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