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So, let me remind you,
yesterday we've defined and

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started to compute line
integrals for work as a vector

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00:00:34,000 --> 00:00:41,000
field along a curve.
So, we have a curve in the

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plane, C.
We have a vector field that

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gives us a vector at every
point.

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And, we want to find the work
done along the curve.

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00:01:04,000 --> 00:01:11,000
So, that's the line integral
along C of F dr,

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or more geometrically,
line integral along C of F.T ds

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where T is the unit tangent
vector,

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00:01:19,000 --> 00:01:23,000
and ds is the arc length
element.

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00:01:23,000 --> 00:01:30,000
Or, in coordinates,
that they integral of M dx N dy

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00:01:30,000 --> 00:01:38,000
where M and N are the components
of the vector field.

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00:01:38,000 --> 00:01:46,000
OK, so -- Let's do an example
that will just summarize what we

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00:01:46,000 --> 00:01:51,000
did yesterday,
and then we will move on to

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00:01:51,000 --> 00:01:57,000
interesting observations about
these things.

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00:01:57,000 --> 00:02:03,000
So, here's an example we are
going to look at now.

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00:02:03,000 --> 00:02:11,000
Let's say I give you the vector
field yi plus xj.

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00:02:11,000 --> 00:02:13,000
So, it's not completely obvious
what it looks like,

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00:02:13,000 --> 00:02:16,000
but here is a computer plot of
that vector field.

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00:02:16,000 --> 00:02:21,000
So, that tells you a bit what
it does.

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00:02:21,000 --> 00:02:24,000
It points in all sorts of
directions.

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00:02:24,000 --> 00:02:31,000
And, let's say we want to find
the work done by this vector

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00:02:31,000 --> 00:02:35,000
field.
If I move along this closed

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00:02:35,000 --> 00:02:40,000
curve, I start at the origin.
But, I moved along the x-axis

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00:02:40,000 --> 00:02:43,000
to one.
That move along the unit circle

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00:02:43,000 --> 00:02:46,000
to the diagonal,
and then I move back to the

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00:02:46,000 --> 00:02:54,000
origin in a straight line.
OK, so C consists of three

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00:02:54,000 --> 00:03:06,000
parts -- -- so that you enclose
a sector of a unit disk -- --

35
00:03:06,000 --> 00:03:17,000
corresponding to angles between
zero and 45�.

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00:03:17,000 --> 00:03:24,000
So, to compute this line
integral, all we have to do is

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00:03:24,000 --> 00:03:33,000
we have set up three different
integrals and add that together.

38
00:03:33,000 --> 00:03:44,000
OK, so we need to set up the
integral of y dx plus x dy for

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00:03:44,000 --> 00:03:54,000
each of these pieces.
So, let's do the first one on

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00:03:54,000 --> 00:03:59,000
the x-axis.
Well, one way to parameterize

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00:03:59,000 --> 00:04:06,000
that is just use the x variable.
And, say that because we are on

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00:04:06,000 --> 00:04:12,000
the, let's see,
sorry, we are going from the

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00:04:12,000 --> 00:04:17,000
origin to (1,0).
Well, we know we are on the

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00:04:17,000 --> 00:04:22,000
x-axis.
So, y there is actually just

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00:04:22,000 --> 00:04:25,000
zero.
And, the variable will be x

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00:04:25,000 --> 00:04:28,000
from zero to one.
Or, if you prefer,

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00:04:28,000 --> 00:04:31,000
you can parameterize things,
say, x equals t for t from zero

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00:04:31,000 --> 00:04:36,000
to one, and y equals zero.
What doesn't change is y is

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00:04:36,000 --> 00:04:41,000
zero, and therefore,
dy is also zero.

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00:04:41,000 --> 00:04:48,000
So, in fact,
we are integrating y dx x dy,

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00:04:48,000 --> 00:04:53,000
but that becomes,
well, zero dx 0,

52
00:04:53,000 --> 00:04:59,000
and that's just going to give
you zero.

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00:04:59,000 --> 00:05:03,000
OK, so there's the line
integral.

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00:05:03,000 --> 00:05:07,000
Here, it's very easy to compute.
Of course, you can also do it

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00:05:07,000 --> 00:05:10,000
geometrically because
geometrically,

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00:05:10,000 --> 00:05:12,000
you can see in the picture
along the x-axis,

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00:05:12,000 --> 00:05:15,000
the vector field is pointing
vertically.

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00:05:15,000 --> 00:05:19,000
If I'm on the x-axis,
my vector field is actually in

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00:05:19,000 --> 00:05:23,000
the y direction.
So, it's perpendicular to my

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00:05:23,000 --> 00:05:24,000
curve.
So, the work done is going to

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00:05:24,000 --> 00:05:26,000
be zero.
F dot T will be zero.

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00:05:48,000 --> 00:05:51,000
OK, so F dot T is zero,
so the integral is zero.

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00:05:51,000 --> 00:05:57,000
OK, any questions about this
first part of the calculation?

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00:05:57,000 --> 00:06:04,000
No? It's OK?
OK, let's move on to more

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00:06:04,000 --> 00:06:11,000
interesting part of it.
Let's do the second part,

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00:06:11,000 --> 00:06:18,000
which is a portion of the unit
circle.

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00:06:18,000 --> 00:06:24,000
OK, so I should have drawn my
picture.

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00:06:24,000 --> 00:06:37,000
And so now we are moving on
this part of the curve that's

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00:06:37,000 --> 00:06:40,000
C2.
And, of course we have to

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00:06:40,000 --> 00:06:43,000
choose how to express x and y in
terms of a single variable.

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00:06:43,000 --> 00:06:46,000
Well, most likely,
when you are moving on a

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00:06:46,000 --> 00:06:49,000
circle, you are going to use the
angle along the circle to tell

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00:06:49,000 --> 00:06:53,000
you where you are.
OK, so we're going to use the

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00:06:53,000 --> 00:06:56,000
angle theta as a parameter.
And we will say,

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00:06:56,000 --> 00:07:02,000
we are on the unit circle.
So, x is cosine theta and y is

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00:07:02,000 --> 00:07:05,000
sine theta.
What's the range of theta?

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00:07:05,000 --> 00:07:11,000
Theta goes from zero to pi over
four, OK?

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00:07:11,000 --> 00:07:15,000
So, whenever I see dx,
I will replace it by,

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00:07:15,000 --> 00:07:19,000
well, the derivative of cosine
is negative sine.

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00:07:19,000 --> 00:07:24,000
So, minus sine theta d theta,
and dy, the derivative of sine

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00:07:24,000 --> 00:07:29,000
is cosine.
So, it will become cosine theta

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00:07:29,000 --> 00:07:34,000
d theta.
OK, so I'm computing the

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00:07:34,000 --> 00:07:41,000
integral of y dx x dy.
That means -- -- I'll be

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00:07:41,000 --> 00:07:52,000
actually computing the integral
of, so, y is sine theta.

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00:07:52,000 --> 00:08:01,000
dx, that's negative sine theta
d theta plus x cosine.

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00:08:01,000 --> 00:08:08,000
dy is cosine theta d theta from
zero to pi/4.

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00:08:08,000 --> 00:08:17,000
OK, so that's integral from
zero to pi / 4 of cosine squared

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00:08:17,000 --> 00:08:22,000
minus sine squared.
And, if you know your trig,

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00:08:22,000 --> 00:08:27,000
then you should recognize this
as cosine of two theta.

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00:08:27,000 --> 00:08:33,000
OK, so that will integrate to
one half of sine two theta from

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00:08:33,000 --> 00:08:36,000
zero to pi over four,
sorry.

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00:08:36,000 --> 00:08:44,000
And, sine pi over two is one.
So, you will get one half.

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00:08:44,000 --> 00:08:49,000
OK, any questions about this
one?

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00:08:49,000 --> 00:09:07,000
No?
OK, then let's do the third one.

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00:09:07,000 --> 00:09:15,000
So, the third guy is when we
come back to the origin along

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00:09:15,000 --> 00:09:18,000
the diagonal.
OK, so we go in a straight line

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00:09:18,000 --> 00:09:20,000
from this point.
Where's this point?

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00:09:20,000 --> 00:09:25,000
Well, this point is one over
root two, one over root two.

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00:09:25,000 --> 00:09:32,000
And, we go back to the origin.
OK, so we need to figure out a

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00:09:32,000 --> 00:09:38,000
way to express x and y in terms
of the same parameter.

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00:09:38,000 --> 00:09:43,000
So, one way which is very
natural would be to just say,

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00:09:43,000 --> 00:09:47,000
well, let's say we move from
here to here over time.

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00:09:47,000 --> 00:09:50,000
And, at time zero, we are here.
At time one, we are here.

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00:09:50,000 --> 00:09:54,000
We know how to parameterize
this line.

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00:09:54,000 --> 00:10:02,000
So, what we could do is say,
let's parameterize this line.

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00:10:02,000 --> 00:10:09,000
So, we start at one over root
two, and we go down by one over

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00:10:09,000 --> 00:10:19,000
root two in time one.
And, same with y.

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00:10:19,000 --> 00:10:24,000
That's actually perfectly fine.
But that's unnecessarily

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00:10:24,000 --> 00:10:27,000
complicated.
OK, why is a complicated?

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00:10:27,000 --> 00:10:30,000
Because we will get all of
these expressions.

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00:10:30,000 --> 00:10:34,000
It would be easier to actually
just look at motion in this

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00:10:34,000 --> 00:10:38,000
direction and then say,
well, if we have a certain work

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00:10:38,000 --> 00:10:42,000
if we move from here to here,
then the work done moving from

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00:10:42,000 --> 00:10:46,000
here to here is just going to be
the opposite,

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00:10:46,000 --> 00:10:48,000
OK?
So, in fact,

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00:10:48,000 --> 00:10:55,000
we can do slightly better by
just saying, well,

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00:10:55,000 --> 00:10:59,000
we'll take x = t,
y = t.

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00:10:59,000 --> 00:11:07,000
t from zero to one over root
two, and take,

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00:11:07,000 --> 00:11:15,000
well, sorry,
that gives us what I will call

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00:11:15,000 --> 00:11:25,000
minus C3, which is C3 backwards.
And then we can say the

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00:11:25,000 --> 00:11:31,000
integral for work along minus C3
is the opposite of the work

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00:11:31,000 --> 00:11:35,000
along C3.
Or, if you're comfortable with

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00:11:35,000 --> 00:11:39,000
integration where variables go
down,

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00:11:39,000 --> 00:11:42,000
then you could also say that t
just goes from one over square

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00:11:42,000 --> 00:11:45,000
root of two down to zero.
And, when you set up your

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00:11:45,000 --> 00:11:49,000
integral, it will go from one
over root two to zero.

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00:11:49,000 --> 00:11:50,000
And, of course,
that will be the negative of

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00:11:50,000 --> 00:11:52,000
the one from zero to one over
root two.

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00:11:52,000 --> 00:11:59,000
So, it's the same thing.
OK, so if we do it with this

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00:11:59,000 --> 00:12:03,000
parameterization,
we'll get that,

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00:12:03,000 --> 00:12:08,000
well of course,
dx is dt, dy is dt.

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00:12:08,000 --> 00:12:16,000
So, the integral along minus C3
of y dx plus x dy is just the

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00:12:16,000 --> 00:12:24,000
integral from zero to one over
root two of t dt plus t dt.

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00:12:24,000 --> 00:12:30,000
Sorry, I'm messing up my
blackboard, OK,

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00:12:30,000 --> 00:12:37,000
which is going to be,
well, the integral of 2t dt,

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which is t2 between these
bounds, which is one half.

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00:12:46,000 --> 00:12:51,000
That's the integral along minus
C3, along the reversed path.

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00:12:51,000 --> 00:13:03,000
And, if I want to do it along
C3 instead, then I just take the

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00:13:03,000 --> 00:13:07,000
negative.
Or, if you prefer,

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00:13:07,000 --> 00:13:11,000
you could have done it directly
with integral from one over root

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00:13:11,000 --> 00:13:14,000
two, two zero,
which gives you immediately the

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00:13:14,000 --> 00:13:19,000
negative one half.
OK, so at the end,

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00:13:19,000 --> 00:13:28,000
we get that the total work --
-- was the sum of the three line

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00:13:28,000 --> 00:13:32,000
integrals.
I'm not writing after dr just

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00:13:32,000 --> 00:13:36,000
to save space.
But, zero plus one half minus

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00:13:36,000 --> 00:13:39,000
one half, and that comes out to
zero.

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00:13:39,000 --> 00:13:44,000
So, a lot of calculations for
nothing.

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00:13:44,000 --> 00:13:49,000
OK, so that should give you
overview of various ways to

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00:13:49,000 --> 00:13:57,000
compute line integrals.
Any questions about all that?

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00:13:57,000 --> 00:14:03,000
No? OK.
So, next, let me tell you about

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00:14:03,000 --> 00:14:07,000
how to avoid computing like
integrals.

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00:14:07,000 --> 00:14:08,000
Well, one is easy:
don't take this class.

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00:14:08,000 --> 00:14:17,000
But that's not,
so here's another way not to do

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00:14:17,000 --> 00:14:20,000
it, OK?
So, let's look a little bit

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00:14:20,000 --> 00:14:24,000
about one kind of vector field
that actually we've encountered

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00:14:24,000 --> 00:14:26,000
a few weeks ago without saying
it.

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00:14:26,000 --> 00:14:30,000
So, we said when we have a
function of two variables,

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00:14:30,000 --> 00:14:32,000
we have the gradient vector.
Well, at the time,

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00:14:32,000 --> 00:14:35,000
it was just a vector.
But, that vector depended on x

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00:14:35,000 --> 00:14:36,000
and y.
So, in fact,

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00:14:36,000 --> 00:14:43,000
it's a vector field.
OK, so here's an interesting

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00:14:43,000 --> 00:14:48,000
special case.
Say that F, our vector field is

163
00:14:48,000 --> 00:14:52,000
actually the gradient of some
function.

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00:14:52,000 --> 00:15:01,000
So, it's a gradient field.
And, so f is a function of two

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00:15:01,000 --> 00:15:08,000
variables, x and y,
and that's called the potential

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00:15:08,000 --> 00:15:12,000
for the vector field.
The reason is,

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00:15:12,000 --> 00:15:16,000
of course, from physics.
In physics, you call potential,

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00:15:16,000 --> 00:15:21,000
electrical potential or
gravitational potential,

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00:15:21,000 --> 00:15:25,000
the potential energy.
This function of position that

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00:15:25,000 --> 00:15:29,000
tells you how much actually
energy stored somehow by the

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00:15:29,000 --> 00:15:33,000
force field, and this gradient
gives you the force.

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00:15:33,000 --> 00:15:37,000
Actually, not quite.
If you are a physicist,

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00:15:37,000 --> 00:15:40,000
that the force will be negative
the gradient.

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00:15:40,000 --> 00:15:44,000
So, that means that physicists'
potentials are the opposite of a

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00:15:44,000 --> 00:15:46,000
mathematician's potential.
Okay?

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00:15:46,000 --> 00:15:48,000
So it's just here to confuse
you.

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00:15:48,000 --> 00:15:50,000
It doesn't really matter all
the time.

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00:15:50,000 --> 00:15:55,000
So to make things simpler we
are using this convention and

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00:15:55,000 --> 00:15:59,000
you just put a minus sign if you
are doing physics.

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00:15:59,000 --> 00:16:13,000
So then I claim that we can
simplify the evaluation of the

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00:16:13,000 --> 00:16:21,000
line integral for work.
Perhaps you've seen in physics,

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00:16:21,000 --> 00:16:24,000
the work done by,
say, the electrical force,

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00:16:24,000 --> 00:16:28,000
is actually given by the change
in the value of a potential from

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00:16:28,000 --> 00:16:30,000
the starting point of the ending
point,

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00:16:30,000 --> 00:16:36,000
or same for gravitational force.
So, these are special cases of

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00:16:36,000 --> 00:16:40,000
what's called the fundamental
theorem of calculus for line

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00:16:40,000 --> 00:16:43,000
integrals.
So, the fundamental theorem of

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00:16:43,000 --> 00:16:46,000
calculus, not for line
integrals, tells you if you

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00:16:46,000 --> 00:16:49,000
integrate a derivative,
then you get back the function.

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00:16:49,000 --> 00:16:52,000
And here, it's the same thing
in multivariable calculus.

191
00:16:52,000 --> 00:16:54,000
It tells you,
if you take the line integral

192
00:16:54,000 --> 00:16:58,000
of the gradient of a function,
what you get back is the

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00:16:58,000 --> 00:16:58,000
function.

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00:17:23,000 --> 00:17:30,000
OK,
so -- -- the fundamental

195
00:17:30,000 --> 00:17:43,000
theorem of calculus for line
integrals -- -- says if you

196
00:17:43,000 --> 00:17:58,000
integrate a vector field that's
the gradient of a function along

197
00:17:58,000 --> 00:18:03,000
a curve,
let's say that you have a curve

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00:18:03,000 --> 00:18:06,000
that goes from some starting
point, P0,

199
00:18:06,000 --> 00:18:15,000
to some ending point, P1.
All you will get is the value

200
00:18:15,000 --> 00:18:21,000
of F at P1 minus the value of F
at P0.

201
00:18:21,000 --> 00:18:25,000
OK, so, that's a pretty nifty
formula that only works if the

202
00:18:25,000 --> 00:18:28,000
field that you are integrating
is a gradient.

203
00:18:28,000 --> 00:18:32,000
You know it's a gradient,
and you know the function,

204
00:18:32,000 --> 00:18:35,000
little f.
I mean, we can't put just any

205
00:18:35,000 --> 00:18:38,000
vector field in here.
We have to put the gradient of

206
00:18:38,000 --> 00:18:41,000
F.
So, actually on Tuesday we'll

207
00:18:41,000 --> 00:18:47,000
see how to decide whether a
vector field is a gradient or

208
00:18:47,000 --> 00:18:49,000
not,
and if it is a gradient,

209
00:18:49,000 --> 00:18:52,000
how to find the potential
function.

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00:18:52,000 --> 00:18:58,000
So, we'll cover that.
But, for now we need to try to

211
00:18:58,000 --> 00:19:05,000
figure out a bit more about
this, what it says,

212
00:19:05,000 --> 00:19:11,000
what it means physically,
how to think of it

213
00:19:11,000 --> 00:19:15,000
geometrically,
and so on.

214
00:19:15,000 --> 00:19:18,000
So, maybe I should say,
if you're trying to write this

215
00:19:18,000 --> 00:19:21,000
in coordinates,
because that's also a useful

216
00:19:21,000 --> 00:19:24,000
way to think about it,
if I give you the line integral

217
00:19:24,000 --> 00:19:27,000
along C,
so, the gradient field,

218
00:19:27,000 --> 00:19:29,000
the components are f sub x and
f sub y.

219
00:19:29,000 --> 00:19:36,000
So, it means I'm actually
integrating f sub x dx plus f

220
00:19:36,000 --> 00:19:38,000
sub y dy.
Or, if you prefer,

221
00:19:38,000 --> 00:19:42,000
that's the same thing as
actually integrating df.

222
00:19:42,000 --> 00:19:46,000
So, I'm integrating the
differential of a function,

223
00:19:46,000 --> 00:19:54,000
f.
Well then, that's the change in

224
00:19:54,000 --> 00:19:56,000
F.
And, of course,

225
00:19:56,000 --> 00:20:02,000
if you write it in this form,
then probably it's quite

226
00:20:02,000 --> 00:20:06,000
obvious to you that this should
be true.

227
00:20:06,000 --> 00:20:11,000
I mean, in this form,
actually it's the same

228
00:20:11,000 --> 00:20:15,000
statement as in single variable
calculus.

229
00:20:15,000 --> 00:20:17,000
OK, and actually that's how we
prove the theorem.

230
00:20:17,000 --> 00:20:27,000
So, let's prove this theorem.
How do we prove it?

231
00:20:27,000 --> 00:20:31,000
Well, let's say I give you a
curve and I ask you to compute

232
00:20:31,000 --> 00:20:34,000
this integral.
How will you do that?

233
00:20:34,000 --> 00:20:38,000
Well, the way you compute the
integral actually is by choosing

234
00:20:38,000 --> 00:20:41,000
a parameter, and expressing
everything in terms of that

235
00:20:41,000 --> 00:20:46,000
parameter.
So, we'll set,

236
00:20:46,000 --> 00:20:57,000
well, so we know it's f sub x
dx plus f sub y dy.

237
00:20:57,000 --> 00:21:03,000
And, we'll want to parameterize
C in the form x equals x of t.

238
00:21:03,000 --> 00:21:09,000
y equals y of t.
So, if we do that,

239
00:21:09,000 --> 00:21:12,000
then dx becomes x prime of t
dt.

240
00:21:12,000 --> 00:21:25,000
dy becomes y prime of t dt.
So, we know x is x of t.

241
00:21:25,000 --> 00:21:31,000
That tells us dx is x prime of
t dt.

242
00:21:31,000 --> 00:21:38,000
y is y of t gives us dy is y
prime of t dt.

243
00:21:38,000 --> 00:21:52,000
So, now what we are integrating
actually becomes the integral of

244
00:21:52,000 --> 00:22:05,000
f sub x times dx dt plus f sub y
times dy dt times dt.

245
00:22:05,000 --> 00:22:09,000
OK, but now,
here I recognize a familiar

246
00:22:09,000 --> 00:22:13,000
guy.
I've seen this one before in

247
00:22:13,000 --> 00:22:15,000
the chain rule.
OK, this guy,

248
00:22:15,000 --> 00:22:19,000
by the chain rule,
is the rate of change of f if I

249
00:22:19,000 --> 00:22:22,000
take x and y to be functions of
t.

250
00:22:22,000 --> 00:22:26,000
And, I plug those into f.
So, in fact,

251
00:22:26,000 --> 00:22:34,000
what I'm integrating is df dt
when I think of f as a function

252
00:22:34,000 --> 00:22:42,000
of t by just plugging x and y as
functions of t.

253
00:22:42,000 --> 00:22:51,000
And so maybe actually I should
now say I have sometimes t goes

254
00:22:51,000 --> 00:22:59,000
from some initial time,
let's say, t zero to t one.

255
00:22:59,000 --> 00:23:03,000
And now, by the usual
fundamental theorem of calculus,

256
00:23:03,000 --> 00:23:07,000
I know that this will be just
the change in the value of f

257
00:23:07,000 --> 00:23:09,000
between t zero and t one.

258
00:23:36,000 --> 00:23:44,000
OK, so integral from t zero to
one of (df /dt) dt,

259
00:23:44,000 --> 00:23:52,000
well, that becomes f between t
zero and t one.

260
00:23:52,000 --> 00:23:55,000
f of what?
We just have to be a little bit

261
00:23:55,000 --> 00:23:58,000
careful here.
Well, it's not quite f of t.

262
00:23:58,000 --> 00:24:02,000
It's f seen as a function of t
by putting x of t and y of t

263
00:24:02,000 --> 00:24:07,000
into it.
So, let me read that carefully.

264
00:24:07,000 --> 00:24:15,000
What I'm integrating to is f of
x of t and y of t.

265
00:24:15,000 --> 00:24:19,000
Does that sound fair?
Yeah, and so,

266
00:24:19,000 --> 00:24:23,000
when I plug in t1,
I get the point where I am at

267
00:24:23,000 --> 00:24:26,000
time t1.
That's the endpoint of my curve.

268
00:24:26,000 --> 00:24:31,000
When I plug t0,
I will get the starting point

269
00:24:31,000 --> 00:24:37,000
of my curve, p0.
And, that's the end of the

270
00:24:37,000 --> 00:24:43,000
proof.
It wasn't that hard, see?

271
00:24:43,000 --> 00:24:56,000
OK, so let's see an example.
Well, let's look at that

272
00:24:56,000 --> 00:25:00,000
example again.
So, we have this curve.

273
00:25:00,000 --> 00:25:03,000
We have this vector field.
Could it be that,

274
00:25:03,000 --> 00:25:05,000
by accident,
that vector field was a

275
00:25:05,000 --> 00:25:08,000
gradient field?
So, remember,

276
00:25:08,000 --> 00:25:12,000
our vector field was y,
x.

277
00:25:12,000 --> 00:25:16,000
Can we think of a function
whose derivative with respect to

278
00:25:16,000 --> 00:25:19,000
x is y, and derivative with
respect to y is x?

279
00:25:19,000 --> 00:25:27,000
Yeah, x times y sounds like a
good candidate where f( x,

280
00:25:27,000 --> 00:25:30,000
y) is xy.
OK,

281
00:25:30,000 --> 00:25:34,000
so that means that the line
integrals that we computed along

282
00:25:34,000 --> 00:25:39,000
these things can be just
evaluated from just finding out

283
00:25:39,000 --> 00:25:44,000
the values of f at the endpoint?
So, here's version two of my

284
00:25:44,000 --> 00:25:50,000
plot where I've added the
contour plot of a function,

285
00:25:50,000 --> 00:25:53,000
x, y on top of the vector
field.

286
00:25:53,000 --> 00:25:57,000
Actually, they have a vector
field is still pointing

287
00:25:57,000 --> 00:26:00,000
perpendicular to the level
curves that we have seen,

288
00:26:00,000 --> 00:26:02,000
just to remind you.
And, so now,

289
00:26:02,000 --> 00:26:05,000
when we move,
now when we move,

290
00:26:05,000 --> 00:26:09,000
the origin is on the level
curve, f equals zero.

291
00:26:09,000 --> 00:26:14,000
And, when we start going along
C1, we stay on f equals zero.

292
00:26:14,000 --> 00:26:17,000
So, there's no work.
The potential doesn't change.

293
00:26:17,000 --> 00:26:21,000
Then on C2, the potential
increases from zero to one half.

294
00:26:21,000 --> 00:26:24,000
The work is one half.
And then, on C3,

295
00:26:24,000 --> 00:26:27,000
we go back down from one half
to zero.

296
00:26:27,000 --> 00:26:40,000
The work is negative one half.
See, that was much easier than

297
00:26:40,000 --> 00:26:47,000
computing.
So, for example,

298
00:26:47,000 --> 00:26:53,000
the integral along C2 is
actually just,

299
00:26:53,000 --> 00:27:00,000
so, C2 goes from one zero to
one over root two,

300
00:27:00,000 --> 00:27:09,000
one over root two.
So, that's one half minus zero,

301
00:27:09,000 --> 00:27:18,000
and that's one half,
OK, because C2 was going here.

302
00:27:18,000 --> 00:27:26,000
And, at this point, f is zero.
At that point, f is one half.

303
00:27:26,000 --> 00:27:29,000
And, similarly for the others,
and of course when you sum,

304
00:27:29,000 --> 00:27:33,000
you get zero because the total
change in f when you go from

305
00:27:33,000 --> 00:27:35,000
here,
to here, to here, to here,

306
00:27:35,000 --> 00:27:37,000
eventually you are back at the
same place.

307
00:27:37,000 --> 00:27:44,000
So, f hasn't changed.
OK, so that's a neat trick.

308
00:27:44,000 --> 00:27:48,000
And it's important conceptually
because a lot of the forces are

309
00:27:48,000 --> 00:27:53,000
gradients of potentials,
namely, gravitational force,

310
00:27:53,000 --> 00:27:57,000
electric force.
The problem is not every vector

311
00:27:57,000 --> 00:28:00,000
field is a gradient.
A lot of vector fields are not

312
00:28:00,000 --> 00:28:02,000
gradients.
For example,

313
00:28:02,000 --> 00:28:08,000
magnetic fields certainly are
not gradients.

314
00:28:08,000 --> 00:28:33,000
So -- -- a big warning:
everything today only applies

315
00:28:33,000 --> 00:28:48,000
if F is a gradient field.
OK, it's not true otherwise.

316
00:29:07,000 --> 00:29:19,000
OK, still, let's see,
what are the consequences of

317
00:29:19,000 --> 00:29:30,000
the fundamental theorem?
So, just to put one more time

318
00:29:30,000 --> 00:29:39,000
this disclaimer,
if F is a gradient field -- --

319
00:29:39,000 --> 00:29:44,000
then what do we have?
Well, there's various nice

320
00:29:44,000 --> 00:29:47,000
features of work done by
gradient fields that are not too

321
00:29:47,000 --> 00:29:53,000
far off the vector fields.
So, one of them is this

322
00:29:53,000 --> 00:30:02,000
property of path independence.
OK, so the claim is if I have a

323
00:30:02,000 --> 00:30:05,000
line integral to compute,
that it doesn't matter which

324
00:30:05,000 --> 00:30:09,000
path I take as long as it goes
from point a to point b.

325
00:30:09,000 --> 00:30:14,000
It just depends on the point
where I start and the point

326
00:30:14,000 --> 00:30:18,000
where I end.
And, that's certainly false in

327
00:30:18,000 --> 00:30:22,000
general, but for a gradient
field that works.

328
00:30:22,000 --> 00:30:25,000
So if I have a point,
P0,

329
00:30:25,000 --> 00:30:28,000
a point, P1,
and I have two different paths

330
00:30:28,000 --> 00:30:33,000
that go there,
say, C1 and C2,

331
00:30:33,000 --> 00:30:39,000
so they go from the same point
to the same point but in

332
00:30:39,000 --> 00:30:44,000
different ways,
then in this situation,

333
00:30:44,000 --> 00:30:55,000
the line integral along C1 is
equal to the line integral along

334
00:30:55,000 --> 00:30:56,000
C2.
Well, actually,

335
00:30:56,000 --> 00:31:02,000
let me insist that this is only
for gradient fields by putting

336
00:31:02,000 --> 00:31:09,000
gradient F in here,
just so you don't get tempted

337
00:31:09,000 --> 00:31:19,000
to ever use this for a field
that's not a gradient field --

338
00:31:19,000 --> 00:31:28,000
-- if C1 and C2 have the same
start and end point.

339
00:31:28,000 --> 00:31:30,000
OK, how do you prove that?
Well, it's very easy.

340
00:31:30,000 --> 00:31:32,000
We just use the fundamental
theorem.

341
00:31:32,000 --> 00:31:35,000
It tells us,
if you compute the line

342
00:31:35,000 --> 00:31:38,000
integral along C1,
it's just F at this point minus

343
00:31:38,000 --> 00:31:41,000
F at this point.
If you do it for C2,

344
00:31:41,000 --> 00:31:45,000
well, the same.
So, they are the same.

345
00:31:45,000 --> 00:31:48,000
And for that you don't actually
even need to know what little f

346
00:31:48,000 --> 00:31:50,000
is.
You know in advance that it's

347
00:31:50,000 --> 00:31:53,000
going to be the same.
So, if I give you a vector

348
00:31:53,000 --> 00:31:56,000
field and I tell you it's the
gradient of mysterious function

349
00:31:56,000 --> 00:31:58,000
but I don't tell you what the
function is and you don't want

350
00:31:58,000 --> 00:32:00,000
to find out,
you can still use path

351
00:32:00,000 --> 00:32:03,000
independence,
but only if you know it's a

352
00:32:03,000 --> 00:32:03,000
gradient.

353
00:32:25,000 --> 00:32:35,000
OK, I guess this one is dead.
So, that will stay here forever

354
00:32:35,000 --> 00:32:40,000
because nobody is tall enough to
erase it.

355
00:32:40,000 --> 00:32:49,000
When you come back next year
and you still see that formula,

356
00:32:49,000 --> 00:32:53,000
you'll see.
Yes, but there's no useful

357
00:32:53,000 --> 00:32:59,000
information here.
That's a good point.

358
00:32:59,000 --> 00:33:06,000
OK, so what's another
consequence?

359
00:33:06,000 --> 00:33:14,000
So, if you have a gradient
field, it's what's called

360
00:33:14,000 --> 00:33:19,000
conservative.
OK, so what a conservative

361
00:33:19,000 --> 00:33:21,000
field?
Well, the word conservative

362
00:33:21,000 --> 00:33:26,000
comes from the idea in physics;
if the conservation of energy.

363
00:33:26,000 --> 00:33:31,000
It tells you that you cannot
get energy for free out of your

364
00:33:31,000 --> 00:33:33,000
force field.
So,

365
00:33:33,000 --> 00:33:36,000
what it means is that in
particular,

366
00:33:36,000 --> 00:33:39,000
if you take a closed
trajectory,

367
00:33:39,000 --> 00:33:44,000
so a trajectory that goes from
some point back to the same

368
00:33:44,000 --> 00:33:52,000
point,
so, if C is a closed curve,

369
00:33:52,000 --> 00:34:03,000
then the work done along C --
-- is zero.

370
00:34:03,000 --> 00:34:06,000
OK, that's the definition of
what it means to be

371
00:34:06,000 --> 00:34:09,000
conservative.
If I take any closed curve,

372
00:34:09,000 --> 00:34:13,000
the work will always be zero.
On the contrary,

373
00:34:13,000 --> 00:34:17,000
not conservative means
somewhere there is a curve along

374
00:34:17,000 --> 00:34:21,000
which the work is not zero.
If you find a curve where the

375
00:34:21,000 --> 00:34:23,000
work is zero,
that's not enough to say it's

376
00:34:23,000 --> 00:34:26,000
conservative.
You have show that no matter

377
00:34:26,000 --> 00:34:30,000
what curve I give you,
if it's a closed curve,

378
00:34:30,000 --> 00:34:34,000
it will always be zero.
So, what that means concretely

379
00:34:34,000 --> 00:34:37,000
is if you have a force field
that conservative,

380
00:34:37,000 --> 00:34:42,000
then you cannot build somehow
some perpetual motion out of it.

381
00:34:42,000 --> 00:34:44,000
You can't build something that
will just keep going just

382
00:34:44,000 --> 00:34:47,000
powered by that force because
that force is actually not

383
00:34:47,000 --> 00:34:50,000
providing any energy.
After you've gone one loop

384
00:34:50,000 --> 00:34:53,000
around, nothings happened from
the point of view of the energy

385
00:34:53,000 --> 00:34:57,000
provided by that force.
There's no work coming from the

386
00:34:57,000 --> 00:34:59,000
force,
while if you have a force field

387
00:34:59,000 --> 00:35:02,000
that's not conservative than you
can try to actually maybe find a

388
00:35:02,000 --> 00:35:04,000
loop where the work would be
positive.

389
00:35:04,000 --> 00:35:07,000
And then, you know,
that thing will just keep

390
00:35:07,000 --> 00:35:08,000
running.
So actually,

391
00:35:08,000 --> 00:35:13,000
if you just look at magnetic
fields and transformers or power

392
00:35:13,000 --> 00:35:15,000
adapters,
and things like that,

393
00:35:15,000 --> 00:35:18,000
you precisely extract energy
from the magnetic field.

394
00:35:18,000 --> 00:35:20,000
Of course, I mean,
you actually have to take some

395
00:35:20,000 --> 00:35:22,000
power supply to maintain the
magnetic fields.

396
00:35:22,000 --> 00:35:25,000
But, so a magnetic field,
you could actually try to get

397
00:35:25,000 --> 00:35:31,000
energy from it almost for free.
A gravitational field or an

398
00:35:31,000 --> 00:35:35,000
electric field,
you can't.

399
00:35:35,000 --> 00:35:41,000
OK, so and now why does that
hold?

400
00:35:41,000 --> 00:35:43,000
Well, if I have a gradient
field,

401
00:35:43,000 --> 00:35:47,000
then if I try to compute this
line integral,

402
00:35:47,000 --> 00:35:50,000
I know it will be the value of
the function at the end point

403
00:35:50,000 --> 00:35:52,000
minus the value at the starting
point.

404
00:35:52,000 --> 00:36:04,000
But, they are the same.
So, the value is the same.

405
00:36:04,000 --> 00:36:09,000
So, if I have a gradient field,
and I do the line integral,

406
00:36:09,000 --> 00:36:13,000
then I will get f at the
endpoint minus f at the starting

407
00:36:13,000 --> 00:36:23,000
point.
But, they're the same point,

408
00:36:23,000 --> 00:36:32,000
so that's zero.
OK, so just to reinforce my

409
00:36:32,000 --> 00:36:38,000
warning that not every field is
a gradient field,

410
00:36:38,000 --> 00:36:45,000
let's look again at our
favorite vector field from

411
00:36:45,000 --> 00:36:49,000
yesterday.
So, our favorite vector field

412
00:36:49,000 --> 00:36:54,000
yesterday was negative y and x.
It's a vector field that just

413
00:36:54,000 --> 00:36:59,000
rotates around the origin
counterclockwise.

414
00:36:59,000 --> 00:37:07,000
Well, we said,
say you take just the unit

415
00:37:07,000 --> 00:37:17,000
circle -- -- for example,
counterclockwise.

416
00:37:17,000 --> 00:37:23,000
Well, remember we said
yesterday that the line integral

417
00:37:23,000 --> 00:37:28,000
of F dr, maybe I should say F
dot T ds now,

418
00:37:28,000 --> 00:37:34,000
because the vector field is
tangent to the circle.

419
00:37:34,000 --> 00:37:43,000
So, on the unit circle,
F is tangent to the curve.

420
00:37:43,000 --> 00:37:49,000
And so, F dot T is length F
times, well, length T.

421
00:37:49,000 --> 00:37:53,000
But, T is a unit vector.
So, it's length F.

422
00:37:53,000 --> 00:37:57,000
And, the length of F on the
unit circle was just one.

423
00:37:57,000 --> 00:38:02,000
So, that's the integral of 1 ds.
So, it's just the length of the

424
00:38:02,000 --> 00:38:07,000
circle that's 2 pi.
And 2 pi is definitely not zero.

425
00:38:07,000 --> 00:38:13,000
So, this vector field is not
conservative.

426
00:38:13,000 --> 00:38:17,000
And so, now we know actually
it's not the gradient of

427
00:38:17,000 --> 00:38:22,000
anything because if it were a
gradient, then it would be

428
00:38:22,000 --> 00:38:28,000
conservative and it's not.
So, it's an example of a vector

429
00:38:28,000 --> 00:38:33,000
field that is not conservative.
It's not path independent

430
00:38:33,000 --> 00:38:38,000
either by the way because,
see, if I go from here to here

431
00:38:38,000 --> 00:38:43,000
along the upper half circle or
along the lower half circle,

432
00:38:43,000 --> 00:38:46,000
in one case I will get pi.
In the other case I will get

433
00:38:46,000 --> 00:38:49,000
negative pi.
I don't get the same answer,

434
00:38:49,000 --> 00:38:54,000
and so on, and so on.
It just fails to have all of

435
00:38:54,000 --> 00:38:59,000
these properties.
So, maybe I will write that

436
00:38:59,000 --> 00:39:06,000
down.
It's not conservative,

437
00:39:06,000 --> 00:39:21,000
not path independent.
It's not a gradient.

438
00:39:21,000 --> 00:39:29,000
It doesn't have any of these
properties.

439
00:39:29,000 --> 00:39:38,000
OK, any questions?
Yes?

440
00:39:38,000 --> 00:39:40,000
How do you determine whether
something is a gradient or not?

441
00:39:40,000 --> 00:39:44,000
Well, that's what we will see
on Tuesday.

442
00:39:44,000 --> 00:39:50,000
Yes?
Is it possible that it's

443
00:39:50,000 --> 00:39:52,000
conservative and not path
independent, or vice versa?

444
00:39:52,000 --> 00:39:54,000
The answer is no;
these two properties are

445
00:39:54,000 --> 00:39:57,000
equivalent, and we are going to
see that right now.

446
00:39:57,000 --> 00:40:12,000
At least that's the plan.
OK, yes?

447
00:40:12,000 --> 00:40:15,000
Let's see, so you said if it's
not path independent,

448
00:40:15,000 --> 00:40:19,000
then we cannot draw level
curves that are perpendicular to

449
00:40:19,000 --> 00:40:23,000
it at every point.
I wouldn't necessarily go that

450
00:40:23,000 --> 00:40:25,000
far.
You might be able to draw

451
00:40:25,000 --> 00:40:27,000
curves that are perpendicular to
it.

452
00:40:27,000 --> 00:40:32,000
But they won't be the level
curves of a function for which

453
00:40:32,000 --> 00:40:34,000
this is the gradient.
I mean, you might still have,

454
00:40:34,000 --> 00:40:36,000
you know,
if you take, say,

455
00:40:36,000 --> 00:40:40,000
take his gradient field and
scale it that in strange ways,

456
00:40:40,000 --> 00:40:42,000
you know, multiply by two in
some places,

457
00:40:42,000 --> 00:40:45,000
by one in other places,
by five and some other places,

458
00:40:45,000 --> 00:40:48,000
you will get something that
won't be conservative anymore.

459
00:40:48,000 --> 00:40:51,000
And it will still be
perpendicular to the curves.

460
00:40:51,000 --> 00:40:56,000
So, it's more subtle than that,
but certainly if it's not

461
00:40:56,000 --> 00:41:01,000
conservative then it's not a
gradient, and you cannot do what

462
00:41:01,000 --> 00:41:04,000
we said.
And how to decide whether it is

463
00:41:04,000 --> 00:41:06,000
or not, they'll be Tuesday's
topic.

464
00:41:06,000 --> 00:41:17,000
So, for now,
I just want to figure out again

465
00:41:17,000 --> 00:41:31,000
actually, let's now state all
these properties -- Actually,

466
00:41:31,000 --> 00:41:41,000
let me first do one minute of
physics.

467
00:41:41,000 --> 00:41:48,000
So, let me just tell you again
what's the physics in here.

468
00:41:48,000 --> 00:42:00,000
So, it's a force field is the
gradient of a potential -- --

469
00:42:00,000 --> 00:42:07,000
so, I'll still keep my plus
signs.

470
00:42:07,000 --> 00:42:13,000
So, maybe I should say this is
minus physics.

471
00:42:13,000 --> 00:42:20,000
[LAUGHTER]
So, the work of F is the change

472
00:42:20,000 --> 00:42:31,000
in value of potential from one
endpoint to the other endpoint.

473
00:42:31,000 --> 00:42:45,000
[PAUSE ] And -- -- so,
you know, you might know about

474
00:42:45,000 --> 00:42:58,000
gravitational fields,
or electrical -- -- fields

475
00:42:58,000 --> 00:43:13,000
versus gravitational -- -- or
electrical potential.

476
00:43:13,000 --> 00:43:16,000
And, in case you haven't done
any 8.02 yet,

477
00:43:16,000 --> 00:43:19,000
electrical potential is also
commonly known as voltage.

478
00:43:19,000 --> 00:43:27,000
It's the one that makes it hurt
when you stick your fingers into

479
00:43:27,000 --> 00:43:33,000
the socket.
[LAUGHTER] Don't try it.

480
00:43:33,000 --> 00:43:45,000
OK, and so now,
conservativeness means no

481
00:43:45,000 --> 00:44:01,000
energy can be extracted for free
-- -- from the field.

482
00:44:01,000 --> 00:44:05,000
You can't just have, you know,
a particle moving in that field

483
00:44:05,000 --> 00:44:08,000
and going on in definitely,
faster and faster,

484
00:44:08,000 --> 00:44:11,000
or if there's actually
friction,

485
00:44:11,000 --> 00:44:25,000
then keep moving.
So, total energy is conserved.

486
00:44:25,000 --> 00:44:29,000
And, I guess,
that's why we call that

487
00:44:29,000 --> 00:44:43,000
conservative.
OK, so let's end with the recap

488
00:44:43,000 --> 00:44:57,000
of various equivalent
properties.

489
00:44:57,000 --> 00:45:04,000
OK, so the first property that
I will have for a vector field

490
00:45:04,000 --> 00:45:12,000
is that it's conservative.
So, to say that a vector field

491
00:45:12,000 --> 00:45:22,000
with conservative means that the
line integral is zero along any

492
00:45:22,000 --> 00:45:26,000
closed curve.
Maybe to clarify,

493
00:45:26,000 --> 00:45:32,000
sorry, along all closed curves,
OK, every closed curve;

494
00:45:32,000 --> 00:45:36,000
give me any closed curve,
I get zero.

495
00:45:36,000 --> 00:45:44,000
So, now I claim this is the
same thing as a second property,

496
00:45:44,000 --> 00:45:53,000
which is that the line integral
of F is path independent.

497
00:45:53,000 --> 00:45:56,000
OK, so that means if I have two
paths with the same endpoint,

498
00:45:56,000 --> 00:45:58,000
then I will get always the same
answer.

499
00:45:58,000 --> 00:46:03,000
Why is that equivalent?
Well, let's say that I am path

500
00:46:03,000 --> 00:46:06,000
independent.
If I am path independent,

501
00:46:06,000 --> 00:46:10,000
then if I take a closed curve,
well, it has the same endpoints

502
00:46:10,000 --> 00:46:13,000
as just the curve that doesn't
move at all.

503
00:46:13,000 --> 00:46:16,000
So, path independence tells me
instead of going all around,

504
00:46:16,000 --> 00:46:21,000
I could just stay where I am.
And then, the work would just

505
00:46:21,000 --> 00:46:24,000
be zero.
So, if I path independent,

506
00:46:24,000 --> 00:46:28,000
tonight conservative.
Conversely, let's say that I'm

507
00:46:28,000 --> 00:46:32,000
just conservative and I want to
check path independence.

508
00:46:32,000 --> 00:46:37,000
Well, so I have two points,
and then I had to paths between

509
00:46:37,000 --> 00:46:39,000
that.
I want to show that the work is

510
00:46:39,000 --> 00:46:43,000
the same.
Well, how I do that?

511
00:46:43,000 --> 00:46:49,000
C1 and C2, well,
I observe that if I do C1 minus

512
00:46:49,000 --> 00:46:54,000
C2, I get a closed path.
If I go first from here to

513
00:46:54,000 --> 00:46:57,000
here, and then back along that
one, I get a closed path.

514
00:46:57,000 --> 00:47:01,000
So, if I am conservative,
I should get zero.

515
00:47:01,000 --> 00:47:05,000
But, if I get zero on C1 minus
C2, it means that the work on C1

516
00:47:05,000 --> 00:47:10,000
and the work on C2 are the same.
See, so it's the same.

517
00:47:10,000 --> 00:47:19,000
It's just a different way to
think about the situation.

518
00:47:19,000 --> 00:47:24,000
More things that are
equivalent, I have two more

519
00:47:24,000 --> 00:47:29,000
things to say.
The third one,

520
00:47:29,000 --> 00:47:38,000
it's equivalent to F being a
gradient field.

521
00:47:38,000 --> 00:47:46,000
OK, so this is equivalent to
the third property.

522
00:47:46,000 --> 00:47:59,000
F is a gradient field.
Why?

523
00:47:59,000 --> 00:48:02,000
Well, if we know that it's a
gradient field,

524
00:48:02,000 --> 00:48:05,000
that we've seen that we get
these properties out of the

525
00:48:05,000 --> 00:48:08,000
fundamental theorem.
The question is,

526
00:48:08,000 --> 00:48:12,000
if I have a conservative,
or path independent vector

527
00:48:12,000 --> 00:48:15,000
field, why is it the gradient of
something?

528
00:48:15,000 --> 00:48:25,000
OK, so this way is a
fundamental theorem.

529
00:48:25,000 --> 00:48:31,000
That way, well,
so that actually,

530
00:48:31,000 --> 00:48:43,000
let me just say that will be
how we find the potential.

531
00:48:43,000 --> 00:48:46,000
So, how do we find potential?
Well, let's say that I know the

532
00:48:46,000 --> 00:48:49,000
value of my potential here.
Actually, I get to choose what

533
00:48:49,000 --> 00:48:51,000
it is.
Remember, in physics,

534
00:48:51,000 --> 00:48:53,000
the potential is defined up to
adding or subtracting a

535
00:48:53,000 --> 00:48:56,000
constant.
What matters is only the change

536
00:48:56,000 --> 00:48:58,000
in potential.
So, let's say I know my

537
00:48:58,000 --> 00:49:01,000
potential here and I want to
know my potential here.

538
00:49:01,000 --> 00:49:04,000
What do I do?
Well, I take my favorite

539
00:49:04,000 --> 00:49:07,000
particle and I move it from here
to here.

540
00:49:07,000 --> 00:49:10,000
And, I look at the work done.
And that tells me how much

541
00:49:10,000 --> 00:49:14,000
potential has changed.
So, that tells me what the

542
00:49:14,000 --> 00:49:18,000
potential should be here.
And, this does not depend on my

543
00:49:18,000 --> 00:49:21,000
choice of path because I've
assumed that I'm path

544
00:49:21,000 --> 00:49:25,000
independence.
So, that's who we will do on

545
00:49:25,000 --> 00:49:29,000
Tuesday.
And, let me just state the

546
00:49:29,000 --> 00:49:36,000
fourth property that's the same.
So, all that stuff is the same

547
00:49:36,000 --> 00:49:42,000
as also four.
If I look at M dx N dy is

548
00:49:42,000 --> 00:49:48,000
what's called an exact
differential.

549
00:49:48,000 --> 00:49:52,000
So, what that means,
an exact differential,

550
00:49:52,000 --> 00:49:56,000
means that it can be put in the
form df for some function,

551
00:49:56,000 --> 00:49:58,000
f,
and just reformulating this

552
00:49:58,000 --> 00:50:02,000
thing, right,
because I'm saying I can just

553
00:50:02,000 --> 00:50:05,000
put it in the form f sub x dx
plus f sub y dy,

554
00:50:05,000 --> 00:50:08,000
which means my vector field was
a gradient field.

555
00:50:08,000 --> 00:50:13,000
So, these things are really the
same.

556
00:50:13,000 --> 00:50:16,000
OK, so after the weekend,
on Tuesday we will actually

557
00:50:16,000 --> 00:50:19,000
figure out how to decide whether
these things hold or not,

558
00:50:19,000 --> 00:50:22,000
and how to find the potential.