# Lec 21 | MIT 18.02 Multivariable Calculus, Fall 2007

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We were looking at vector fields last time.
Last time we saw that if a vector field happens to be a
gradient field -- -- then the line integral can be computed
actually by taking the change in value of the potential between
the end point and the starting point of the curve.
If we have a curve c, from a point p0 to a point p1
then the line integral for work depends only on the end points
and not on the actual path we chose.
We say that the line integral is path independent.
And we also said that the vector field is conservative
because of conservation of energy which tells you if you
start at a point and you come back to the same point then you
haven't gotten any work out of that force.
If we have a closed curve then the line integral for work is
just zero. And, basically,
we say that these properties are equivalent being a gradient
field or being path independent or being conservative.
And what I promised to you is that today we would see a
criterion to decide whether a vector field is a gradient field
or not and how to find the potential function if it is a
gradient field. So, that is the topic for today.
The question is testing whether a given vector field,
let's say M and N compliments, is a gradient field.
For that, well, let's start with an
observation. Say that it is a gradient field.
That means that the first component of a field is just the
partial of f with respect to some variable x and the second
component is the partial of f with respect to y.
Now we have seen an interesting property of the second partial
derivatives of the function, which is if you take the
partial derivative first with respect to x,
then with respect to y, or first with respect to y,
then with respect to x you get the same thing.
We know f sub xy equals f sub yx, and that means M sub y
equals N sub x. If you have a gradient field
then it should have this property.
You take the y component, take the derivative with
respect to x, take the x component,
differentiate with respect to y,
you should get the same answer. And that is important to know.
So, I am going to put that in a box.
It is a broken box. The claim that I want to make
is that there is a converse of sorts.
This is actually basically all we need to check.
Conversely, if, and I am going to put here a
condition, My equals Nx, then F is a gradient field.
What is the condition that I need to put here?
Well, we will see a more precise version of that next
week. But for now let's just say if
our vector field is defined and differentiable everywhere in the
plane. We need, actually,
a vector field that is well-defined everywhere.
You are not allowed to have somehow places where it is not
well-defined. Otherwise, actually,
you have a counter example on your problem set this week.
If you look at the last problem on the problem set this week,
it gives you a vector field that satisfies this condition
everywhere where it is defined. But, actually,
there is a point where it is not defined.
And that causes it, actually, to somehow -- I mean
everything that I am going to say today breaks down for that
example because of that. I mean, we will shed more light
on this a bit later with the notion of simply connected
regions and so on. But for now let's just say if
it is defined everywhere and it satisfies this criterion then it
is a gradient field. If you ignore the technical
condition, being a gradient field means essentially the same
thing as having this property. That is what we need to check.
Let's look at an example. Well, one vector field that we
have been looking at a lot was - yi xj.
Remember that was the vector field that looked like a
rotation at the unit speed. I think last time we already
decided that this guy should not be allowed to be a gradient
field and should not be conservative because if we
integrate on the unit circle then we would get a positive
answer. But let's check that indeed it
fails our test. Well, let's call this M and
let's call this guy N. If you look at partial M,
partial y, that is going to be a negative one.
If you take partial N, partial x, that is going to be
one. These are not the same.
So, indeed, this is not a gradient field.
Any questions about that? Yes?
Your question is if I have the property M sub y equals N sub x
only in a certain part of a plane for some values of x and
y, can I conclude these things?
And it is a gradient field in that part of the plane and
conservative and so on. The answer for now is,
in general, no. And when we spend a bit more
time on it, actually, maybe I should move that up.
Maybe we will talk about it later this week instead of when
I had planned. There is a notion what it means
for a region to be without holes.
Basically, if you have that kind of property in a region
that doesn't have any holes inside it then things will work.
The problem comes from a vector field satisfying this criterion
in a region but it has a hole in it.
Because what you don't know is whether your potential is
actually well-defined and takes the same value when you move all
around the hole. It might come back to take a
different value. If you look carefully and think
hard about the example in the problem sets that is exactly
what happens there. Again, I will say more about
that later. For now we basically need our
function to be, I mean,
I should still say if you have this property for a vector field
that is not quite defined everywhere,
you are more than welcome, you know,
you should probably still try to look for a potential using
methods that we will see. But something might go wrong
later. You might end up with a
potential that is not well-defined.
Let's do another example. Let's say that I give you this
vector field. And this a here is a number.
The question is for which value of a is this going to be
possibly a gradient? If you have your flashcards
then that is a good time to use them to vote,
assuming that the number is small enough to be made with.
Let's try to think about it. We want to call this guy M.
We want to call that guy N. And we want to test M sub y
versus N sub x. I don't see anyone.
I see people doing it with their hands, and that works very
well. OK.
The question is for which value of a is this a gradient?
I see various people with the correct answer.
OK. That a strange answer.
That is a good answer. OK.
The vote seems to be for a equals eight.
Let's see. What if I take M sub y?
That is going to be just ax. And N sub x?
That is 8x. I would like a equals eight.
By the way, when you set these two equal to each other,
they really have to be equal everywhere.
You don't want to somehow solve for x or anything like that.
You just want these expressions, in terms of x and
y, to be the same quantities. I mean you cannot say if x
equals z they are always equal. Yeah, that is true.
But that is not what we are asking.
Now we come to the next logical question.
Let's say that we have passed the test.
We have put a equals eight in here.
Now it should be a gradient field.
The question is how do we find the potential?
That becomes eight from now on. The question is how do we find
the function which has this as gradient?
One option is to try to guess. Actually, quite often you will
succeed that way. But that is not a valid method
on next week's test. We are going to see two
different systematic methods. And you should be using one of
these because guessing doesn't always work.
And, actually, I can come up with examples
where if you try to guess you will surely fail.
I can come up with trick ones, but I don't want to put that on
the test. The next stage is finding the
potential. And let me just emphasize that
we can only do that if step one was successful.
If we have a vector field that cannot possibly be a gradient
then we shouldn't try to look for a potential.
It is kind of obvious but is probably worth pointing out.
There are two methods. The first method that we will
see is computing line integrals. Let's see how that works.
Let's say that I take some path that starts at the origin.
Or, actually, anywhere you want,
but let's take the origin. That is my favorite point.
And let's go to a point with coordinates (x1,
y1). And let's take my favorite
curve and compute the line integral of that field,
you know, the work done along the curve.
Well, by the fundamental theorem, that should be equal to
the value of the potential at the end point minus the value at
the origin. That means I can actually write
f of (x1, y1) equals -- -- that line integral plus the value at
the origin. And that is just a constant.
We don't know what it is. And, actually,
we can choose what it is. Because if you have a
potential, say that you have some potential function.
And let's say that you add one to it.
It is still a potential function.
Adding one doesn't change the gradient.
You can even add 18 or any number that you want.
This is just going to be an integration constant.
It is the same thing as, in one variable calculus,
when you take the anti-derivative of a function it
is only defined up to adding the constant.
We have this integration constant, but apart from that we
know that we should be able to get a potential from this.
And this we can compute using the definition of the line
integral. And we don't know what little f
is, but we know what the vector field is so we can compute that.
Of course, to do the calculation we probably don't
want to use this kind of path. I mean if that is your favorite
path then that is fine, but it is not very easy to
compute the line integral along this,
especially since I didn't tell you what the definition is.
There are easier favorite paths to have.
For example, you can go on a straight line
from the origin to that point. That would be slightly easier.
But then there is one easier. The easiest of all,
probably, is to just go first along the x-axis to (x1,0) and
then go up parallel to the y-axis.
Why is that easy? Well, that is because when we
do the line integral it becomes M dx N dy.
And then, on each of these pieces, one-half just goes away
because x, y is constant. Let's try to use that method in
our example.
Let's say that I want to go along this path from the origin,
first along the x-axis to (x1,0) and then vertically to
(x1, y1). And so I want to compute for
the line integral along that curve.
Let's say I want to do it for this vector field.
I want to find the potential for this vector field.
Let me copy it because I will have to erase at some point.
4x squared plus 8xy and 3y squared plus 4x squared.
That will become the integral of 4x squared plus 8 xy times dx
plus 3y squared plus 4x squared times dy.
To evaluate on this broken line, I will,
of course, evaluate separately on each of the two segments.
I will start with this segment that I will call c1 and then I
will do this one that I will call c2.
On c1, how do I evaluate my integral?
Well, if I am on c1 then x varies from zero to x1.
Well, actually, I don't know if x1 is positive
or not so I shouldn't write this.
I really should say just x goes from zero to x1.
And what about y? y is just 0.
I will set y equal to zero and also dy equal to zero.
I get that the line integral on c1 -- Well, a lot of stuff goes
away. The entire second term with dy
goes away because dy is zero. And, in the first term,
8xy goes away because y is zero as well.
I just have an integral of 4x squared dx from zero to x1.
By the way, now you see why I have been using an x1 and a y1
for my point and not just x and y.
It is to avoid confusion. I am using x and y as my
integration variables and x1, y1 as constants that are
representing the end point of my path.
And so, if I integrate this, I should get four-thirds x1
cubed. That is the first part.
Next I need to do the second segment.
If I am on c2, y goes from zero to y1.
And what about x? x is constant equal to x1 so dx
becomes just zero. It is a constant.
If I take the line integral of c2, F dot dr then I will get the
integral from zero to y1. The entire first term with dx
goes away and then I have 3y squared plus 4x1 squared times
dy. That integrates to y cubed plus
4x1 squared y from zero to y1. Or, if you prefer,
that is y1 cubed plus 4x1 squared y1.
Now that we have done both of them we can just add them
together, and that will give us the formula for the potential.
F of x1 and y1 is four-thirds x1 cubed plus y1 cubed plus 4x1
squared y1 plus a constant. That constant is just the
integration constant that we had from the beginning.
Now you can drop the subscripts if you prefer.
You can just say f is four-thirds x cubed plus y cubed
plus 4x squared y plus constant. And you can check.
If you take the gradient of this, you should get again this
vector field over there. Any questions about this method?
Yes? No.
Well, it depends whether you are just trying to find one
potential or if you are trying to find all the possible
potentials. If a problem just says find a
potential then you don't have to use the constant.
This guy without the constant is a valid potential.
You just have others. If your neighbor comes to you
and say your answer must be wrong because I got this plus
18, well, both answers are correct.
By the way. Instead of going first along
the x-axis vertically, you could do it the other way
around. Of course, start along the
y-axis and then horizontally. That is the same level of
difficulty. You just exchange roles of x
and y. In some cases,
it is actually even making more sense maybe to go radially,
start out from the origin to your end point.
But usually this setting is easier just because each of
these two guys were very easy to compute.
But somehow maybe if you suspect that polar coordinates
will be involved somehow in the answer then maybe it makes sense
to choose different paths. Maybe a straight line is better.
Now we have another method to look at which is using
anti-derivatives. The goal is the same,
still to find the potential function.
And you see that finding the potential is really the
multivariable analog of finding the anti-derivative in the one
variable. Here we did it basically by
hand by computing the integral. The other thing you could try
to say is, wait, I already know how to take
anti-derivatives. Let's use that instead of
computing integrals. And it works but you have to be
careful about how you do it. Let's see how that works.
Let's still do it with the same example.
We want to solve the equations. We want a function such that f
sub x is 4x squared plus 8xy and f sub y is 3y squared plus 4x
squared. Let's just look at one of these
at a time. If we look at this one,
well, we know how to solve this because it is just telling us we
have to integrate this with respect to x.
Well, let's call them one and two because I will have to refer
to them again. Let's start with equation one
and lets integrate with respect to x.
Well, it tells us that f should be,
what do I get when I integrate this with respect to x,
four-thirds x cubed plus, when I integrate 8xy,
y is just a constant, so I will get 4x squared y.
And that is not quite the end to it because there is an
integration constant. And here, when I say there is
an integration constant, it just means the extra term
does not depend on x. That is what it means to be a
constant in this setting. But maybe my constant still
depends on y so it is not actually a true constant.
A constant that depends on y is not really a constant.
It is actually a function of y. The good news that we have is
that this function normally depends on x.
We have made some progress. We have part of the answer and
we have simplified the problem. If we have anything that looks
like this, it will satisfy the first condition.
Now we need to look at the second condition.
We want f sub y to be that. But we know what f is,
so let's compute f sub y from this.
From this I get f sub y. What do I get if I
differentiate this with respect to y?
Well, I get zero plus 4x squared plus the derivative of
g. I would like to match this with
what I had. If I match this with equation
two then that will tell me what the derivative of g should be.
If we compare the two things there, we get 4x squared plus g
prime of y should be equal to 3y squared by 4x squared.
And, of course, the 4x squares go away.
That tells you g prime is 3y squared.
And that integrates to y cubed plus constant.
Now, this time the constant is a true constant because g did
not depend on anything other than y.
And the constant does not depend on y so it is a real
constant now. Now we just plug this back into
this guy. Let's call him star.
If we plug this into star, we get f equals four-thirds x
cubed plus 4x squared y plus y cubed plus constant.
I mean, of course, again, now this constant is
optional. The advantage of this method is
you don't have to write any integrals.
The small drawback is you have to follow this procedure
carefully. By the way, one common pitfall
that is tempting. After you have done this,
what is very tempting is to just say, well,
let's do the same with this guy.
Let's integrate this with respect to y.
You will get another expression for f up to a constant that
depends on x. And then let's match them.
Well, the difficulty is matching is actually quite
tricky because you don't know in advance whether they will be the
same expression. It could be you could say let's
just take the terms that are here and missing there and
combine the terms, you know, take all the terms
that appear in either one. That is actually not a good way
to do it, because if I put sufficiently
complicated trig functions in there then you might not be able
to see that two terms are the same.
Take an easy one. Let's say that here I have one
plus tangent square and here I have a secan square then you
might not actually notice that there is a difference.
But there is no difference. Whatever.
Anyway, I am saying do it this way, don't do it any other way
because there is a risk of making a mistake otherwise.
I mean, on the other hand, you could start with
integrating with respect to y and then differentiate and match
with respect to x. But what I am saying is just
take one of them, integrate,
get an answer that involves a function of the other variable,
then differentiate that answer and compare and see what you
get. By the way, here,
of course, after we simplified there were only y's here.
There were no x's. And that is kind of good news.
I mean, if you had had an x here in this expression that
would have told you that something is going wrong.
g is a function of y only. If you get an x here,
maybe you want to go back and check whether it is really a
gradient field. Yes?
Yes, this will work with functions of more than two
variables. Both methods work with more
than two variables. We are going to see it in the
case where more than two means three.
We are going to see that in two or three weeks from now.
I mean, basically starting at the end of next week,
we are going to do triple integrals, line integrals in
space and so on. The format is first we do
everything in two variables. Then we will do three variables.
And then what happens with more than three will be left to your
imagination. Any other questions about
either of these methods? A quick poll.
Who prefers the first method? Who prefers the second method?
Wow. OK.
Anyway, you will get to use whichever one you want.
And I would agree with you, but the second method is
slightly more effective in that you are writing less stuff.
You don't have to set up all these line integrals.
On the other hand, it does require a little bit
more attention. Let's move on a bit.
Let me start by actually doing a small recap.
We said we have various notions. One is to say that the vector
field is a gradient in a certain region of a plane.
And we have another notion which is being conservative.
It says that the line integral is zero along any closed curve.
Actually, let me introduce a new piece of notation.
To remind ourselves that we are doing it along a closed curve,
very often we put just a circle for the integral to tell us this
is a curve that closes on itself.
It ends where it started. I mean it doesn't change
anything concerning the definition or how you compute it
or anything. It just reminds you that you
are doing it on a closed curve. It is actually useful for
various physical applications. And also, when you state
theorems in that way, it reminds you,oh..
I need to be on a closed curve to do it.
And so we have said these two things are equivalent.
Now we have a third thing which is N sub x equals M sub y at
every point. Just to summarize the
discussion. We have said if we have a
gradient field then we have this.
And the converse is true in suitable regions.
We have a converse if F is defined in the entire plane.
Or, as we will see soon, in a simply connected region.
I guess some of you cannot see what I am writing here,
but it doesn't matter because you are not officially supposed
to know it yet. That will be next week.
Anyway, I said the fact that Nx equals
My implies that we have a gradient field and is only if a
vector field is defined in the entire plane or in a region that
is called simply connected. And more about that later.
Now let me just introduce a quantity that probably a lot of
you have heard about in physics that measures precisely fairly
ought to be conservative. That is called the curl of a
vector field.
For the definition we say that the curl of F is the quantity N
sub x - M sub y. It is just replicating the
information we had but in a way that is a single quantity.
In this new language, the conditions that we had over
there, this condition says curl F equals zero.
That is the new version of Nx equals My.
It measures failure of a vector field to be conservative.
The test for conservativeness is that the curl of F should be
zero. I should probably tell you a
little bit about what the curl is, what it measures and what it
does because that is something that is probably useful.
It is a very strange quantity if you put it in that form.
Yes? I think it is the same as the
physics one, but I haven't checked the physics textbook.
I believe it is the same. Yes, I think it is the same as
the physics one. It is not the opposite this
time. Of course, in physics maybe you
have seen curl in space. We are going to see curl in
space in two or three weeks. Yes?
Yes. Well, you can also use it. If you fail this test then you
know for sure that you are not gradient field so you might as
well do that. If you satisfy the test but you
are not defined everywhere then there is still a bit of
ambiguity and you don't know for sure.
OK. Let's try to see a little bit
what the curl measures. Just to give you some
intuition, let's first think about a velocity field.
The curl measures the rotation component of a motion.
If you want a fancy word, it measures the vorticity of a
motion. It tells you how much twisting
is taking place at a given point.
For example, if I take a constant vector
field where my fluid is just all moving in the same direction
where this is just constants then,
of course, the curl is zero. Because if you take the
partials you get zero. And, indeed,
that is not what you would call swirling.
There is no vortex in here. Let's do another one where this
is still nothing going on. Let's say that I take the
radial vector field where everything just flows away from
the origin. That is f equals x, y.
Well, if I take the curl, I have to take partial over
partial x of the second component,
which is y, minus partial over partial y of
the first component, which is x.
I will get zero. And, indeed,
if you think about what is going on here,
there is no rotation involved. On the other hand,
if you consider our favorite rotation vector field -- --
negative y and x then this curl is going to be N sub x minus M
sub y, one plus one equals two.
That corresponds to the fact that we are rotating.
Actually, we are rotating at unit angular speed.
The curl actually measures twice the angular speed of a
rotation part of a motion at any given point.
Now, if you have an actual motion,
a more complicated field than these then no matter where you
are you can think of a motion as a combination of translation
effects, maybe dilation effects,
maybe rotation effects, possibly other things like that.
And what a curl will measure is how intense the rotation effect
is at that particular point. I am not going to try to make a
much more precise statement. A precise statement is what a
curl measures is really this quantity up there.
But the intuition you should have is it measures how much
rotation is taking place at any given point.
And, of course, in a complicated motion you
might have more rotation at some point than at some others,
which is why the curl will depend on x and y.
It is not just a constant because how much you rotate
depends on where you are. If you are looking at actual
wind velocities in weather prediction then the regions with
high curl tend to be hurricanes or tornadoes or things like
that. They are not very pleasant
things. And the sign of a curl tells
you whether you are going clockwise or counterclockwise.
Curl measures twice the angular velocity of the rotation
component of a velocity field. Now, what about a force field?
Because, after all, how we got to this was coming
from and trying to understand forces and the work they do.
So I should tell you what it means for a force.
Well, the curl of a force field -- -- measures the torque
exerted on a test object that you put at any point.
Remember, torque is the rotational analog of the force.
We had this analogy about velocity versus angular velocity
and mass versus moment of inertia.
And then, in that analogy, force divided by the mass is
what will cause acceleration, which is the derivative of
velocity. Torque divided by moment of
inertia is what will cause the angular acceleration,
namely the derivative of angular velocity.
Maybe I should write that down.
Torque divided by moment of inertia is going to be d over dt
of angular velocity. I leave it up to your physics
teachers to decide what letters to use for all these things.
That is the analog of force divided by mass equals
acceleration, which is d over dt of velocity.
And so now you see if the curl of a velocity field measure the
angular velocity of its rotation then,
by this analogy, the curl of a force field
should measure the torque it exerts on a mass per unit moment
of inertia. Concretely, if you imagine that
you are putting something in there,
you know, if you are in a velocity field the curl will
tell you how fast your guy is spinning at a given time.
If you put something that floats, for example,
in your fluid, something very light then it is
going to start spinning. And the curl of a velocity
field tells you how fast it is spinning at any given time up to
a factor of two. And the curl of a force field
tells you how quickly the angular velocity is going to
increase or decrease. OK.
Well, next time we are going to see Green's theorem which is
actually going to tell us a lot more about curl and failure of
conservativeness. 556 00:50:11,000 --> NaN:NaN:NaN,NaN