TTU Math2450 Calculus3 Secs 13.1 - 13.2

Vector Fields and Line Integrals

PROFESSOR: I'll
go over the exam.

It's good review for
the final, and it's

a good feedback for you in
case you have questions.

I do not change grades,
I do not curve your exam.

I do not make adjustments
after I give you the grade.

Therefore, it's very
important for me

to explain why you
got what you got.

Not everybody did
well on this exam.

Most people did
pretty good, and I'm

quite happy with what I see
as a average for the class.

However, there are
many open questions

from many people, things
they didn't quite understand,

and I would like
to discuss those.

First of all, the midterm
exam was 11 questions.

10 were mandatory, so
the maximum possible

percentage-wise was 110%.

So for somebody who
did perfectly fine,

they would have 110%.

There is one person
only who got the high.

I didn't disclose
his name, but I would

like to say congratulations.

And I'm going to go ahead and
solve each problem with you,

for you.

So you have the
function f of x, y,

to be x squared minus y squared.

And the differential was f
sub x dx plus f sub y dy,

whihc is 2x dx, minus 2y dy.

That was something very easy.

It was not supposed to give you
any headache, and most of you

did a fine job on this one.

What created some
problems to most students

was the second problem, though.

And I sorry to hear
that, sorry to see that.

Find the directional
derivative of a function,

the same function as before.

So I have taken advantage
of the previous problem,

in order to make
your do time shorter.

At the point p of coordinates
x equals 0, y equals 1,

you will have a direction
given by the vector.

What does it mean
direction given?

Analyze that direction given
by the vector means what?

Not the vector i minus j is y,
because it's not a unit vector.

What is the corresponding
direction given by it?

A corresponding
direction given by it

is 1 over square root of 2i
minus 1 over square root of 2j.

So it's a collinear vector--
that is, unit varies.

Say it again, [INAUDIBLE]?

The direction u represents
a collinear vector.

So pointing in the
same direction as v,

but it has to be unitary.

Why?

Because the definition of
the directional derivative

is a function along the
direction u at the point p,

was given by the formula
partial derivative at p

and at 1, plus partial
derivative at p times 2.

Did I expect to
write all this down?

Yes, I did, as I
showed you last time.

So you have 2x evaluated
at-- what is x?

0.

1 times u1.

This is u1 minus 2y,
evaluated at 0, 1 times

minus 1 over root
2, which is u2.

Well that means the
first term goes away,

because this is going to be 0.

And after the second
term, you have a plus.

y is 1, thank god, that's easy.

2 over square root of
2, the answer is root 2.

So any other answer would
normally receiving a 0.

The answer was b, square root 2.

Now, on number three, the
function given is different.

f of x, y equals e to the xy.

And they say, the
gradient of this function

is at an arbitrary point.

Say is again?

The gradient of this function
is at an arbitrary point.

That was only part
of the problem.

A little bit of credit for
just writing the gradient.

This is actually
easy, a piece of cake.

y is that.

You have f sub x
i plus f sub y j.

It equals y to the xy
i plus x to the xyj.

That's very good.

Alright, OK?

Then, which direction-- it's
just the gradient right?

The direction corresponds
to the gradient.

They don't ask you for the u.

Actually, you don't need the u.

You just need the tangent
plane in this case.

And if you know the equation
of the tangent plane,

as I told you to remember that,
that would be very helpful.

Write your answer in
the space provided.

So what did I expect you to do?

First this, and then write
the equation z minus z0

equals f sub x times x minus
x0, plus f sub y, y minus y0.

Here at the point p,
evaluate it at the point p.

But attention, what
is the point p?

Well, p is the origin,
because we say at the origin.

Oh, so that makes things easier.

I'm not done.

Half of the problem
is still coming.

If you did until this point, I
can only give you 5 out of 10

or something like that.

Many people made
a mistake at z0.

Attention guys, you plus
in that 0, you don't get 0.

For god's sake, it's 1, right?

So z0 is 1.

Now you're getting
the sense that you

have z minus 1 equals
f sub x, computed as 0,

0 will be 0, lucky you.

f sub y computed at 0, 0, you
were expected to say that.

So the answer for this
problem was z equals 1.

Still, if you messed up, I
gave some partial credit,

because I didn't want to punish
you too much, too harshly.

On number four, find
the direction u-- now

you're using number
three, so I should not

erase number three completely.

On number four, you
use number three, so

the same type of function.

But it says find the
direction in which

this function increases most
rapidly, at the point 1, 1.

OK.

So what do you do? you compute
the gradient at the point 1,1,

and you say, this
is a piece of cake.

It's going to be ei plus ej, ee.

Wonderful.

So what you do is a u is the
gradient f over the length

of the gradient of f at p.

Which is ee divided
by the length of it.

But you say, I don't have
to compute the length of it.

I know what is pulling
your two e's is what?

And no matter what you have here
you get the same unique result.

Remember we talked
about that uniqueness?

This is what I
tried to emphasize,

that you can have 77,
ee, 99, 55, 100 and 100.

If you divide by the norm,
you still get the same answer.

Not 11, but 1 over
root 2, 1 over root 2.

So no matter what
you had there--

it could have had a million,
or something instead of e,

you still have the same u.

Yes, put it back.

Give yourself
points, modify that.

OK, so let me tell you.

Normally I should penalize,
because I say write the answer

in the space provided.

And thank god you had
enough space, right?

Look, this person wrote--
I shouldn't show you who

he is, he's not in here anyway.

He has space and he
provided last year

with no square root
of 2, because only two

rows are enough to write that.

It's OK, I understand
you forgot to copy.

My son did the same thing.

He got a scantron at the UIL.

Come to visit my son, I wanted
to kill him, but it's OK.

He got all the answers
right, and then

the teacher-- that reminds
me of a movie with Mr. Bean.

So the teacher comes
to him and says,

wow your scantron is blank.

So what was I supposed to do?

Adjust for all the answers
you got in the box,

put them in the scantron.

Oh, really?

So he goes quickly.

And then he got only
75% of them transferred.

The rest of them
were not transferred.

I don't know what they did.

I have no idea.

But the professor would
have given full credit,

even for the answers
that he had in the box.

From what I understood,
the rule for scantrons,

exams like you I only say,
if you don't have them

on the scantron,
they don't count.

This is very harsh,
because we don't do that.

For example, the
final-- if you--

that's why I'm trying
to read everything.

Suppose you box you answer and
it's 1 over square root of 2.

If that's the right answer.

Then if have the
multiple choice,

and they forgot to circle
1 over square root of 2.

I still give you
100% on that problem.

Some professor do not.

So this is at the latitude
at whoever makes the rules,

or whoever writes the exam.

OK.

So again for the final, even for
the multiple choice problems,

I still need the solutions.

I'm going to ask you
to use a bluebook.

Some professors do not
ask you to use a bluebook.

They say, as long as you can
write on the sheet, circle

the answer, I'm fine.

I'm not fine.

I want to keep what's
in the bluebook.

So buy-- how much is it?

Like a dollar?

Buy the books ahead of time,
make sure you have them.

Now, number five was a piece of
cake once you did number four.

You have a question?

STUDENT: What size bluebook
do you need for the final?

PROFESSOR: The big one.

Bigger than that, right?

The direction u for five.

With the problem four was i
plus j over root 2, right?

This is what you remember
that you did in problem four.

If you didn't do problem four,
you cannot do problem five.

Problem five says, this
is parallel to one line.

This is parallel to--
what is i plus j?

Of course, you don't
have to draw that.

I'm not expecting
you to draw that.

y equals x is the
first bisection.

All you had to do was
circle C, and that

was-- once you circled
C, you get full credit.

If you don't do that, you
don't get credit for anything.

Now six.

What is the maximum rate
of increase of the function

z the same of your friend,
your fellow z equals

e to the xy at p0,
coordinates 1, 1?

Then the value of the
maximum rate of change is?

A noun.

What's the simplest
way to do it?

There are two ways to do it.

One is the long way,
one is the short way.

What's the short way, guys?

Just compute the
length of the gradient.

The length of the
gradient at the point P.

So you have whatever
that was, ee in length.

So the answer was e root 2.

Am I right?

What was the long way?

I saw somebody do it.

This is a lot more
work, but of course,

would be to compute the
directional derivative

at the point p
for this function.

In the direction of u,
where u is the gradient

divided by the length.

at the point p.

And you get, of course,
the same answer.

Why?

Because we proved that actually
the maximum rate of change

represented directional
derivative exactly

in the direction
given by the gradient.

This is something we proved.

One of the few things
we proved in this class.

Alright.

So the answer was e root 2.

Let's move on to number seven.

Number seven-- and remind
me of your five points.

Can you email me, so
I have an Excel sheet,

and I'll put it in.

Consider the function f of x,
y e to the negative x squared,

y squared.

What can you tell me about
this type of function?

It's the headache function.

If I would ask you to do
an anti-derivative of each

of the negative squares,
you would say Magdalene,

didn't you say that
this is impossible?

While the anti-derivative
exists, it cannot be expressed.

It cannot be expressed as
an elementary function.

And that's a big headache.

This problem is beautiful,
why is it beautiful?

Because in the end,
it becomes magic.

So it's a positive function.

It's like a bell on top
of the church something.

And then, you have to
compute double integral

over the unit disk of
centers of 0 and radius 1.

Of e to the negative x
squared minus y squared dx/dy.

Well then you say, well
I've done this kind of thing

before, but not with
Cartesian coordinates.

We did it with the Jacobian
r, that changes everything

into polar coordinates.

So this guy becomes e
to the minus r squared.

Each of the numbers are
squared dr, d theta.

D on the unit
[INAUDIBLE] disk means

the radius goes from 0 to 1.

This is a blessing for us,
because it's easy data.

Then we have 0 to 2 pi.

You could have put
it in any order.

For u, it's easier to close your
eyes when it comes to theta.

Say, theta is independent.

He is like a partition
that has to do nothing

with what's inside here.

So let's pull him
out of this picture.

And he wants to live by himself.

An integral from 0 to 2 pi of
d theta was of course 2 pi.

He's happy to go
out, having fun.

This guy inside has to
be thoroughly computed.

In the sense that you
perform the substitution.

I was actually amused that half
of you did u equals r squared,

and half of you did u
equals minus r squared.

It really doesn't
matter which one.

But the problem is
that some of you

made a mess when you put the
limit points back in place,

and you made mistakes.

Somebody even got
negative answers,

I was about to
fall off the chair.

Of course, I was in a good
mood because it was a holiday,

I graded them.

Fortunately, I graded
them over the break.

So after I came
back from Georgia.

I have minus r dr. rdr
was minus a half du.

This fellow is just
into the u, and he's

a blessing because
the [INAUDIBLE]

So into the u, however,
take it between 1 and what?

Not 0 and 1.

But when you have 0
here, you have 0 here.

When you have 1,
you have minus 1.

So pay attention to
that, otherwise, you

get something that
makes no sense.

Times minus a half.

That, you will have
to be careful about.

Why?

Because there will be a minus
from here and here, in the end,

the answer will be positive.

And that's reminding me
of that city plumber joke

when he doesn't pay attention
to the limits of integration.

And you can get a minus
volume, or a minus area.

So e to the minus 1 minus 1.

But that leaves a
negative number,

but when you multiply it by a
minus, you have 1 minus 1/e.

1 minus 1/e.

Good, thank god.

This is a nice guy, less than 1.

And this is key to your
answer, because 2 goes away

and pi stays in place
and this is less than pi.

So the answer to this question
was an answer less than pi.

And if you didn't get
it, I'm very sorry,

if you didn't get less than
pi, you didn't get any points.

But, there are enough chances
for you to get another point.

I was brokenhearted for 10
people or more out of 25

did not remember what I
taught in class about the area

of a collateral triangle.

And it broke my heart,
and I was about to cry,

but I said, c'mon, they'll
do it better in the final.

Honestly, I was
so brokenhearted.

So this is 1, 0, 0.

This was 0, 1, 0.

This was 0, 0, 1.

On number eight.

Thank you.

Beautiful.

It's an equilateral
triangle, and the l side

of that equilateral triangle
is the square root of 2.

I even taught you how to cheat.

That's why I was mad.

I taught you how to cheat,
and you didn't take advantage.

So the area was l squared,
square root of 2, 4.

Which we did this together
in fifth or sixth grade

by multiplying that height and
the width and divided by 2.

And then we came up with this
formula with the Pythagorean

theorem in the classroom.

If eligible to, you can
very quickly get an answer.

So that's going to be 2 root
2 over 4, just root 3 over 2.

And when I saw that people got
something else except root 3

over 2, that broke my heart.

Really.

You have plenty of
time to catch up

with that being on your final.

Did I expect you to really
do the surface integral?

Some people again, need to write
integral over the shaded domain

d a square root
of f sub x squared

plus f sub y squared plus 1.

That was the right track,
because this is root 3.

And then the area, you get the
area of the 1 times 1 over 2,

right?

1/3 is the area.

Root 3 gets out of this, so
you have-- when you integrate,

you have the area of the
shaded base that I have.

And you get the same answer.

No matter how you do
it, with calculators

or without calculators, you
still could have passed.

Am I if you didn't
get the answer?

No.

Absolutely.

But it hurts me as if, I
don't know, a relative of mine

messed up some task.

That's why it's better that
you don't know your students,

because when you
know your students,

you know that they
could have done better,

because you know them.

So we can say, OK,
it really hurts

when you know that they messed
up, not because they are not

smart or educated, but because
they just either didn't

pay attention or they
were stressed out.

However, my substitute,
the guy who came here,

was my Ph.D. student.

He got a doctoral degree
mathematics with me last year.

And he told me you were
not stressed out at all.

And I said, thank god.

I'm glad that they were calm.

And he said, I didn't
look at the exam,

but it seemed like they did very
well and they were comfortable.

And I was so happy.

I was in Athens, Georgia.

And reading this
email I said, yay!

Everybody's going to get an A!

So I come home and
I start grading it.

I was sad to see that my
prediction was not correct.

But anyway, [INAUDIBLE]
with an average of B.

For an honors class, it's OK.

I just expected a lot better.

And I know it's going to be
a lot better in the final.

Number nine.

This was done by
almost everybody,

except for a few people who
messed up on the limits.

I don't know why.

When they compute-- when they
drew, they drew x squared,

and they drew square root of xn.

Of course, you were supposed--
the answer was 0 to 1,

integral of.

Now, if you do first
x, you have x from y

squared to square root of y.

You guys with me?

Because this is
smaller than that.

OK?

So you have 1 and dx dy equals
to integral from 0 to 1,

integral x squared to
square root of x1 dy dx.

Now, what a few people did--
and I just forgave them.

They just-- one
put this like that.

And here, he put root 2.

Root y and y squared.

Don't do that.

It's like chasing that
a positive number equals

a negative number, which
is all complete nonsense.

So the correct answer was
we put y squared down,

and square root of y
because this guy is

bigger than this guy for
something between 0 and 1.

Because I told you.

Square root of 0.04 is bigger
than the square of that.

OK.

Now am I happy with that?

I'm quite happy.

In general, people understood
the vertical strip method

compared to the
horizontal strip method.

And why am I happy?

Because I was asked by three
people from other classes

to help them, over
there, on the corridor.

And I asked them,
who is your teacher?

This and that.

But we did not understand
reversing the order

of integration in class.

And I said, how come?

Well, they didn't
explain it very well.

So I started
explaining it to them.

And then I realized that
it's a conflict of interest.

I'm not allowed to do that.

And then I go, oh my god, I
cannot do the homework for you.

I'm not allowed.

But I was already talking.

So I said, guys, can you do it?

I don't know.

I said, do you draw?

Why would we draw?

They didn't teach
us how to draw.

I said, but how do you
know about vertical strips

and horizontal strips?

No.

And how do you do this?

We don't know.

We felt like we have
to figure it out.

Without drawing, without
understanding how the vertical

strips are drawn
between two functions,

and how you switch
the horizontal strips,

you cannot do this
problem, period.

So if you don't have--
maybe some people have

enough imagination--
but that's very rare--

That they can close
their eyes and they

can see a picture
with their eyes closed

and they can solve that.

But that's not the way to learn.

The way to learn is a very
visual learning thing.

So that's why we
draw all the time.

STUDENT: Professor, you
can cheat these with Cal 2.

PROFESSOR: Yes.

You can do that with Cal 2.

What's the problem?

You have integral from 0 to 1.

Square root of y
minus y squared.

Well, they learn to
do the other one.

The one with square root x
minus x squared, 0,1 and so on.

But they were told explicitly
to write-- the professor even

left these empty and put
spaces, fill in the spaces.

And they say, how the heck
do we fill in those spaces?

Plus the whiteboard problems
have the empty spaces.

And they couldn't
believe that at all.

And one of them went to the
tutoring center and was lucky.

Because he got--
this is like when

you go to a medical
doctor, sometimes you

are lucky and get a good
doctor who takes care of you,

figures out what
your problem is.

And sometimes, they give
you the wrong medicine.

So one of them got the right
tutor who knew how to explain

and sort of knew something.

But the other one got a tutor
who never took Calculus 3

and said, I don't know what the
heck these multiple snakes are.

So I'm not going to
be able to help you.

So he was very disappointed.

OK.

Compute the area of the domain
D from the previous problem.

This was something that
nobody's telling you,

hey, you have to do it
with the double snakes.

You can do it with just
with a simple snake

and you're still fine.

So in Calc 1-- this
Calc 1, whatever it is.

In Calc 1, you learn that
you have to integrate this

and you'll get 2/3 x
to the 3/2 minus 1/3 x

cubed at x equals 1 minus
whatever you have with 0.

But at 0, you have 0, so
you say, forget about it.

And you have 2/3 minus 1/3
equals 1/3, then you're done.

OK?

Did I expect you
to show me work?

No.

For everybody who
wrote 1.3-- and there

were many people who
did this mentally,

and they came up with 1/3.

They got 10 pionts
on the problem.

Finally, number 11.

Without computing the
volume inside the sphere,

x squared plus y squared
plus z squared equals 2.

Set up a triple integral
corresponding to it

in the space provided below.

Some people, a few
people, messed up.

They forgot the Jacobian.

So they put the 1 instead of
r squared [? side-side. ?]

When you work in
three components,

they do fine setting
up the limits.

[INAUDIBLE] 1 here.

Don't look at it in the final.

You can ruin your life this way.

So we have r squared sine phi.

Phi was the latitude
from the North Pole.

it doesn't matter in
which order you do it.

But I would do to
er b phi b theta.

You tell me what the end
points are, and we are done.

STUDENT: From 0 to 5.

PROFESSOR: 0 to--

STUDENT: No, on the first one.

PROFESSOR: 0 to--

STUDENT: Dr?

It's the square root of 2.

PROFESSOR: Mm-hmm.

STUDENT: And b theta--

PROFESSOR: 0.

2pi.

And theta, all around.

STUDENT: 2pi.

PROFESSOR: Longitude
360 meridian degrees.

OK.

0 to 2pi.

So good.

So we are done.

Did I expect you
to write it down?

No.

I had three people who
were nice and wrote down 4.

I mean, they actually
did the work.

Maybe they had
nothing better to do.

I have no idea why.

4pi i cubed over 3, right?

And then they proved the formula
in general using the Jacobian.

Using the formula, they got
the correct formula for r

equals square root of 2.

And I was very happy.

But did I ask you to do that?

No.

Did I give you extra credit.

No.

So all the extra credit
was just one problem to

asked to do exactly what
you were told to do.

I don't know about how
you feel about this exam,

but it wasn't a hard exam.

It was not an easy exam.

It was an exam that
was supposed to test

what you learned until now
all through the course.

And that was the whole idea.

I think you've
learned very much,

and I think you did fine,
the majority of you.

And that should ease
the pressure on you

when it comes to
preparing for the final.

I was thinking last night, I'm
going to send you, probably

by email or in-person
in class, two

or three samples of the
final from old finals

that inspire us when
we write the final.

A few of us will provide
problems and comments

and suggestions when we write
out the departmental final.

But the final will be
departmental for all sections.

I don't expect more than
15 problems on the final.

I have yet to think and
decide if I want to [? lift ?]

probably the same policy.

I mean, the final is
the same for everybody.

But the policy about how
to give partial credit

or not give partial
credit. [INAUDIBLE].

And I already decided that
I'm going to read everything,

so in case that you mess up
at the end with your miracle

answer, you still
get partial credit

for your integrals
[INAUDIBLE] shown.

Also, one of those 15 problems.
might be for extra credit.

I have to think a
little bit better

how-- what is the maximum
weight I want to put.

What I would say, since I never
[INAUDIBLE] open a homework,

and I never curve
exams, I would think

I could make 110% as
the possible maximum.

In this case, you
have some cushion

to make a mistake or two and
still get a perfect score.

OK.

I'm going to move
on to a new chapter.

I have actually
moved on already,

but nobody believed me.

Last time, I started Chapter 13.

Chapter 13 is a mixture of
mathematics and physics.

You will be surprised
how many things

are coming from solid
mechanics, fluid mechanics.

Yes, Regan.

STUDENT: [INAUDIBLE]

PROFESSOR: For a job?

You want me to come with you?

[LAUGHTER]

STUDENT: Because I tried
to talk to you [INAUDIBLE].

PROFESSOR: Yes, yes.

Yes.

Yeah.

And you have to sign up.

Start a [? sheet, ?] attend
[? the sheet, ?] and sign

your name and good luck
with the interview.

You should have told me before!

I could have said
a prayer for you.

This things are very stressful!

I remember my own interviews.

There were several.

I didn't know anything about it,
and my hands were all sweaty.

And you know you should never
shake hands with somebody

when your hands are sweaty.

You have to do like this first.

Be confident and
don't be nervous.

Don't sweat or anything.

Because they can see that.

All right.

You just be yourself.

Do you have earrings?

Because after my
several job interviews--

those are good earrings-- I
was told that I should never

wear dangling earrings at the
interviews, which I did not,

because I didn't have any.

But I love dangling earrings.

And I was asking some academics
why that was [? our ?] problem.

And they say they
are distracting.

Because mathematicians
are like cats.

[LAUGHTER]

PROFESSOR: --pendulum,
and then they

get hypnotized by the dangling.

So I don't know.

I think most of the
interviewers have some problems

and they find some things
distracting or annoying.

Otherwise, I think you are fine.

You're dressed fine
for an interview.

OK.

So now serious job.

We have to remember some of
the things we don't remember.

Which are the gradient for
a function of let's say

three variables.

Let's grow up a little bit.

And that was what
the vector field

F sub xi plus F sub
[? I j ?] plus F sub z k.

Right?

At an arbitrary point
xyz in your domain.

So where xyz is in some
domain, you are in a potato.

And the meaning of the gradient,
the geometric meaning of this,

doesn't look like a
theta [INAUDIBLE].

It's some sort of solid
that it corresponds

to a closed surface.

And this closed surface
that closes up on its own

is having a hard
time [INAUDIBLE].

It has a normal.

And this normal is given by
the gradient of this function,

we can increase
[? it ?] like that.

You remember that.

And that was a long time ago.

But you should
still master that.

Last time, I gave you
the z equals f of xy,

z equals little f
of xy, as a graph

of the function of two variables
over a domain in plane.

We computed the
gradient of that.

But that's what we did all
through the [? meter ?].

So that's no fun.

We know that too well.

On this problem, I
gave you some new piece

of information last time.

So I said, if you have
a vector field that

looks F 1i plus F
2j plus F 3k, where

Fi is C1, that means
that the differentiable

and the derivatives
are continuous,

what was the divergence of it?

Well, that was before
the Easter break.

And I know we had a long break.

I cannot recover from this break
so easily, because it was long.

And I also traveled last week.

But before I traveled, I
remember that I gave you this.

And you memorized it.

Most of you memorised it.

How was it?

The first component
differentiated with respect

to the first variable
plus the second component

differentiated with respect
to the second variable.

Plus the third component
differentiated with respect

to the third variable.

So I'm asking you, as
an exercise, like I

did last time, the same thing.

Exercise one for this section.

Compute divergence
of the gradient

of F, where F is a
C1 function of xyz.

That means F is
[? like this ?] differentiable

and with continuous derivatives.

What does it mean?

It means that you have to
compute divergence of F sub xi

plus F sub yj plus F sub zk.

And you're thinking,
I can do that!

By definition, I take the
first component-- who was that?

Hmm?

STUDENT: Brian.

PROFESSOR: Oh, right.

I thought that somebody
wanted to come in

and then he heard me
and changed his mind.

[LAUGHTER]

PROFESSOR: F sub x
parentheses [INAUDIBLE]

x plus F sub-- like when
you go on a blind date

and you see, change your mind.

OK.

F sub y y plus F sub z z.

Do you remember that
I gave away 95 cents

for this type of question?

So what was this operator?

We can write it better.

We can write it using the
second partial derivatives

with respect to z, y, and z.

And we gave a name to this one.

We called this names--

STUDENT: Laplacian.

PROFESSOR: Laplacian.

Laplace operator.

Laplace.

Laplace.

Laplacian.

That's how you spell it.

Laplac-ian.

OK?

Of F. And then what do you have?

You have to introduce
a new notation in.

When you see this
triangle that looks

like an equilateral
triangle, this

means Laplacian of something.

So if you have a function
of two variables-- so

let's say z equals F of xy.

What is the Laplacian
of this little f?

Little f x x plus little f y y.

So we could be second
partial with respect

to x plus the second
partial with respect to y.

What if I have something else?

Like let me give you a
more general function.

Let's say I have a
differentiable function

of N variables with
continuous derivatives.

And it looks like crazy.

It looks like that.

x1, x2, x n minus what?

Well, the Laplace operator in
this case will be F sub x1 x1

plus [? A of ?] sub x2 x2.

Which means the partial of
F, the second derivative

with respect to x2.

And plus the last derivative
with respect-- two [INAUDIBLE]

with respect to
the same variable.

The last variable is xm minus 1.

This could be one million and 1.

I don't know.

You can have this as many
variables as you want.

Now, actually in
engineering, there

are functions that
have many parameters.

You have three
special opponents.

Then you have time.

Then you have temperature,
then you have pressure,

then you have god knows what.

The surface tension
of the membrane.

Many things.

You really have a
million parameters.

Actually, it's impossible.

It's even hard to work
with 10 parameters.

Imagine always
working with equations

that have lots of variables
and having do deal with that.

In fluid flows,
hydrodynamical problems,

most the time in
3D turbulent flows,

for example, then you have
xyz spatial coordinates

and time T. So even
with four variables,

once you get those operators,
you could have something like F

sub x x x t plus g sub
x x t plus and so on.

All sorts of ugly components.

Sometimes you'll have
equations of fluid flows

in dynamic software.

Fluid flows with
turbulence are really

an area of
mathematics in itself,

of really complicated equations
with most of the operators.

I was looking at
them in Georgia,

where I went to this conference.

Most of those
equations were order 4.

Of course, most of them you
cannot even think about solving

by hand, or with
any known methods.

You can solve them numerically
with computational software.

That is the only [INAUDIBLE]
that modern mathematics

has in some areas right now.

The right software, in
order to find solutions

to a fluid flow with turbulence.

That is the solution to
this type of equation.

Like [INAUDIBLE], for example.

Now we are going to see--
well, you are going to see.

I'm too old and I saw
that 20 years ago.

When you're going
3350 [INAUDIBLE]

differential equations.

And then, if you do PD
3350 one in engineering,

You're going to see
lots of equations

that are hard to solve.

But in many of them, you're
going to see partials,

like that.

And you're going to
say, oh, thank god

that I like partials
in Calc Three

so they became my friends.

And you'll never have
headaches-- [? you know what ?]

would be easy, if you understood
that notion of differential

well, the notion of partial
derivatives very well.

So I'm going to erase this one.

OK.

And then I'll say, I
don't how many of you--

I'll try to make this
formula more visible.

Some of you maybe, who
are engineering majors

know about curl.

Have you heard about curl?

Curl of a vector value function.

No.

You haven't.

Suppose that you have a
vector value function.

That is F of coordinates
x, y, z, the coordinates.

The C1 of over
seven domain omega.

Omega is the domain that your
special coordinates live in.

Xyz living some potato.

That's it.

Whose solid body enclosed
by a closed surface.

In that potato, F is a
differentiable function

with respect to xyz, and the
derivatives are continuous.

Now, in most cases, if
you work with Laplacian,

this is not enough C1.

If you work with Laplacian,
what do you want?

What do you need?

You have F sub x
x plus F sub y1.

So you need C2.

You work with at least C2.

Many examples have C infinity.

That means you're having
really beautiful functions that

are elementary.

Some of them even
polynomial approximations.

And then you really
can differentiate

them ad infinitum and all
the derivatives [INAUDIBLE],

and then you can
call yourself lucky.

How do you introduce the
notion of curl of it?

And it sounds funny, and this
is why they made this fun.

And my hair used to be
curly, but I shaved my head

over the holiday,
and now it's between.

So curl of F is something
that looks horrible

when you try to memorize it.

So you say, OK, if I'm going
to get this on the final,

you better wear this T-shirt.

No, there is something
better than that.

One time I was the wearing-- OK.

My students got no permission
from the [INAUDIBLE]

to come in with a cheat sheet.

But I was wearing a T-shirt
that had Green's theorem.

I don't know how
many of you have

heard about Green's theorem.

We are going to learn
it in two weeks.

And I was wearing that T-shirt.

And it was by accident, OK?

I didn't do it on purpose
to help my students cheat.

So one student at some
point goes like, well, I

don't remember Green's theorem.

And then he looked my T-shirt.

Oh, all right.

Never mind.

So I had Green's theorem
on my shirt, [INAUDIBLE].

But it's hard to wear like
10 T-shirts, one for the-- I

have one for the formula of the
curvature of a curve in space.

Remember that one,
how it is so nasty?

OK, I have this one.

I have Green's theorem.

I have [INAUDIBLE], all the
important formulas actually.

I have 10 T-shirts.

And then I was
thinking, how will I

be if I were like taking ten
T-shirts on top of the other

and taking them one off at
a time during the final.

There is no cheat sheet.

There are no formula
sheets, no nothing.

But I would look like
Joey from "Friends."

Remember Joey, when he was
dressed in many layers.

So rather than
that, I say ask me.

Say oh, you know,
I'm freaking out.

I'm taking this final,
and I forgot curl.

Rather than not attempting
the complex problem at all,

ask me before the exam,
and I will remind everybody

how to set up the curl formula.

So you simply have
to think in terms

of operators-- ddx, ddy, ddz.

What are these?

These are derivative operators.

So if you take this and
multiply it by a function,

that means df, ds-- [INAUDIBLE].

All right, so in this
case, if F is-- I'll

go by my T-shirt-- PI plus QJ
plus RK, where PQ and R are all

scalar functions of xyz.

STUDENT: Then we
will not forget it.

PROFESSOR: Then we are
no longer forget it,

and you'll no longer
need my T-shirt.

All right, so how
do you do that?

You go expand along
your first row,

I times whoever the minor
will be, which is this guy.

How do you do the
[? cowboy ?] problem?

These guys multiply each other.

So you go dr, dy.

Plus or minus?

Minus dq, dz.

Close times I. So the I is
the corresponding element

to the minor that
I just completed.

This minor is the determinant,
which is exactly this guy.

And this is exactly
what my T-shirt says.

Right, precisely.

OK.

The second term, if
we put the minus-- no,

they changed the signs.

That's the thing.

I would put minus, because I am
expanding along the first row.

And the second that I'm in
minus something minor times

J. Which minor?

Let me make in the lime.

Lime is a nice color.

And then I'll take this,
this, this, and that-- dr,

dx shooting [? cowboys ?]
there-- minus dq, dz.

And of course they wrote
dq, dz minus dr, dx.

So I would leave it like that.

It doesn't matter.

You can put the
minus in if you want.

Plus the k dot.

k goes at the end.

All right, now k
goes at the end.

And then k multiplies this
determinant-- dq, dx minus dp,

dy.

dq, dx minus dp, dy.

Is it hard?

No.

It is not going to
be hard to memorize.

So then how did we do that?

We set up the first row to
be I, J, K, the second row

to be ddx, ddy, and ddz.

And then all in order the
components of your vector value

function in the exact
order they are with respect

to the standard basis i j k.

All right, now there
are other names

and other symbols for
curl of F. They use

curl because it's in English.

Well actually, in
Great Britain I

saw that they used [INAUDIBLE],
or else they use both.

In my language, in Romanian,
we call it [? rotore. ?]

And I saw that in French
it's very similar.

They use the same.

Now in the mechanical
engineering notation

it's funny.

They use another symbol and a
cross [? broad dot ?] symbol F.

And by that they mean
curl F. So if you

talk to a professor who's
in mechanical engineering,

or fluid mechanics,
or something,

when they talk about curl,
they will use this notation.

When they use this
other notation,

what do you think this is again?

Divergence, yes.

I told you last time
that is divergence of F.

So make the distinction
between-- again,

when are you leaving?

Huh?

OK, so you have
been in [INAUDIBLE].

And then we have
this distinction

we use here, like
for dot product

and you use here
as a cross product.

Now you have to understand
the conceptual difference is

huge between these guys.

This is a scalar function.

This is a vector function--
vector, scalar-- vector,

scalar, vector scalar.

Because I've had to do
it on so [INAUDIBLE].

It makes [INAUDIBLE].

And I heard of
colleagues complaining

while grading the final
that the students did not

understand that this is a
vector, and this is a scalar.

OK, a few simple exercises--
I'm going to go ahead and do

some of them.

We tried to make the
data on the final exam

very accessible and very
easy to apply in problems.

And one of the problems that--
we'll start with example 2--

would be this one.

And you may think, why?

Sometimes we put it in disguise.

And we said assume you
have a sphere-- that's

the unit sphere-- of
origin O. And say compute.

What is the equation of
the unit sphere, guys?

X squared plus y squared plus
z squared equals one, right?

From [INAUDIBLE], F
equals normal-- external

normal-- to the unit sphere
pointing out, [? through ?]

than N is the same at a
different point as the position

vector.

Then compute.

[? Now follow. ?] Gradient
of F, divergence of F,

and curl of F. Now that
should be a piece of cake.

Now one is not [INAUDIBLE]
so much of a piece of cake

if you don't understand what
the problem wants from you.

It is to actually graph
the expression of this one.

So you're going to
say what is the normal

to a function like that?

First of all, we just
talked today about it.

If you have a function, even
if it's implicitly as F of x,

y, z equals c, in that case N
is your friend from the past.

If it's a unit normal,
unit normal to a surface

happens all the
time in engineering.

Whether you do solid
mechanics or fluid mechanics,

you always have to
complete these things.

This is going to be hard.

The gradient of F
divided by the length

of-- but here I have a problem.

I have to put G here, because
G will be my position vector.

This is the point x,y,z.

Or you prefer big R. But
I think I prefer big G,

because big R looks
like a scalar radius,

and I don't like that.

So the position vector
will be the circle middle

that starts at the origin and
whose N is on the surface,

right?

And this is the equation,
xy equals yj plus [? ek1. ?]

Because my point x,y,z has a
corresponding vector xi plus yj

plus zk-- big deal.

Now I'm trying to convince
you that, for the unit

normal for the sphere, I
have the same kind of thing.

So how do we compute
this normally?

I take the function F that
implicitly defines the surface.

All right, so in my case
F is something else.

What is it? x squared plus
y squared plus z squared.

Let's compute it.

N is going to be [INAUDIBLE].

It's very nice.

2x comma 2y comma 2z divided
by the square root of the sums.

Do I like this?

Uh, no, but I'll have to do
it whether I like it or not.

I want to simplify
up and down via 2.

Can I do that?

Of course I can.

I'm going to get x,y,z divided
by square root of x squared

plus y squared plus z squared.

And this was 1.

STUDENT: Wouldn't
there still be a 2

there, because it's 2
squared [INAUDIBLE]?

PROFESSOR: No, I pulled it out.

That's exactly what I said.

There was a 4 inside.

I pulled out with the forceps.

I put it up here,
square root of 4.

And I have a 2 here,
and that cancels out.

So I got something much
simpler than you guys

expected at first.

I got xi plus yj plus
zk as being the normal.

Did you expect this?

And you were supposed to
expect that this is y,

because this is the position
vector that has one length.

The length of a
root vector is 1,

and the point is on the sphere.

The normal will be
exactly the continuation.

Take your root
vector, and continue

in the same direction--
this is the beauty

of the normal to a surface,
that it continues the radius.

It continues the radius of the
sphere in the same direction.

So you copy and paste
your vector here.

Position vector G will be
the same as the normal N.

All you do is you shift,
but it's the same vector

at the different point.

Instead of starting at
O, it starts at P. So

[? that ?] is the same vector.

So you take the radius vector
from inside the sphere--

the position vector--
and you shift it out,

and that's the
normal to the sphere.

So the equation is still
xi plus yj plus zk.

Yes, sir.

STUDENT: Does it remain the
same for any other functions,

like [INAUDIBLE]?

PROFESSOR: For the
unit sphere, yes it is.

But for a general sphere, no.

For example, what
if my sphere will be

of center origin and radius R?

And its position vector
v is x,y,z-- like that.

[INAUDIBLE] I don't know.

G, right?

That's the position normal.

STUDENT: [INAUDIBLE] just
divide them by the R.

PROFESSOR: You just
divide by the R.

So instead of
radius being big R,

your unit vector
will be this one.

And you take this one
and shift it here,

and that's all you have.

For the sphere, it's beautiful.

For any surface in general, no.

Let me show you.

You have a bunch of [INAUDIBLE],
and your position vectors

look like crazies like that.

And the normals could
be-- they don't have

to continue their position.

They could be-- it depends how
the tangent planes look like.

And the tangent
plane at the point

has to be perpendicular
to the normal.

So the normal field is the
N of [INAUDIBLE] vectors.

But the little thingies
that look like rectangles

or whatever they are--
those are the tangent planes

of those points.

So in general there is
no obvious relationship

between the position and
the normal for the surface.

You are really lucky
for this [? field. ?]

And for many reasons, like
how beautiful the sphere is,

these functions will
be easy to compute.

Can you tell me what they
are without computing?

Because that should
be a piece of cake.

What is the gradient field?

STUDENT: [INAUDIBLE]
to that one?

That's the x, y, and z.

PROFESSOR: For the sphere.

STUDENT: 2x, 2y--

PROFESSOR: Actually, let's do it
for both divergence G and curl

G. And you say wait, they
will be-- so gradient-- no,

I meant here.

You don't have gradient.

When F is a scalar function,
then you have gradient.

Then for that gradient you're
going to have divergence.

And for that-- I changed
notations, that's shy

I have to fix it.

Because F used to be that,
and it's not a vector anymore.

So big F is not
a vector anymore.

It's a scalar function, and now
I have to change the problem.

What is the gradient there?

What's divergence
of the gradient?

[INAUDIBLE] gradient of F.
And for the G that I gave you,

I want the divergence
in the curve?

So I made the problem
fluffier that it was before.

More things to
confuse for practice.

What's the gradient?

We did it before.

2x--

STUDENT: 2xi, 2--

PROFESSOR: 2y, 2z-- we are at a
[? 93 ?] point p on the sphere.

It could be anywhere--
anywhere in space.

What's the divergence
of this individual?

So remember guys,
what I told you?

First component differentiated
with a straight 2x

plus second component
differentiated with respect

to y plus third
component differentiated

with respect to z.

2 plus 2 plus 2 equals
6-- piece of cake.

And curl of the gradient
of F-- is that hard?

[? STUDENT: Yeah. ?]

PROFESSOR: No, but we have
to know the definition.

And without looking at the
T-shirt, how do we do that?

The determinant-- I, J, K.
Operators-- ddx, ddy, and ddz.

STUDENT: [INAUDIBLE]
2x, 2y, 2z, correct?

PROFESSOR: And we copy and
paste the three components.

[INAUDIBLE] in the trash.

I'll take the blue.

So we put 2x, 2y, 2z.

Do you think it's going
to be easy or hard?

Do you see the answer?

Some of are very sharp,
and you may see the answer.

For example, when the
cowboys shoot at each other

like this, dz, dy is here.

dy, dz is here.

So this, as a minor, is
0-- 0I, an eye for an eye.

And what else?

dz, dx-- dx dz, same
thing, minus 0j.

Is this meant to say minus 0j?

Yes it is.

But I did it because I want
you to have the good habit

of saying plus minus plus.

And that's finally
the same kind of thing

that'll give you 0k
if you think that when

you do partial derivative of
y with respect to [? f ?],

you get 0.

You have 0.

So some student of mine asked,
so this is the 0 vector,

how in the world do I
write a 0 vector on short?

Let me show you how.

You're going to laugh at me.

Some people write 0 bar,
which means the 0 vector.

Some other people don't
like it, it's silly.

Some people write O with
double like that, meaning that,

hey, this is a vector element,
the vector with its components

of 0, 0, 0-- to distinguish
that vector from the number 0,

which is not in bold-- So the
notations for the vector are 0.

So I'm going to
write here 0, 0, 0.

How about Mr. G?

Mr. G will act similarly.

When you do the divergence
it's going to be-- 1

plus 1 plus 1 equals 3.

You should remember this thing.

We are going to do
the divergence 3,

and they will ask you to do a
triple integral of a divergence

of a vector field.

And when you do
that, you are going

to get a triple
integer of something

like 3, which is a custom, which
will make your life very easy.

So you will very easily
compute those triple integrals

of constants.

Curl of G, G being
of [? a. ?] OK?

I should make the distinction
between a scalar function

and a vector function by putting
a G bar on the vector function.

How about this?

Is it hard?

No, because it's
the same fellow.

Instead of that, I
have just x, y, z.

The answer will be the same.

So I still want to get
0, 0, 0-- the vector 0.

So the point was that
we will give you enough.

You may expect them
to be very hard,

but they are not
going to be very hard.

Let's do one more like the
ones we have in the book.

What do you think
this one will be?

I'm making you a new
vector value function.

That's maybe two
little exercises

we can do just working
exercise three, four,

I don't know what they are.

Let me give you R
vector of x, y, z

equals yzI plus xzj plus xyk.

Compute the curl.

Let me write it like engineers
do just for fun-- [INAUDIBLE]

cross.

R is the same as
curl R, which is I,

J, K-- oh my god--
ddx, ddy, ddz.

Why is z-- xz-- xy.

Are you saying oh, that's
not so easy anymore.

You-- you will see
that it becomes easy,

OK? i times what is the minor?

This times-- x, right?

Minus x plus minus j
times 1 minus what?

Minor will be the red thingie.

And the red thingie
is beautiful,

because it's gonna be y
minus y plus k times--

who do you think it's
gonna be? z z minus.

So it's still 0.

Do we expect something
like that on the final?

An easy computation.

Somebody says, find me the
curve of this function.

And the functions
usually we give

you are nice and significant.

Something where the
result will be pretty.

OK.

Let me see what else I wanted.

I'm gonna-- I have space here.

So compute the curl and
Laplace operator of f of xyz

equals x squared yzi plus x y
squared zj plus xy z squared k.

Of divergence.

Sorry, guys.

This is not a-- it's
not a scalar function.

I want the divergence
and the curl.

The curl will be a vector.

The divergence will
be a scalar function.

Later on I'll give you a
nice function where you can

compute the Laplace operator.

That's gonna have to
be a scalar function.

And although the
Laplace operator

can be generalized
to vector functions,

and I'll tell you later
how-- what that is.

It's very easy.

It's practically the Laplace
operators in every direction.

OK.

So let's see the curl.

i j k, d dx, d dy, d dz.

Today I'm gonna cook
up the homework.

And with all the practice
that we are doing now,

you should have absolutely
no problem doing the homework

for the first two sections.

At least for the section--
today's section, 13.1.

x squared yz, x y
squared z, xy z squared.

You see there is some
sort of symmetry.

I'm playing a game here.

So I have i.

I want you to tell me
[INAUDIBLE], see now,

I don't work much in groups.

I don't make you
work in groups, but I

want you to answer my question.

So what is going to be this
minor that I-- the first thing

is gonna be?

STUDENT: x z squared.

PROFESSOR: Very good.

STUDENT: Minus x y squared.

PROFESSOR: Minus j.

Potential [? plus or ?] minus.

OK, what is the next guy?

STUDENT: y z squared.

PROFESSOR: y z
squared, thank you.

STUDENT: x squared.

PROFESSOR: x squared, right?

Plus k times--

STUDENT: y squared z.

PROFESSOR: So, you
see what I'm doing?

I'm doing this from
respect to x. y squared z.

You said it, right.

Minus this guy, x squared z.

Can I write it more-- I don't
really like the way I wrote it.

But I'll write like that.

How about x times z squared
minus y squared i minus j.

Or maybe better plus j.

I'll change this up.

Plus j at the end.

Because it's the
vector. y times x

squared minus z squared
j plus-- who gets out?

z-- z times y squared
minus x squared k.

OK.

Good There is some
symmetry in there.

The break in the symmetry
is in the middle.

Because, as you
see, x is separate,

and then z is followed
by y, and then

x squared-- x is followed
by z and y is followed by x.

So I have some-- some
symmetry of some sort.

What else did I want?

Divergence operator.

And that will be the
last example of the kind.

[INAUDIBLE]

How do you write the divergence?

Is this hard?

Very easy?

I'm going go ask you to simplify
because I don't like it,

like, as a sum.

2xyz--

STUDENT: 2xyz plus
2xyz plus 2xyz.

PROFESSOR: And now you see
why I don't like it as a sum.

Because it's 6xyz and it's
very pretty like that.

I'd like you-- on
the exam, I'd like

you to take the function
and box the answer,

and that's all I want you to do.

All right.

I'm gonna go ahead and erase.

I'm going to move
on to 13.2, but I'd

like to review some physics
a little bit with you

and see what you
remember from physics.

It's a little bit messy.

I'll use this instead because
I like the board to be clean.

If I were to ask you to
remember work in physics,

I would say-- I'm changing a
little bit the order in 13.2.

I'd like you to go
back in time and see

what work was in physics class.

STUDENT: [INAUDIBLE] force dx.

PROFESSOR: What if you
didn't know any calculus?

Let's go a long time.

The section is
13.2 preliminaries.

STUDENT: Force
multiplied distance.

PROFESSOR: Very good.

Preliminary work.

[INAUDIBLE] The notion of work
from physics-- hey, come on.

Physics, or engineering,
mechanics, whatever you study,

work.

Imagine that you're
taking a-- this

is your body that you're playing
with-- not your own body,

but the body you are
acting on in physics.

And you are dragging this
object from a place A

to a place B, another position.

A B is the distance.

And the force is parallel
to the trajectory.

This is very important.

This is a simpler case.

In general, it's not so simple.

So this force is acting,
and it's a constant force.

And you pull the object
from one place to another.

That's case one.

In case two, life is harder.

You actually pull
the poor object

with the force in
this direction.

Actually, most of
us do that, right?

If I were to have a gliding
object on the surface,

I would actually act on
that object in the direction

of my arm by pulling it.

So when I displace this body
from point a to point b,

I still travel the
distance d, but-- so d

is a displacement vector
that can be written like--

or it can be drawn like that.

I have to be smart
in both cases,

figure out what
I want from life,

because it's not so clear.

When they taught me I
think the first time

I was in-- oh my
God-- eighth grade,

and they-- that was
a long time ago.

In this case, w is
gonna be a scalar

and I'm gonna have the magnitude
of the force F. F is a vector,

but to indicate it in Newtons
or whatever I measure it in,

it's gonna be in magnitude.

Times the little d,
but instead of little d

I should be a little
smarter and say,

Magdalena, this is the magnitude
of the vector A B, which

is a displacement vector.

So this is called work.

And if the force is 10
Newtons, and the distance

is 10 meters, because we
want to go international.

We want to be global, right,
at Texas Tech, so good.

So we have 100 Newton-meters.

Now you can measure-- well,
you can have another example.

I'm thinking gravity and then
you can say it in pounds,

and that measures force.

And you have other units
that are not international.

I'm not gonna mess up.

When you have the work
in this case, though,

it's more complicated.

And I'm not gonna be mad
at you. [INAUDIBLE] is

trying to tell me what it is.

I'm not going to be mad at
the people who don't know

what the work is in this case.

Although, I was looking
at-- I am the person

who has run a
committee to oversee

the finals for different
math classes, all the math

classes we offer here.

And every semester I see the
[? streak ?] pre-calculus,

calc 1, calc 2, calc 3.

In trig and
pre-calculus, they'll

always have a work there.

And I was wondering how many
of you took pre-calculus,

and how many of you
remember that you

studied this in pre-calculus.

It's a little bit awkward.

I'm thinking, how do they do
it, but I gave you the formula.

And they say the force in
itself as a vector dot product

the displacement vector.

So they are both
forces in dot product.

And I was surprised to see
that they gave you [INAUDIBLE]

If I were to express it,
how would I express it?

I'll say the magnitude
of F, of course,

in Newtons, whatever it
is, times the magnitude

of the displacement vector--

STUDENT: Multiply those cosines.

PROFESSOR: Cosine of
the angle between.

And I'm too lazy, I
don't know, theta.

Let's call it angle theta.

Because I don't want to
include that in locations, OK?

It really doesn't matter
in which direction

I'm going, because cosine theta,
thank God, is an even function.

It's equal to cosine
of minus theta.

So whether I go this way or
that way, it's the same cosine.

All right.

So the cosine of the
angle between the two.

It's very easy when you
don't need calculus.

But when you use calculus,
because your trajectory is

no longer a line, life is
becoming more complicated.

So we have to come up
with a different formula,

with a different notion of work.

I'm gonna erase-- are
you guys done with that?

Is it visible?

You're done.

OK.

So again, life is not so
easy in reality anymore.

I have a particle in physics-- a
photon enters a four-star hotel

and says-- talks to the
bellboy, and the bellboy,

can I help you
with your luggage.

No, I'm traveling light.

[LAUGHTER]

So the particle, the
photon-- whatever.

A particle is moving-- is
moving on a trajectory.

Suppose that
trajectory is planar,

just to make your
life easier at first.

It's in the plane x y.

And this is the little
particle that's moving.

And this is R. And
that is x i plus yj.

Good.

And this is the point x, y.

And that's the position,
the current position

of my particle, right now.

Not in the past,
not in the future.

My particle is moving,
and this is now.

Suppose time doesn't even exist.

We think of the movies
that we saw lately,

in The Theory of Everything.

So then, they say OK, we only
care about now. x, y is now

and that is the current
position vector.

Well, what would be
the work between now--

whatever now-- and the next,
let's say, this is gonna be x1,

y1.

And this is x0, y0.

That's the general formula,
will be x i plus yj.

So I actually cannot
forget about time.

Not as much as I want.

So x and y-- x and y are
both changing in time.

We're gonna have x equals
x sub t, y equals y sub t.

Do you guys remember what we
call that kind of equation

for a curve from here to here?

Para--

STUDENT: Parametrization.

PROFESSOR: Parametrization,
or parametric equations.

Parametric equations.

Good.

So I have may the
force be with you.

I have a force.

I have a force, and I have
some sort of displacement.

But I cannot express that
displacement linearly anymore.

I'm moving along a
[INAUDIBLE] of a curve.

So I have to think differently.

And the work will be defined,
whether you like it or not,

as F vector field, dot
product with dR over c.

And you say, what in
the world is that?

What would c be?

How do I integrate along a path?

And I will tell you in a
second what we mean by that.

Meaning that-- this is by
definition if you want.

This is like an
application of calculus 1.

It can be proved, but we
don't-- we do a rigorous job

in the book about introducing
and proving that along

a curvilinear path.

This is gonna be-- where
am I here, at time t0?

And this is at time t1.

That means x0 is x of t0.

y0 is y of t0.

And at the finish point,
I'm at x1, which is x of t1,

and y1 equals y of t1.

So between t0 and
t1, I have traveled

and I have F where
measure at x of t y of t,

where t is between--
moving between t0 and t1.

I'm done with this
is the F part.

What is the dR?

Now you guys know
about differential.

Thank God you know
about differential,

because this is gonna
help you very much.

OK.

So instead of dR, I'm going
to write dot, and let's

see how I write-- what I write
in terms of dR. You may say,

well, what does she mean?

dR was dxi plus dyj.

And you say, why is that?

I don't understand.

Because R itself is x i plus yj.

And x is a function of t,
and y is a function of t.

That means that when you
apply the differential,

you are gonna apply the
differentials to dx and dy,

and these are gonna be
infinitesimal displacement.

Infinitesimal displacement.

Infinitesimal displacement.

What is an infinitesimal
displacement in terms of time?

Well, we have our
parametric equations.

So Mr. dx as a differential
is just x prime dt.

It's like in the
[INAUDIBLE] substitution.

dy is just y prime dt.

So let me write this down again.

This is x prime of t i plus
y prime of t j times dt.

So Mr. dt is like
a common factor.

If he wants to go
out for a walk,

he says, I'm gonna
go out for a walk.

I go out for a walk.

So dR is actually x
prime of t times i

plus y prime of t times j dt.

And this will represent
the derivative of R

with respect to pi.

So that will be what?

The differential.

Differential of R
with respect to pi.

This is the same as
writing dx i plus dy j.

And it's the same as writing
dR. Why is this happening?

Because it's [INAUDIBLE]
Because x and y

themselves are functions of one
variable only, which is time.

This is why it happens.

Oh, so we will
simply have to do--

to learn new things, right?

We are gonna have
to learn new things,

like integral from a time--
fixed time 0 to t1, which

is 10 seconds, of a dot product
between a certain vector that

depends on time and another
vector that depends on time,

and dt.

So we are gonna
have to learn how

to compute the work through this
type of curvilinear integral.

And this is-- this is called
either path integral-- path

integral along the curve c, or
curvilinear integral along c.

Yes.

STUDENT: Let's say if I
move the force this is

[INAUDIBLE] function, correct?

So if I can find
[? arc length ?] that

is between the x--

PROFESSOR: Yeah--

STUDENT: [INAUDIBLE] points.

PROFESSOR: Yeah, we will do
the one with that length next.

The [? reason ?] so--

STUDENT: Is it harder?

PROFESSOR: No.

No, you can pass
through a plane.

And you can-- we'll
do that next time.

You will have an integral
with respect to S.

So the integration will
be with respect to dS,

they are correct.

And then you will have a
function that depends on S

[INAUDIBLE] So I'll-- for
today, I'll only teach you that.

Next time I'll teach
you that with respect

to arc length, which is also
very-- it's not hard at all.

STUDENT: OK.

PROFESSOR: So I will work
with you on [INAUDIBLE]

Now assume that we have-- I
will spray all this thing.

Assume that I have a problem.

I have a parabola-- arc
of a parabola, all right?

Between-- let's
say the parabola is

y equals x squared
between two points.

And I'll ask you to
compute some work,

and I'll tell you in a second
what [INAUDIBLE] to do.

So exercise-- assume
the parabola y

equals x squared between
points A of coordinates 0, 0

and point B of coordinates 1, 1.

a, Parametrized this parabola
in the simplest way you can.

And b, compute the work
along this arc of a parabola,

arc AB of this parabola.

For [INAUDIBLE] the
function big F of t,

you see that-- I'm going to say,
no, big F of the point x, y,

because you haven't parametrized
that yet, big F of x,

y being xi plus yg.

So you say, OK, wait a minute.

W will be integral over
the arc of a parabola.

Do you want to draw that first?

Yes, I need to draw that first.

So I have this parabola from A
to B. A is of coordinates 0, 0.

B is of coordinates 1, 1.

And this is y equals x squared.

So what kind of
parametrization is

the simplest one, the
regular one that people take?

Take x to be t?

And of course, take y in that
case. y will be t squared.

And for 1 you have 1.

For 0 you have 0.

When you have that work by
definition, what was that?

It was written as integral
or on the graph C. Let's call

this path C a curvilinear path.

Look, script C-- so beautiful.

Let me [INAUDIBLE] red and
draw the C of what is work?

F force-- may the force be with
us-- dot dR. All righty, that's

a little bit of a headache.

This F is going to be-- can I
write an alternative formula

that I have not written yet
but I will write in a second?

dR will be dxi plus dyj.

So I can also
write that F dot dR

as the dot product will seem to
be-- what was the dot product

guys, do you remember?

First component times
first component, F1dx

plus second scalar component
times second scalar component,

F2dy.

I'll write it down.

Along the path C I'll
have F1dx plus F2dy.

But god knows what it's
going to be in terms of time.

So I have to change
variable thinking.

Okey-dokey, Mr.
dx by substitution

was x prime to T. Mr. dy by
substitution was y prime dt.

So I'd rather write
this in a simpler way.

This is a new object,
path integral.

But we know this
object from Calc I

as being a simple
integral from time t0--

I'll write it down-- time
t1, F1x prime of t plus F2y

prime of t.

This is the integral dt.

This would be a
piece of cake for us

to apply in this problem.

Equals-- now you tell me
what I'm supposed to write.

Because if you don't, I'm
going to not write anything.

t0 for me is what time?

When did we leave this?

0.

And when did we arrive?

At 1 o'clock.

We arrived when t is 1, or
every one second or whatever

depending on [INAUDIBLE].

OK, from 0 to 1, now who is F1?

F1 is this.

But it drives me crazy.

Because I need this
to be expressed in t.

So I think of x and
y as functions of t.

So if 1 is not x,
not [INAUDIBLE]

right here, but t, which is the
same thing in parametrization--

this is t, t times.

Who is x prime?

1, thank god.

That is easy, times 1, plus F2.

Who is F2?

t squared.

I'll have to write it down.

Times who is y prime?

2t.

y prime is t2.

So I write it down-- 2t, dt.

So that's how I compute
this integral back.

Is it hard?

No, because it's just a simple
integral from Calculus I.

So I have to integrate
what function?

A polynomial, 2t cubed
plus t with respect

to t between 0,
time 0 and time 1.

Good, let's do it.

Because that's a piece of
cake-- 2 times t to the 4 over 4

plus t squared over 2.

I take the whole thing between,
I apply the fundamental theorem

of calculus, and I have between
t equals 1 up and t equals 0

down.

What's the final answer?

It's a single final answer.

And again, on the
exam, on the final,

do not expect a
headache computation.

Do expect something
simple like that

where you don't
need a calculator.

You just have either integers
only or simple fractions

to add, and you
should get the answer.

What is the answer, guys?

1-- 1/2 plus 1/2 equals 1.

So 1 is the value
of the work in what?

Measured in newtons
times meters,

whatever your units are.

When you drag the
object from this point

to this point, on which the
acting force is the only

acting force-- it
could be the result

that there are several forces.

That is that force
that you have here.

Is it useful?

It's very useful for engineers.

It's very useful for physicists.

It's very useful for
anybody who works

in applied mathematics, this
notion of work given like that.

I'm going to go ahead and erase.

And I'll ask you one
thing here that is not

in the book I think
as far as I remember.

Can you guys prove that this
sophisticated formula becomes

your formula of the
one you claimed,

the first formula you gave me?

Is it hard?

Do you think it's
hard to prove this?

OK, what if we have the
simplest possible case.

Let's think of--

STUDENT: [INAUDIBLE]

PROFESSOR: Yeah,
I'm thinking maybe I

should do, well, A to B, right?

AB, what kind of expression
do I have [INAUDIBLE]?

If I take this to be-- I
could have any line, right?

I could have any line.

But if I have any line,
I can pick my frame

according to my preference.

Nobody's going to
tell me, well, you

have to take the
frame like that,

and then your line will be of
the form ax plus by equals.

No, I'll just take the
frame to be this one, where

AB will be x axis, and A
will be of coordinates 0, 0

and B will be of
coordinates B and 0.

And this is just my line.

So x will be moving between
0 and B. And y is 0, right?

It should be, at least.

And then F, let's say, should
be this function, this.

I'll assume the
angle is constant,

just like I had it with theta.

And then it's acting all
the way on your object.

You have the same
angle here always.

So F is F1i plus F2j.

dR will be dxi plus dyj.

But then you say, wait a minute,
but didn't you say, Magdalena,

that you are along this line?

Didn't you say that y is 0?

So which y?

So there is no y.

So this is 0, right?

OK, Mr. x, I want to
parametrize my trajectory.

How do I parametrize
it the simplest way?

I'll take x to be t.

And time will be
exactly between 0 and d.

And y will be 0.

And thank you god,
because that's easy.

And so all you
need to give me is

W is integral of F
dR C in that case.

So what am I going
to have in that case?

I'll have this formula.

I'll skip a step, and
I'll have that formula.

And that means I have integral
from t0 equals 0 to t1

equals B.

F1-- now you have to tell
me what F1 will be. x prime

[? noted ?] is 1.

The second guy is 0, thank you
very much, and [INAUDIBLE].

F1 will be what?

Well, life is nice.

F1 will be the projection of
the vector F on my x-axis.

So F1 is the length of
this blue vector, I'll say.

So F1 is a scalar.

Let's say F1 is a
scalar component.

That means it's F
length cosine theta.

Because it's hypotenuse
times cosine theta.

So it's easy.

So you have length
of F, how much it is,

how big this vector is, times
cosine theta, times what

when you integrate it, guys?

When you integrate 1 with
respect to t, what do you get?

t between d and 0.

So you have t between d and 0.

We got the formula.

So we got that F length
times [INAUDIBLE]

times cosine theta times d,
this is the displacement.

This is the cosine.

This is the magnitude of
the force that I'm-- look,

this is the force.

My force is along my arm.

I'm just dragging
this poor object.

The force I'm
acting with, suppose

it's always the same parallel
to that that I can feel.

So that's what I have, F
cosine theta, and it was easy.

So as a particular case
of this nasty integral,

I have my old work from
school that I had to believe.

I tell you guys, I did
not believe a word.

Because my teacher
in eighth grade

came up with this
out of nothing,

and we were supposed to
be good students preparing

for a high school like this
kind of scientific-- back home,

there are different
kinds of high school.

There is scientific high
school with emphasis

in math and physics.

There is one for
chemistry/biology.

There is one for language,
linguistics, [INAUDIBLE],

and so on.

And I was for the
math and physics one.

And I had to solve this formula
without understanding it.

And it took me many
other years to understand

that it's just a little
piece of a big picture,

and that there's
something bigger than what

we were taught in eighth grade.

STUDENT: [INAUDIBLE]

PROFESSOR: Yeah,
yeah, it's true.

Now I want to ask
you a question.

So do you think that I would get
any kind of conservation laws

in physics that
apply to calculus?

I mean, how hard is it
really to compute the work?

And I'm making an
announcement now.

Since I have not
given you a break,

I have to let you
go in a few minutes.

But I'm making a
big announcement

without proving it.

So we will, in about
one week at the maximum,

in maximum one week, study the
independence of path of work

if that work is performed
by a conservative force.

And you're going to
say, wait a minute,

what the heck is a conservative
force and what does she mean?

Well, I just showed you that
the work is a path integral.

We don't know what that is.

I'll introduce more.

I just introduced the definition
of a path integral with respect

to parametrization, general
parametrization with respect

to t.

So that becomes an integral
with respect to dt,

like the one in
Calc I. This is how

you have to view it at first.

But guys, if this force
is not just any force,

it's something magic, if F comes
from a scalar potential that

is F represents the gradient
of a scalar function F--

this is called scalar
potential-- then

F is called-- now let's
see how much money I

have for just the last two or
three minutes that I have left.

I don't have money
or I have money?

Come on, big money.

No, I have $5.

I was looking for $1.

Here, I'll give you $5
if you give me $4 back

if you guess-- I don't know.

So maybe in your
engineering courses-- maybe

I give you some candy instead.

So if there is a scalar function
little f of coordinates x,

y, whatever you
have in the problem,

so that big F will be the nabla.

F nabla means the gradient.

We say that F comes
from a scalar potential.

But it has also a name,
which is called-- god.

It starts with a C, ends
with an E. In that case,

if this is going to be equal
to nabla F, in that case,

there is a magic theorem
that I'm anticipating.

I'm not proving.

I'm doing exercises right now.

We'll see it in two sections,
that the work does not depend

on the path you are taking.

So you can go from A to B like
that, or you can go like this.

You can go like this.

You can go like this.

You can go like that.

You can go on a parabola,
on a line, on anything.

The result is always the same.

And it's like the
fundamental theorem

of Calc III in
plane for the work.

So you have little f endpoint.

STUDENT: Is that [INAUDIBLE].

PROFESSOR: Little
f of [INAUDIBLE].

So all that matters is
computing this scalar potential

here and here, making
the difference,

and that will be your work.

It's a magic thing.

In mechanical
engineering maybe you

met it, in physics-- in
mechanical engineering,

because that's where you guys
drag all sorts of objects

around.

STUDENT: Conservative.

PROFESSOR: Ah, thank god.

Rachel, you're a
math major I think.

You're an engineering major.

STUDENT: [INAUDIBLE]

PROFESSOR: Wow, OK, and
who else said conservative?

And were there other people
who said conservative?

I'm sorry I don't have.

Well, next time I'll
bring a bunch of dollars,

and I'll start giving
prizes as dollar bills

like I used to give in
differential equations.

Everybody was so
happy in my class.

Because for everything that
they got quickly and right,

they got $1.

So conservative-- very good.

Remember that for
the next few lessons.

We will show that when this f is
magical, that is conservative,

you guys don't have to
compute the integral at all.

There's no parametrization,
no nothing.

It really doesn't depend
on what path you take.

All you would need is to figure
who this little f will be,

this scalar potential.

Our future work can do that.

And then you compute the values
of that scalar potential here

and here, make the difference.

And for sure you'll have
such a problem in the final.

So I'm just anticipating
it, because I

want this to be absorbed
in time into your system.

When we will do the
final exam review,

you should be baptized
in this kind of problem

so that everybody will get
100% on that for the final.

OK, now I'll let you go.

Sorry I didn't give you a break.

But now I give you more time.

And enjoy the day.

I'll see you Thursday.

I'm moving to my office.

If you have questions,
you can come to my office.

Maybe you were getting close.

How did-- did you know,
or it just came to you?

[BACKGROUND CHATTER]

STUDENT: Do you know
what section it would be?

Because I don't even think
he's listed or anything.

PROFESSOR: Send me an email
if you don't figure it out.

But for sure [INAUDIBLE].

STUDENT: OK, because I was
going to do the honors,

but it was with [INAUDIBLE].

I don't know if he's
good, or she's good.

PROFESSOR: She's good.

But he's fantastic in the
sense that he will help you

whenever you stumble.

He's an extremely good teacher.

He explains really well.

He has a talent.

STUDENT: I'll look for him.

Thank you.

PROFESSOR: And if
you don't get him,

she is good as well-- not
exceptional like he is.

He's an exceptional teacher.

STUDENT: I'll go to the office.

PROFESSOR: Yes,
yes, [INAUDIBLE].