PROFESSOR: We will pick
up from where we left.

I hope the attendance will
get a little bit better today.

It's not even Friday,
it's Thursday night.

So last time we talked a
little bit about chapter 13,

we started 13-1.

I wanted to remind you that we
revisited the notion of work.

Now, if you notice
what the book does,

it doesn't give
you any specifics

about the force field.

May the force be with you.

They don't say what kind
of animal this f is.

We sort of informally
said I'm going to have

some sort of path integral.

And I didn't say what conditions
I was assuming about f.

And I just said that r
is the position vector.

It's important for us to
imagine that is plus c1, what

does that mean, c1?

It means that this function,
let's write it R of t, equals.

Let's say we are implying not
in space, so we have x of t,

y of t, the parametrization
of this position vector.

Of course we wrote that
last time as well, we

said x is x of t, y is y of t.

But why I took c1
and not continuous?

Could anybody tell me?

If I'm going to go ahead
and differentiate it,

of course I'd like it
to be differentiable.

And its derivatives
should be continuous.

But that's actually not
enough for my purposes.

So if I want R of t
to be c1, that's good,

I'm going to smile.

But when we did that in
chapter-- was it chapter 10?

It was chapter 10,
Erin, am I right?

We assumed this was
a regular curve.

A regular curve is not
just as differentiable

with the derivative's
continuous with respect to time.

x prime of t, y prime
of t, both must exist

and must be continuous.

We wanted something
else about the velocity.

Do you remember the drunken bug?

The drunken but was
fine and he was flying.

As long as he was flying,
everything was fine.

When did the drunken
bug have a problem?

When the velocity
field became 0,

at the instant where the bug
lost his velocity, right?

So we said regular means c1 and
R prime of t at any value of t

should be different from 0.

We do not allow the particle
to stop on it's way.

We don't allow it, whether it
is a photon, a drunken bug,

an airplane, or whatever it is.

We don't want it to
stop in it's trajectory.

Is that good for
other reasons as well?

Very good for the
reason that we want

to think later in arc length.

[INAUDIBLE] came up with
this idea last time.

I didn't want to tell you
the truth, but he was right.

One can define certain
path integrals with respect

to s, with respect to
arc length parameter.

But as you remember
very well, he

had this correspondence between
an arbitrary parameter type t

and s, and this is s of t.

And also going back
and forth, that

means from s you head back to t.

So here's s of t and
this is t of s, right?

So we have this correspondence
and everything worked fine

in terms of being
able to invert that.

And having some sort
of equal morphisms

as long as the
velocity was non-0.

OK, do you remember
who s of t was?

S of t was defined--
It was a long time ago.

So I'm reminding you
s of t was integral

from 0 to t-- or from t0 to t.

Your favorite initial
moment in time.

Of the speed, uh-huh, and
what the heck was the speed?

The speed was the norm or the
length of the R prime of t.

This is called speed, that
we assume different from 0,

for a good purpose.

We can go back and forth
between t and s, t to s,

s to t, with
differentiable functions.

Good, so now we can apply the
inverse mapping through them.

We can do all sorts
of stuff with that.

On this one we did not
quite define it rigorously.

What did they say is?

We said f would be a
good enough function,

but know that I do
not need f to be c1.

This is too strong, too strong.

So in Calc 1 when you had to
integrate a function of one

variable you just
assumed that- in Calc 1

I remember- you assume
that continuous.

It doesn't even have
to be continuous

but let's assume that
f would be continuous.

OK, so you have, in one sense,
that the composition with R,

if you have f of x
of t, y of t, z of t.

In terms of time will be a
functions of one variable,

and this will be continuous.

All right?

OK, now what if
it's not continuous?

Can't I have a piecewise,
continuous function?

Like in Calc 1, do you guys
remember we had some of this?

And from here like that
and from here like this

and from here like that.

And we had these continuities,
and this was piecewise

continuous.

Yeah, for god sake,
I can integrate that.

Why do we assume integral
of a continuous function?

Just to make our
lives easier and also

because we are in freshman
and sophomore level Calculus.

If we were in advanced
Calculus we would say,

I want this function
to be integrable.

This is a lot weaker
than continuous,

maybe the set of discontinuities
is also very large.

Who told you that you
have finitely many jumps

these continuities?

Maybe you have a
much larger set.

And this is what you learn
in advanced Calculus.

But you are not at
the level of a senior

yet so we'll just assume,
for the time being,

that f is continuous.

All right, and we say,
what is this animal?

We called it w and
be baptised it.

We said, just give
it some sort of name

and we say that is work.

And by definition,
by definition,

this is going to be
integral from-- Now,

the thing is, we define this as
a simple integral with respect

to time as a definition.

That doesn't mean
that I introduced

the notion of path
integral the way I should,

I was cheating on that.

So the way we
introduced it was like,

let f be a function of
the spatial coordinates

in terms of time.

x, y, z are space
coordinates, t is time.

So I have f of R of t here.

Dot Product, who
the heck is the R?

This is nothing but
a vector art drawing.

These are both
vectors, sometimes

I should put them in bold
like they do in the book.

To make it clear I can
put a bar on top of them,

they are free vectors.

So, f of R times
R prime of t dt.

And your favorite moments of
time are-- let's say on my arc

that I'm describing from time,
t, equals a, to time equals b.

Therefore, I'm going to
take time for a to b.

And this is how we define
the work of a force.

The work of a
force that's acting

on a particle that is moving
between time, a, and time, b,

on this arc of a curve
which is called c.

Do you like this c?

Okay, and the
force is different.

So we have a force field.

So I cheated, I knew
a lot in the sense

that I didn't tell
you how you actually

introduce the path integral.

Now this is more or less
where I stop and [INAUDIBLE].

But couldn't we actually
introduce this integral

and even define it with respect
to some arc length grammar?

Maybe if everything goes
fine in terms of theory?

And the answer is yes.

And I'm going to show
you how one can do that.

I'm going to go ahead and
clean here a little bit.

I'm going to leave this on
by comparison for awhile.

And then I will assume
something that we have not

defined whatsoever, which is
an animal called path integral.

So the path integral of a vector
field along a trajectory, c.

I don't know how to draw.

I will draw some skewed
curve, how about that?

Some pretty skewed curve, c,
it's not self intersecting,

not necessarily.

You guys have to imagine
this is like the trajectory

of an airplane in
the sky, right?

OK, and I have it on d
equals a, to d equals b.

But I said forget
about the time, t,

maybe I can do everything
in arc length forever.

So if that particle,
or airplane,

or whatever it is has
a continuous motion,

that's also differentiable.

And the velocity
never becomes zero.

Then I can parametrize
an arc length

and I can say, forget about
it, I have integral over c.

See, this is c,
it's not f, okay?

But f of x of s, y
of s, z of s, okay?

And this is going to be a ds.

And you'll say, yes Magdelina--
this is little s, I'm sorry.

Yes, Magdelina, but what
the heck is this animal,

you've never introduced it.

I have not introduced it because
I have to discuss about it.

When we introduced Riemann
sums, then we took the limit.

We always have to think how
to partition our domains.

So this curve can be partitioned
in as many as n, this is s k.

S k, this is s1, and this is
s n, the last of the Mohicans.

I have n sub intervals,
pieces of the art.

And how am I going
to introduce this?

As the limit, if it exists.

Because I can be in
trouble, maybe this limit

is not going to exist.

The sum of what?

For every [? seg ?] partition
I will take a little arbitrary

point inside the subarc.

Subarc?

STUDENT: Yeah.

PROFESSOR: Subarc,
it's a little arc.

Contains a-- let's take it here.

What am I going to
define in terms of wind?

s k, y k, and z k, some
people put a star on it

to make it obvious.

But I'm going to
go ahead and say

x star k, y star k, z star k,
is my arbitrary point in the k

subarc.

Times, what shall I multiply by?

A delta sk, and then
I take k from one to n

and I press to the
limit with respect n.

But actually I could also
say in some other ways

that the partitions length goes
to 0, delta s goes to zero.

And you say but,
now wait a minute,

you have s1, s2 s3, s4,
s k, little tiny subarc,

what the heck is delta s?

Delta s is the largest subarc.

So the length of the largest
subarc, length of the largest

subarc in the partition.

So the more points I take,
the more I refine this.

I take the points closer
and closer and closer

in this partition.

What happens to the
length of this partition?

It shrinks to-- it goes to 0.

Assuming that this
would be the largest

one, well if the
largest one goes to 0,

everybody else goes to 0.

So this is a Riemann
sum, can we know for sure

that this limit exists?

No, we hope to god
that this limit exists.

And if the limit exists then
I will introduce this notion

of integral around the back.

And you said, OK I
believe you, but look,

what is the connection
between the work- the way

you introduced it as a simple
Calculus 1 integral here-

and this animal that looks like
an alien coming from the sky.

We don't know how to look at it.

Actually guys it's not so
bad, you do the same thing

as you did before.

In a sense that, s is connected
to any time parameter.

So Mr. ds says, I'm
your old friend,

trust me, I know who I am.
ds was the speed times dt.

Who can tell me if we are in R
three, and we are drunken bugs,

ds will become what?

A long square root times
dt, and what's inside here?

I want to see if
you guys are awake.

[INTERPOSING VOICES]

PROFESSOR: Very good, x prime
of t squared, I'm so lazy

but I'll write it down. y prime
of t squared plus z prime of t

squared.

And this is going
to be the speed.

So I can always do
that, and in this case

this is going to become always
some-- let's say from time, t0,

to time t1.

Some in the integrals
of-- some of the limit

points for the time.

I'm going to have
f of R of s of t,

in the end everything
will depend on t.

And this is my face being happy.

It's not part of the integral.

Saying what?

Saying that, guys, if
I plug in everything

back in terms of t- I'm
more familiar to that type

of integral- then I have what?

Square root of-- that's
the arc length element

x prime then t squared,
plus y prime then t

squared, plus z prime
then t squared, dt.

So in the end it is-- I think
the video doesn't see me

but it heard me, presumably.

This is our old
friend from Calc 1,

which is the simple integral
with respect to t from a to b.

OK, all right, and we
believe that the work

can be expressed like that.

I introduced it
last time, I even

proved it on some
particular cases

last time when Alex wasn't
here because, I know why.

Were you sick?

ALEX: I'll talk to
you about it later.

I'm a bad person.

PROFESSOR: All right, then
I'm dragging an object like,

the f was parallel to the
direction of displacement.

And then I said
the work would be

the magnitude of f times the
magnitude of the displacement.

And then we proved that is
just a particular case of this,

we proved that last time,
it was a piece of cake.

Actually, we proved
the other one.

It proved that if force is going
to be oblique and at an angle

theta with the displacement
direction, then

the work will be the
magnitude of the force times

cosine of theta, times the
magnitude of displacement,

all right?

And that was all an application
of this beautiful warp formula.

Let's see something more
interesting from an application

viewpoint.

Assume that you are
looking at the washer,

you are just doing laundry.

And you are looking at
this centrifugal force.

We have two forces, one is
centripetal towards the center

of the motion, circular
motion, one is centrifugal.

I will take a
centrifugal force f,

and I will say I want
to measure at the work

that this force is producing
in the circular motion

of my dryer.

My poor dryer died so I
had to buy another one

and it cost me a lot of money.

And I was thinking,
such a simple thing,

you pay hundreds
of dollars on it

but, anyway, we take
some things for granted.

I will take the washer because
the washer is a simpler

case in the sense that the
motion-- I can assume it's

a circular motion of
constant velocity.

And let's say this
is the washer.

And centrifugal
force is acting here.

Let's call that--
what should it be?

Well, it's continuing
the position vector

so let's call that lambda
x I, plus lambda y j.

In the sense that it's
collinear to the vector that

starts at origin, and here
is got to be x of t, y.

X and y are the
special components

at any point on my
circular motion.

If it's a circular
motion I have x

squared plus y squared
equals r squared,

where the radius is the
radius of my washer.

You have to compute
the work produced

by the centrifugal force
in one full rotation.

It doesn't matter, I can have
infinitely many rotations.

I can have a hundred rotations,
I couldn't care less.

But assume that the
motion has constant speed.

So if I wanted, I could
parametrize in our things

but it doesn't
bring a difference.

Because guys, when this
speed is already constant,

like for the circular motion
you are familiar with.

Or the helicoidal case
you are familiar with,

you also saw the case when
the speed was constant.

Practically, you were
just rescaling the time

to get to your speed, to your
time parameter s, arc length.

So whether you work with
t, or you work with s,

it's the same thing if
the speed is a constant.

So I'm not going to use my
imagination to go and do it

with respect to s.

I could, but I
couldn't give a damn

because I'm going to have
a beautiful t that you

are going to help me recover.

From here, what is
the parametrization

that comes to mind?

Can you guys help me?

I know you can after all
the review of chapter 10

and-- this is what?

R what?

You should whisper cosine t.

Say it out loud, be
proud of what you know.

This is R sine t.

And let's take t
between 0 and 2 pi.

One, the revolution only,
and then I say, good.

The speed is what?

Speed, square root of
x prime the t squared

plus y prime the t squared.

Which is the same as writing
R prime of the t in magnitude.

Thank God we know that.

How much is this?

R, very good, this
is R, very good.

So life is not so hard, it's--
hopefully I'll be able to do

the w.

What is the w?

It's the path integral
all over the circle

I have here, that I traveled
counterclockwise from any point

to any point.

Let's say this would be
the origin of my motion,

then I go back.

And I have this
force, F, that I have

to redistribute in terms
of R. So this notation

is giving me a little
bit of a headache,

but in reality it's
going to be very simple.

This is the dot product, R
prime dt, which was the R.

Which some other people asked
me, how can you write that?

Well, read the
review, R of x equals

x of t, i plus y of t, j.

Also, read the
next side plus y j.

It short, the dR
differential out of t,

sorry, I'll put R.
dR is dx i plus dy j.

And if somebody wants
to be expressing

this in terms of speeds,
we'll say this is x prime dt,

this is y prime dt.

So we can rewrite this x prime
then t i, plus y prime then t

j, dt.

All right?

OK, which is the same
thing as R prime of t, dt.

[INAUDIBLE]

This looks awfully theoretical.

I say, I don't like it, I
want to put my favorite guys

in the picture.

So I have to think, when
I do the dot product

I have the dot product
between the vector that

has components f1 and f2.

How am I going to do that?

Well, if I multiply with this
guy, dot product, the boss guy,

with this boss guy.

Are you guys with me?

What am I going to do?

First component times
first component,

plus second component times
second component of a vector.

So I have to be smart and
understand how I do that.

Lambda is a constant.

Lambda, you're my
friend, you stay there.

x is x of t, x of
t, but I multiplied

with the first
component here, so I

multiplied by x prime of t.

Plus lambda times y of
t, times y prime of t.

And who gets out of the
picture is dt at the end.

I have integrate with
respect to that dt.

This would be incorrect, why?

Because t has to move
between some specific limits

when I specify what
a path integral is.

I cannot leave a c-- very good,
from 0 to 2 pi, excellent.

Is this hard?

No, It's going to
be a piece of cake.

Why is that a piece of cake?

Because I can keep writing.

You actually are faster than me.

STUDENT: Your chain rule
is already done for you.

PROFESSOR: Right,
and then lambda

gets out just because--
well you remember

you kick the lambda out, right?

And then I've put R cosine
t times minus R sine t.

I'm done with who?

I'm done with this
fellow and that fellow.

And plus y, R sine
t, what is R prime?

R cosine, thank you guys, dt.

And now I'm going to ask
you, what is this animal?

Stare at that, what
is the integrand?

Is a friend of
yours, he's so cute.

He's staring at you and
saying you are done.

Why are you done?

What happens to the integrand?

It's zero, it's a
blessing, it's zero.

How come it's zero?

Because these two terms
simplify, they cancel out.

They cancel out, thank god
they cancel out, I got a zero.

So we discover something
that a physicist

or a mechanical engineer
would have told you already.

And do you think he would
have actually plugged

in the path integral?

No, they wouldn't
think like this.

He has a simpler explanation for
that because he's experienced

with linear experiments.

And says, if I drag this like
that I know what to work with.

If I drag in, like
at an angle, I

know that I have the
magnitude of this,

cosine theta, the angle.

So, he knows for linear
cases what we have.

For a circular case he
can smell the result

without doing the path integral.

So how do you think the guy,
if he's a mechanical engineer,

would think in a second?

Say well, think of
your trajectory, right?

It's a circle.

The problem is that centrifugal
force being perpendicular

to the circle all the time.

And you say, how can a line
be perpendicular to a circle?

It is, it's the
normal to the circle.

So when you say this
is normal to the circle

you mean it's normal to
the tangent of the circle.

So if you measure the angle made
by the normal at every point

to the trajectory of a
circle, it's always lines.

So he goes, gosh, I got
cosine of 90, that's zero.

So if you have some
sort of work produced

by something perpendicular
to your trajectory,

that must be zero.

So he or she has
very good intuition.

Of course, how do we prove it?

We are mathematicians, we
prove the path integral,

we got zero for the
work, all right?

But he could sense that kind
of stuff from the beginning.

Now, there is another example
where maybe you don't have

90 degrees for your trajectory.

Well, I'm going to just take--
what if I change the force

and I make a difference problem?

Make it into a
different problem.

I will do that later, I
won't go and erase it.

Last time we did one
that was, compute

the work along a parabola
from something to something.

Let's do that again.

For some sort of
a nice force field

I'll take your vector valued
function to be nice to you.

I'll change it, y i plus x j.

And then we are in
plane and we move

along this parabola between
0, 0 and-- what is this guys?

1, 1-- well let make it
into a one, it's cute.

And I'd like you to measure
the work along the parabola

and also along the arc of a--
along the segment of a line

between the two points.

So I want you to compute
w1 along the parabola,

and w2 along this
thingy, the segment.

Should it be hard?

No, this was old
session for many finals.

I remember, I think it
was 2003, 2006, 2008,

and very recently, I
think a year and 1/2 ago.

A problem like that was given.

Compute the path
integrals correspond

to work for both parametrization
and compare them.

Is it hard?

I have no idea, let me think.

For the first one we
have parametrization

that we need to distinguish
from the other one.

The first parametrization
for a parabola,

we discussed it last
time, was of course

the simplest possible
one you can think of.

And we did this
last time but I'm

repeating this because I
didn't want Alex to miss that.

And I'm going to say integral
from some time to some time.

Now, if I'm between 0
and 1, time of course

will be between 0 and
1 because x is time.

All right, good,
that means what else?

This Is f1 and this is f2.

So I'm going to have f1 of t,
x prime of t, plus f2 of t,

y prime of t, all the
[? sausage ?] times dt.

Is this going to be hard?

Hopefully not, I'm going to
have to identify everybody.

Identify this guys prime
of t with respect to t

is 1, piece of cake, right?

This fellow is-- you told
me last time you got 2t

and you got it right.

This guy, I have to be
a little bit careful

because y is the fourth guy.

This is t squared and this is t.

So my integral will be a joke.

0 to 1, 2t squared plus t
squared equals 3t squared.

Is it hard to integrate?

No, for God's sake,
this is integral-- this

is t cubed between 0
and 1, right, right?

So I should get
1, and if I get, I

think I did it right, if I get
the other parametrization you

have to help me write it again.

The parametrization
of this straight line

between 0, 0 and 1, 1.

Now on the actual
exam, I'm never

going to forgive you if you
don't know how to parametrize.

Now you know it but two months
ago you didn't, many of you

didn't.

So if somebody gives you 2
points, OK, in plane I ask you,

how do you write that symmetric
equation of the line between?

You were a little
bit hesitant, now

you shouldn't be hesitant
because it's a serious thing.

So how did we write that?

We memorized it. x minus
x1, over x2 minus x1

equals y minus y1,
over y2 minus y1.

This can also be written
as-- you know, guys,

that this over that
is the actual slope.

This over that, so
it can be written

as a [INAUDIBLE] formula.

It can be written in many ways.

And if we put a, t, we transform
it into a parametric equation.

So you should be
able, on the final,

to do that for any segment of
a line with your eyes closed.

Like, you see the
numbers, you plug them in,

you get the
parametric equations.

We are nice on the exams
because we usually give you

a line that's easy to write.

Like in this case you would
have x equals t and y equal,

let's see if you
are asleep yet, t.

Why is that?

Because the line that joins
0, 0 and 1, 1 is y equals x.

So y equals x is called
also, first bicycle.

It's the old friend of
yours from trigonometry,

from Pre-Calc, from algebra,
I don't know where, college

algebra.

Alrighty, is this hard to do?

No, it's easier than before.
w2 is integral from 0 to 1,

this is t and this is t, this
is t and this is t, good.

So we have t times
1 plus t times 1,

it's like a funny, nice, game
that's too simple, 2t, 2t.

So the fundamental
theorem of Calc

says t squared between 0
and 1, the answer is 1.

Am I surprised?

Look at me, do I
look surprised at all

that I got the same answer?

No, I told you a
secret last time.

I didn't prove it.

I said that R times happy times.

When depending on the
force that is with you,

you have the same work
no matter what path

you are taking between a and b.

Between the origin
and finish line.

So I'm claiming that if I
give you this zig-zag line

and I asked you what-- look,
it could be any crazy path

but it has to be a nice
differentiable path.

Along this differentiable
path, no matter

how you compute the work,
that's your business,

I claim I still get 1.

Can you even think why,
some of you remember maybe,

the force was key?

STUDENT: It's a
conservative force.

PROFESSOR: It had to
be good conservative.

Now this is conservative but
why is that conservative?

What the heck is a
conservative force?

So let's write it
down on the-- we

say that the vector
valued function, f,

valued in R2 or R3, is
conservative if there exists

a smooth function.

Little f, it actually
has to be just c1,

called scalar potential.

Called scalar potential,
such that big F as a vector

field will be not little f.

That means it will be the
gradient of the scalar

potential.

Definition, that
was the definition,

and then criterion for a f
in R2 to be conservative.

I claim that f equals f1 i,
plus f two eyes, no, f2 j.

I'm just making
silly puns, I don't

know if you guys follow me.

If and only if f sub 1
prime, with respect to y,

is f sub 2 prime,
with respect to x.

Can I prove this?

Prove, prove Magdelina, don't
just stare at it, prove.

Why would that be
necessary and sufficient?

Well, for big F
to be conservative

it means that it has
to be the gradient

of some little function, little
f, some scalar potential.

Alrighty, so let me
write it down, proof.

f conservative if
and only if there

exists f, such that gradient
of f is F. If and only

if-- what does it
mean about f1 and f2?

f1 and f2 are the f
sub of x and the f sub

y of some scalar potential.

So if f is the gradient, that
means that the first component

has to be little f sub x And
the second component should

have to be little f sub y.

But that is if and only if f
sub 1 prime, with respect to y,

is the same as f sub 2
prime, with respect to x.

What is that?

The red thing here is called
a compatibility condition

of this system.

This is a system of two OD's.

You are going to
study ODEs in 3350.

And you are going
to remember this

and say, Oh, I know that because
she taught me that in Calc 3.

Not all instructors will
teach you this in Calc 3.

Some of them fool you
and skip this material

that's very important
to understand in 3350.

So guys, what's going to
happened when you prime this

with respect to y?

You get f sub x prime,
with respect to y.

When you prime this
with respect to x

you get f sub y prime,
with respect to x.

Why are they the same thing?

I'm going to remind
you that they

are the same thing
for a smooth function.

Who said that?

A crazy German mathematician
whose name was Schwartz.

Which means black, that's
what I'm painting it in black.

Because is the Schwartz
guy, the first criterion

saying that no
matter in what order

you differentiate the
smooth function you

get the same answer for
the mixed derivative.

So you see we prove if
and only if that you

have to have this
criterion, otherwise

it's not going to
be conservative.

So I'm asking you,
for your old friend, f

equals- example
one or example two,

I don't know- y i plus x j.

Is the conservative?

You can prove it in two ways.

Prove in two differently
ways that it is conservative.

a, find the criteria.

What does this criteria say?

Take your first component,
prime it with respect to y.

So y prime with respect to y.

Take your second component,
x, prime it with respect to x.

Is this true?

Yes, and this is me,
happy that it's true.

So this is 1 equals
1, so it's true.

So it must be conservative,
so it must be conservative.

Could I have done
it another way?

By definition, by definition,
to prove that a force field

is conservative by
definition, that it a matter

of the smart people.

There are people
who- unlike me when

I was 18- are able to see
the scalar potential in just

about any problem I give them.

I'm not going to make this
experiment with a bunch of you

and I'm going to reward you
for the correct answers.

But, could anybody
see the existence

of the scalar potential?

So these there, this exists.

That's there exists a little
f scalar potential such

that nabla f equals F.

And some of you may see
it and say, I see it.

So, can you see
a little function

f scalar function so
that f sub of x i is y

and f sub- Magdelena--
x of y j is this x j?

STUDENT: [INAUDIBLE]?

PROFESSOR: No, you need to
drink some coffee first.

You can get this, x times what?

Why is that?

I'll teach you how to get it.

Nevertheless, there
are some people

who can do it with
their naked eye

because they have a little
computer in their head.

But how did I do it?

It's just a matter of
experience, I said,

if I take f to be x y,
I sort of guessed it.

f sub x would be y and f sub y
will be x so this should be it,

and this is going to do.

And f is a nice function,
polynomial in two variables,

it's a smooth function.

I'm very happy I'm
over the domain,

open this or whatever,
open domain in plane.

I'm very happy, I have
no problem with it.

So I can know that this is
conservative in two ways.

Either I get to the
source of the problem

and I find the little
scalar potential whose

gradient is my force field.

Or I can verify the
criterion and I say,

the derivative of
this with respect to y

is the derivative
of this with respect

to x is-- one is the same.

The same thing, you're going
to see it again in math 3350.

All right, that I
taught many times,

I'm not going to teach
that in the fall.

But I know of some
very good people

who teach that in the fall.

In any case, they
would reteach it to you

because good teachers don't
assume that you know much.

But when you will see
it you'll remember me.

Hopefully fondly, not cursing
me or anything, right?

OK, how do we actually
get to compute f by hand

if we're not experienced
enough to guess it like I was

experienced enough to guess?

So let me show you how you solve
a system of two differential

equations like that.

So how I got-- how
you are supposed

to get the scalar potential.

f sub x equals F1,
f sub y equals F2.

So by integration, 1 and 2.

And you say, what
you mean 1 and 2?

I'll show you in a second.

So for my case,
example 2, I'll take

my f sub x must be y, right?

Good.

My f sub y must be x, right?

Right.

Who is f?

Solve this property.

Oh, I have to start
integrating from the first guy.

What kind of information
am I going to squeeze?

I'm going to say I
have to go backwards,

I have to get-- f
is going to be what?

Integral of y with respect to
x, say it again, Magdelina.

Integral of y with respect
to x, but attention,

this may come because, for
me, the variable is x here,

and y is like, you cannot
stay in this picture.

So I have a constant
c that depends on y.

Say what?

Yes, because if you go backwards
and prime this with respect

to x, what do you
get? f sub x will

be y because this is
the anti-derivative.

Plus this prime with
respect to x, zero.

So this c of y may, a
little bit, ruin your plans.

I've had students
who forgot about it

and then they got in trouble
because they couldn't get

the scalar potential correctly.

All right?

OK, so from this one you
say, OK I have some--

what is the integral of y dx?

xy, plus some guy c
constant that depends on y.

From this fellow
I go, but I have

to verify the second condition,
if I don't I'm dead meat.

There are two coupled
equations, these

are coupled equations
that have to be

verified at the same time.

So f sub y will be prime with
respect to y. x plus prime

with respect to y. c prime
of y, God gave me x here.

So I'm really lucky in that
sense that c prime of y

will be 0 because I have
an x here and an x here.

So c of y will simply be
any constant k. c of y

is just a constant k, it's
not going to depend on y,

it's a constant k.

So my answer was not correct.

The best answer would have
been f of xy must be xy plus k.

But any function like xy will
work, I just need one to work.

I just need a scalar
potential, not all of them.

This will work, x2, xy plus 7
will work, xy plus 3 will work,

xy minus 1,033,045 will work.

But I only need one
so I'll take xy.

Now that I trained
your mind a little bit,

maybe you don't need
to actually solve

the system because your
brain wasn't ready before.

But you'd be amazed, we
are very trainable people.

And in the process of doing
something completely new,

we are learning.

And your brain next,
will say, I think

I know how to function
a little bit backwards.

And try to integrate and
see and guess a potential

because it's not so hard.

So let me give you example 3.

Somebody give you over a
domain in plane x i plus y j,

and says, over D, simply
connected domain in plane,

open, doesn't matter.

Is this conservative?

Find a scalar potential.

This is again, we
do section 13-2,

so today we did 13-1
and 13-2 jointly.

Find the scalar potential.

Do you see it now?

STUDENT: [INAUDIBLE]?

PROFESSOR: Excellent,
we teach now, got it.

He says, I know where
this comes from.

I've got it, x squared
plus y squared over 2.

How did he do it?

He's a genius.

No he's not, he's just
learning from the first time

when he failed.

And now he knows what he has
to do and his brain says,

oh, I got it.

Now, [INAUDIBLE] could
have applied this method

and solved the coupled
system and do it slowly

and it would have taken
him another 10 minutes.

And he's in the final, he
doesn't have time to spare.

If he can guess the potential
and then verify that,

it's going to be easy for him.

Why is that?

This is going to be 2x over 2
x i, and this is 2y over 2 y j.

So yeah, he was right.

All right, let me
give you another one.

Let's see who gets this one.

F is a vector valued
function, maybe a force field,

that is this.

Of course there are many
ways-- maybe somebody's

going to ask you
to prove this is

conservative by the criterion,
but they shouldn't tell you

how to do it.

So show this is conservative.

If somebody doesn't want the
scalar potential because they

don't need it, let's say.

Well, prime f1 with respect to
y, I'll prime this with respect

to x.

f1 prime with respect to
y equals 2x is the same

as f2 prime with respect with.

Yeah, it is conservative, I
know it from the criterion.

But [INAUDIBLE] knows
that later I will ask him

for the scalar potential.

And I wonder if he can find
it for me without computing it

by solving the system.

Just from his
mathematical intuition

that is running in the
background of your--

STUDENT: x squared,
multiply y [INAUDIBLE].

PROFESSOR: x squared
y, excellent.

Zach came up with
it and anybody else?

Alex?

So all three of you, OK?

Squared y, very good.

Was it hard?

Yeah, it's hard for most people.

It was hard for me
when I first saw

that in the first 30
minutes of becoming familiar

with the scalar
potential, I was 18 or 19.

But then I got it in
about half an hour

and I was able to
do them mentally.

Most of the examples I
got were really nice.

Were on purpose made nice for
us for the exam to work fast.

And now let's see why
would the work really not

depend on the trajectory
you are taking

if your force is conservative.

If the force is
conservative there

is something magic
that's going to happen.

And we really don't
know what that is,

but we should be able to prove.

So theorem, actually
this is funny.

It's called the fundamental
theorem of path integrals

but it's the fundamental
theorem of calculus 3.

I'm going to write it like
this, the fundamental theorem

of calc 3, path integrals.

It's also called- 13.3,
section- Independence of path.

So remember you have a
work, w, over a path, c.

F dot dR where there R is the
regular parametrized curve

overseen.

This is called a
supposition vector.

Regular meaning c1, and
never vanishing speed,

the velocity never vanishes.

Velocity times 0 such
that f is continuous,

or a nice enough integral.

If F is conservative of
scalar potential, little f,

then the work, w, equals
little f at the endpoint

minus little f at the origin.

Where, by origin and endpoint
are those for the path,

are those for the arc, are
those for the curve, c.

So the work, the w, will
be independent of time.

So w will be independent of f.

And you saw an
example when I took

a conservative function
that was really nice,

y times i plus x times j.

That was the force field.

Because that was
conservative, we got w being 1

no matter what path we took.

We took a parabola, we
took a straight line,

and we could have
taken a zig-zag

and we still get w equals 1.

So no matter what
path you are taking.

Can we prove this?

Well, regular classes don't
prove anything, almost nothing.

But we are honor students
so lets see what we can do.

We have to understand
what's going on.

Why do we have this fundamental
theorem of calculus 3?

The work, w, can be expressed--
assume f is conservative

which means it's going to come
from a potential little f.

Where f is [INAUDIBLE] scalar
function over my domain, omega.

Now, the curve, c,
is part of this omega

so I don't have any
problems on the curve.

w will be rewritten beautifully.

So I'm giving you a
sketch of a proof.

But you would be able to do
this, maybe even better than me

because I have taught
you what you need to do.

So this is going to
be f1 i, plus f2 j.

And I'm going to write it.
f1 times-- what is this guys?

dR, I taught you, you taught
me, x prime of t, right?

Plus f2 times y prime of t,
all dt, and time from t0 to t1.

I start my motion along
the curve at t equals t0

and I finished my
motion at t equals t1.

Do I know where f1 and f2 are?

This is the point,
that's the whole point,

I know who they are, thank God.

And now I have to again
apply some magical think,

I'll ask you in a
minute what that is.

So, what is f1?

df dx, or f sub base.

If you don't like f sub base,
if you don't like my notation,

you put f sub x, right?

And this df dy.

Why?

Because it's
conservative and that

was the gradient of little f.

Of course I'm using the fact
that the first component would

be the partial of little
f with respect to x.

The second component would
be the partial of little

f with respect to y.

Have you seen this
formula before?

What in the world
is this formula?

STUDENT: It's the chain rule?

PROFESSOR: It's the chain rule.

I don't have a dollar but I
will give you a dollar, OK?

Imagine a virtual dollar.

This is the chain rule.

So by the chain
rule we can write

this to be the
derivative with respect

to t of little f of
x of t, and y of t.

Alrighty, so I know
what I'm doing.

I know that by chain rule I had
little f evaluated at x of t,

y of t, and time t,
prime with respect to t.

Now when we take the fundamental
theorem of calculus, FTC.

That reminds me, I was
teaching calc 1 a few years ago

and I said, that's the
Federal Trade Commission.

Federal Trade Commission,
fundamental theorem

of calculus.

So coming back to
what I have, I prove

that w is the Federal
Trade Commission, no.

w is the application
of something

that we knew from calc
1, which is beautiful.

f of xt, y of t dt, this
is nothing but what?

Little f evaluated at-- I'm
going to have to write it down,

this whole sausage.

f of x of t1, y of t1,
minus f of x of t0, y of t0.

For somebody as lazy as
I am, that they effort.

How can I write It better?

f at the endpoint
minus f at the origin.

And of course, we are trying to
be quite rigorous in the book.

We would never say
that in the book.

We actually denote
the first point

with p, the origin, and
the endpoint with q.

So we say, f of q minus f of p.

And we proved q e d, we
proved the fundamental theorem

of path integrals, the
independence of that.

So that means the work
is independent of path

when the force is conservative.

Now attention, if f is not
conservative you are dead meat.

You cannot say what I just said.

So I'll give you two
separate examples

and let's see how we
solve each of them.

A final exam type of
problem-- every final exam

contains an
application like that.

Even the force field, f,
or the vector value, f.

Is it conservative?

Prove what is proved and
after that-- so the path

integral in any way you can.

If it's conservative you're
really lucky because you're

in business.

You don't have to do any work.

You just find the little
scalar potential evaluated

at the endpoints and subtract,
and that's your answer.

So I'm going to give you an
example of a final exam problem

that happened in the past year.

So, a final type exam problem.

f of xy equals 2xy
i plus x squared j,

over r over r squared.

STUDENT: Didn't we just
do [INAUDIBLE] that?

PROFESSOR: Well, I just did
that but I changed the problem.

I wanted to keep the
same force field.

STUDENT: Alright.

PROFESSOR: OK.

Compute the work, w,
performed by f along the arc

of the circle in the picture.

And they draw a picture.

And they do a picture for you,
and you stare at this picture

and-- So you say,
oh my God, if I

were to parametrize it
would be a little bit of-- I

could, but it would be
a little bit of work.

I would have x equals 2
cosine t, y equals sine t.

I would have to plug in and
do that whole work, definition

with parametrization.

Do you have to parametrize?

Not in this case, why?

Because the f is conservative.

If they ask you- some professors
give hints, most of them

are nice and give hints-
show f is conservative.

So that's a big
hint in the sense

that you see it
immediately, how you do it.

You have 2xy prime
with respect to y,

is 2x, which is x squared
prime with respect to x.

So it is conservative.

But he or she told you more.

He said, I'm selling
you something here,

you have to get your
own scalar potential.

And you did, and
you got x squared y.

Now, most of the
scalar potentials

that we are giving
you on the exam

can be seen with naked eyes.

You wouldn't have to do
all the integration of that

coupled system with respect
to x, with respect to y,

integrate backwards,
and things like that.

What do I need to do
in that case guys?

Say in words.

Since the force is
conservative, just two lines.

I'm applying the fundamental
theorem of calculus.

I'm applying the fundamental
theorem of path integrals.

I know the work is
independent of that.

So w, in this case
is already there,

is going to be f at the point.

I didn't say how I'm going to
travel it, in what direction.

f of q minus f of p.

And you'll say, well
what does it mean?

How do we do that?

That means x squared
y, evaluated at q,

who the heck is q?

Attention, negative 2,
0, Matt, you got it?

OK, right?

So, are you guys with me?

Right?

And p is 2, 0.

So at negative 2, 0,
minus x squared y at 2, 0.

So, if you set 0, big, as
you knew that you got 0,

that is the answer.

Now if somebody would give you
a wiggly look that this guy's

wrong, then he took this past.

He's going to do
exactly the same work

if he's under influence the
same conservative force.

If the force acting
on it is the same.

No matter what path you
are taking-- yes, sir?

STUDENT: Is it still working
for your self-intersecting.

PROFESSOR: Yeah, because
you're not stopping,

it works for any
parametrization.

So if you're able to parametrize
that as a differentiable

function so that the
derivative would never vanish,

it's going to work, right?

All right, it can work also
for this piecewise contour

or any other path point.

As long as it starts and
it ends at the same point

and as long as your
conservative force is the same.

So that force is very
[INAUDIBLE], yes?

STUDENT: Do [INAUDIBLE] graphs
but the endpoints the same.

And if they are conservative
then [INTERPOSING VOICES].

PROFESSOR: If everything
is conservative

the work along this
path is the same.

There were people
who played games

like that to catch if
the student knows what

they are talking-- yes, sir?

STUDENT: So, in a
problem, if they wanted

to find the work
couldn't we simplify it

by just saying, we
need to find-- because,

since we're in R2, couldn't we
just say it's a straight line?

Because that's the--
Like, instead of a curve

we could just set the straight
line [INTERPOSING VOICES].

PROFESSOR: If it
were moving from here

to here in a straight
line you would still

get the same answer.

And you could have-- if
you did that, actually,

if you were to compute this
kind of work on a straight line

it has-- let me
show you something.

You see, when you compute
dx y dx, plus x squared dy.

If y is 0 like Matthew said,
I'm walking on-- I'm not drunk.

I'm walking straight.

y will be 0 here, and 0 here,
and the integral will be 0.

So he would have noticed
that from the beginning.

But unless you know the
force is conservative,

there is no guarantee
that on another path

you don't have a
different answer, right?

So, let me give you another
example because now Matthew

brought this up.

A catchy example
that a professor

gave just to make the
students life miserable,

and I'll show it
to you in a second.

He said, for the picture--
very similar to this one,

just to make people confused.

Somebody gives you
this arc of a circle

and you travel from a to b.

And this the thing.

And he says, compute w for
the force given by y i plus j.

And the students
said OK, I guess

I'm going to get 0 because
I'm going to get something

like y dx plus x dy.

And if y is 0, I get 0, and
that way you wouldn't be 0

and I'm done.

No, it's not how you
think because this is not

conservative.

So you cannot say I can change
my path and it's still going

to be the same.

No, why is this
not conservative?

Quickly, this prime
with respect to y

is 1, this prime with
respect to x is 0.

So 1 different from
0 so, oh my God, no.

In that case, why do we do?

We have no other choice but
say, x equals 2 cosine t,

y equals 2 sine t, and
t between 0 and pi.

And then I get integral of
f1, y, what the heck is y?

2 sine t, times x prime of t.

Yeah, minus 2 sine
t, this is x prime.

Plus 1, are you guys with me?

Times y prime, which
is 2 cosine t, dt.

And t between 0 and pi.

And you get something
ugly, you get 0 to pi.

What is the nice thing?

When you integrate this with
respect to t, you get sine t.

And thank God, sine, whether you
are at 0 or if pi is still 0.

So this part will disappear.

So all you have left is
minus 4 sine squared dt.

But you are not done
so compute this at home

because we are out of time.

So don't jump to
conclusions unless you know

that the force is conservative.

If your force is
not conservative

then things are going
to look very ugly

and your only chance is to go
back to the parametrization,

to the basics.

So we are practically
done with 13.3

but I want to watch
more examples next time.

And I'll send you the homework.

Over the weekend you would be
able to start doing homework.

Now, when shall I
grab the homework?

What if I closed it right
before the final, is that?

STUDENT: Yeah.

PROFESSOR: Yeah?