000 --> 00:23:35.000
intervals.
For each time interval,
00:23:35.000 --> 00:23:39.000
I have a small motion which is,
essentially,
00:23:39.000 --> 00:23:43.000
velocity times delta t,
and then I dot that with a
00:23:43.000 --> 00:23:50.000
force and I sum them.
Let's do an example.
00:23:50.000 --> 00:23:58.000
Let's say that we want to find
the work of this force.
00:23:58.000 --> 00:24:00.000
I guess that was the first
example we had.
00:24:00.000 --> 00:24:04.000
It is a force field that tries
to make everything rotate
00:24:04.000 --> 00:24:08.000
somehow.
Your first points along these
00:24:08.000 --> 00:24:11.000
circles.
And let's say that our
00:24:11.000 --> 00:24:16.000
trajectory, our particle is
moving along the parametric
00:24:16.000 --> 00:24:24.000
curve.
x = t, y = t^2 for t going from
00:24:24.000 --> 00:24:28.000
zero to one.
What that looks like -- Well,
00:24:28.000 --> 00:24:31.000
maybe I should draw you a
picture.
00:24:31.000 --> 00:24:43.000
Our vector field.
Our trajectory.
00:24:43.000 --> 00:24:48.000
If you try to plot this,
when you see y is actually x
00:24:48.000 --> 00:24:53.000
squared, so it a piece of
parabola that goes from the
00:24:53.000 --> 00:24:58.000
origin to (1,1).
That is what our curve looks
00:24:58.000 --> 00:25:01.000
like.
We are trying to get the work
00:25:01.000 --> 00:25:04.000
done by our force along this
trajectory.
00:25:04.000 --> 00:25:07.000
I should point out;
I mean if you are asking me how
00:25:07.000 --> 00:25:10.000
did I get this?
That is actually the wrong
00:25:10.000 --> 00:25:13.000
question.
This is all part of the data.
00:25:13.000 --> 00:25:15.000
I have a force and I have a
trajectory, and I want to find
00:25:15.000 --> 00:25:17.000
what the work done is along that
trajectory.
00:25:17.000 --> 00:25:24.000
These two guys I can choose
completely independently of each
00:25:24.000 --> 00:25:29.000
other.
The integral along C of F dot
00:25:29.000 --> 00:25:34.000
dr will be -- Well,
it is the integral from time
00:25:34.000 --> 00:25:41.000
zero to time one of F dot the
velocity vector dr over dt times
00:25:41.000 --> 00:25:46.000
dt.
That would be the integral from
00:25:46.000 --> 00:25:50.000
zero to one.
Let's try to figure it out.
00:25:50.000 --> 00:25:58.000
What is F?
F, at a point (x,
00:25:58.000 --> 00:26:01.000
y), is <- y,
x>.
00:26:01.000 --> 00:26:06.000
But if I take the point where I
am at time t then x is t and y
00:26:06.000 --> 00:26:09.000
is t squared.
Here I plug x equals t,
00:26:09.000 --> 00:26:15.000
y equals t squared,
and that will give me negative
00:26:15.000 --> 00:26:19.000
t squared, t.
Here I will put negative t
00:26:19.000 --> 00:26:24.000
squared, t dot product.
What is the velocity vector?
00:26:24.000 --> 00:26:33.000
Well, dx over dt is just one,
dy over dt is 2t.
00:26:33.000 --> 00:26:43.000
So, the velocity vector is 1,2t
dt.
00:26:43.000 --> 00:26:47.000
Now we have to continue the
calculation.
00:26:47.000 --> 00:26:51.000
We get integral from zero to
one of, what is this dot
00:26:51.000 --> 00:26:54.000
product?
Well, it is negative t squared
00:26:54.000 --> 00:26:57.000
plus 2t squared.
I get t squared.
00:26:57.000 --> 00:27:01.000
Well, maybe I will write it.
Negative t squared plus 2t
00:27:01.000 --> 00:27:06.000
squared dt.
That ends up being integral
00:27:06.000 --> 00:27:12.000
from zero to one of t squared
dt, which you all know how to
00:27:12.000 --> 00:27:18.000
integrate and get one-third.
That is the work done by the
00:27:18.000 --> 00:27:23.000
force along this curve.
Yes?
00:27:23.000 --> 00:27:29.000
Well, I got it by just taking
the dot product between the
00:27:29.000 --> 00:27:36.000
force and the velocity.
That is in case you are
00:27:36.000 --> 00:27:47.000
wondering, things go like this.
Any questions on how we did
00:27:47.000 --> 00:27:51.000
this calculation?
No.
00:27:51.000 --> 00:28:01.000
Yes?
Why can't you just do F dot dr?
00:28:01.000 --> 00:28:05.000
Well, soon we will be able to.
We don't know yet what dr means
00:28:05.000 --> 00:28:10.000
or how to use it as a symbol
because we haven't said yet,
00:28:10.000 --> 00:28:14.000
I mean, see,
this is a d vector r.
00:28:14.000 --> 00:28:17.000
That is kind of strange thing
to have.
00:28:17.000 --> 00:28:21.000
And certainly r is not a usual
variable.
00:28:21.000 --> 00:28:24.000
We have to be careful about
what are the rules,
00:28:24.000 --> 00:28:28.000
what does this symbol mean?
We are going to see that right
00:28:28.000 --> 00:28:30.000
now.
And then we can do it,
00:28:30.000 --> 00:28:33.000
actually, in a slightly more
efficient way.
00:28:33.000 --> 00:28:35.000
I mean r is not a scalar
quantity.
00:28:35.000 --> 00:28:39.000
R is a position vector.
You cannot integrate F with
00:28:39.000 --> 00:28:47.000
respect to r.
We don't know how to do that.
00:28:47.000 --> 00:28:52.000
OK.
Yes?
00:29:10.000 --> 00:29:12.000
The question is if I took a
different trajectory from the
00:29:12.000 --> 00:29:15.000
origin to that point (1,1),
what will happen?
00:29:15.000 --> 00:29:19.000
Well, the answer is I would get
something different.
00:29:19.000 --> 00:29:21.000
For example,
let me try to convince you of
00:29:21.000 --> 00:29:23.000
that.
For example,
00:29:23.000 --> 00:29:31.000
say I chose to instead go like
this and then around like that,
00:29:31.000 --> 00:29:35.000
first I wouldn't do any work
because here the force is
00:29:35.000 --> 00:29:38.000
perpendicular to my motion.
And then I would be going
00:29:38.000 --> 00:29:40.000
against the force all the way
around.
00:29:40.000 --> 00:29:42.000
I should get something that is
negative.
00:29:42.000 --> 00:29:46.000
Even if you don't see that,
just accept it at face value
00:29:46.000 --> 00:29:50.000
that I say now.
The value of a line integral,
00:29:50.000 --> 00:29:54.000
in general, depends on how we
got from point a to point b.
00:29:54.000 --> 00:29:57.000
That is why we have to compute
it by using the parametric
00:29:57.000 --> 00:30:00.000
equation for the curve.
It really depends on what curve
00:30:00.000 --> 00:30:01.000
you choose.
00:30:14.000 --> 00:30:22.000
Any other questions.
Yes?
00:30:22.000 --> 00:30:25.000
What happens when the force
inflects the trajectory?
00:30:25.000 --> 00:30:28.000
Well, then, actually,
you would have to solve a
00:30:28.000 --> 00:30:31.000
differential equation telling
you how a particle moves to find
00:30:31.000 --> 00:30:35.000
what the trajectory is.
That is something that would be
00:30:35.000 --> 00:30:39.000
a very useful topic.
And that is probably more like
00:30:39.000 --> 00:30:42.000
what you will do in 18.03,
or maybe you actually know how
00:30:42.000 --> 00:30:45.000
to do it in this case.
What we are trying to develop
00:30:45.000 --> 00:30:49.000
here is a method to figure out
if we know what the trajectory
00:30:49.000 --> 00:30:52.000
is what the work will be.
It doesn't tell us what the
00:30:52.000 --> 00:30:55.000
trajectory will be.
But, of course,
00:30:55.000 --> 00:30:57.000
we could also find that.
But here, see,
00:30:57.000 --> 00:30:59.000
I am not assuming,
for example,
00:30:59.000 --> 00:31:02.000
that the particle is moving
just based on that force.
00:31:02.000 --> 00:31:04.000
Maybe, actually,
I am here to hold it in my hand
00:31:04.000 --> 00:31:06.000
and force it to go where it is
going,
00:31:06.000 --> 00:31:10.000
or maybe there is some rail
that is taking it in that
00:31:10.000 --> 00:31:14.000
trajectory or whatever.
I can really do it along any
00:31:14.000 --> 00:31:16.000
trajectory.
And, if I wanted to,
00:31:16.000 --> 00:31:18.000
if I knew that was the case,
I could try to find the
00:31:18.000 --> 00:31:20.000
trajectory based on what the
force is.
00:31:20.000 --> 00:31:36.000
But that is not what we are
doing here.
00:31:36.000 --> 00:31:41.000
Let's try to make sense of what
you asked just a few minutes
00:31:41.000 --> 00:31:45.000
ago, what can we do directly
with dr?
00:31:45.000 --> 00:31:50.000
dr becomes somehow a vector.
I mean, when I replace it by dr
00:31:50.000 --> 00:31:55.000
over dt times dt,
it becomes something that is a
00:31:55.000 --> 00:32:01.000
vector with a dt next to it.
In fact -- Well,
00:32:01.000 --> 00:32:07.000
it is not really new.
Let's see.
00:32:07.000 --> 00:32:15.000
Another way to do it,
let's say that our force has
00:32:15.000 --> 00:32:21.000
components M and N.
I claim that we can write
00:32:21.000 --> 00:32:27.000
symbolically vector dr stands
for its vector whose components
00:32:27.000 --> 00:32:31.000
are dx, dy.
It is a strange kind of vector.
00:32:31.000 --> 00:32:38.000
I mean it is not a real vector,
of course, but as a notion,
00:32:38.000 --> 00:32:45.000
it is a pretty good notation
because it tells us that F of dr
00:32:45.000 --> 00:32:51.000
is M dx plus N dy.
In fact, we will very often
00:32:51.000 --> 00:32:58.000
write, instead of F dot dr line
integral along c will be line
00:32:58.000 --> 00:33:03.000
integral along c of M dx plus N
dy.
00:33:03.000 --> 00:33:06.000
And so, in this language,
of course, what we are
00:33:06.000 --> 00:33:08.000
integrating now,
rather than a vector field,
00:33:08.000 --> 00:33:11.000
becomes a differential.
But you should think of it,
00:33:11.000 --> 00:33:12.000
too, as being pretty much the
same thing.
00:33:12.000 --> 00:33:17.000
It is like when you compare the
gradient of a function and its
00:33:17.000 --> 00:33:20.000
differential,
they are different notations
00:33:20.000 --> 00:33:24.000
but have the same content.
Now, there still remains the
00:33:24.000 --> 00:33:27.000
question of how do we compute
this kind of integral?
00:33:27.000 --> 00:33:30.000
Because it is more subtle than
the notation suggests.
00:33:30.000 --> 00:33:35.000
Because M and N both depend on
x and y.
00:33:35.000 --> 00:33:38.000
And, if you just integrate it
with respect to x,
00:33:38.000 --> 00:33:40.000
you would be left with y's in
there.
00:33:40.000 --> 00:33:41.000
And you don't want to be left
with y's.
00:33:41.000 --> 00:33:46.000
You want a number at the end.
See, the catch is along the
00:33:46.000 --> 00:33:51.000
curve x and y are actually
related to each other.
00:33:51.000 --> 00:33:54.000
Whenever we write this,
we have two variables x and y,
00:33:54.000 --> 00:33:58.000
but, in fact,
along the curve C we have only
00:33:58.000 --> 00:34:00.000
one parameter.
It could be x.
00:34:00.000 --> 00:34:02.000
It could be y.
It could be time.
00:34:02.000 --> 00:34:04.000
Whatever you want.
But we have to express
00:34:04.000 --> 00:34:06.000
everything in terms of that one
parameter.
00:34:06.000 --> 00:34:11.000
And then we get a usual single
variable integral.
00:34:11.000 --> 00:34:17.000
How do we evaluate things in
this language?
00:34:17.000 --> 00:34:21.000
Well, we do it by substituting
the parameter into everything.
00:34:50.000 --> 00:35:06.000
The method to evaluate is to
express x and y in terms of a
00:35:06.000 --> 00:35:17.000
single variable.
And then substitute that
00:35:17.000 --> 00:35:20.000
variable.
Let's, for example,
00:35:20.000 --> 00:35:25.000
redo the one we had up there
just using these new notations.
00:35:25.000 --> 00:35:31.000
You will see that it is the
same calculation but with
00:35:31.000 --> 00:35:40.000
different notations.
In that example that we had,
00:35:40.000 --> 00:35:52.000
our vector field F was negative
.
00:35:52.000 --> 00:35:55.000
What we are integrating is
negative y dx plus x dy.
00:35:55.000 --> 00:35:58.000
And, see, if we have just this,
we don't know how to integrate
00:35:58.000 --> 00:35:59.000
that.
I mean, well,
00:35:59.000 --> 00:36:01.000
you could try to come up with
negative x, y or something like
00:36:01.000 --> 00:36:03.000
that.
But that actually doesn't make
00:36:03.000 --> 00:36:08.000
sense.
It doesn't work.
00:36:08.000 --> 00:36:13.000
What we will do is we will
actually have to express
00:36:13.000 --> 00:36:18.000
everything in terms of a same
variable,
00:36:18.000 --> 00:36:21.000
because it is a single integral
and we should only have on
00:36:21.000 --> 00:36:25.000
variable.
And what that variable will be,
00:36:25.000 --> 00:36:30.000
well, if we just do it the same
way that would just be t.
00:36:30.000 --> 00:36:34.000
How do we express everything in
terms of t?
00:36:34.000 --> 00:36:37.000
Well, we use the parametric
equation.
00:36:37.000 --> 00:36:45.000
We know that x is t and y is t
squared.
00:36:45.000 --> 00:36:46.000
We know what to do with these
two guys.
00:36:46.000 --> 00:36:49.000
What about dx and dy?
Well, it is easy.
00:36:49.000 --> 00:37:01.000
We just differentiate.
dx becomes dt, dy becomes 2t dt.
00:37:01.000 --> 00:37:03.000
I am just saying,
in a different language,
00:37:03.000 --> 00:37:05.000
what I said over here with dx
over dt equals one,
00:37:05.000 --> 00:37:09.000
dy over dt equals 2t.
It is the same thing but
00:37:09.000 --> 00:37:17.000
written slightly differently.
Now, I am going to do it again.
00:37:17.000 --> 00:37:20.000
I am going to switch from one
board to the next one.
00:37:20.000 --> 00:37:30.000
My integral becomes the
integral over C of negative y is
00:37:30.000 --> 00:37:39.000
minus t squared dt plus x is t
times dy is 2t dt.
00:37:39.000 --> 00:37:44.000
And now that I have only t
left, it is fine to say I have a
00:37:44.000 --> 00:37:49.000
usual single variable integral
over a variable t that goes from
00:37:49.000 --> 00:37:51.000
zero to one.
Now I can say,
00:37:51.000 --> 00:37:55.000
yes, this is the integral from
zero to one of that stuff.
00:37:55.000 --> 00:38:00.000
I can simply it a bit and it
becomes t squared dt,
00:38:00.000 --> 00:38:04.000
and I can compute it,
equals one-third.
00:38:04.000 --> 00:38:08.000
I have negative t squared and
then I have plus 2t squared,
00:38:08.000 --> 00:38:11.000
so end up with positive t
squared.
00:38:11.000 --> 00:38:20.000
It is the same as up there.
Any questions?
00:38:20.000 --> 00:38:32.000
Yes?
dy is the differential of y,
00:38:32.000 --> 00:38:38.000
y is t squared,
so I get 2t dt.
00:38:38.000 --> 00:38:46.000
I plug dt for dx,
I plug 2t dt for dy and so on.
00:38:46.000 --> 00:38:51.000
And that is the general method.
If you are given a curve then
00:38:51.000 --> 00:38:56.000
you first have to figure out how
do you express x and y in terms
00:38:56.000 --> 00:38:58.000
of the same thing?
And you get to choose,
00:38:58.000 --> 00:39:00.000
in general, what parameter we
use.
00:39:00.000 --> 00:39:14.000
You choose to parameterize your
curve in whatever way you want.
00:39:14.000 --> 00:39:28.000
The note that I want to make is
that this line integral depends
00:39:28.000 --> 00:39:40.000
on the trajectory C but not on
the parameterization.
00:39:40.000 --> 00:39:43.000
You can choose whichever
variable you want.
00:39:43.000 --> 00:39:49.000
For example,
what you could do is when you
00:39:49.000 --> 00:39:52.000
know that you have that
trajectory,
00:39:52.000 --> 00:39:56.000
you could also choose to
parameterize it as x equals,
00:39:56.000 --> 00:40:03.000
I don't know, sine theta,
y equals sine square theta,
00:40:03.000 --> 00:40:09.000
because y is x squared where
theta goes from zero to pi over
00:40:09.000 --> 00:40:11.000
two.
And then you could get dx and
00:40:11.000 --> 00:40:15.000
dy in terms of d theta.
And you would be able to do it
00:40:15.000 --> 00:40:19.000
with a lot of trig and you would
get the same answer.
00:40:19.000 --> 00:40:22.000
That would be a harder way to
get the same thing.
00:40:22.000 --> 00:40:27.000
What you should do in practice
is use the most reasonable way
00:40:27.000 --> 00:40:31.000
to parameterize your curve.
If you know that you have a
00:40:31.000 --> 00:40:35.000
piece of parabola y equals x
squared, there is no way you
00:40:35.000 --> 00:40:38.000
would put sine and sine squared.
You could set x equals,
00:40:38.000 --> 00:40:40.000
y equals t squared,
which is very reasonable.
00:40:40.000 --> 00:40:43.000
You could even take a small
shortcut and say that your
00:40:43.000 --> 00:40:47.000
variable will be just x.
That means x you just keep as
00:40:47.000 --> 00:40:50.000
it is.
And then, when you have y,
00:40:50.000 --> 00:40:53.000
you set y equals x squared,
dy equals 2x dx,
00:40:53.000 --> 00:40:56.000
and then you will have an
integral over x.
00:40:56.000 --> 00:41:09.000
That works.
So, this one is not practical.
00:41:09.000 --> 00:41:11.000
But you get to choose.
00:41:45.000 --> 00:41:52.000
Now let me tell you a bit more
about the geometry.
00:41:52.000 --> 00:41:55.000
We have said here is how we
compute it in general,
00:41:55.000 --> 00:41:59.000
and that is the general method
for computing a line integral
00:41:59.000 --> 00:42:00.000
for work.
You can always do this,
00:42:00.000 --> 00:42:04.000
try to find a parameter,
the simplest one,
00:42:04.000 --> 00:42:07.000
express everything in terms of
its variable and then you have
00:42:07.000 --> 00:42:11.000
an integral to compute.
But sometimes you can actually
00:42:11.000 --> 00:42:14.000
save a lot of work by just
thinking geometrically about
00:42:14.000 --> 00:42:21.000
what this all does.
Let me tell you about the
00:42:21.000 --> 00:42:29.000
geometric approach.
One thing I want to remind you
00:42:29.000 --> 00:42:34.000
of first is what is this vector
dr?
00:42:34.000 --> 00:42:41.000
Well, what is vector delta r?
If I take a very small piece of
00:42:41.000 --> 00:42:47.000
the trajectory then my vector
delta r will be tangent to the
00:42:47.000 --> 00:42:50.000
trajectory.
It will be going in the same
00:42:50.000 --> 00:42:53.000
direction as the unit tangent
vector t.
00:42:53.000 --> 00:42:58.000
And what is its length?
Well, its length is the arc
00:42:58.000 --> 00:43:02.000
length along the trajectory,
which we called delta s.
00:43:02.000 --> 00:43:09.000
Remember, s was the distance
along the trajectory.
00:43:09.000 --> 00:43:21.000
We can write vector dr equals
dx, dy, but that is also T times
00:43:21.000 --> 00:43:24.000
ds.
It is a vector whose direction
00:43:24.000 --> 00:43:29.000
is tangent to the curve and
whose length element is actually
00:43:29.000 --> 00:43:35.000
the arc length element.
I mean, if you don't like this
00:43:35.000 --> 00:43:41.000
notation, think about dividing
everything by dt.
00:43:41.000 --> 00:43:45.000
Then what we are saying is dr
over dt, which is the velocity
00:43:45.000 --> 00:43:47.000
vector.
Well, in coordinates,
00:43:47.000 --> 00:43:51.000
the velocity vector is dx over
dt, dy over dt.
00:43:51.000 --> 00:43:57.000
But, more geometrically,
the direction of a velocity
00:43:57.000 --> 00:44:03.000
vector is tangent to the
trajectory and its magnitude is
00:44:03.000 --> 00:44:10.000
speed ds over dt.
So, that is really the same
00:44:10.000 --> 00:44:14.000
thing.
If I say this,
00:44:14.000 --> 00:44:20.000
that means that my line
integral F to dr,
00:44:20.000 --> 00:44:28.000
well, I say I can write it as
integral of M dx plus N dy.
00:44:28.000 --> 00:44:32.000
That is what I will do if I
want to compute it by computing
00:44:32.000 --> 00:44:36.000
the integral.
But, if instead I want to think
00:44:36.000 --> 00:44:42.000
about it geometrically,
I could rewrite it as F dot T
00:44:42.000 --> 00:44:45.000
ds.
Now you can think of this,
00:44:45.000 --> 00:44:50.000
F dot T is a scalar quantity.
It is the tangent component of
00:44:50.000 --> 00:44:53.000
my force.
I take my force and project it
00:44:53.000 --> 00:44:58.000
to the tangent direction to a
trajectory and the I integrate
00:44:58.000 --> 00:45:04.000
that along the curve.
They are the same thing.
00:45:04.000 --> 00:45:07.000
And sometimes it is easier to
do it this way.
00:45:07.000 --> 00:45:15.000
Here is an example.
This is bound to be easier only
00:45:15.000 --> 00:45:18.000
when the field and the curve are
relatively simple and have a
00:45:18.000 --> 00:45:20.000
geometric relation to each
other.
00:45:20.000 --> 00:45:24.000
If I give you an evil formula
with x cubed plus y to the fifth
00:45:24.000 --> 00:45:28.000
or whatever there is very little
chance that you will be able to
00:45:28.000 --> 00:45:32.000
simplify it that way.
But let's say that my
00:45:32.000 --> 00:45:42.000
trajectory is just a circle of
radius a centered at the origin.
00:45:42.000 --> 00:45:52.000
Let's say I am doing that
counterclockwise and let's say
00:45:52.000 --> 00:46:00.000
that my vector field is xi yj.
What does that look like?
00:46:00.000 --> 00:46:07.000
Well, my trajectory is just
this circle.
00:46:07.000 --> 00:46:11.000
My vector field,
remember, xi plus yj,
00:46:11.000 --> 00:46:15.000
that is the one that is
pointing radially from the
00:46:15.000 --> 00:46:18.000
origin.
Hopefully, if you have good
00:46:18.000 --> 00:46:20.000
physics intuition here,
you will already know what the
00:46:20.000 --> 00:46:26.000
work is going to be.
It is going to be zero because
00:46:26.000 --> 00:46:34.000
the force is perpendicular to
the motion.
00:46:34.000 --> 00:46:41.000
Now we can say it directly by
saying if you have any point of
00:46:41.000 --> 00:46:49.000
a circle then the tangent vector
to the circle will be,
00:46:49.000 --> 00:46:51.000
well, it's tangent to the
circle,
00:46:51.000 --> 00:46:54.000
so that means it is
perpendicular to the radial
00:46:54.000 --> 00:47:00.000
direction,
while the force is pointing in
00:47:00.000 --> 00:47:09.000
the radial direction so you have
a right angle between them.
00:47:09.000 --> 00:47:16.000
F is perpendicular to T.
F dot T is zero.
00:47:16.000 --> 00:47:22.000
The line integral of F dot T ds
is just zero.
00:47:22.000 --> 00:47:26.000
That is much easier than
writing this is integral of x
00:47:26.000 --> 00:47:29.000
over dx plus y over dy.
What do we do?
00:47:29.000 --> 00:47:33.000
Well, we set x equals a cosine
theta, y equals a sine theta.
00:47:33.000 --> 00:47:37.000
We get a bunch of trig things.
It cancels out to zero.
00:47:37.000 --> 00:47:42.000
It is not much harder but we
saved time by not even thinking
00:47:42.000 --> 00:47:45.000
about how to parameterize
things.
00:47:45.000 --> 00:47:50.000
Let's just do a last one.
That was the first one.
00:47:50.000 --> 00:47:55.000
Let's say now that I take the
same curve C,
00:47:55.000 --> 00:48:03.000
but now my vector field is the
one that rotates negative yi
00:48:03.000 --> 00:48:11.000
plus xj.
That means along my circle the
00:48:11.000 --> 00:48:23.000
tangent vector goes like this
and my vector field is also
00:48:23.000 --> 00:48:28.000
going around.
So, in fact,
00:48:28.000 --> 00:48:34.000
at this point the vector field
will always be going in the same
00:48:34.000 --> 00:48:41.000
direction.
Now F is actually parallel to
00:48:41.000 --> 00:48:47.000
the tangent direction.
That means that the dot product
00:48:47.000 --> 00:48:52.000
of F dot T, remember,
if it is the component of F in
00:48:52.000 --> 00:48:57.000
this direction that will be the
same of the length of F.
00:48:57.000 --> 00:49:02.000
But what is the length of F on
this circle if this length is a?
00:49:02.000 --> 00:49:05.000
It is just going to be a.
That is what we said earlier
00:49:05.000 --> 00:49:08.000
about this vector field.
At every point,
00:49:08.000 --> 00:49:11.000
this dot product is a.
Now we know how to integrate
00:49:11.000 --> 00:49:13.000
that quite quickly.
00:49:25.000 --> 00:49:30.000
Because it becomes the integral
of a ds, but a is a constant so
00:49:30.000 --> 00:49:34.000
we can take it out.
And now what do we get when we
00:49:34.000 --> 00:49:38.000
integrate ds along C?
Well, we should get the total
00:49:38.000 --> 00:49:43.000
length of the curve if we sum
all the little pieces of arc
00:49:43.000 --> 00:49:47.000
length.
But we know that the length of
00:49:47.000 --> 00:49:52.000
a circle of radius a is 2pi a,
so we get 2(pi)a squared.
00:49:52.000 --> 00:50:01.000
If we were to compute that by
hand, well, what would we do?
00:50:01.000 --> 00:50:08.000
We would be computing integral
of minus y dx plus x dy.
00:50:08.000 --> 00:50:13.000
Since we are on a circle,
we will probably set x equals a
00:50:13.000 --> 00:50:16.000
times cosine theta,
y equals a times sine theta for
00:50:16.000 --> 00:50:23.000
theta between zero and 2pi.
Then we would get dx and dy out
00:50:23.000 --> 00:50:27.000
of these.
So, y is a sine theta,
00:50:27.000 --> 00:50:33.000
dx is negative a sine theta d
theta, if you differentiate a
00:50:33.000 --> 00:50:40.000
cosine, plus a cosine theta
times a cosine theta d theta.
00:50:40.000 --> 00:50:45.000
Well, you will just end up with
integral from zero to 2pi of a
00:50:45.000 --> 00:50:51.000
squared time sine squared theta
plus cosine square theta times d
00:50:51.000 --> 00:50:53.000
theta.
That becomes just one.
00:50:53.000 --> 00:50:56.000
And you get the same answer.
It took about the same amount
00:50:56.000 --> 00:50:59.000
of time because I did this one
rushing very quickly,
00:50:59.000 --> 00:51:03.000
but normally it takes about at
least twice the amount of time
00:51:03.000 --> 00:51:06.000
to do it with a calculation.
That tells you sometimes it is
00:51:06.000 --> 00:51:08.000
worth thinking geometrically.