Mathematics - Multivariable Calculus - Lecture 19

Uploaded by UCBerkeley on 12.11.2009

We are studying line integrals.

Very exciting, right.
Line integrals -- actually today we will prove a theorem,
a result about line integrals, which is a first of a series of
results that I promised you last time which relate
integration in different dimension.
So, let's talk a little more about line integrals.

Last time we talked about line integrals of vector fields, but
actually this is not the only kind of a line integral that
you can have, and so today we'll talk about
other kinds as well.
Let me remind you what we learned last time.

Suppose that you have a vector field on the plane.

As always, the vector field can be written in two different
ways -- you can write it like this or you can write it like
this, and you can choose whatever way you prefer,
whichever way you like.
So you're given a -- this is a vector field, and in
addition you are given a curve on the plane.
Here's the curve with some -- in this case with two boundary
points, but in general we could actually also look at closed
curves, which don't have a boundary.
So, a line integral is a combination, is a pairing, if
you will, it's a marriage of a vector field and a curve it
takes two objects to produce an integral.
You need a vector field and you need a curve.
So that integral we write like this.
We write it f dot di over c.
And we write it in the following way. dr here, also
stands for vector which has two components, dx and dy, and when
you take the dot product of these two vectors -- this one
having these two components and that one having those
two components.
You will produce P dx plus Q dy.

And then we can write it down more concretely in a way
suitable for calculation by parameterizing the curve.
So we write x as a function of some auxiliary variable t, and
y is the function of some auxiliary variable t, where t
is between some values a and b where, say, this point will
correspond to t equal a, and this will point will
correspond to t equal b.
If you do that, you can actually write it as an
ordinary integral, just an ordinary 1-dimensional
integral from a to b with respect to dt now.
But as always there's sort of a price to pay, and the price to
pay is that we have to replace dx by dt, but we can't just
replace dx by dt, we write it as dx dt times dt.
So this quantity is reminiscent of Jacobian's.
Or even earlier in one variable calculus we always had to
factor when we made the change of variables.
And this kind of factors show up here as well.
So the end result that get P dx dt plus Q dy dt, dt.
And that's already an ordinary integral in t.
So, notice the difference between these two.
Here I write this integral as an integral over a curve, which
is on the plane, but here I rewrite it as an integral
over just the basic interval from a to b.
Just the basic line segment from a to b.
Certainly this is much simpler than this, and we've been able
to pass from this form to this form by using this
This permutization, as I already explained many times,
what it does it really identifies this complicate
curve with this simple line segment.
Those are the formulas which describe how
they're identified.

So this is sequel a again, this is sequel b, and all the values
in between correspond to some points on the curve.
All right, is that clear?
Could I do one simple example?
I did an example last time.
So last time I had an example, which was actually
slightly more complicated.
In my example last time I had vector fields defined in three
space, but let me just simplify that example a little bit and
explain this in the 3-dimensional case.
So, example, let's say you have a curve which x is equal to
t squared and y is equal to p cubed.
And so you have -- that's the curve [UNINTELLIGIBLE]
from 0 to 1.
And then you have the vector field which is let's just say x
and i plus xy squared times j.
It doesn't matter, it's just to illustrate this.
OK, so in this case what is a line integral?

You're going to have an integral from 0 to 1,
which has these limits.
And then you simply substitute, just substitute all of this
stuff into the formula.
So this is P, this is P, this is Q.
So you get this is P, and P written in terms of t variable
is t squared, then you have to write dxdt -- dx dt is 2 t dt
plus -- well, if I want to write it exactly in that form
I'll just put the dt outside.
Plus, if you put xy squared, xy squared is, what, t
to the eighth, right.
dy dt is 3t squared and then dt.
And so then you get a very simple integral.
A function in t, dt from 0 to 1.
Is that clear?
So I will not do this calculation of dt,
I'll just stop here.
All right, so, what do these integrals represent?
The last time we discussed, last time we discussed that
this integral, this integral represents the work done by
this vector field viewed as a force vector field, work done
by a force field f in moving an object or particle
along this curve c, right.
So, and when we were deriving it, the way we derived it is
as usual with integrals -- we break the curve
into small pieces.
We then approximate each piece by a line segment.
So let's say that's is -- I blew up one of those segments,
so that I say this is, that's this segment, which I blew up
on this board to make it easier to keep track of things.
On a small scale because the curve is smooth, we can
approximate it well by a line segment, by a linear thing, by
a piece of straight line.
And so let's say this is the piece number i.
So it will be let's say, in this case it might be a dozen
pieces, but eventually we'll make this partition finer and
finer for the thousand, million, billion and so on.
So let's focus on one of those piece.
When we focus on one of those pieces, we have -- here we have
with respect to the x-axis we have the distance as delta x,
delta xi, and here it will be delta yi, and then we will
have our force vector field.
And because we are in a very small scale, in a very small
piece of the curve, it would be a good approximation to think
that the vector field is just constant along this segment
because it's very small.
So it is going to look like this and so on.
So you will have some -- let's call this vector fi.

And so the work done, as I explained last time -- remember
I was erasing things -- the work done here is easy to find,
it's just the dot product of fi and delta r.
So the work, let's call it work number i, work done on the i
segment is just a dot product of this vector, which is red
one, and this one which is a yellow one.
And this yellow vector is, we'll call it delta ri, has two
components, one is delta xi, and the other one is delta yi.
So I want to dot those two those things -- fi dot delta
ri, both are vectors.

And more concretely then, this is going to be the first
component which will be some pi times delta xi plus a second
component ui times delta wi.
So that's the work done approximately,
approximate work.
I should say this is really approximate because we are
making two kinds of approximation here.
First of all, we're approximating the actual
segment of this curve.
It is curvy, even though it is very small, but it's curvy in
principle, but we are approximating it by
something linear.
That's the first approximation we make.
And the second approximation we make is that the vector field
also varies in general a little bit along this segment, but we
are assuming that actually doesn't vary, so we pick a
particular value somewhere in the middle and we just assume
that i stays the same along this segment.
So this is a kind of approximation that we
make at this point.
But the idea is that as our partition becomes finer and
finer, this approximation, the error of this approximation
will go away.
So we'll actually get the exact result.
So the next step, in other words, is to take the sum, so
take the total work, which is the sum of the work done on
each of those segments, and that's approximately
now equal to what?
To f dot fi dot delta ri, which can also be written as the
sum of P i delta xi plus Q i delta yi.

That's a formula we get.
I'm essentially repeating what I said last time, but you'll
see why -- you will see in a minute why I want
to repeat this.
So, always with integrals, this sum, as the partition becomes
finer, actually trends to the integral, what's
called the intregral.
So that's how we end up with this integral
and what integral?
Well, it will help f dot what you call dr, so delta r under
this limit becomes differential dr.
So we get fdr, that's this line integral which we are talking
about now precisely as a total work done along the curve.
And also you can think in terms of -- you can also think in
terms of this expression which is a more detailed expression
for this dot product, so that's how you get this second formula
on the left board, which is a pdx plus qdy.
It's the same thing, it's just writing the dot
product and component.
So that's what we've done so far and that's called line
integral -- line integral over vector field.
And I think the terminology's a little bit misleading because
it suggests, the line suggests that we are integrating over
something linear, but we're not.
I guess line here means to represent the fact that we're
doing a 1-dimensional integral, so something which is also -- a
line is 1-dimensional, so that's why we call it line
integral, but in fact I would rather call it a curvy integral
or a curve integral because we're really integrating
over a curve.
So, what I would like to say now is that this is not the
only type of line integral -- let's follow the standard
terminology, the only type of line integral that we can have.
As you see in this line integral, what we are doing is
at the end of the day we are calculating the expression like
this where there are two types of terms, one has this form,
some function times ds, and the second one is some
function times dy.
So this already makes you wonder what are this dx and dy
that what is it that allows us to insert them under the
integral and make the integration -- that's
the first question.
And a second natural question is whether there is something
else like dxdy that we could use potentially also to produce
line integral or integrals over the curves.
And the answer is that yes, there are actually other ways
to produce line integrals, which are also meaningful from
the point of view of applications, which have a
different interpretation though, not as work
done by force.
And In trying to generalize it, we simply have to go back and
look at this picture one more time.
So what exactly, what is the structure of the
integral that we obtain?
Well, the structure is really -- the structure of the
integral is really dictated by what we've done at one of
the segments, of the segment number i.
What we've done is we've taken the dot product between vector
fi and the displacement vector along this segment
of the curve.
But that expression came from physics, that was because we
tried to find the work done by the force.
But let's look at the result.
The result is really the sum of these two expressions, di
delta xi and qi delta yi.
These are really the basic expressions that we have
written down for this i segment, which have eventually
led us to consider those integrals.
So if you want to generalize this integral we should
do it at this step.
Once we write that i's term, essentially there is no choice,
there will be a particular integral as the limit when
we sum up over all i.
So we see what are we doing?
So we have this segment, and let's focus on
this term for example.
So this term is a product of two factors.
The first factor is a value of some function, so it's a number
essentially, but this number depends on the i segment, so at
the end of the day when we take the limit this will
become a function.
So it's some particular quantity which we
attach to the segment.
But the second factor actually has geometric meaning.
The second factor has something to do with the shape
of the segment.
It actually is the length of the projection of this line
segment onto the x-coordinate.
Of course, somewhere in the background I imagine
the coordinate system, xy-coordinate system.
And so this is a projection of this vector on to the
x-axis where delta wy is a projection onto this factor.
So the way to produce this integral in other words is
to combine two things.
One is a function, like p, and the other one is some length,
and this length could be the length of the projection onto
the x-axis or it could be the projection onto the y-axis.
And if somebody comes and chooses another coordinate
system, for example, polar coordinate system, which
would have instead of xy would have r and theta.
They could also take the displacement with respect
to one of the coordinates of that system.
For example, instead of delta x and delta y, you could have
taken delta r or delta theta if you were working with
polar coordinates.
So in other words there are many choices here.
We choose delta x and delta y for two reasons.
First of all, there is a very, in this case, there is a very
nice physical interpretation of the result, of the resulting
integral, and second of all, our xy-coordinate system is
somehow the most basic coordinate system
that we can have.
There is, however, one quantity, one geometric
quantity, one length attached to this picture which is
somehow intrinsic and doesn't depend on any coordinate, and
that's just the length of this integral.
So in retrospect, even though we've got now these very nice
integrals which have this very nice interpretation of the work
done by force and so on, we can now go back and in retrospect
see that actually perhaps the most natural thing to do, now
that we're starting to analyze it in more detail, would be
take instead of delta x or delta y just delta f, namely
the length of the segment.
So this way we get a different type of line integral, a
different type of line integral is obtained.
If we take instead of delta x or delta y -- let's even say
delta xi or delta yi, we take delta si, which is the
length of the segment, the length of the segment.
All right, so what do we get in this case?
So in this case, let's call this now, let's call this delta
si and of course, delta si is nothing but the square root
of detla xi squared plus delta wi squared, right.
So how about taking instead of this expression, take the
expression -- in fact, maybe let me go and this board so
that we can follow the analogy more closely.
So, take the expression, instead of this one, take the
expression some -- put some f i, where f is some function.
Some number -- well, in this case some number, because it's
attached to the i segment.
Times delta si, which you can also write as fi square
root of delta xi squared plus delta yi squared.
So if we take the sum now of this expression, we take the
sum of this expression, this will tend to a different kind
of integral, which it will still be an integral over c,
which we'll have f times dx squared plus dy squared.
So then, of course, we ask what does it mean to
have such an integral?
Well, to do justice to such an integral we really should
parameterize the curve the way we did here.
So let's go back to this permatization, when we
parameterize it like this we get this auxiliary variable t,
so now our curve gets replaced for all intensive purposes by a
simple line interval -- it makes things much easier.
So we will end up with an integral again from a to b
over this auxiliary variable.
We'll have here f where we'll substitute x as a function of
t and y as a function of t.
And then here I will make a trick.
I will divide everything by dt and multiply by dt.
But if I divide by dt, that means that under the square
root I divide by tt squared.
So I will end up with dxzt squared, dydt squared
-- you had a question.
Why did I change it to delta s?
Why are we doing all these integrals in the first place,
because we have to, no, because this is what we have to do in
this class, but no, because this integral makes sense and
then you need it in applications, for example, if
you want to calculate the work done by force.
By the way, this is not the only application
of line integrals.
We'll talk more about them today about the fundamental
theory, but for instance, if you look at the end of that
section, 16.2 in the exercises, there is discussion of the
Ampere Law, and Ampere's law is about line integrals of
magnetic field caused by current going -- say you have a
current going through a wire and this causes magnetic field,
so you would like to know the force of that magnetic field,
and you can find what the line integrals at magnetic field is
by knowing the current along this wire.
I'm getting to the answer of your question.
This is an introduction to the answer.
So this was the reason.
I'm explaining now a different type of integral, and I will
explain now that this too has application to
something meaningful.
But you can think about it, you could think about
these integrals from two points of view.
One is the kind of utilitarian point of view, which means why
do I need this, what is situation where I actually need
this integral, and for this we have answers.
of vector fields you have work done by force, for example.
But more than that, for instance, Ampere's law can
be understood in terms of line integral.
This type of integrals, as I will explain now, will allow
you to calculate things like masses of wires.
I'll get to this in a minute.
So this is sort of utilitarian point of view.
A second point of view is the more, is a more
abstract point of view.
We certainly know the utility of integrals in general, but so
far in one variable calculus, we would only calculate the
integral of a function over dt.
What I tried to explain in the last lecture is that you can
only understand the most beautiful results in calculus
by developing the most general possible integrals as you can
take, and the theory of most general integral.
So you can stand on the point of view of why are we doing --
so, I would turn your question around -- your question was why
don't we just stay with this, why do we take delta xi, is it
just to make our life more complicated?
I would turn it around and I say why only this?
Now that we understand this, why not do something more
general, you see, and then see if that has some
So, the integrals I'm talking about are the most general
integrals over a general curve on the plane.
That's the answer.
And this is one example, or perhaps -- I mean this example
leads to this integral, this is one example.
This is an example over integral of a vector field,
and this is another example.
You had a question.
That's right, you're right.
So delta si is approximately equal to the length
of the segment.
When I wrote length of a segment, I was implicitly
referring to this line segment, which was approximating my
curve, but if you would like to think of delta si as the
original one, then it would be approximate.
No, no, it's always that ds is always the differential, so it
is a linear increment, it is a linear increment.
So the question is about approximation.
Well, it's -- in some sense it's true.
It's depends on how exactly you introduce an implementation,
but if you introduce in a certain way, you're right,
that's what will happen.
So, I've given you this sort of a pep talk about the utility of
integrals, but now I have to explain what is
this useful for.
And this is useful for calculating things like
length, like the total length, for example.
What is the total length of this curve?
The total length of this curve is such an integral where
the function f is 1.
So this is the point.

So this integral, if f is equal to 1, for example -- by the
way, there is notation for this, which we'll
call, let's call ds.
So then if f is 1, fds is just the integral of ds.
This is just the length, arc length of c, which, of
course, something which we discussed before.
What we are doing now is we are generalizing that definition of
the arc length by allowing an arbitrary function
instead of 1.
And this is something which is useful because let's suppose
that you have a wire, a thin wire in the shape of this
curve, and let's suppose you want to find the
mass of this wire.
So mass of the wire is a funny thing?

I guess it was something else.
So we want to find mass of this wire.
Now, but if the wire, if it is homogeneous, that's going to be
the same as the length or is going to be proportional to the
length of some coefficient.
But suppose that it's heavier on this side and it's lighter
on this side, so it's not homogeneous.
What you have is a density function.
This is very similar to when we discussed masses
of 2-dimensional, 3-dimensional object.
So you're dealing with some density function, and that will
be your function f, and to compute the total mass you'd
have to integrate this function f with respect to this ds,
with respect to this measure.
In mathematics these kind of objects are called measures.
What I'm trying to explain to you is that the structure of
the integral, of this line integral, the basic structure
of this integral is as follows.
The integrand is always the product, which can be thought
of as a product, of something which is external like mass
density or vector field and something which is internal,
something which is intrinsic to that curve, and something
that has to do with the measurements along the curve.
It could be, it could the length of the projection
onto the x-axis.
It could be the length to the projection to the y-axis or it
could be the length itself.
Each time you get a particular integral.
So these are the two types of integral, and I know that this
is a little bit confusing because it looks like you have
two different things, and especially if you read in the
book, I think it is very confusing.
So what I was trying to explain to you is why there is such a
variety of integrals and what is the relation between them.
That's right.
So, in other words, what you're saying is you can observe the
analogy between the following terms: P dx, Q dy, or F ds.
So, you have the three different types of quantities
which you're allowed to integrate.
And the first factor here is something which is external, it
has to do with vector field or function, like mass density
function of the wire and things like that.
The second one is intrinsic to the curve itself.
It has to do with measuring things along this curve.
Here you measure something along the x direction, here you
measure along the y direction, here you just take the length
of this little piece, for each piece.
And these are not all possible ways, because -- well, in some
sense this is the most intrinsic one, because the
length is a length, it doesn't know about coordinates.
This is already a little bit arbitrary because we have
chosen a particular coordinate system, and as we know there
are many coordinate systems and essentially there is
no preferred one.
Because, for example, I could rotate the x and y coordinate
system, so that's already new coordinate system.
And we've learned to appreciate other coordinate system, for
example, when we did double integral.
Amongst the coordinate systems we know is the
polar coordinate system.
So, in fact, you could also have expressions like
P dr or Q d theta.
That's also allowed.
And each time you have such an integral, the way you compute
it is you parametrize your curve and you reduce the
integral to the integral over a line segment on the
auxiliary t line.
So in the case of a line integral of a vector field, you
end up with this integral, and in the case of a second type of
integral you end up with this type of integral.
But in both cases you get an integral just along a line
segment on the t line.

So, this was to explain the different types of integrals,
because I know that this is something which drives people
crazy because the ways it's explained in the book I think
is not really, how should I say, there's not enough
motivation given.
So what I've tried to do is to motivate them and
explain the difference.
But I have to say that we mostly will be interested in
line integrals of vector fields, not integrals of
functions so much with respect to the measure ds, but line
integrals of vector field -- this will be our main concern.
And here now, we will talk about the first example of a
kind of a fundamental theory of calculus applied to
this type of integral.
So to explain this, I have to introduce a particular kind of
vector field, which is actually something we've learned before.
So, let's remember the notion of gradient, gradient.
If you have a function -- again, I work on the plane.
So it's a function of two variables.
We can take partial derivatives of this function -- we have f
sub x and x sub y, and we can put them together in what we
call the gradient vector.

All right, this is the gradient vector.
When we started the gradient, we talked about the
gradient attached to a particular point, xy.
So it would be a particular point xy, and we would evaluate
the gradient, so we'll have a vector for that
particular point.
So it was kind of a static option -- it was attached
to a particular point.
But if you look at this formula, it looks like
actually we have a vector value function.
We actually have a vector attached to each point.
So this is precisely the kind of thing that we
now call a vector field.
So we can now look at the gradient and recognize that
the gradient of a function is actually a vector field.
This is a vector field.
Why, because at each point you have a vector.
At each point, xy, you have of vector, namely the vector
combining the two partial derivatives of this
function at this point.
And such vector fields are called conservative.
And conservative here is not meant to be like a conservative
liberal, but it's due to the different etymology
of the word.
Conservative because it conserves something -- it has
to do with conservation.
It's something we'll talk about later.
So it's called conservative, easy to remember.

Now, at Berkeley we don't like conservative thing, right.
But this vector fields are good, no problem.
So what can we say about line integrals of conservative
vector fields?
So this will be the main subject of the rest
of this lecture.
It turns out for such vector fields, we actually have an
analog of the fundamental theory of calculus, of the
one variable calculus, which looks at follows.

So, I preserve, I keep the same picture.
I have my curve c and I can see the line integral of a vector
field, it's just that now my vector field is going to be a
gradient vector field or conservative vector field.
It is a gradient of a function f.
And what can we learn about line integrals
of such vector fields?

So it turns out that we have the following
beautiful formula.
So on the left hand side, we have just the line integral
of this vector field over this curve.
And on the right hand side, we just evaluate the function
at the end point.

So you see, this is the first type of result that
I promised last time.
I promised last time that we will have results of
the following nature.
On the left hand side, we'll have an integral over some
domain d of something which is a derivative of another object.
And on the right hand side, we'll have an integral over
the boundary of this domain of that object itself.
And so you see in this particular case -- perhaps I
should explain that this is point a and this is point b.
So in this particular case, your domain z is this curve.
And the boundary of is domain consists of
two points, b and a.
So indeed on the left we get an integral over c, and on the
right we just get the values at the two end points.
This is a boundary, right, this is a boundary.

So what is going omega here?
Omega is, in this particular case, is just our function f.
So this integral just means evaluating this function f
at the point b minus value at point a.
You have a question.
Shouldn't be a double integral on the left hand side.
I'm sorry.
So, the question is whether we should put some double
integrals on the right.
But see, this is a very schematic form, so I'm
not keeping track of how many integration signs.
In fact, well, up to now we have an agreement that if
it's a double integral we put two integral signs.
If it's a three integral put three integral signs.
But in general you can just say all of them are integrals.
And so you -- we will know what kind of integral it is by
looking at the domain of integration, 2-dimensional,
3-dimensional, or 1-dimensional, right.
So this is schematic, but you're right, to be consistent
with all notation, the number of integration signs on the
left would have to be greater by one than the number of
integration signs on right.
the right.
So, maybe this minus 1, the number minus 1, and here
will be the certain number of integration sign.
But in this particular case, we do have a single integral on
the left, right, because it's an integral over the curve.
And on the right we have an integral over zero-dimensional
object, and usually integral over zero-dimensional
object, we never use the integration sign for that.
There will be 0 integration sign, because the
zero-dimensional object.
And integration over zero-dimensional set is just
evaluating your function at the points of that set, with signs
though, so it's important which signs we are taking.
And let me explain that -- that's actually a
very important point.
You see, when I parametrize my curve, this gives it
orientation because when I identify this curve with this
line segment, I'm using on this line segment I have a
natural orientation.
I'm going from lower value to higher value -- from
lower to greater value.
So here, a is smaller, a is less than d, of course.
Because I write it like this.
So when I identify my curve, c, with that line interval, this
also, it allows me to move the orientation to the curve.
In other words, the curve inherits the orientation of
that interval, which if this point corresponds to this and
this point corresponds to that, the orientation
would be like this.
But if it was the other way around the orientation would
be in the opposite direction.
So in this equation, we are taking the curve
oriented from a to b.
So it goes like this from a to b.
And on the right hand side with the end point evaluated the
function at that point minus a value at the initial point.
So it has to be consistent, because if it's not consistent
you will get an equation with the wrong sign -- you'll
have an overall sign.

What does f stand for.
It was on this board but I could close it. f is
the function. f of xy.
What does the big F mean?
What is the big F.
There is no big F here.
So the question is we see big F there and we see big F here,
but we don't see big F there.
So this is what I explained.
I said this is the most general line integral.
We take a most general vector field and we
take the line integral.
Amongst all vector fields, there are special ones which
are called conservative.
A vector field f is called conservative if it can be
written as nabla f, as gradient of some function.
Not all vector fields are like this.
But from now on or at this moment, we focus just on this
conservative vector field, vector fields which have
this shape, nabla f.
In this case, the line integral will look like this because the
vector field is nabla f, so we just put this nabla f in
the formula so that's the line integral.
The statement of this theorem, which is called the theorem
about -- elemental theorem for line integral is that in this
particular case, the line integral of this vector field
can be written in a much simpler way.
It's just the difference of the values of the function
at the end points.
So this is -- first of all, I want to say that this is a
natural generalization of the fundamental theorem of
calculus, which we learned for functions in one variable.
Very good.
So the question is does it mean that the path, going from a
to b, doesn't really matter.
This is absolutely correct, and this is one of the most
important corollaries of this, because on the left hand
side, it looks like the integral depends on the
choice of the path.
We could go like this or we could go like this or maybe
just take the most boring one is just a straight line.
But on the right, nothing uses the shape of a curve,
only the end points.
So this formula immediately tells us that actually this
line integral, which Apriori -- which a priori depends on
the choice of the path of c.
Actually it does not, only depends on the end points.
And that's one of the most beautiful things about it,
which you don't see -- which without really analyzing this
in detail, we wouldn't be able to see that.
It's a surprising fact about line integral.
And I'm getting to it in a moment.
But before I get to this I wanted to compare, I want to
compare this formula to the formula which we started with
last time, which is the fundamental theorem of calculus
for functions in one variable, which is that if you have a
function in one variable, then we have the formula like this
where you integrate over line segment from a to b, and
that's f of b minus f of a.
So you see there's a clear connection between the two
formulas, there's a clear analogy between
the two formulas.
In both cases on the left we have a single integral
of something which is a derivative.
Here is a function one variable so there's only one derivative.
Here we have a function in two variables, so there isn't a
single derivative, but there is a gradient, and what we can do
is we can take the line integral, the gradient
vector field.
And on the right hand side is exactly the same, it's just
evauating at the end points and taking the difference.
So, in fact, this is a special case of this more general
formula, this is a special case.
We have now generalized the fundamental theorem of
1-dimensional calculus to the case of line integrals
over arbitrary curves.
Now, let's look at this in more detail and let's try to figure
out some consequences of this.
What is this good for?
And the first consequence is the one which has already been
observed, is the independence of the line integral
of the path.
Which is really striking, if you think about it.
Because earlier today I wrote some formula for line integral,
right, some specific example.
And this example, it was sort of a painstaking process
because I have first of all parameterize my curve.
Then I had to express my vector field in terms of
this parameterization.
Then I had to take the derivative of x and y and I had
to put all this together and then finally integrate, right,
so it's a lot of work.
But now what this formula is telling that is that instead of
doing all of that, instead of going through this complicated
process, what I could do is just take some function --
there is some function, f, so that the result, the result is
just the difference between the values of function
at the end points.
So this is really striking fact.
So it's a shortcut.
It's a shortcut, it is a shortcut, so now you're happy.

OK, so everybody's happy when things become easier I guess,
but, you know, sometimes to get things easier, you have to
first make things more complicated to
make them easier.
So I think it was -- I forgot who it was who said that.
It was a physicist.
I think it was Edward Taylor who said what
is your goal in life?
He said my goal in life is to make complicated things easier
and easy things complicated.
So, in a sense -- that's not necessarily my
goal in this class.
But I agree that this is certainly simplification.
The right hand side, we will all agree that the right
hand side is much simpler than the left hand side.
But there's a caveat, there's a caveat to this, because
this does not apply to all vector fields.
This only applies to vector fields which are conservative.
So if you vector field is not conservative
you're out of luck.
There's no simplification, you have to go
through this process.
But there is a huge class of vector fields for which this
applies, and now we will discuss the which class, how to
find out, how to see whether your vector field is a good one
or allows for such simplification.
But before I do that I would like to derive the consequences
of this, or is this called corollaries.
So, corollary one is that if f is conservative, if f is a
conservative vector field, which is -- let's just
say is conservative.
When I say it's conservative it means that there is some
function f small such that f is equal to nabla of that
function, the gradient of that function.
Then the line integral of this vector field is independent of
the curve, depends only on the end points.

In other words, this is equal to the integral over a
different curve of any other curve provided that that other
curve has the same end points.
What they mean by this, of course, is just what I have
already drawn here, but let me draw it one more time.
So let's say you have these two end points. but so that's
this, this, this and so on.
I don't want to leave the blackboard, but you could -- I
could walk all over this classroom and then go outside
and come back.
And as long as I start here and come back here, the line
integral, no matter how complicated it would be to
calculate it, the result would be the same.
So this is really a striking fact, which enables us if the
question on the would be calculate over this curve, it
looks complicated but you know right away that instead you
could calculate over the straight line, which
is much easier.
So this really allows for great simplification.
Another way to phrase this, another way to phrase
it is the following.
Again, we'll be under the assumption that after the
conservative vector field, that if you integrate over a closed
curve -- a closed curve is one which does not have influence,
which does not have a boundary.
I'll draw it like this.
So this is called -- such a curve a closed curve.
Closed curves does not have a boundary.
We talked about this last time, like a circle, for example.
It's boundaries is empty.

So if you integrate over this, you will always get 0.
So in this case, you don't even need to choose a simple path,
you just know that you get 0.
How to see that.
Well, let's pick up a point here.
So you can think of this point, you can think of this curve as
a curve with a boundary, but it's boundary, in this
boundary, this one point appears in two different ways.
First of all, it appears as the initial point, something that
I have previously called a.
But second of all, the same point appears also as the
end point, a point which I previously called b.
It's one in the same point.
If you do this integral according to the fundamental
theorem for line integrals, and we are allowed to do this
because I'm under the assumption that my vector field
is the gradient of a function.
Then we know that this integral is equal to the difference
between the values at the point b and point a.
But point b and a are the same in this case.
So when I take the difference I get 0, you see.
So this is equal to -- this is explanation. f of b minus f of
a is 0, because it's the same point.
In fact, these two statements are equivalent to each other
for the following reason.
If you have two paths -- in this statement, which I call
corollary one, the emphasis always on the fact that if you
have two curves connecting two points, connecting the
same point, then the answer is the same.
But suppose I know that for any closed curve, for a closed
curve the integral is 0.
This is also implied -- this fact will be implied.
Because -- let me draw it on a separate board.
So, suppose I already know that the integral over a
closed curve, any closed curve, of fdr, is 0.
Then I claim that this means that integral over c -- and I
mean any, here I mean really any, any closed curve anywhere.
If I have an integral over some curve c, this will be equal to
the integral of the, of course, c-prime, as long as they
have the same end points.
How would I prove this?
Well, so here is my curve c, and here is my curve c-prime.

So to prove this is the same as to prove this minus this is 0.

But what is this minus this?
It means that I go like this, but then I change
orientation on this one.
So I go -- so, this is minus c-prime.

So it means I go like this, and I go like this.
And these two curves together combine into closed curve for
which the integral is 0.
You see, that's how I get at this 0.
So, c minus c-prime is a closed curve.
Is this clear?
Any questions about this?

How many people follow this argument?
How many people have not followed this argument?
Don't be afraid.
You didn't?
Not quite.
OK, so what's the second point?
Now, so this argument was clear -- this argument was clear why
the integral over the closed curve is 0 because you have the
same initial and end points.
So what I'm trying to explain now is that suppose I know,
suppose I have an integral -- suppose I have a vector field
such that for any closed curve, for any closed
cruve, the integral is 0.
I want to say, I want to demonstrate that in that case,
the integral over any curve, even with boundary like this,
would necessarily be equal to the integral over any other
curve with the same boundary.

In other words,
I would like to derive this formula from this formula, and
the way I do it is I say OK -- so, I wanted to show that this
is equal to this, but it's the same as saying that
this minus this is 0.
And now I explain that the difference between the two
integrals actually could be thought of as an integral over
the curve, which is c, the original one, this one, minus
c-prime, but c minus c-prime is actually a closed curve.
So since I had assumed that for all closed curves it was 0,
this implies that the difference is 0.
So this was just to say that these two properties are
actually equivalent to each other, so you should not think
of this corollary one and corollary two as two
separate statements.
One is past independent, one is past independence -- the
integral's the same over all paths with the same end point.
The other one is a statement that the integral over any
closed curve is 0, and the point is that these
two statements are equivalent each other.

so where does this leave us?
So, we see now that if you have a conservative vector field,
things simplify -- let's just look this way.
Things simplify a lot.
So of course then, the next question is how do we find out
if a given vector field is conservative or not, because if
it is conservative we can apply this and we have all this nice
properties, and if it's not then we have to use
other methods.
So this now is really the question.
How to determine whether f is conservative.

And if it is, how to find the function, function f, such that
this vector field is the gradient of that function.
That's the question.
As a warm-up, let's observe one property of
conservative vector field.

So, we are given -- so let's suppose that f is
convservative, so f is PQ, and let's suppose that it is
conservative, so it is nabla f.
So, in other words, let's start with, from the other end.
Let's suppose it is conservative, what
can we say about it?
What kind of properties does it have?
If it is conservative then f of B minus f of A is
a line integral, right.
That's the formula to have, but this is still way too abstract.
First of all, we talk about some curves and
things like that.
I would like some very simple algebraic criteria where I
don't do any integration, but just look at the components,
dQ, and I would like to have a criterion to say whether
it's conservative or not.

Does it have a proof.
Are we going to prove it in this class.
Well, I was hoping to prove it in this class, but I
may run out of time today.
So, the proof is very simple and it's the book.
The proof is follows almost immediately from mental theorem
of 1-dimensional calculus, when you just spell out
what this means.
In other words, when you parameterize the curve and
substitute everything, you just -- you get on the
knows the mental theorem of 1-dimensional calculus.
So I would like an algebraic criterion.
You see, I'd like an algebraic criterion which does
not involve any curves or any integration.
So see, what does this mean?
This means that p is f sub x.
And Q is f sub y, so some function y.
So here's what I can do.
I cannot compare these two guys, because a priori, the two
parts of derivatives that a priori have nothing to
do with each other.
But what I can do is, I can kind of equalize them
in the following way.
I can take P sub y.
I can take the partial derivative of P
with respect to y.
If I do that, I get f x, y.
On the other hand, I can take Q sub x.
That's a partial derivative of Q with respect to
x, and that's f y,x.

So now remember our good friend Clairaut told us that if you
have a function which has continuous first partial
derivatives, then it doesn't matter in which order you take
second partial derivatives.
Now these two thing are actually equal to each other.
So, what does it mean?
It means that if your vector field is conservative, it
will -- this property will necessarily have to hold.
In other words, P sub y will have to be equal to Q sub x.
That's a very simple--.

So, we conclude that in this case, P sub y is equal to
Q sub x, or if you will, dP dy is equal to dQ dx.
Here is an example.
So let's say the vector field is ye to the x plus sign y
times I plus e to the x plus cosine y plus 2yj.

So this is P and this is Q.
So let me ask you this question, is this vector
field conservative?

So, in other words, conservative would mean that
there is some function above it such that you get this by
taking f sub x and you get this by taking f sub y.
It's very difficult to guess right away just by looking at
those two functions, it's very difficult to guess whether such
a function f exists, because it's like you should be able to
extract in some sense the anti-derivative in x and y
and see that the result is somehow the same.
But now we have a very simple crterion as to whether it is
conservative or not, because we can argue as follows.
We can say well, if it is conservative, then dP y would
have to be equal to dQ x.
Are they equal?
So, the natural question is are they equal.
So let's calculate.
What is dP dy?

dP dy is e to the x plus cosine y.
Everybody agrees?
And now dQ dx is e to the x plus cosine y.
It's the same.
So there is a chance that it is conservative.
I have not yet -- so far what I've said is that if the vector
field is conservative, then this has to hold, and we see
that they are equal, so there is a potential, there is a
possibility that this that this vector field is, in
fact, conservative.
But it turns out, in fact, that in this particular case, this
is not only a necessary condition, but it is a
sufficient condition, and this vector field is, in
fact, conservative.
So what is the precise statement and how do we
find this function f in this particular case.
So, by the way, for this particular vector field then,
it would be very easy to calculate its line integral.
You don't need to parameterize anything, any curves and so on.
You just find this function f and you evaluate that
function at the end points.
For example, if there are no end points, if it's a closed
curve, you know the answer is zero right away, so it's a--.
How to find the function f, I'm getting to this.
No, no you haven't lost anything.
OK, what we've established so far is that if f is
conservative, then dpdy equals dqdx.
In other words, this statement implies this statement.

But actually the claim is that the converse is also true.
That if this formula is true, dP dy is equal to dQ dx, then
there exists a function f such that the vector field is
a gradient vector field.
Under one assumption, the converse is also true under the
assumption that the vector field is defined on the region
which is simply connected.
The region in our two.
Vector fields are defined on the plane, right.
So as defined on the region in r2, which is simply connected.

So I have to explain what simply connected means.
Simply connected simply means -- it doesn't simply mean
disconnected, it's a different property.

So, you see that connected, connected is easy, connected
means that -- this is disconnected, right, this is
disconnected, this is connected.
So this we know, that's easy.
Connected means there's only one piece, you can not
separate the object.
But simply connected is a slightly more
sophisticated property.
So, well, I drew it for -- I drew it in the case of
2-dimensional regions, but if you want, since we talk about
1-dimensional object curves, maybe it's better to just talk
about this curve as opposed -- the curve which is the union of
these two, as opposed to curve which is just this one.
So this is connected and this is disconnected.
But simply connected region -- well, it's true in both
settings, it's true for curves so it's going to be true for
regions like this, as you like.
But simply connected is something more complicated.
Simply connected means that any curve that you can draw in your
region, so let's suppose you have a region like this.
Simply connected means that if you draw any curve in this
region, like any closed curve like this, you can deform it so
as to under the deformation it all collapses into one
point within the region.
So in this particular case, I can deform it, I can
make it smaller, smaller, smaller, and like this.
Now contrast this to the case of a region which has a hole.
So it's like [? analise. ?]
So in this case, if I take this loop, I can not deform it
continuously to a point.
I can not collapse it into one point, because I can make it
smaller, all right, I can make it even more smaller, but I can
not pass through this hole.
So simply connected, put in the most down the earth terms,
it is a region which doesn't have any holes.
And holes, actually even if it's just the region was out
with one point thrown away, it's already
non-simply connected.
So this is simply connected and this not, and this is not,
because even here because this point is thrown out, I can
not pass by this point.
And this turns out to be the only obstacle to the converse
statement, which is a mild -- it's really a mild restriction,
because to begin with, most of our vector fields are actually
defined on the entire plane.
For example, this vector field is surely defined
on the entire plane.
In other words, you can put any x and y and it makes sense,
the formulas make sense.
So, and the entire plane is certainly simply connected.
So any disk on the plane is simply connected.
It's only when the vector field is not defined at a particular
point or it's not defined on the -- there is some hole like
this that you can not apply this result.
In fact, in the homework, in one of the exercise of the
homework, there is an example of a vector field, which is
defined on a non-simply connected domain for which,
what we are going to do now does not work, for which
this implication this way does not work.
Are the regions with holes [UNINTELLIGIBLE]
Very good point.
So these are actually connected, these are all
connected, right, because it's only one piece.
I can not separate it into two pieces, it's only
one piece, right.
So that's why I talked about connectivity first because I
wanted to emphasize that actually these are two
different properties.
You can have a connected region, but not
simply connected.
Let us assume though that our region, our vector field is
defined on a simply connected domain.
For example, of this vector field is, because it's defined
on the entire plane and the plane for sure is
simply connected.
Then I claim that actually the converse statement is true,
which means that if you know that dpdy is equal to dQ dx,
which is what we actually found to be the case for our vector
field, then there exists a function f so that our vector
field is a gradient of that.
So, now only that question remains is how do we find this
f, and that's done by a very simple algorithm where we
simply take anti-derivatives of this component of the vector
field -- an anti-derivative of the component of
the vector field.
So let's find f in our example.
So what I'm going to do is I'm going to take this, I'm going
to take this p and I'm going to take its anti-derivative.
Because p is -- so, we're looking for a function f such
that this is derivative with respect to x and this is a
derivative with respect to y.
But if so, we should be able to find f by taking
anti-derivative of this with respect to x, right.
So take anti-derivative of p with respect to x.

What do we get?
Well, so in other words, with respect to x means that we
treat y as a constant, not as a variable but as a
constant, x is a variable.
So when we take that anti-derivative we get ye
to the x plus x sign y.
Don't be surprised that you get x sign y, because b though it's
a sign y, but it's a sign of y which is now treated
as a constnat.
So, this is treated for all intensive purposes as a
constant with respect to x, so we just put an x in front,
that's what gives us the anti-derivative.
Now, if we were in one variable calculus, we actually wouldn't
stop here, we would write a constant.
This is very important to always remember that you can
always write a constant y because when you take the
derivative constant disappears.
So, there isn't a unique anti-derivative, there is
an anti-derivative up to addition of a constant.
But now we have two variables.
So this should be a constant with respect to the x
variable, but it may very well depend on y.
So what we should do, the only difference from old
calculations is that you should put dependence on y.
And now we find this function of y by comparing to
the result for Q.
I mean calculating the -- so, we sort of go
zig-zagging like this.
We start here, we take anti-derivative.
but we get some indeterminancy.
So, we resolve the indeterminancy by going
back to the second derivative, to f sub y.
So let's take, so the first step is like this, second is
differentiate with respect to y.
What do you get?
You get e to x plus x cosine y plus c-prime of y.
That's what we get.
But this is what we should get.
We should get e to the x plus x cosine y plus 2y.
So we recognize that the first terms have appeared.
And now we want to find what is c-prime.
Well, comparing to the formula for Q, we find that c-prime is
2y, and that means that c of y is y squared plus a constant,
which is now an honest constant because now we are dealing with
a function in one variable.
So the answer is this function plus y squared
plus al constant.
And that's how you find the function f for a
conservative vector field.
All right, so we'll continue next time.
Have a good weekend.