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So, let me remind you,
yesterday we've defined and
00:00:28.000 --> 00:00:34.000
started to compute line
integrals for work as a vector
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.
00:00:55.000 --> 00:01:04.000
And, we want to find the work
done along the curve.
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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
00:01:16.000 --> 00:01:19.000
where T is the unit tangent
vector,
00:01:19.000 --> 00:01:23.000
and ds is the arc length
element.
00:01:23.000 --> 00:01:30.000
Or, in coordinates,
that they integral of M dx N dy
00:01:30.000 --> 00:01:38.000
where M and N are the components
of the vector field.
00:01:38.000 --> 00:01:46.000
OK, so -- Let's do an example
that will just summarize what we
00:01:46.000 --> 00:01:51.000
did yesterday,
and then we will move on to
00:01:51.000 --> 00:01:57.000
interesting observations about
these things.
00:01:57.000 --> 00:02:03.000
So, here's an example we are
going to look at now.
00:02:03.000 --> 00:02:11.000
Let's say I give you the vector
field yi plus xj.
00:02:11.000 --> 00:02:13.000
So, it's not completely obvious
what it looks like,
00:02:13.000 --> 00:02:16.000
but here is a computer plot of
that vector field.
00:02:16.000 --> 00:02:21.000
So, that tells you a bit what
it does.
00:02:21.000 --> 00:02:24.000
It points in all sorts of
directions.
00:02:24.000 --> 00:02:31.000
And, let's say we want to find
the work done by this vector
00:02:31.000 --> 00:02:35.000
field.
If I move along this closed
00:02:35.000 --> 00:02:40.000
curve, I start at the origin.
But, I moved along the x-axis
00:02:40.000 --> 00:02:43.000
to one.
That move along the unit circle
00:02:43.000 --> 00:02:46.000
to the diagonal,
and then I move back to the
00:02:46.000 --> 00:02:54.000
origin in a straight line.
OK, so C consists of three
00:02:54.000 --> 00:03:06.000
parts -- -- so that you enclose
a sector of a unit disk -- --
00:03:06.000 --> 00:03:17.000
corresponding to angles between
zero and 45�.
00:03:17.000 --> 00:03:24.000
So, to compute this line
integral, all we have to do is
00:03:24.000 --> 00:03:33.000
we have set up three different
integrals and add that together.
00:03:33.000 --> 00:03:44.000
OK, so we need to set up the
integral of y dx plus x dy for
00:03:44.000 --> 00:03:54.000
each of these pieces.
So, let's do the first one on
00:03:54.000 --> 00:03:59.000
the x-axis.
Well, one way to parameterize
00:03:59.000 --> 00:04:06.000
that is just use the x variable.
And, say that because we are on
00:04:06.000 --> 00:04:12.000
the, let's see,
sorry, we are going from the
00:04:12.000 --> 00:04:17.000
origin to (1,0).
Well, we know we are on the
00:04:17.000 --> 00:04:22.000
x-axis.
So, y there is actually just
00:04:22.000 --> 00:04:25.000
zero.
And, the variable will be x
00:04:25.000 --> 00:04:28.000
from zero to one.
Or, if you prefer,
00:04:28.000 --> 00:04:31.000
you can parameterize things,
say, x equals t for t from zero
00:04:31.000 --> 00:04:36.000
to one, and y equals zero.
What doesn't change is y is
00:04:36.000 --> 00:04:41.000
zero, and therefore,
dy is also zero.
00:04:41.000 --> 00:04:48.000
So, in fact,
we are integrating y dx x dy,
00:04:48.000 --> 00:04:53.000
but that becomes,
well, zero dx 0,
00:04:53.000 --> 00:04:59.000
and that's just going to give
you zero.
00:04:59.000 --> 00:05:03.000
OK, so there's the line
integral.
00:05:03.000 --> 00:05:07.000
Here, it's very easy to compute.
Of course, you can also do it
00:05:07.000 --> 00:05:10.000
geometrically because
geometrically,
00:05:10.000 --> 00:05:12.000
you can see in the picture
along the x-axis,
00:05:12.000 --> 00:05:15.000
the vector field is pointing
vertically.
00:05:15.000 --> 00:05:19.000
If I'm on the x-axis,
my vector field is actually in
00:05:19.000 --> 00:05:23.000
the y direction.
So, it's perpendicular to my
00:05:23.000 --> 00:05:24.000
curve.
So, the work done is going to
00:05:24.000 --> 00:05:26.000
be zero.
F dot T will be zero.
00:05:48.000 --> 00:05:51.000
OK, so F dot T is zero,
so the integral is zero.
00:05:51.000 --> 00:05:57.000
OK, any questions about this
first part of the calculation?
00:05:57.000 --> 00:06:04.000
No? It's OK?
OK, let's move on to more
00:06:04.000 --> 00:06:11.000
interesting part of it.
Let's do the second part,
00:06:11.000 --> 00:06:18.000
which is a portion of the unit
circle.
00:06:18.000 --> 00:06:24.000
OK, so I should have drawn my
picture.
00:06:24.000 --> 00:06:37.000
And so now we are moving on
this part of the curve that's
00:06:37.000 --> 00:06:40.000
C2.
And, of course we have to
00:06:40.000 --> 00:06:43.000
choose how to express x and y in
terms of a single variable.
00:06:43.000 --> 00:06:46.000
Well, most likely,
when you are moving on a
00:06:46.000 --> 00:06:49.000
circle, you are going to use the
angle along the circle to tell
00:06:49.000 --> 00:06:53.000
you where you are.
OK, so we're going to use the
00:06:53.000 --> 00:06:56.000
angle theta as a parameter.
And we will say,
00:06:56.000 --> 00:07:02.000
we are on the unit circle.
So, x is cosine theta and y is
00:07:02.000 --> 00:07:05.000
sine theta.
What's the range of theta?
00:07:05.000 --> 00:07:11.000
Theta goes from zero to pi over
four, OK?
00:07:11.000 --> 00:07:15.000
So, whenever I see dx,
I will replace it by,
00:07:15.000 --> 00:07:19.000
well, the derivative of cosine
is negative sine.
00:07:19.000 --> 00:07:24.000
So, minus sine theta d theta,
and dy, the derivative of sine
00:07:24.000 --> 00:07:29.000
is cosine.
So, it will become cosine theta
00:07:29.000 --> 00:07:34.000
d theta.
OK, so I'm computing the
00:07:34.000 --> 00:07:41.000
integral of y dx x dy.
That means -- -- I'll be
00:07:41.000 --> 00:07:52.000
actually computing the integral
of, so, y is sine theta.
00:07:52.000 --> 00:08:01.000
dx, that's negative sine theta
d theta plus x cosine.
00:08:01.000 --> 00:08:08.000
dy is cosine theta d theta from
zero to pi/4.
00:08:08.000 --> 00:08:17.000
OK, so that's integral from
zero to pi / 4 of cosine squared
00:08:17.000 --> 00:08:22.000
minus sine squared.
And, if you know your trig,
00:08:22.000 --> 00:08:27.000
then you should recognize this
as cosine of two theta.
00:08:27.000 --> 00:08:33.000
OK, so that will integrate to
one half of sine two theta from
00:08:33.000 --> 00:08:36.000
zero to pi over four,
sorry.
00:08:36.000 --> 00:08:44.000
And, sine pi over two is one.
So, you will get one half.
00:08:44.000 --> 00:08:49.000
OK, any questions about this
one?
00:08:49.000 --> 00:09:07.000
No?
OK, then let's do the third one.
00:09:07.000 --> 00:09:15.000
So, the third guy is when we
come back to the origin along
00:09:15.000 --> 00:09:18.000
the diagonal.
OK, so we go in a straight line
00:09:18.000 --> 00:09:20.000
from this point.
Where's this point?
00:09:20.000 --> 00:09:25.000
Well, this point is one over
root two, one over root two.
00:09:25.000 --> 00:09:32.000
And, we go back to the origin.
OK, so we need to figure out a
00:09:32.000 --> 00:09:38.000
way to express x and y in terms
of the same parameter.
00:09:38.000 --> 00:09:43.000
So, one way which is very
natural would be to just say,
00:09:43.000 --> 00:09:47.000
well, let's say we move from
here to here over time.
00:09:47.000 --> 00:09:50.000
And, at time zero, we are here.
At time one, we are here.
00:09:50.000 --> 00:09:54.000
We know how to parameterize
this line.
00:09:54.000 --> 00:10:02.000
So, what we could do is say,
let's parameterize this line.
00:10:02.000 --> 00:10:09.000
So, we start at one over root
two, and we go down by one over
00:10:09.000 --> 00:10:19.000
root two in time one.
And, same with y.
00:10:19.000 --> 00:10:24.000
That's actually perfectly fine.
But that's unnecessarily
00:10:24.000 --> 00:10:27.000
complicated.
OK, why is a complicated?
00:10:27.000 --> 00:10:30.000
Because we will get all of
these expressions.
00:10:30.000 --> 00:10:34.000
It would be easier to actually
just look at motion in this
00:10:34.000 --> 00:10:38.000
direction and then say,
well, if we have a certain work
00:10:38.000 --> 00:10:42.000
if we move from here to here,
then the work done moving from
00:10:42.000 --> 00:10:46.000
here to here is just going to be
the opposite,
00:10:46.000 --> 00:10:48.000
OK?
So, in fact,
00:10:48.000 --> 00:10:55.000
we can do slightly better by
just saying, well,
00:10:55.000 --> 00:10:59.000
we'll take x = t,
y = t.
00:10:59.000 --> 00:11:07.000
t from zero to one over root
two, and take,
00:11:07.000 --> 00:11:15.000
well, sorry,
that gives us what I will call
00:11:15.000 --> 00:11:25.000
minus C3, which is C3 backwards.
And then we can say the
00:11:25.000 --> 00:11:31.000
integral for work along minus C3
is the opposite of the work
00:11:31.000 --> 00:11:35.000
along C3.
Or, if you're comfortable with
00:11:35.000 --> 00:11:39.000
integration where variables go
down,
00:11:39.000 --> 00:11:42.000
then you could also say that t
just goes from one over square
00:11:42.000 --> 00:11:45.000
root of two down to zero.
And, when you set up your
00:11:45.000 --> 00:11:49.000
integral, it will go from one
over root two to zero.
00:11:49.000 --> 00:11:50.000
And, of course,
that will be the negative of
00:11:50.000 --> 00:11:52.000
the one from zero to one over
root two.
00:11:52.000 --> 00:11:59.000
So, it's the same thing.
OK, so if we do it with this
00:11:59.000 --> 00:12:03.000
parameterization,
we'll get that,
00:12:03.000 --> 00:12:08.000
well of course,
dx is dt, dy is dt.
00:12:08.000 --> 00:12:16.000
So, the integral along minus C3
of y dx plus x dy is just the
00:12:16.000 --> 00:12:24.000
integral from zero to one over
root two of t dt plus t dt.
00:12:24.000 --> 00:12:30.000
Sorry, I'm messing up my
blackboard, OK,
00:12:30.000 --> 00:12:37.000
which is going to be,
well, the integral of 2t dt,
00:12:37.000 --> 00:12:46.000
which is t2 between these
bounds, which is one half.
00:12:46.000 --> 00:12:51.000
That's the integral along minus
C3, along the reversed path.
00:12:51.000 --> 00:13:03.000
And, if I want to do it along
C3 instead, then I just take the
00:13:03.000 --> 00:13:07.000
negative.
Or, if you prefer,
00:13:07.000 --> 00:13:11.000
you could have done it directly
with integral from one over root
00:13:11.000 --> 00:13:14.000
two, two zero,
which gives you immediately the
00:13:14.000 --> 00:13:19.000
negative one half.
OK, so at the end,
00:13:19.000 --> 00:13:28.000
we get that the total work --
-- was the sum of the three line
00:13:28.000 --> 00:13:32.000
integrals.
I'm not writing after dr just
00:13:32.000 --> 00:13:36.000
to save space.
But, zero plus one half minus
00:13:36.000 --> 00:13:39.000
one half, and that comes out to
zero.
00:13:39.000 --> 00:13:44.000
So, a lot of calculations for
nothing.
00:13:44.000 --> 00:13:49.000
OK, so that should give you
overview of various ways to
00:13:49.000 --> 00:13:57.000
compute line integrals.
Any questions about all that?
00:13:57.000 --> 00:14:03.000
No? OK.
So, next, let me tell you about
00:14:03.000 --> 00:14:07.000
how to avoid computing like
integrals.
00:14:07.000 --> 00:14:08.000
Well, one is easy:
don't take this class.
00:14:08.000 --> 00:14:17.000
But that's not,
so here's another way not to do
00:14:17.000 --> 00:14:20.000
it, OK?
So, let's look a little bit
00:14:20.000 --> 00:14:24.000
about one kind of vector field
that actually we've encountered
00:14:24.000 --> 00:14:26.000
a few weeks ago without saying
it.
00:14:26.000 --> 00:14:30.000
So, we said when we have a
function of two variables,
00:14:30.000 --> 00:14:32.000
we have the gradient vector.
Well, at the time,
00:14:32.000 --> 00:14:35.000
it was just a vector.
But, that vector depended on x
00:14:35.000 --> 00:14:36.000
and y.
So, in fact,
00:14:36.000 --> 00:14:43.000
it's a vector field.
OK, so here's an interesting
00:14:43.000 --> 00:14:48.000
special case.
Say that F, our vector field is
00:14:48.000 --> 00:14:52.000
actually the gradient of some
function.
00:14:52.000 --> 00:15:01.000
So, it's a gradient field.
And, so f is a function of two
00:15:01.000 --> 00:15:08.000
variables, x and y,
and that's called the potential
00:15:08.000 --> 00:15:12.000
for the vector field.
The reason is,
00:15:12.000 --> 00:15:16.000
of course, from physics.
In physics, you call potential,
00:15:16.000 --> 00:15:21.000
electrical potential or
gravitational potential,
00:15:21.000 --> 00:15:25.000
the potential energy.
This function of position that
00:15:25.000 --> 00:15:29.000
tells you how much actually
energy stored somehow by the
00:15:29.000 --> 00:15:33.000
force field, and this gradient
gives you the force.
00:15:33.000 --> 00:15:37.000
Actually, not quite.
If you are a physicist,
00:15:37.000 --> 00:15:40.000
that the force will be negative
the gradient.
00:15:40.000 --> 00:15:44.000
So, that means that physicists'
potentials are the opposite of a
00:15:44.000 --> 00:15:46.000
mathematician's potential.
Okay?
00:15:46.000 --> 00:15:48.000
So it's just here to confuse
you.
00:15:48.000 --> 00:15:50.000
It doesn't really matter all
the time.
00:15:50.000 --> 00:15:55.000
So to make things simpler we
are using this convention and
00:15:55.000 --> 00:15:59.000
you just put a minus sign if you
are doing physics.
00:15:59.000 --> 00:16:13.000
So then I claim that we can
simplify the evaluation of the
00:16:13.000 --> 00:16:21.000
line integral for work.
Perhaps you've seen in physics,
00:16:21.000 --> 00:16:24.000
the work done by,
say, the electrical force,
00:16:24.000 --> 00:16:28.000
is actually given by the change
in the value of a potential from
00:16:28.000 --> 00:16:30.000
the starting point of the ending
point,
00:16:30.000 --> 00:16:36.000
or same for gravitational force.
So, these are special cases of
00:16:36.000 --> 00:16:40.000
what's called the fundamental
theorem of calculus for line
00:16:40.000 --> 00:16:43.000
integrals.
So, the fundamental theorem of
00:16:43.000 --> 00:16:46.000
calculus, not for line
integrals, tells you if you
00:16:46.000 --> 00:16:49.000
integrate a derivative,
then you get back the function.
00:16:49.000 --> 00:16:52.000
And here, it's the same thing
in multivariable calculus.
00:16:52.000 --> 00:16:54.000
It tells you,
if you take the line integral
00:16:54.000 --> 00:16:58.000
of the gradient of a function,
what you get back is the
00:16:58.000 --> 00:16:58.000
function.
00:17:23.000 --> 00:17:30.000
OK,
so -- -- the fundamental
00:17:30.000 --> 00:17:43.000
theorem of calculus for line
integrals -- -- says if you
00:17:43.000 --> 00:17:58.000
integrate a vector field that's
the gradient of a function along
00:17:58.000 --> 00:18:03.000
a curve,
let's say that you have a curve
00:18:03.000 --> 00:18:06.000
that goes from some starting
point, P0,
00:18:06.000 --> 00:18:15.000
to some ending point, P1.
All you will get is the value
00:18:15.000 --> 00:18:21.000
of F at P1 minus the value of F
at P0.
00:18:21.000 --> 00:18:25.000
OK, so, that's a pretty nifty
formula that only works if the
00:18:25.000 --> 00:18:28.000
field that you are integrating
is a gradient.
00:18:28.000 --> 00:18:32.000
You know it's a gradient,
and you know the function,
00:18:32.000 --> 00:18:35.000
little f.
I mean, we can't put just any
00:18:35.000 --> 00:18:38.000
vector field in here.
We have to put the gradient of
00:18:38.000 --> 00:18:41.000
F.
So, actually on Tuesday we'll
00:18:41.000 --> 00:18:47.000
see how to decide whether a
vector field is a gradient or
00:18:47.000 --> 00:18:49.000
not,
and if it is a gradient,
00:18:49.000 --> 00:18:52.000
how to find the potential
function.
00:18:52.000 --> 00:18:58.000
So, we'll cover that.
But, for now we need to try to
00:18:58.000 --> 00:19:05.000
figure out a bit more about
this, what it says,
00:19:05.000 --> 00:19:11.000
what it means physically,
how to think of it
00:19:11.000 --> 00:19:15.000
geometrically,
and so on.
00:19:15.000 --> 00:19:18.000
So, maybe I should say,
if you're trying to write this
00:19:18.000 --> 00:19:21.000
in coordinates,
because that's also a useful
00:19:21.000 --> 00:19:24.000
way to think about it,
if I give you the line integral
00:19:24.000 --> 00:19:27.000
along C,
so, the gradient field,
00:19:27.000 --> 00:19:29.000
the components are f sub x and
f sub y.
00:19:29.000 --> 00:19:36.000
So, it means I'm actually
integrating f sub x dx plus f
00:19:36.000 --> 00:19:38.000
sub y dy.
Or, if you prefer,
00:19:38.000 --> 00:19:42.000
that's the same thing as
actually integrating df.
00:19:42.000 --> 00:19:46.000
So, I'm integrating the
differential of a function,
00:19:46.000 --> 00:19:54.000
f.
Well then, that's the change in
00:19:54.000 --> 00:19:56.000
F.
And, of course,
00:19:56.000 --> 00:20:02.000
if you write it in this form,
then probably it's quite
00:20:02.000 --> 00:20:06.000
obvious to you that this should
be true.
00:20:06.000 --> 00:20:11.000
I mean, in this form,
actually it's the same
00:20:11.000 --> 00:20:15.000
statement as in single variable
calculus.
00:20:15.000 --> 00:20:17.000
OK, and actually that's how we
prove the theorem.
00:20:17.000 --> 00:20:27.000
So, let's prove this theorem.
How do we prove it?
00:20:27.000 --> 00:20:31.000
Well, let's say I give you a
curve and I ask you to compute
00:20:31.000 --> 00:20:34.000
this integral.
How will you do that?
00:20:34.000 --> 00:20:38.000
Well, the way you compute the
integral actually is by choosing
00:20:38.000 --> 00:20:41.000
a parameter, and expressing
everything in terms of that
00:20:41.000 --> 00:20:46.000
parameter.
So, we'll set,
00:20:46.000 --> 00:20:57.000
well, so we know it's f sub x
dx plus f sub y dy.
00:20:57.000 --> 00:21:03.000
And, we'll want to parameterize
C in the form x equals x of t.
00:21:03.000 --> 00:21:09.000
y equals y of t.
So, if we do that,
00:21:09.000 --> 00:21:12.000
then dx becomes x prime of t
dt.
00:21:12.000 --> 00:21:25.000
dy becomes y prime of t dt.
So, we know x is x of t.
00:21:25.000 --> 00:21:31.000
That tells us dx is x prime of
t dt.
00:21:31.000 --> 00:21:38.000
y is y of t gives us dy is y
prime of t dt.
00:21:38.000 --> 00:21:52.000
So, now what we are integrating
actually becomes the integral of
00:21:52.000 --> 00:22:05.000
f sub x times dx dt plus f sub y
times dy dt times dt.
00:22:05.000 --> 00:22:09.000
OK, but now,
here I recognize a familiar
00:22:09.000 --> 00:22:13.000
guy.
I've seen this one before in
00:22:13.000 --> 00:22:15.000
the chain rule.
OK, this guy,
00:22:15.000 --> 00:22:19.000
by the chain rule,
is the rate of change of f if I
00:22:19.000 --> 00:22:22.000
take x and y to be functions of
t.
00:22:22.000 --> 00:22:26.000
And, I plug those into f.
So, in fact,
00:22:26.000 --> 00:22:34.000
what I'm integrating is df dt
when I think of f as a function
00:22:34.000 --> 00:22:42.000
of t by just plugging x and y as
functions of t.
00:22:42.000 --> 00:22:51.000
And so maybe actually I should
now say I have sometimes t goes
00:22:51.000 --> 00:22:59.000
from some initial time,
let's say, t zero to t one.
00:22:59.000 --> 00:23:03.000
And now, by the usual
fundamental theorem of calculus,
00:23:03.000 --> 00:23:07.000
I know that this will be just
the change in the value of f
00:23:07.000 --> 00:23:09.000
between t zero and t one.
00:23:36.000 --> 00:23:44.000
OK, so integral from t zero to
one of (df /dt) dt,
00:23:44.000 --> 00:23:52.000
well, that becomes f between t
zero and t one.
00:23:52.000 --> 00:23:55.000
f of what?
We just have to be a little bit
00:23:55.000 --> 00:23:58.000
careful here.
Well, it's not quite f of t.
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
00:24:02.000 --> 00:24:07.000
into it.
So, let me read that carefully.
00:24:07.000 --> 00:24:15.000
What I'm integrating to is f of
x of t and y of t.
00:24:15.000 --> 00:24:19.000
Does that sound fair?
Yeah, and so,
00:24:19.000 --> 00:24:23.000
when I plug in t1,
I get the point where I am at
00:24:23.000 --> 00:24:26.000
time t1.
That's the endpoint of my curve.
00:24:26.000 --> 00:24:31.000
When I plug t0,
I will get the starting point
00:24:31.000 --> 00:24:37.000
of my curve, p0.
And, that's the end of the
00:24:37.000 --> 00:24:43.000
proof.
It wasn't that hard, see?
00:24:43.000 --> 00:24:56.000
OK, so let's see an example.
Well, let's look at that
00:24:56.000 --> 00:25:00.000
example again.
So, we have this curve.
00:25:00.000 --> 00:25:03.000
We have this vector field.
Could it be that,
00:25:03.000 --> 00:25:05.000
by accident,
that vector field was a
00:25:05.000 --> 00:25:08.000
gradient field?
So, remember,
00:25:08.000 --> 00:25:12.000
our vector field was y,
x.
00:25:12.000 --> 00:25:16.000
Can we think of a function
whose derivative with respect to
00:25:16.000 --> 00:25:19.000
x is y, and derivative with
respect to y is x?
00:25:19.000 --> 00:25:27.000
Yeah, x times y sounds like a
good candidate where f( x,
00:25:27.000 --> 00:25:30.000
y) is xy.
OK,
00:25:30.000 --> 00:25:34.000
so that means that the line
integrals that we computed along
00:25:34.000 --> 00:25:39.000
these things can be just
evaluated from just finding out
00:25:39.000 --> 00:25:44.000
the values of f at the endpoint?
So, here's version two of my
00:25:44.000 --> 00:25:50.000
plot where I've added the
contour plot of a function,
00:25:50.000 --> 00:25:53.000
x, y on top of the vector
field.
00:25:53.000 --> 00:25:57.000
Actually, they have a vector
field is still pointing
00:25:57.000 --> 00:26:00.000
perpendicular to the level
curves that we have seen,
00:26:00.000 --> 00:26:02.000
just to remind you.
And, so now,
00:26:02.000 --> 00:26:05.000
when we move,
now when we move,
00:26:05.000 --> 00:26:09.000
the origin is on the level
curve, f equals zero.
00:26:09.000 --> 00:26:14.000
And, when we start going along
C1, we stay on f equals zero.
00:26:14.000 --> 00:26:17.000
So, there's no work.
The potential doesn't change.
00:26:17.000 --> 00:26:21.000
Then on C2, the potential
increases from zero to one half.
00:26:21.000 --> 00:26:24.000
The work is one half.
And then, on C3,
00:26:24.000 --> 00:26:27.000
we go back down from one half
to zero.
00:26:27.000 --> 00:26:40.000
The work is negative one half.
See, that was much easier than
00:26:40.000 --> 00:26:47.000
computing.
So, for example,
00:26:47.000 --> 00:26:53.000
the integral along C2 is
actually just,
00:26:53.000 --> 00:27:00.000
so, C2 goes from one zero to
one over root two,
00:27:00.000 --> 00:27:09.000
one over root two.
So, that's one half minus zero,
00:27:09.000 --> 00:27:18.000
and that's one half,
OK, because C2 was going here.
00:27:18.000 --> 00:27:26.000
And, at this point, f is zero.
At that point, f is one half.
00:27:26.000 --> 00:27:29.000
And, similarly for the others,
and of course when you sum,
00:27:29.000 --> 00:27:33.000
you get zero because the total
change in f when you go from
00:27:33.000 --> 00:27:35.000
here,
to here, to here, to here,
00:27:35.000 --> 00:27:37.000
eventually you are back at the
same place.
00:27:37.000 --> 00:27:44.000
So, f hasn't changed.
OK, so that's a neat trick.
00:27:44.000 --> 00:27:48.000
And it's important conceptually
because a lot of the forces are
00:27:48.000 --> 00:27:53.000
gradients of potentials,
namely, gravitational force,
00:27:53.000 --> 00:27:57.000
electric force.
The problem is not every vector
00:27:57.000 --> 00:28:00.000
field is a gradient.
A lot of vector fields are not
00:28:00.000 --> 00:28:02.000
gradients.
For example,
00:28:02.000 --> 00:28:08.000
magnetic fields certainly are
not gradients.
00:28:08.000 --> 00:28:33.000
So -- -- a big warning:
everything today only applies
00:28:33.000 --> 00:28:48.000
if F is a gradient field.
OK, it's not true otherwise.
00:29:07.000 --> 00:29:19.000
OK, still, let's see,
what are the consequences of
00:29:19.000 --> 00:29:30.000
the fundamental theorem?
So, just to put one more time
00:29:30.000 --> 00:29:39.000
this disclaimer,
if F is a gradient field -- --
00:29:39.000 --> 00:29:44.000
then what do we have?
Well, there's various nice
00:29:44.000 --> 00:29:47.000
features of work done by
gradient fields that are not too
00:29:47.000 --> 00:29:53.000
far off the vector fields.
So, one of them is this
00:29:53.000 --> 00:30:02.000
property of path independence.
OK, so the claim is if I have a
00:30:02.000 --> 00:30:05.000
line integral to compute,
that it doesn't matter which
00:30:05.000 --> 00:30:09.000
path I take as long as it goes
from point a to point b.
00:30:09.000 --> 00:30:14.000
It just depends on the point
where I start and the point
00:30:14.000 --> 00:30:18.000
where I end.
And, that's certainly false in
00:30:18.000 --> 00:30:22.000
general, but for a gradient
field that works.
00:30:22.000 --> 00:30:25.000
So if I have a point,
P0,
00:30:25.000 --> 00:30:28.000
a point, P1,
and I have two different paths
00:30:28.000 --> 00:30:33.000
that go there,
say, C1 and C2,
00:30:33.000 --> 00:30:39.000
so they go from the same point
to the same point but in
00:30:39.000 --> 00:30:44.000
different ways,
then in this situation,
00:30:44.000 --> 00:30:55.000
the line integral along C1 is
equal to the line integral along
00:30:55.000 --> 00:30:56.000
C2.
Well, actually,
00:30:56.000 --> 00:31:02.000
let me insist that this is only
for gradient fields by putting
00:31:02.000 --> 00:31:09.000
gradient F in here,
just so you don't get tempted
00:31:09.000 --> 00:31:19.000
to ever use this for a field
that's not a gradient field --
00:31:19.000 --> 00:31:28.000
-- if C1 and C2 have the same
start and end point.
00:31:28.000 --> 00:31:30.000
OK, how do you prove that?
Well, it's very easy.
00:31:30.000 --> 00:31:32.000
We just use the fundamental
theorem.
00:31:32.000 --> 00:31:35.000
It tells us,
if you compute the line
00:31:35.000 --> 00:31:38.000
integral along C1,
it's just F at this point minus
00:31:38.000 --> 00:31:41.000
F at this point.
If you do it for C2,
00:31:41.000 --> 00:31:45.000
well, the same.
So, they are the same.
00:31:45.000 --> 00:31:48.000
And for that you don't actually
even need to know what little f
00:31:48.000 --> 00:31:50.000
is.
You know in advance that it's
00:31:50.000 --> 00:31:53.000
going to be the same.
So, if I give you a vector
00:31:53.000 --> 00:31:56.000
field and I tell you it's the
gradient of mysterious function
00:31:56.000 --> 00:31:58.000
but I don't tell you what the
function is and you don't want
00:31:58.000 --> 00:32:00.000
to find out,
you can still use path
00:32:00.000 --> 00:32:03.000
independence,
but only if you know it's a
00:32:03.000 --> 00:32:03.000
gradient.
00:32:25.000 --> 00:32:35.000
OK, I guess this one is dead.
So, that will stay here forever
00:32:35.000 --> 00:32:40.000
because nobody is tall enough to
erase it.
00:32:40.000 --> 00:32:49.000
When you come back next year
and you still see that formula,
00:32:49.000 --> 00:32:53.000
you'll see.
Yes, but there's no useful
00:32:53.000 --> 00:32:59.000
information here.
That's a good point.
00:32:59.000 --> 00:33:06.000
OK, so what's another
consequence?
00:33:06.000 --> 00:33:14.000
So, if you have a gradient
field, it's what's called
00:33:14.000 --> 00:33:19.000
conservative.
OK, so what a conservative
00:33:19.000 --> 00:33:21.000
field?
Well, the word conservative
00:33:21.000 --> 00:33:26.000
comes from the idea in physics;
if the conservation of energy.
00:33:26.000 --> 00:33:31.000
It tells you that you cannot
get energy for free out of your
00:33:31.000 --> 00:33:33.000
force field.
So,
00:33:33.000 --> 00:33:36.000
what it means is that in
particular,
00:33:36.000 --> 00:33:39.000
if you take a closed
trajectory,
00:33:39.000 --> 00:33:44.000
so a trajectory that goes from
some point back to the same
00:33:44.000 --> 00:33:52.000
point,
so, if C is a closed curve,
00:33:52.000 --> 00:34:03.000
then the work done along C --
-- is zero.
00:34:03.000 --> 00:34:06.000
OK, that's the definition of
what it means to be
00:34:06.000 --> 00:34:09.000
conservative.
If I take any closed curve,
00:34:09.000 --> 00:34:13.000
the work will always be zero.
On the contrary,
00:34:13.000 --> 00:34:17.000
not conservative means
somewhere there is a curve along
00:34:17.000 --> 00:34:21.000
which the work is not zero.
If you find a curve where the
00:34:21.000 --> 00:34:23.000
work is zero,
that's not enough to say it's
00:34:23.000 --> 00:34:26.000
conservative.
You have show that no matter
00:34:26.000 --> 00:34:30.000
what curve I give you,
if it's a closed curve,
00:34:30.000 --> 00:34:34.000
it will always be zero.
So, what that means concretely
00:34:34.000 --> 00:34:37.000
is if you have a force field
that conservative,
00:34:37.000 --> 00:34:42.000
then you cannot build somehow
some perpetual motion out of it.
00:34:42.000 --> 00:34:44.000
You can't build something that
will just keep going just
00:34:44.000 --> 00:34:47.000
powered by that force because
that force is actually not
00:34:47.000 --> 00:34:50.000
providing any energy.
After you've gone one loop
00:34:50.000 --> 00:34:53.000
around, nothings happened from
the point of view of the energy
00:34:53.000 --> 00:34:57.000
provided by that force.
There's no work coming from the
00:34:57.000 --> 00:34:59.000
force,
while if you have a force field
00:34:59.000 --> 00:35:02.000
that's not conservative than you
can try to actually maybe find a
00:35:02.000 --> 00:35:04.000
loop where the work would be
positive.
00:35:04.000 --> 00:35:07.000
And then, you know,
that thing will just keep
00:35:07.000 --> 00:35:08.000
running.
So actually,
00:35:08.000 --> 00:35:13.000
if you just look at magnetic
fields and transformers or power
00:35:13.000 --> 00:35:15.000
adapters,
and things like that,
00:35:15.000 --> 00:35:18.000
you precisely extract energy
from the magnetic field.
00:35:18.000 --> 00:35:20.000
Of course, I mean,
you actually have to take some
00:35:20.000 --> 00:35:22.000
power supply to maintain the
magnetic fields.
00:35:22.000 --> 00:35:25.000
But, so a magnetic field,
you could actually try to get
00:35:25.000 --> 00:35:31.000
energy from it almost for free.
A gravitational field or an
00:35:31.000 --> 00:35:35.000
electric field,
you can't.
00:35:35.000 --> 00:35:41.000
OK, so and now why does that
hold?
00:35:41.000 --> 00:35:43.000
Well, if I have a gradient
field,
00:35:43.000 --> 00:35:47.000
then if I try to compute this
line integral,
00:35:47.000 --> 00:35:50.000
I know it will be the value of
the function at the end point
00:35:50.000 --> 00:35:52.000
minus the value at the starting
point.
00:35:52.000 --> 00:36:04.000
But, they are the same.
So, the value is the same.
00:36:04.000 --> 00:36:09.000
So, if I have a gradient field,
and I do the line integral,
00:36:09.000 --> 00:36:13.000
then I will get f at the
endpoint minus f at the starting
00:36:13.000 --> 00:36:23.000
point.
But, they're the same point,
00:36:23.000 --> 00:36:32.000
so that's zero.
OK, so just to reinforce my
00:36:32.000 --> 00:36:38.000
warning that not every field is
a gradient field,
00:36:38.000 --> 00:36:45.000
let's look again at our
favorite vector field from
00:36:45.000 --> 00:36:49.000
yesterday.
So, our favorite vector field
00:36:49.000 --> 00:36:54.000
yesterday was negative y and x.
It's a vector field that just
00:36:54.000 --> 00:36:59.000
rotates around the origin
counterclockwise.
00:36:59.000 --> 00:37:07.000
Well, we said,
say you take just the unit
00:37:07.000 --> 00:37:17.000
circle -- -- for example,
counterclockwise.
00:37:17.000 --> 00:37:23.000
Well, remember we said
yesterday that the line integral
00:37:23.000 --> 00:37:28.000
of F dr, maybe I should say F
dot T ds now,
00:37:28.000 --> 00:37:34.000
because the vector field is
tangent to the circle.
00:37:34.000 --> 00:37:43.000
So, on the unit circle,
F is tangent to the curve.
00:37:43.000 --> 00:37:49.000
And so, F dot T is length F
times, well, length T.
00:37:49.000 --> 00:37:53.000
But, T is a unit vector.
So, it's length F.
00:37:53.000 --> 00:37:57.000
And, the length of F on the
unit circle was just one.
00:37:57.000 --> 00:38:02.000
So, that's the integral of 1 ds.
So, it's just the length of the
00:38:02.000 --> 00:38:07.000
circle that's 2 pi.
And 2 pi is definitely not zero.
00:38:07.000 --> 00:38:13.000
So, this vector field is not
conservative.
00:38:13.000 --> 00:38:17.000
And so, now we know actually
it's not the gradient of
00:38:17.000 --> 00:38:22.000
anything because if it were a
gradient, then it would be
00:38:22.000 --> 00:38:28.000
conservative and it's not.
So, it's an example of a vector
00:38:28.000 --> 00:38:33.000
field that is not conservative.
It's not path independent
00:38:33.000 --> 00:38:38.000
either by the way because,
see, if I go from here to here
00:38:38.000 --> 00:38:43.000
along the upper half circle or
along the lower half circle,
00:38:43.000 --> 00:38:46.000
in one case I will get pi.
In the other case I will get
00:38:46.000 --> 00:38:49.000
negative pi.
I don't get the same answer,
00:38:49.000 --> 00:38:54.000
and so on, and so on.
It just fails to have all of
00:38:54.000 --> 00:38:59.000
these properties.
So, maybe I will write that
00:38:59.000 --> 00:39:06.000
down.
It's not conservative,
00:39:06.000 --> 00:39:21.000
not path independent.
It's not a gradient.
00:39:21.000 --> 00:39:29.000
It doesn't have any of these
properties.
00:39:29.000 --> 00:39:38.000
OK, any questions?
Yes?
00:39:38.000 --> 00:39:40.000
How do you determine whether
something is a gradient or not?
00:39:40.000 --> 00:39:44.000
Well, that's what we will see
on Tuesday.
00:39:44.000 --> 00:39:50.000
Yes?
Is it possible that it's
00:39:50.000 --> 00:39:52.000
conservative and not path
independent, or vice versa?
00:39:52.000 --> 00:39:54.000
The answer is no;
these two properties are
00:39:54.000 --> 00:39:57.000
equivalent, and we are going to
see that right now.
00:39:57.000 --> 00:40:12.000
At least that's the plan.
OK, yes?
00:40:12.000 --> 00:40:15.000
Let's see, so you said if it's
not path independent,
00:40:15.000 --> 00:40:19.000
then we cannot draw level
curves that are perpendicular to
00:40:19.000 --> 00:40:23.000
it at every point.
I wouldn't necessarily go that
00:40:23.000 --> 00:40:25.000
far.
You might be able to draw
00:40:25.000 --> 00:40:27.000
curves that are perpendicular to
it.
00:40:27.000 --> 00:40:32.000
But they won't be the level
curves of a function for which
00:40:32.000 --> 00:40:34.000
this is the gradient.
I mean, you might still have,
00:40:34.000 --> 00:40:36.000
you know,
if you take, say,
00:40:36.000 --> 00:40:40.000
take his gradient field and
scale it that in strange ways,
00:40:40.000 --> 00:40:42.000
you know, multiply by two in
some places,
00:40:42.000 --> 00:40:45.000
by one in other places,
by five and some other places,
00:40:45.000 --> 00:40:48.000
you will get something that
won't be conservative anymore.
00:40:48.000 --> 00:40:51.000
And it will still be
perpendicular to the curves.
00:40:51.000 --> 00:40:56.000
So, it's more subtle than that,
but certainly if it's not
00:40:56.000 --> 00:41:01.000
conservative then it's not a
gradient, and you cannot do what
00:41:01.000 --> 00:41:04.000
we said.
And how to decide whether it is
00:41:04.000 --> 00:41:06.000
or not, they'll be Tuesday's
topic.
00:41:06.000 --> 00:41:17.000
So, for now,
I just want to figure out again
00:41:17.000 --> 00:41:31.000
actually, let's now state all
these properties -- Actually,
00:41:31.000 --> 00:41:41.000
let me first do one minute of
physics.
00:41:41.000 --> 00:41:48.000
So, let me just tell you again
what's the physics in here.
00:41:48.000 --> 00:42:00.000
So, it's a force field is the
gradient of a potential -- --
00:42:00.000 --> 00:42:07.000
so, I'll still keep my plus
signs.
00:42:07.000 --> 00:42:13.000
So, maybe I should say this is
minus physics.
00:42:13.000 --> 00:42:20.000
[LAUGHTER]
So, the work of F is the change
00:42:20.000 --> 00:42:31.000
in value of potential from one
endpoint to the other endpoint.
00:42:31.000 --> 00:42:45.000
[PAUSE ] And -- -- so,
you know, you might know about
00:42:45.000 --> 00:42:58.000
gravitational fields,
or electrical -- -- fields
00:42:58.000 --> 00:43:13.000
versus gravitational -- -- or
electrical potential.
00:43:13.000 --> 00:43:16.000
And, in case you haven't done
any 8.02 yet,
00:43:16.000 --> 00:43:19.000
electrical potential is also
commonly known as voltage.
00:43:19.000 --> 00:43:27.000
It's the one that makes it hurt
when you stick your fingers into
00:43:27.000 --> 00:43:33.000
the socket.
[LAUGHTER] Don't try it.
00:43:33.000 --> 00:43:45.000
OK, and so now,
conservativeness means no
00:43:45.000 --> 00:44:01.000
energy can be extracted for free
-- -- from the field.
00:44:01.000 --> 00:44:05.000
You can't just have, you know,
a particle moving in that field
00:44:05.000 --> 00:44:08.000
and going on in definitely,
faster and faster,
00:44:08.000 --> 00:44:11.000
or if there's actually
friction,
00:44:11.000 --> 00:44:25.000
then keep moving.
So, total energy is conserved.
00:44:25.000 --> 00:44:29.000
And, I guess,
that's why we call that
00:44:29.000 --> 00:44:43.000
conservative.
OK, so let's end with the recap
00:44:43.000 --> 00:44:57.000
of various equivalent
properties.
00:44:57.000 --> 00:45:04.000
OK, so the first property that
I will have for a vector field
00:45:04.000 --> 00:45:12.000
is that it's conservative.
So, to say that a vector field
00:45:12.000 --> 00:45:22.000
with conservative means that the
line integral is zero along any
00:45:22.000 --> 00:45:26.000
closed curve.
Maybe to clarify,
00:45:26.000 --> 00:45:32.000
sorry, along all closed curves,
OK, every closed curve;
00:45:32.000 --> 00:45:36.000
give me any closed curve,
I get zero.
00:45:36.000 --> 00:45:44.000
So, now I claim this is the
same thing as a second property,
00:45:44.000 --> 00:45:53.000
which is that the line integral
of F is path independent.
00:45:53.000 --> 00:45:56.000
OK, so that means if I have two
paths with the same endpoint,
00:45:56.000 --> 00:45:58.000
then I will get always the same
answer.
00:45:58.000 --> 00:46:03.000
Why is that equivalent?
Well, let's say that I am path
00:46:03.000 --> 00:46:06.000
independent.
If I am path independent,
00:46:06.000 --> 00:46:10.000
then if I take a closed curve,
well, it has the same endpoints
00:46:10.000 --> 00:46:13.000
as just the curve that doesn't
move at all.
00:46:13.000 --> 00:46:16.000
So, path independence tells me
instead of going all around,
00:46:16.000 --> 00:46:21.000
I could just stay where I am.
And then, the work would just
00:46:21.000 --> 00:46:24.000
be zero.
So, if I path independent,
00:46:24.000 --> 00:46:28.000
tonight conservative.
Conversely, let's say that I'm
00:46:28.000 --> 00:46:32.000
just conservative and I want to
check path independence.
00:46:32.000 --> 00:46:37.000
Well, so I have two points,
and then I had to paths between
00:46:37.000 --> 00:46:39.000
that.
I want to show that the work is
00:46:39.000 --> 00:46:43.000
the same.
Well, how I do that?
00:46:43.000 --> 00:46:49.000
C1 and C2, well,
I observe that if I do C1 minus
00:46:49.000 --> 00:46:54.000
C2, I get a closed path.
If I go first from here to
00:46:54.000 --> 00:46:57.000
here, and then back along that
one, I get a closed path.
00:46:57.000 --> 00:47:01.000
So, if I am conservative,
I should get zero.
00:47:01.000 --> 00:47:05.000
But, if I get zero on C1 minus
C2, it means that the work on C1
00:47:05.000 --> 00:47:10.000
and the work on C2 are the same.
See, so it's the same.
00:47:10.000 --> 00:47:19.000
It's just a different way to
think about the situation.
00:47:19.000 --> 00:47:24.000
More things that are
equivalent, I have two more
00:47:24.000 --> 00:47:29.000
things to say.
The third one,
00:47:29.000 --> 00:47:38.000
it's equivalent to F being a
gradient field.
00:47:38.000 --> 00:47:46.000
OK, so this is equivalent to
the third property.
00:47:46.000 --> 00:47:59.000
F is a gradient field.
Why?
00:47:59.000 --> 00:48:02.000
Well, if we know that it's a
gradient field,
00:48:02.000 --> 00:48:05.000
that we've seen that we get
these properties out of the
00:48:05.000 --> 00:48:08.000
fundamental theorem.
The question is,
00:48:08.000 --> 00:48:12.000
if I have a conservative,
or path independent vector
00:48:12.000 --> 00:48:15.000
field, why is it the gradient of
something?
00:48:15.000 --> 00:48:25.000
OK, so this way is a
fundamental theorem.
00:48:25.000 --> 00:48:31.000
That way, well,
so that actually,
00:48:31.000 --> 00:48:43.000
let me just say that will be
how we find the potential.
00:48:43.000 --> 00:48:46.000
So, how do we find potential?
Well, let's say that I know the
00:48:46.000 --> 00:48:49.000
value of my potential here.
Actually, I get to choose what
00:48:49.000 --> 00:48:51.000
it is.
Remember, in physics,
00:48:51.000 --> 00:48:53.000
the potential is defined up to
adding or subtracting a
00:48:53.000 --> 00:48:56.000
constant.
What matters is only the change
00:48:56.000 --> 00:48:58.000
in potential.
So, let's say I know my
00:48:58.000 --> 00:49:01.000
potential here and I want to
know my potential here.
00:49:01.000 --> 00:49:04.000
What do I do?
Well, I take my favorite
00:49:04.000 --> 00:49:07.000
particle and I move it from here
to here.
00:49:07.000 --> 00:49:10.000
And, I look at the work done.
And that tells me how much
00:49:10.000 --> 00:49:14.000
potential has changed.
So, that tells me what the
00:49:14.000 --> 00:49:18.000
potential should be here.
And, this does not depend on my
00:49:18.000 --> 00:49:21.000
choice of path because I've
assumed that I'm path
00:49:21.000 --> 00:49:25.000
independence.
So, that's who we will do on
00:49:25.000 --> 00:49:29.000
Tuesday.
And, let me just state the
00:49:29.000 --> 00:49:36.000
fourth property that's the same.
So, all that stuff is the same
00:49:36.000 --> 00:49:42.000
as also four.
If I look at M dx N dy is
00:49:42.000 --> 00:49:48.000
what's called an exact
differential.
00:49:48.000 --> 00:49:52.000
So, what that means,
an exact differential,
00:49:52.000 --> 00:49:56.000
means that it can be put in the
form df for some function,
00:49:56.000 --> 00:49:58.000
f,
and just reformulating this
00:49:58.000 --> 00:50:02.000
thing, right,
because I'm saying I can just
00:50:02.000 --> 00:50:05.000
put it in the form f sub x dx
plus f sub y dy,
00:50:05.000 --> 00:50:08.000
which means my vector field was
a gradient field.
00:50:08.000 --> 00:50:13.000
So, these things are really the
same.
00:50:13.000 --> 00:50:16.000
OK, so after the weekend,
on Tuesday we will actually
00:50:16.000 --> 00:50:19.000
figure out how to decide whether
these things hold or not,
00:50:19.000 --> 00:50:22.000
and how to find the potential.