what we discussed
last time in 11.6

which was-- do you remember
the topics we discussed?

We discussed the
valuation of a derivative.

What else have we
discussed about them?

I'm gonna split the fields,
although they are related.

Gradient and the steepest
ascent and descent.

So in which direction
will z equals

f of x and y,
differential cofunction,

and with a derivative
that is continuous?

So will this have the
maximum rate of change?

This is just review.

Why am I doing review?

Well, after talking to
you on a personal basis,

like one-on-one basis by email
and a little bit in person,

I realized that you like
very much when I review.

When I briefly review
some of the notions.

I will give you the essentials
for the section, that

was 11.5 is embedded in this.

Embedded.

So 11.5 is embedded in
11.6, and in one shot

we can talk of them.

In 11.7 you're gonna see
some extrema of functions

of two variables,
like max and min,

and then we have 11.8, which
is Lagrange multipliers.

What have we done about
this differentiable function

with continuous
partial derivatives?

Let's assume that
it's smooth, OK?

And in that case, we
define the partial--

the directional derivative
in the direction of u at u,

v with unit vector at the
point x, y, but let's fix it

x0, y0, using a limit of
a difference quotient,

just like any derivative
should be introduced.

But I'm not gonna
repeat that definition.

Why is that?

Because I want you to give
me an alternative definition,

which is something
that is simpler

to use in applications,
which is what?

Who remembers?

The partial derivative
of f with respect to x

measured at the point x0, y0.

I'm gonna use this color, and
then I'll change the color.

And I'll say times u1.

Plus-- change the color again--
f sub y at x0, y0, times u2.

And this is just a
times-- multiplication

between real values.

Because u1 and u2 will be
nobody but the components--

the real value components
of the unit vector direction

that we have.

So, guys, remember, direction
in this books means unit vector.

Every time I say
direction, I should

say that's a unit vector.

Is there any way, any
other way, to express this?

Maybe with a vector
multiplication of some sort--

multiplication between vectors.

See, what if I had a pink
vector with pink components,

and a blue vector
with blue components?

I'm getting somewhere.

And I'm sneaky.

And you feel--
you know where I'm

getting because you had
chapter nine fresh in your mind

and a certain product
between vectors.

So what is this?

It's a dot product, excellent.

Rachel, right?

So it's a dot product.

And I'm gonna run
it, and it's going

to be just the gradient
of f at the point p0.

But I'm going to write
it again, x0, y0,

although it drives
me crazy to have

to write that all the time.

Dot product or scalar
product with what vector u?

I'm not gonna put a bar on
that because the-- that would

be like an oxymoron.

The gradient is a bar thing.

It's always a vector, so
I'm not gonna write a bar.

But I write a bar
on u, reminding you

that u is a free vector.

All right, do you like this?

Yes.

It's a compactified form
of the fluffy expression

we had before.

It's much easier to remember
than the limit definition.

Of course, it's
equivalent to it.

And in applications, I-- well,
we ask about that all the time.

What was the gradient?

So see, in mathematics
everything is related.

And talking about--
speaking about the devil--

I mean, not you, but we
weren't talking about you--

just this gradient.

Gradient f at the
point p will simply

be the vector whose components
in the direction of i and j

are the partials.

The partial derivative
with respect

to x, plus the partial
derivative with respect to y.

OK, last time we dealt even
with gradients of-- functions

of more variables.

If I have n
variables, so x1, x2,

x3, xn, then this vector will
have n components-- f sub x1,

f sub x2, f sub x3, f sub x4.

Somebody stop me.

F sub xn.

So I have n of them, and
it's an n [INAUDIBLE].

It's a sum ordered [INAUDIBLE].

OK.

A set of n elements
in this order.

It's an order set,
[INAUDIBLE] n.

All right.

So the order matters.

Now, steepest
ascent and descent.

In which direction do I have
the maximum rate of change?

And the answer is--
this is Q and this

is A-- the maximum
rate of change

happens in the direction of
the gradient at every point--

every point of
the domain where I

have smoothness or [INAUDIBLE]
or [INAUDIBLE] whatever.

All right.

And what else?

I claimed last
time, but I didn't

prove, that in that
direction that I claimed c

from [INAUDIBLE].

The directional derivative
is maximized exactly

in the direction
of the gradient.

Can we prove that?

Now we can prove it.

Before I couldn't prove it
because you couldn't see it,

because I didn't look
at it as a dot product.

And we were all blind, like
guiding each other in the dark.

Me blind, you blind.

We couldn't see.

Now we can see.

So now we can see
how to prove my claim

that the directional
derivative, which

is measuring the
maximum-- measuring

the instantaneous rate of
change in a direction-- compass

direction, like,
what is that, east?

East, northeast,
southwest, whatever.

Those are the compass
directions like i, j, i

plus j over square
root 2 and so on,

those are called
compass directions

because you hold the compass
in your hand as horizontal

as you can, and you
refer to the floor.

Even if you are on a slope.

Maybe you imagine me on a
slope, hiking or whatever.

Going down, going up.

But the compass should
always be kept horizontal.

Do you hike, Alex?

STUDENT: Yeah.

PROFESSOR: I'm sorry,
I've put you on the spot.

So whatever you do when you
think of a path up or down

is measured in the direction
of the horizontal plane,

the compass direction.

And that is the gradient
I was talking about.

You see, it's a function.

It's a vector in plane.

That means a geographic
compass direction.

Prove it.

Let's prove the claim.

Let's prove the claim,
because this is Tuesday,

and it's almost weekend, and
we have to prove something

this week, right?

Now, do you like what you see?

Well, I have no idea.

What if I want to measure this.

This could be a negative
number, but it doesn't matter.

Assume that I take
the dot product,

and I think, well, it's a scalar
product, it must be positive.

What do I get?

This guy-- if I'm a physicist,
I'm going to say this guy

is the length of the
first vector at p0.

Gradient at p0.

That's the length of u.

Duh, that is 1.

Last time I checked, u was
unitary, so that's silly of me,

but I'll write it anyway.

Times the cosine of the
angle between-- oh, cosine

of the parenthesis angle
between nabla f and u.

OK.

When is this maximized?

STUDENT: When the angle
between nabla f and u is 0.

PROFESSOR: Exactly.

When this is pi, so this
quantity is maximized-- gosh,

I hate writing a lot.

I had to submit homework
in the past two days,

and one was about biological
research and the other one

about stress management.

The stress management class
stresses me out the most.

I shouldn't make it public.

Really, because we have to write
these essays of seven or eight

pages every Tuesday,
every end of the week.

Twice a week.

, So I realized how much
I hate writing down a lot,

and what a blessing it
is to be a mathematician.

You abbreviate everything.

You compress everything.

I love formulas.

So what we have here is
maximized when the cosine is 1.

And if you have
become 0 to 2 pi open,

then theta 0 is
your only option.

Well, if you take
an absolute value,

you could also have it
in the other direction,

cosine pi, but in that
case you change the sign.

So what you get--
you get a maximum.

Let's say you hike, right?

I'm just hiking in my brain.

The maximum rate of
change in this direction,

climbing towards the peak.

And then the steepest descent
is the exactly minus gradient

direction.

So I could have 0
or pi for the angle.

That's the philosophical
meaning of that.

All right.

So the directional
derivative, which is this guy,

when does it become maximum?

When the angle is 0.

So I'm done.

QED.

What does it mean?

That I know this happens
when-- when direction u is

the direction of the gradient.

Can I write u equals
gradient of f?

Not quite.

I should say divided
by norm or magnitude.

Why is that?

And you say, Magdalena,
didn't you say like 10 times

that u is a unit vector?

You want u to be a
unit vector direction.

So the direction should be
the direction of the gradient

in order to maximize this
directional derivative.

But then you have
to take the gradient

and divide it by its magnitude.

Let's compute it.

Let's see what we get.

Let's see what we get.

Now, I'm sorry about my
beautiful handwriting,

but I'm-- well, I'm gonna
have to-- I have room here.

And actually I can
use this formula.

So in the direction
of the gradient,

when u is the gradient.

Let's take u to be-- what
did I say over there?

Gradient of f over the
magnitude of gradient of f.

I'll take this guy
and drag him here.

[INAUDIBLE]

What will the value of the
directional derivative be?

We've done last time that--
the same thing on an example.

We've done it on a function,
beautiful f of x, y

equals x squared plus y squared.

This is the type
of function I like,

because they are the
fastest to deal with.

But anyway, we'll have all
sorts of other functions.

What am I going to write here?

I have to write.

Well, by definition,
now, by my new way

to look at the
definition, I'm gonna

have a gradient at
the point, the vector,

dot-- who is u again?

The gradient this time,
that special value,

divided by absolute-- by
magnitude of the gradient.

But what in the world is that?

This animal is--
what if you take

a vector multiplied by itself?

Dot product.

No, not dot product.

What you get is the
magnitude squared.

All right.

So although the pinkie
guy and the pinkie guy

are magnitude of f squared
divided by magnitude of f.

And Alex said, but wait,
that's just-- I know.

I didn't want to just
jump ahead too fast.

We get gradient
of f in magnitude.

So, beautiful.

So we know who that maximum is.

The maximum of
the rate of change

will be for-- this
equals f of x, y.

The max of the rate of change
is-- what is that again?

Magnitude of lambda f, in
the direction nabla-- nabla f

divided by its magnitude.

This is what we discovered.

And now I'm going
to ask you, what

is the minimum rate of
change at the same point?

STUDENT: [INAUDIBLE].

PROFESSOR: Just
parallel opposite.

So it's gonna-- I'm gonna
have the so-called highest--

steepest, not highest,
highest means maximum.

The lowest value.

So I'm going to have
the lowest value, which

indicates the steepest descent.

Me going down on-- in the snow,
I'm dreaming, on a sleigh,

or on a plastic bag.

That would give me
the steepest descent,

and the steepest descent
will correspond to what?

I'm going to make an NB.

Nota bene.

In Latin.

Note the minimum will
be minus magnitude

of f, [INAUDIBLE] nabla f,
in the opposite direction.

Shall I write it in words?

Let me write it in as
O-P-P from opposite-- no,

from opposite--
opposite direction.

What do I mean
opposite direction?

Opposite direction
to the gradient.

Which is the same direction,
if you think about,

because it's the same line.

So it's going to be minus nabla
f over magnitude of nabla f.

It's like when we were
in [INAUDIBLE], which

was 1 minus x squared minus
y squared, whatever it was--

we had i plus j for the
descent, and minus i minus

j after the ascent--
the steepest descent

and the steepest ascent.

We started with examples
because it's easier

to understand
mathematics-- actually

it's easier to understand
anything on an example.

And then-- if the
example is good.

If the example is
bad, it's confusing.

But if the example is
good, you understand just

about any concept, and then
you move on to the theory,

and this is the theory.

And it looks very abstract.

When somebody steps
in this classroom

and they haven't taken more than
calc 2 they will get scared,

and they will never
want to take calc 3.

Well, that's why-- I didn't
want to scare you off yet.

OK, so this is what you have
to remember from section 11.6

with 11.5 embedded in it.

Now, one thing that
I would like to see

would be more examples and
connections to other topics.

So one example that I picked--
and I think it's a nice one.

I just copied from the book.

I usually don't
bring cheat sheets.

I don't like professors who
bring books to the class

and start reading
out of the book.

I think that's ridiculous.

I mean, as if you guys couldn't
read your own book at home.

And I try to make up examples
that are easier than the ones

from the book to start with.

But here's example 4, which
is not so easy to deal with,

but it's not hard either.

And I picked it because
I saw this browsing

through the previous
finals, I saw it

as a pattern coming
every now and then.

Find the direction in which
f increases or decreases

most rapidly.

I have to write
down beautifully.

Find the directions in which
f increases or decreases most

rapidly at p0 coordinates 2, 1.

And, what is the
maximum-- that's

a question-- what is the maximum
rate of change or of increase?

This type of problem is also
covered in the Khan Academy

videos and also the MIT
library, but I don't

feel they do a very good job.

They cover it just
lightly, as if they

were afraid to speak too much
about-- give too many examples

and talk too much
about the subject.

So what do you guys want
to do with this one?

Help me solve the problem.

That was kind of the idea.

So we start with
computing the what?

The animal called--
what's the animal?

STUDENT: Gradient.

PROFESSOR: Gradient.

Thank you very much.

So we do that and we
start differentiating.

With respect to x, we get
a product [INAUDIBLE].

So your product [INAUDIBLE] 1--
it's differentiated-- times e

to the 2y minus x [INAUDIBLE]
plus x undifferentiated.

The second guy, prime.

Copy and paste, Magdalena.

Times-- don't
forget the minus 1,

because-- I'm talking to myself.

Because if you do, you get
a 0 on this in the final.

I'm talking to myself.

OK?

All right.

Plus, parenthesis-- the same
procedure with respect to y.

When I do it with
respect to y, this

is good review of
the whole chapter.

Yes, sir.

STUDENT: [INAUDIBLE].

PROFESSOR: This primed
with respect to x.

Am I doing something wrong?

No.

Are you with me?

So this guy primed
with respect to x, I'm

going to write it es copied.

And take that out, you maintain
and differentiate with respect

to x.

So I did it right.

OK, the second one, x is a
constant for me right now.

Who is the variable y?

So I copy and paste e
to the whole argument

times-- I cover everything
else with my hand,

and I differentiate the
argument with respect to y.

And I get prime sub 2
and a j, and I say, thank

god, this was a little bit long.

You realize that if you make
the slightest algebra mistake,

it's all over for you.

In that case, I
ask my colleagues,

what do you guys do when guy
missed that or missed this?

0, 0, no points-- OK,
maybe a little bit,

maybe a tiny bit
of extra credit.

But pay attention to your math.

So you know what you need to do.

Now I'm going to go on and
say, but I am at the point 0.

By the way, I really don't
like what we did in the book.

OK, I should not say that
out loud, but it's too late.

The book denotes that
sometimes, well, we

try not to do that too
often, but not by F sub 0.

Because some other
books use that.

I don't like that.

So every time-- you
should never do that.

Because it gives
the feeling that

you're differentiating
a constant or something.

OK, so I always try to
say, gradient [? up ?]

is-- which means, I
have a fixed value.

But I don't fix the value
before I took the gradient.

This is too confusing
as a notation.

Don't do it.

Close your eyes
when you get to it

when you're reading the book.

OK, now I have to plug
in instead of x sub 2--

I tried to remember
that-- y equals 1.

So I go 1 times e to
the 2 times 1 minus 2

plus 2 times e to
the 2 times 1 minus--

I wish I had Data with me, I
mean the guy from Star Trek.

Because he could do this in
just a fraction of a second

without me having to bother
with this whole thing.

STUDENT: But then
if Data existed,

why would we be math majors?

PROFESSOR: Exactly,
so we do this

so that we can program
people, I mean androids,

and eventually learn
how to clone ourselves.

So let's see what we have.

e to the 0 is 1.

e to the 0 is 1, 1 minus
2-- are you guys with me?

I'm going too fast?

STUDENT: No, the
[? board ?] says

the y component is equal to
1, x component is equal to 2.

PROFESSOR: x
component equals to 2.

x0 is 2, and y0 is 1.

And I plug in x0
equals [INAUDIBLE].

So 2 times e to the 2 times 1
minus 2-- you think I like it?

I don't like it.

But anyway, it's my life.

I have to go on.

So we have 1 minus 2
plus minus 1, right?

Minus 1i-- minus
1i sounds scary.

OK, plus 4j-- minus i plus 4j.

Is it lovely?

No, I hate it.

The magnitude is going
to be square root of 17.

But that's life.

I mean, you as
engineering majors

see that all the time,
and even worse than that.

So I'm going to say the
gradient of F at P0 in magnitude

will be square root of 17.

What is that?

That is the maximum rate
of change, right guys?

But in which direction
does that happen?

My beloved book says, in the
direction of minus i plus 4j.

That's our answer, so
in the direction of--

STUDENT: Over root 17.

PROFESSOR: No, let
me tell you what.

We fought about that as the
authors when we wrote the book.

So I said, if you're
saying, in the direction of,

then you have to
say, over root 17.

And my coauthor said,
no, actually, Magdalena,

as a matter of English--
since you're not a native,

you don't understand.

In the direction
of a certain vector

means the direction
of a certain vector,

the direction could be that.

But I say equivalently, in
the direction double dot

minus i plus 4j over
square root of 17,

meaning that this
is the direction.

It's a matter of interpretation.

I don't understand it,
but it's your language.

Yes, sir.

STUDENT: [INAUDIBLE].

PROFESSOR: I'm saying--

STUDENT: Which one do you like?

PROFESSOR: Which one do I like?

This one.

STUDENT: [INAUDIBLE].

PROFESSOR: Thank you.

So if we define direction
to be a unit vector,

let's be consistent and not say,
of me, of you, of my cousin,

of whatever.

All right, was this hard?

No.

Do I have a caveat about
this kind of problem?

Yeah, that was kind of the idea.

If I put that in
the midterm, guys,

please do it two or
three times, make

sure you didn't make
any algebra mistakes.

Because if you know the
theory, I will still

give you like 30% or something.

But if you mess up
with the algebra,

I have no choice but giving
up the 70%, whatever that is.

I try to be fair and give
you something for everything

you do and know.

But try not to mess
it up too badly,

because it's very
easy to mess it up.

Yes, sir.

STUDENT: So which
way is increasing?

PROFESSOR: OK,
find the direction

in which r increases or
decreases most rapidly.

OK, the direction in
which I could draw it,

this is increasing in
the direction of that.

At what rate?

At the rate of
square root of 17.

Good question.

How about the other one?

In the direction of plus i
minus 4j over square root of 17,

I get the rate of change
minus square root of 17.

And that's it.

Do we have to say that?

Ehh, I give you extra
credit if you say that.

But at this point, I'm saying
I'm happy with what I just

wrote on the board.

STUDENT: [INAUDIBLE].

PROFESSOR: Yeah, so
it's like finding

on which path are you
going to get to the top

the fastest when you
climb a mountain.

It's the same kind of question.

Because it's all about
z being an altitude.

z equals F of x.

I will go ahead and
erase the whole thing.

And I wish you good luck now.

I wish you good luck because
I was asked by [INAUDIBLE]

to solve a problem like the one
you gave me in the web work,

but I don't remember it.

But it's OK.

it was a review of
what we did last time.

And we said, instead of that,
where the tangent plane--

we know what that is.

At P0 was-- guys, by the
final I want this memorized.

There is no question.

So z minus z0 is
like Taylor's formula

in the linear approximation.

You truncate.

You throw away the
second order, and so on.

So what is that?

f sub x at delta x.

Oh, x equals 0.

I'm lazy.

Times x minus x0--
this is the delta x.

This is the delta z.

And this is not the round
surface, curvy and everything.

This is the plane approximation,
the planar approximation--

A-P-P-R. I'm not done--
plus f sub y y minus y0.

So this is the equation of the
plane that's tangent pi at 0.

And let me draw
the surface pink.

Because I'm a girl, and because
I want to draw this in pink.

Let's call it some
S, for Surface.

Can we paint an S here?

OK, but if I'm giving
the same picture

for a different equation--
so I have F of x, y, z

equals constant, that's
the implicit form.

I make this face.

Why do I make this face?

Because I've got
three confessions.

I'm more like a
priestess in mathematics.

People don't like
implicit differentiation.

We will do a little
bit of that today,

because you told me your stories
from implicit differentiation,

and I got scared.

Those were horror stories.

And I don't want those to repeat
in the final or the midterm.

So I'm going to do something
with implicit differentiation

as well.

What did I want to say?

If we apply this, we'd be wrong.

But we have to remember
what the normal would be.

And the normal was
the gradient of big F.

So those will be F sub xi plus
F sub yj plus big F sub zk.

Then force, such a
surface, even implicitly,

the tangent plane looks
a little bit differently.

But it's the same story.

And I prove that.

You may not believe it,
or may not remember.

But I proved that, that
it's one and the same thing.

So I get F sub x.

I get 0i0.

This was the A dot,
times x minus x plus.

What's next?

From the coefficients
coming from the gradient--

that was the gradient.

So this is F of
x, y, z equals C.

And the gradient was-- this is
the normal, F sub x, F sub y,

F sub z angular bracket
equals gradient of F, which

is more or less than normal.

The unit normal will be just
gradient of F over length of F.

So let's continue-- F sub
y at x0y0 times delta y.

This is B. I had the problem
with memorizing these,

especially since when I was
18 I did not understand them

whatsoever first
year of college.

I had to use markers,
put them in markers

and glue them to my closet.

Because when I
was 18, of course,

I was looking in the
mirror all the time.

So whenever I got
into the closet,

opened the door to the
mirror, next to the mirror

there was this formula.

So whether I liked it or not--
I didn't-- I memorized it just

by seeing it every day
when I opened the door.

Yes, sir.

STUDENT: That green one, should
that be x minus 0 or z minus 0?

PROFESSOR: z minus 0.

That is my mistake.

Thank you.

So again, you have delta
x, delta y, delta z.

Thank you.

Is it hard to memorize?

No, not if you
put it in markers.

I think you will do just fine.

What have we done as time?

Let me review it really quickly.

We said, wait a minute.

How come they are
one and the same?

You said, oh I'm
getting a headache.

I don't understand why
they are one and the same.

And we said, yes, but you
see, this guy is nothing but z

minus F of x, y.

So from this form, you can make
it implicit it by pulling out F

to the left and creating this
big F of x, y, and z equals 0.

And what does it mean?

It means that F sub x will be
oh my god, this has nothing

to do with us minus F sub x.

F sub y will be minus F sub y.

Am I right?

And F sub z, big F sub z, is
simply-- there's no z here-- 1.

So coming back to the guide's
A, B, C, forget about A,

B, C. I'll take the A, and
I'll replace it with minus F

sub x, which doesn't write.

And it's time for him to
go away-- minus F sub x.

And this is F sub
y with the minus.

And finally, Mr. C,
who is happy, he is 1.

So he says, I'm happy.

You're going to separate me.

We are going to separate him.

So this equation is
nothing but what?

Let's write it from
the left to the right.

Let's keep the green guy
in the left hand side.

And everybody else
goes for a moving sale.

The blue and the pink go away.

And when they go
to the other side,

they have a minus,
pick up a minus sign.

But with the minus, the
minus here is a plus sign.

Are you guys with me?

So the blue guy has
moved and changed sign.

Where's the pink?

The pink guy will also move.

And he picked up this.

So practically, this and that
formula are one and the same.

They're both used for
the same tangent plane.

It depends how you
introduce the tangent plane.

And [INAUDIBLE], before class
started, 20 or 25 minutes

before class, when I was in a
hurry, I answered you briefly.

You made some algebra
mistake in the-- you got it?

I would like to make
up one like those.

But I forgot what
your surface was.

Was it an ellipsoid?

STUDENT: Ellipsoid.

PROFESSOR: OK, I'll
make up an ellipsoid.

STUDENT: [INAUDIBLE].

PROFESSOR: Yeah, mhmm,
if you have it with you.

If you don't have it
with you, that's fine.

So I'm going to
go ahead and keep

just the implicit equation.

Because the ellipsoid is given
by the implicit equation.

And everything
else I will erase.

And that was problem 24.

How many problems?

Well, you still have
a lot, up to 49.

STUDENT: 42.

PROFESSOR: 42, OK, I reduced it.

Now, when I sent you
an email on Sunday,

I said I was giving
you an extension.

STUDENT: Till March.

PROFESSOR: Till a lot of March.

Because I thought March the
2nd, and I gave a few more days.

So you have one more week.

STUDENT: When's spring break?

PROFESSOR: Spring
break is the 14th,

but this is due
on the 9th right?

STUDENT: The 10th.

PROFESSOR: The 10th--
maybe I don't remember.

So what was your problem?

What's your problem?

What?

Can I take it?

So you have an
ellipsoid, which comes

from the ellipse 3x
squared [INAUDIBLE].

Then you have a 2z
squared equals 9.

Then it says, you
look at the point P

of coordinates minus
1-- I haven't even

checked if it's correct,
but it should be.

So 3 plus-- I did not
program the problem.

3 plus 4, 7, plus 2, 9,
so he did a good job.

We want the tangent plane.

I'll put it here.

We want the tangent plane.

How do we compute
the tangent plane?

You say, this is F
of x, y, z, right?

So F sub x equals 6x.

F sub y equals 2y.

F sub z equals 4z.

Computing it at P0,
what do we have?

x is minus 1, y is
minus 2, z is minus 1.

I should get negative 6,
negative 4, and minus 4.

And then I should plug
in and get minus 6 times

x minus minus 1.

I have to pay attention myself.

It's not easy to get the
algebra right-- minus 4 times y

plus 2 minus 4 times
z plus 1 equals 0.

And I hope I get what you got--
minus 6x minus 4y minus 4z,

so many of those.

You've got to divide by 2.

I'm not getting that.

And then minus 6-- I'm
going to write it down.

So it's even, right?

The whole thing minus 18, divide
by 2 should be-- divide by 2,

and did you change the
signs, [INAUDIBLE]?

What's your password?

No.

[LAUGHING] Check if I'm
getting the same thing you got.

So I get 3x plus 2y plus 2z.

I divide by minus
2, right, plus 9?

Did you both get the same thing?

STUDENT: [INAUDIBLE].

PROFESSOR: You didn't?

Well, I'm trying to
simplify all my answers

with this simplification.

So I guess of course
if you enter it

like that, it's going to work.

Now, I need you guys
to help me on this one.

Find the parametric
form-- it's so easy,

but this is a problem
session-- of the line passing

through the same point
that's perpendicular

to the tangent plane.

Express your answer in the
parametric form of the type

the one that you know that
I don't like very much,

but I will write it down-- a2t
[INAUDIBLE] b2 a3t plus b3.

This is how he wants me to
write it, which I don't like.

I would've even preferred
it to be in symmetric form.

It's the same.

I'll put the t, and I'm fine.

So x minus x0 over
l, y minus y0 over m,

z minus z0 over n, that
was the symmetric form.

I make it parametric
by saying equal to t.

So what were the
parametric equations?

x equals lt plus x0.

That's the normal.

y equals mt plus y0.

z equals nt plus z0.

So finally, my answer-- I'll
check with [INAUDIBLE] answer

in a second-- should be take the
normal from the tangent plane,

3, 2, 2, right?

2t plus whatever,
this is 2t plus z

equals-- first it's 3, 3, 2,
2, 3, 2, 2, plus x0 y0 z0.

So erase the pluses and minuses.

And those should
be the equations.

And I should write them down.

It's OK, you have the same.

Now you just have to take
them, this, this, and that,

and put it in this form.

This is a combination problem.

Why do I say
combination problem?

It's combining Chapter
11 with Chapter 9.

This was the review
from Chapter 9.

Did you have trouble
understanding

the radiant or
the tangent planes

or anything like that,
implicit form, explicit form?

Let me do an application,
since I'm doing review anyway.

I'm done with the Section 11.6.

But before I want
to go further, I

want to do some
review of Chapter

11 sections 11.1 through 11.6.

I said something about
implicit differentiation

being a headache
for many of you.

One person asked me, how do
you compute z sub x and/or z

sub y based on the equation x
squared plus y squared plus z

squared equals 5?

And of course this is
implicit differentiation.

Why implicit?

OK, because this is an implicit
equation of the type F of x, y,

z equals constant.

When do we call it explicit?

When one of the
variables, x or y or z,

is given explicitly in
terms of the other two.

So if this would be-- well,
here it's hard to pull it out.

But whether it be upper
part then lower hemisphere,

z would be plus or minus.

So you have two caps,
two hemispheres,

plus/minus square root 5 minus
x squared minus y squared.

Well, that's two functions.

We don't like that.

We want to be able to do
everything in one shot

without splitting it into
two different graphs.

So how do we view z
to be a function of x,

you're going to ask yourself.

You imagine inside
this thing that x and y

are independent variables--
independent variables.

They can take
whatever they want.

One is temperature.

One is time.

They run like crazies.

But z depends on both
temperature and time, like us

unfortunately.

It's so cold outside.

I hate it.

OK, you promised me,
and it came true.

Who promised me?

Matthew, I give you a
brownie point for that.

Because you said last week
it's going to be 80 degrees.

And it was.

So the prophecy came true.

On the other hand,
it came back too bad.

And of course it's
not Matthew's fault.

He didn't say what's
going to happen this week.

All right, in this case,
implicit differentiation

is just a philosophical thing.

It's a very important
philosophical step

that you're taking-- think.

Think of z being a
function of x and y.

And two, differentiate
z with respect to x.

So what do you
mean, differentiate

with respect to x?

By differentiating
the entire equation,

both sides of an equation
with respect to x.

So for you, x is
the wanted variable.

y is like a constant.

z is a function of x
is not hard at all.

So what is going to happen
actually if you were to do it?

Theoretically, you
would go like that.

If I'm going to differentiate
this guy with respect to x,

what is the philosophy?

The chain rule tells
me, differentiate F

with respect to the first
variable, and then times dx/dx.

And you say, god, now
that was silly, right?

Differentiate with respect to x.

That's the chain rule.

Plus differentiate
F with respect

to the second place,
second variable, and then

say, dy with respect to x.

But are dx and dy married?

Do they depend on one another?

Do they file an income
tax return together?

They don't want to have
anything to do with one another.

Thank god, so x and y are
independent variables.

If you're taking
statistics or researching

any other kind of
physics, chemistry,

you know that these are
called independent variables,

and this is called the
dependent variable.

And then you have what's
called the constraint.

In physics and engineering and
mechanics, F of some variable

equals C. It's called
constraint usually.

OK, so this guy is all silly.

These guys don't want to have
to do anything with one another.

And then you get plus.

Finally, dF with respect
to the third place,

and then that third place, z,
is occupied by a function that's

a function of x.

So you go, dz/dx.

Why del and not d?

Because poor z is a function
of two variables, x and y.

So you cannot say, dz/dx.

You have to say
del z dx, equals 0.

Thank god, I got to the
end where I wanted to get.

Now, if I want to see what's
going on, it's a piece of cake.

That's 1.

And I get that Mr. z sub x,
which other people write dz/dx,

but I don't, because I don't
like it-- I keep mixing x.

Equals-- how do I
pull this guy out?

How do I substitute for that?

I get Mr. First Fellow
here to the other side.

He's going to pick up
a minus at whatever d

I have divided by-- so this
guy divided by this guy.

STUDENT: What happened
to dF/dy, dy/dx?

PROFESSOR: So again, that
is a very good thing.

So dF/dy was behaving.

He was nice.

But when we got to dy with
respect to dx, y said,

I'm not married to dx.

I have nothing to do with dx.

I'm independent from this.

So dy/dx is 0.

And so this guy
disappears. dx/dx is 1.

Duh, that's a piece of cake.

So I'm done.

This is actually a formula
that looks sort of easy.

But there is a lot
hidden behind it.

This is the implicit
function theorem.

where you of course assume
that these partial derivatives

exist, are continuous,
everything is nice.

It's a beautiful result.
People actually get

to learn it only when they are
big, I mean big mathematically,

mature, in graduate school,
first or second year

of graduate school.

We call that
intermediate analysis

or advanced-- very
advanced-- calculus.

Because this calculus
is advanced enough,

but I'm talking about
graduate level calculus.

And this is the so-called
implicit function theorem.

So if you will ever be even not
necessarily a graduate student

in mathematics, but a graduate
student in physics or something

related to pure science,
remember this result. So let me

see what's going to
happen in practice.

In practice, do we
have to learn this?

No, in practice we can build
everything from scratch, again,

just the way we did
it with the formula.

So for the example
I gave you, it

should be a piece of cake
to do the differentiation.

But I'm going to step by step.

Step one, think.

You have to think.

If you don't think,
you cannot do math.

So you have x squared
plus y squared,

the independent guys, and z,
who is married to both of them.

Or maybe z is the baby.

These are two spouses that are
independent from one another.

And z is their baby.

Because he depends
on both of them.

So you thought you had
a different approach

to the problem, different
vision of what's going on.

Now finally, step two,
differentiate with respect

to one only, x only.

You could of course do the
same process with respect to y.

And in some of the
final exam problems,

we are asking, compute
z sub x and z sub y.

The secret is that--
maybe I shouldn't

talk too much again-- when
I grade those finals, if you

do z sub x, I give you 100%.

Because z sub y is the same.

So I really don't care.

Sometimes there
are so many things

to do that all I care
is, did he or she cover

the essential work?

So with respect to x, x squared
differentiated with respect

to x-- 2x.

Good first step, now, y squared
differentiated with respect

to x.

0-- am I going to write 0?

Yes, because I'm silly.

But I don't have to.

2 times z of x, y.

2 jumps down, z of x, y-- I'm
not done with the chain rule.

STUDENT: z sub x.

PROFESSOR: It's z
sub x, very good.

This is dz/dx.

I'm not going to hide
it completely like that.

That is the same thing.

y prime is 0, thank god.

So you say, if I were to
keep in mind that that's

the derivative of big
F with respect to x,

I could plug in
everything in here.

I could plug in the formula.

But why memorize the
formula and plug it

in when you can do everything
from scratch all over again?

Math is not about memorization.

If you are good, for example,
some people here-- I'm

not going to name them--
are in sciences that involve

a lot of memorization.

More power to them.

I was not very good at that.

So I'm going to go ahead and
write z sub x pulled down

minus 2x divided by 2z.

I'm too lazy to remind
you that z is the baby,

and he depends on
his parents x and y.

I'm not going to write that.

And that's the answer.

So you have minus x/z.

So for example, if somebody
says, compute z sub x

at the point on the sphere,
that is 0, root 5, and 0,

what do you have to do?

You have to say,
z sub x equals--

and now I'm asking you
something that is minus 0/0.

Assuming that the expressions,
the derivatives, are defined

and the denominator one
is different from 0-- so

whenever you do the
implicit function theorem,

you can apply with the
condition that you are away

from points where derivative
of F with respect to z are 0.

So this is a problem
that's not well posed.

So to give you a
well-posed problem, what

do I need to do on the final?

I have to say the
same-- 2, 1, and 0.

STUDENT: z can't be 0.

PROFESSOR: No, I know.

So I go, z is 0 is too easy.

Let's have y to be 0.

STUDENT: 2, 0, 1.

PROFESSOR: Very good, x equals
2, z equals 1, excellent.

So z sub x at the
point 2, 0, 1 will

be by the implicit
function theorem minus 2/1

equals negative.

You see, that's a slope
in a certain direction

if you were to look at z with
respect to x in the plane x, z.

OK, what else?

Nothing-- that was review
of chain rule and stuff.

And you have to
review chain rule.

Make yourself a note.

Before the midterm, I have
to memorize the chain rule.

Yes, sir.

STUDENT: [INAUDIBLE].

PROFESSOR: I will do that either
in the review session today

or in the review
for the midterm, OK?

And I'm thinking about that.

In March, I want to
dedicate at least 10 days

for the review for the midterm.

Yes, sir.

STUDENT: When is the midterm?

PROFESSOR: The midterm
is on the 2nd of April.

Several people asked me--
OK, I forgot about that.

I have to tell you guys.

Several people
asked me questions

by email about the midterm.

So the midterm-- write
down for yourselves-- will

be over the following chapters.

Chapter 10, no Chapter 9.

Chapter 9 is [INAUDIBLE].

Chapter 11 all,
Chapter 10 only what

we have required-- 10.1, 10.2,
and 10.3-- and Chapter 12,

all but Section 12.6.

Because I see that some of
you study ahead of time.

More power to you.

You know what to read.

Skip Section 12.6.

And I'm planning to not give
you anything after Chapter 13

on the midterm.

But of course, Chapter 13 will
be on the final emphasized

in at least six problems
out of the 15 problems

you'll have on the
final, all right?

We still have plenty of time.

Chapter 9, guys, you
were concerned about it.

It's some sort of
embedded, you see?

Wherever you go,
wherever you turn,

you bump into some parametric
equations of a line

or bump into a tangent line.

That's the dot product that
you dealt with, delta F dot N.

So it's like an obsession,
repetitive review of Chapter 9

at ever step.

Vector spaces are
very important.

Vectors in general
are very important.

I'm going to move
onto 11.7 right now.

We'll take a break.

Why don't we take a short
break now, five minutes.

And then we have to
go on until 2:50.

So practically we
have one more hour.

Take a break, eat,
drink something.

I don't want a big break.

Because then a big break
we'll just fall asleep.

I'm tired as well.

So we have to keep going.

[BACKGROUND CHATTER]

PROFESSOR: All right.

I will start with a
little bit of a review

of some friend of yours.

And since we are
in Texas, of course

this counts as an obsession.

This is going to be extrema of
functions of several variables.

Do I draw better lately?

I think I do.

That's why I stopped
drinking coffee.

I'm drinking white tea.

It's good for you.

All right.

White tea.

For some reason, the
black tea was giving me

the shaking and all that.

Too much black tea.

I don't know, maybe
it has less caffeine.

Jasmine is good,
green, or white.

STUDENT: I think green
has less [INAUDIBLE].

PROFESSOR: OK.

So above this
saddle is a function

of two variables-- you
know a lot already,

but I'm asking you to compute
the partial derivatives

and the gradient.

And you're going to
jump on it and say

I'm doing [INAUDIBLE] anyway.

So I've got 2x, and
this is minus 2y.

If I want to ask
you the differential

on the final or
midterm, you will say

that f sub xdx plus x of ygy.

Everybody knows that.

Don't break my heart.

Don't say 2x minus y,
because I'll never recover.

Every time I see that,
I die 100 deaths.

So don't forget about
the x and the y,

which are the important guys
of infinitesimal elements.

This is a 1 form.

In mathematics, any
combination of a dx and dy

in a linear combination
in the 1 form.

It's a consecrated terminology.

But I'm not asking you
about the differential.

I'm asking you
about the gradient.

All righty, and that is a f
sub xi plus f sub yk, which

is exactly 2xi minus 2yj.

And you say all right,
but I want to take a look,

I always have started
with examples.

Hopefully they are good.

Let's look at the tangent
vectors to the surface.

We discussed about the
notion of tangent vector

before, remember, when
we had r sub u and r sub

v form the parametrization.

Now look at the tangent
vectors for this graph

along the x direction
going this way,

and along the y
direction going this way.

We see that both of them are
horizontal at the origin.

And that's a beautiful thing.

And so this origin is a
so-called critical point.

Critical point for a
differentiable function.

Z equals f of xy is
a point in plane x0i0

where the partial
derivatives vanish.

And according to the book, and
many books, all don't exist.

Well I don't like that.

Even our book says if you
have a function in calc 1--

let's say b equal g
of u, critical point.

Do you remember what
a critical point was?

U0, in calc 1 we said either a
point where g prime of u was 0,

or g prime of u 0 doesn't exist.

Although u is 0
is in the domain.

I don't like that.

You say wait a minute,
why don't you like that?

I don't like that for
many reasons practically.

If you have the
absolute value function,

you'll say yeah, yeah, but
look, I considered the corner

to be a point of
non-differentiability,

but it's still an extreme
value, a critical point.

According to our book
in Calculus 1, yeah.

We extended this
definition to ugly points,

points where you don't have
a [? pick ?] or a value

or an inflection, but you
have something ugly like

a [? cusp, ?] a
corner, the ugliness.

I don't like that
kind of ugliness,

because I want to have
more information there.

I maybe even have a point with
a bigger problem than that.

First of all, when I
say critical point,

I have to assume the point is
in the domain of the function.

But then what kind of
ugliness I can have there?

I don't even want
to think about it.

So in the context of
my class-- in context

of my class-- calc 3 honors, I
will denote a critical point.

Is the x0y0 such that
f sub x at x0y0 is 0.

One slope is 0, the
other slope is 0.

f sub y is x0y0, of course.

And no other are the points.

What am I going to call the
[INAUDIBLE] points where

derivatives don't exist?

I simply say I
have a singularity.

I have a singularity.

What type of singularity we can
discuss in an advanced calculus

setting.

If you're math majors, you're
going to have the chance

to discuss that later on.

So remember that I would
prefer both in the context

of calculus 1 and calculus
3 to say critical value

is where the derivative
becomes zero.

Not undefined, plus, minus,
infinity, or something

really crazy, one on the
left, one on the right.

So I don't want to have
any kind of complications.

Now you may say, but I
thought that since you

have those slopes
both zero, that

means that the tangent plane
at the point is horizontal.

And that's exactly what it is.

I agree with you.

If somebody would draw the
tangent plane to the surface,

S-- S is for surface,
but it's funny,

S is also coming from saddles.

So that's a saddle
point, saddle surface.

Origin is so-called
saddle point.

We don't know yet why.

The tangent plane
at 0, at the origin,

will be 0.0 in this case.

Why?

Well, it's easy to see.

z minus 0 equals f sub x,
x minus x0 plus f sub y,

y minus y0.

But this is 0 and that's
0, so z equals zero.

So thank you very much.

Poor horse.

I can take a horizontal
plane, imaginary plane

and make it be tangent to
the saddle in all directions

at the point in the middle.

All right.

STUDENT: So you're saying
[? the critical ?] point

is where both--

PROFESSOR: Where both
partial derivatives vanish.

They have to both vanish.

In case of calculus
1, of course there

is only one derivative that
vanishes at that point.

What if I were
in-- now, you see,

the more you ask me
questions, the more I think

And it's a dangerous thing.

What if I had z equals
f of x1, x2, x3, xn?

Critical point would be where
all the partial derivatives

will be zero.

And then the situation
becomes more complicated,

but it's doable.

The other is the classification
of special points.

Classification of
critical points

based on second
partial derivatives.

The objects you want to study
in this case are several.

One of the most important ones
is the so-called discriminant.

What is the discriminant?

You haven't talked about
discriminants since a long time

ago.

And there is a relationship
between discriminant

in high school algebra and
discriminant in calculus 3.

The discriminant
the way we define

it is D, or delta--
some people denote it

like this, some
people by delta--

and that is the following.

This is the determinant.

f sub xx, f sub xy,
f sub yx, f sub yy,

computed at the point
p0, which is critical.

So p0 first has to satisfy
those two equations,

and then I'm going to have
to compute the [INAUDIBLE]

at that point.

But you say wait a
minute, Magdalena,

what the heck is this?

Well this is the second
partial with respect

of x, one after the other,
second partial with respect

to y, one after the other.

These guys are equal.

Remember that there was
a German mathematician

whose name was Schwartz, the
black cavalier, the black man.

Schwartz means black in German.

And he came up with
this theorem that it

doesn't matter in which order
you differentiate, f sub xy

or f sub yx is the same thing as
long as the function is smooth.

So I'm very happy about that.

Now there are these
other guys, A, B,

C. It's very easy to
remember, it's from the song

that you all learned
in kindergarten.

Once you know your ABC, you
come back to the discriminant.

So f sub xx at the point p0,
f sub xy at the point p0,

and f sub yy at the point p0.

Second partial with respect to
x, second partial with respect

to x and y, mixed one,
mixed derivative, and second

partial with respect to y.

You have to plug in the values
for the p0 will be x0, y0.

The critical point
you got from what?

From solving this system.

So you got x0y0 by
solving that system.

Come back, plug in, compute
those, get ABC as numbers.

And who is D going
to be based on ABC?

According to the diagram that
I drew, it's easy for you guys

to see that A and
B and C are what?

Related to D. So D will
simply be A, B, B, and C,

computed at the point p0.

So it's going to
be now-- now that

will remind you of something.

AC minus B-squared.

OK?

When we had the quadratic
formula in school--

I'm not going to write it.

I'm going to write it here.

So what was the
quadratic formula?

ax-squared plus bx
plus c equals 0.

That was algebra.

Baby algebra.

What do we call that?

High school algebra?

x12 plus minus b plus minus
square root of b-squared minus

4ac divided by 2a.

Now don't don;t know what
kind of professors you had.

But I had a teacher when
I was in high school.

Every time she taught me
something and I did not

absorb it, she was all over me.

She was preparing me for
some math competitions,

and she taught me a trick.

She said look,
Magdalena, pay attention.

If b would be an even number--
take b to be 2b prime,

2-- give me another letter.

2 big B. Right?

Then, the quadratic formula
would be easier to use.

Because in that case, you get
x1 2 equals minus-- b is 2b.

So you have just 2b like that.

Plus minus square
root 4b squared

minus 4ac divided by 2a.

She explained this
to me once and then

she expected me to remember
it for the rest of my life.

And then she said minus big B
plus minus square root of bb

squared minus ac.

Do you see why?

It's because you pull out the
factor of 4, square root of 4

is 2.

2, 2, and 2 simplify.

And then she gave me
to solve problems.

STUDENT: What about the a?

STUDENT: How about the a?

STUDENT: Because you
divide it by [INAUDIBLE].

PROFESSOR: Divided by.

I forgot to write it down.

Because I didn't have space.

I said, I'm not going
to bend and doodle.

So when you have x-squared
plus 2x-- let's say minus 3.

And she gave me that.

And I said OK, let me do it.

Let me do it.

x1 2 minus 2 plus minus square
root b-squared minus 4ac,

which is 12, divided by 2.

And she started screaming.

And she started
screaming big time.

Do you know why?

She said, I just told
you the half formula.

By half formula, I mean
she meant this one.

So when-- and I said OK,
OK, the half formula.

But then for maybe
another seven years,

I did this with the
formula-- with the formula

that everybody knows.

And at the end, I
would remember I

could have done
the half formula,

but I didn't do it
because I'm in a routine.

So the way she wanted
me to do this was what?

Who is the half of 2?

1.

So put minus 1 plus minus
square root of big B

is 1-squared minus
a times c, which

is plus 3, divided by
1 divided by nobody.

This way you don't have
to simplify it further,

and you do it faster.

So you get minus 1 plus minus
root 4, which is minus 5 and 3.

But of course, you could
have done this by factoring.

So you could have
said wait a minute.

Two numbers that multiply-- um--

STUDENT: [INAUDIBLE]
square root.

PROFESSOR: I didn't do right.

So it's 4--

STUDENT: [INAUDIBLE].

PROFESSOR: Yeah.

So you get x plus 5 times--

STUDENT: It's x minus 1--

[INTERPOSING VOICES]

PROFESSOR: Oh, I think I--

STUDENT: Square root.

PROFESSOR: Square root.

I'm sorry, guys.

OK.

Thank you for that.

1 and minus 3.

So x plus 3 times
x minus 1, which

is the same-- the exact same
as x-squared plus 2x minus 3

equals.

All right?

So just the way she insisted
that I learn the half formula.

I'm not insisting that you learn
the half formula, god forbid.

But see here there is
some more symmetry.

The four doesn't appear anymore.

b-squared minus 4ac appeared
here, but here it doesn't.

Here you're going to
have b-squared minus ac.

There is a reason.

This comes from a
discriminant just like that.

And this is why I told
you the whole secret

about the half
quadratic formula.

Not because I wanted
you to know about it,

but because I wanted you to see
that there is a pattern here.

You have-- for the
half formula, you

have plus minus square root
of a new type of discriminant.

People even call this
discriminant b-squared

minus 4ac.

b-squared minus ac.

So for us, it is
ac minus b-squared.

It's just the opposite of
that discriminant you have.

Now depending on the sign
of this discriminant,

you can go ahead and classify
the critical values you have.

So classification
is the following.

Classification of
special critical points.

If delta at p0 is negative,
then p0 is a saddle point.

If delta at p0 is 0,
nothing can be said yet

about the nature of the point.

So I make a face, a sad face.

If delta at p0 is greater than
0, then I have to ramify again.

And I get if a is positive,
it's going to look like a smile.

Forget about this side.

It's going to look like a smile.

So it's going to be a valley
point, what do we call that?

Relative minimum,
or valley point.

Don't say valley
point on the exam, OK?

Relative minimum.

If a is less than
0 at the point,

then locally the
surface will look

like I have a peak-- a relative
maximum Peaks and valleys.

Just the way you
remember them in Calc 1.

Now it's a little
bit more complicated

because the functions
have two variables.

But some of the patterns
can be recognized.

Let's go back to
our original example

and say wait a
minute, Magdalena.

You just gave us a
saddle, but we didn't

do the whole classification.

Yes, we didn't, because I
didn't go over the next steps.

z equals x-squared
minus y-squared.

Again, we computed the gradient.

We computed the
partial derivatives.

And then what was that one in
finding the critical points?

So f sub x equals
0, f sub y equals 0.

Solve for x and y.

And that's good,
because that's going

to give me a lot of
information, a lot that

will give me exactly where
the critical points may be.

So that is if and only if I need
to solve 2x equals 0 minus 2y

equals 0.

Is this system hard to solve?

No.

That's exactly why I picked it.

Because it's easy to solve.

The only solution is
x0 equals y0 equals 0.

So the origin--
that's exactly where

you put your butt on the
saddle when you ride the horse.

That is the only
critical point you have.

The only one.

Now if we want to classify that,
what kind of-- is it a valley?

No.

It looks like a valley
in the direction

of the axis of the horse,
Because the saddle's

going to look like that.

This is the horse.

That's the head of
the horse I'm petting.

And this is the
tail of the horse.

So in this direction, the saddle
will be shaped like a parabola,

like a valley.

But in the
perpendicular direction,

it's going to be
shaped going down,

like a parabola going down.

So it's neither a
valley nor a peak.

It's a valley in one
direction, and a peak

in another direction.

And that's the saddle point.

So say it again.

What is that?

It looks like a valley in
one principle direction

and the peak in the other
principle direction.

And then that's going
to be a saddle point.

Indeed, how do we figure this
out by the method I provided?

Well, who is A?

A is f sub xx at the point.

2x goes primed one time.

f sub x was 2x.

f sub y was 2y.

f sub xx is 2.

f sub xB is f sub xy.

What is that?

0.

Good, that makes my life easier.

C equals f sub yy.

What is that?

2.

OK, this is looking beautiful.

Because I don't have
to plug in any values.

The D is there for me to see it.

And it's going to consist of the
determinant having the elements

2, 0, 0, 2-- minus 2, minus 2.

I'm sorry, guys, I
missed here the minus.

And it cost me my life-- 2x
and minus 2y, and here minus 2.

STUDENT: It didn't
cost you your life,

because you caught it before
you were done with the problem.

PROFESSOR: I caught it up there.

I'm taking the final exam.

I still get 100%,
because I caught it up

at the last minute.

So 2, 0, 0, minus
2-- I knew that I

had to get something negative.

So I said, for god's sake, I
need to get a saddle point.

That's why it's the
horse in the saddle.

So I knew I should
get minus 4, negative.

All right, so the
only thing I have

to say as a final
answer is the only

critical point of this surface
that I'm too lazy to write

about-- don't write that.

So the only critical
point on the surface z

equals x squared
minus y squared will

be at the origin O
of corner 0, 0, 0

where the discriminant
being negative

indicates it's going
to be a saddle point.

And that's it-- nothing else.

You don't need more.

But there are more examples.

Because life is hard.

And I'm going to give
you another example.

Well, OK, this one.

Suppose we have the
surface-- that's

still going to be very easy.

But I want to make the
first examples easy.

I have a reason why.

This is a function of
two variables, right?

It's still a polynomial in
two variables of order 2.

And how do I solve for the
classification of the extrema?

I'm looking for local extrema,
not absolute-- local extrema.

I'm not constrained.

I'm saying, what do you
mean, no constraint?

Constrained would have
been, let's say that x and y

are in the unit disc.

Or let's say x and y are
on the circle x squared

plus y squared equals 1.

That would be a constraint.

But they're not
constrained about anything.

x and y are real numbers.

They can take the whole
plane as a domain.

So I get f sub x equals 0, f sub
y equals 0, solve for x and y,

get the critical values.

I get very nice 2x.

I have to pay attention.

Because now this is
not so easy anymore--

plus prime with respect to x,
2y, prime with respect to x, 0,

prime with respect to x, plus
3, prime of this, OK, equals 0.

f sub y-- 0 plus
prime with respect

to y, 2x, plus prime
with respect to y, 2y,

plus nothing, prime with
respect to y equals 0.

And now you have
to be very smart.

Well, you have to be perceptive
and tell me what I got.

What is this that we mean?

Look at this system.

It looks like crazy.

STUDENT: [INAUDIBLE] the
origin or-- because can't you

just subtract it down?

PROFESSOR: Is this possible?

And what does this mean?

What do we call such a system?

Inconsistent system--
we call it inconsistent.

How can I make this
problem to be possible,

to have some critical points?

STUDENT: If you add 3x.

PROFESSOR: How about
that, just remove the 3x

and see what's going to happen.

Oh, in that case,
I have something

that's over-determined, right?

I have something that
tells me the same thing.

So I'm priming
with respect to x.

I get that.

I'm priming with respect to y.

I get this.

I get 0.

So I don't even need
the second equation.

And that means
the critical point

is any point of
the form-- shall I

put a Greek letter alpha minus
alpha or lambda minus lambda?

What shall I do?

So any point that is situated
on the second bisector,

I mean the x, y plane.

And this is the x,
and this is the y.

And I say, what does it
mean, x plus y equals 0?

Not this line-- don't draw it.

That is x equals y.

The other one, called
the second bisector-- y

equals negative x, so not
this one, the diagonal,

but the diagonal that's
on the corridor, this one.

All right, so any point of
the form alpha minus alpha,

here's the critical point.

The question is, how am I going
to get to the classification

for such points?

Can anybody help me?

So step two--

STUDENT: Solve the equation.

STUDENT: Solve alpha for one
of the two variables first.

PROFESSOR: Take alpha minus
alpha-- could be anything.

And then I'll say, f
sub-- this is f sub x.

And this is f sub y.

What is f sub x?

f sub xx, I'm sorry.

STUDENT: [INAUDIBLE].

PROFESSOR: Huh?

2.

OK, are you with me?

So you know what it is.

f sub xy equals?

STUDENT: 2.

PROFESSOR: 2.

f sub yy equals?

STUDENT: 2

PROFESSOR: 2-- that's
the mystery man.

The book doesn't
give this example,

and it drives me crazy.

And I wanted to give you
some bad example where

the classification doesn't work.

Because we always cook
up nice examples for you

and claim everything
is beautiful.

Life is not always beautiful.

So you get 0.

In that case, nothing can be
said with this classification.

I make a face, sad face.

So what do I hope?

To get to Maple or MATLAB and
be able to draw that, or a TI-92

if my mother would
give me $200 and some.

I told her.

She asked me what to
buy for my birthday.

I have a TI-83 or something.

And it was cheap.

I bought it on eBay, and
then I stopped using it.

And then I saw this TI-92
that can draw surfaces

in three dimensions.

And I said, this is like MATLAB.

You just carry it
in your pocket.

It's only a little
bit too expensive.

All right, how
about another kind?

Look at this one.

You cannot tell
with the naked eye.

But you can go ahead
and do this step one

looking for critical values.

So the system, f sub x will
be 6x plus 2y equals 0.

f sub y will be--
who's going to tell me?

2x plus 2y equals 0.

Now, by elimination or by
substitution or by anything

I want, I subtract the
second from the first.

What do I get?

I get 4x equals 0.

And that gives me the only
possibility is x0 equals 0.

And then I say, OK, if my
only one is 0, then y is 0.

0 is 0.

So I only have one critical
point, which is the origin.

Now, do I know, what
am I going to get?

Not unless I'm a
genius and I can

see two steps ahead of time.

I would need to do ABC
quickly in my head.

Some of you are able, thank god.

But some of you,
like me, are not.

So I have to take a few
seconds to see what's going on.

A-- f sub xx at the point is 0.

B-- f sub xy.

C-- f sub yy.

What do we do?

We get 6.

Are we happy about it?

We don't know yet,
to be happy or not.

f sub xy or f sub yx,
you see, Mr. Schwarz

is now happy that
he proved to you

that it doesn't matter
which order you're

taking for a polynomial
that's a smooth function.

You always have the same.

And finally, C is 2.

And you are ready to do the
D. And I could smell that D,

but I didn't want
to say anything.

6, 2, 2, and 2-- is
that a nice thing?

Yeah, we haven't encountered
this example yet.

Because according to
the classification,

this is greater than 0.

Does it really matter
what value it is?

No, it only matters
that it is positive.

And if it's positive, that
means I can move on with my life

and look at the classification.

From this point where
delta or v is positive,

I'm going to get a ramification
into separate cases.

And who is going to
tell me next what to do?

Look at A. Oh, by
the way, talking

about the quadratic
formula from school,

from kindergarten,
when we computed

the-- I'll use the general one,
minus b plus minus square root

of b squared minus 4ac over 2a.

We were afraid of
some special cases

when we were looking at that.

Especially when
delta was negative,

that was really
imaginary and so on.

But one thing we remember
from ninth grade--

was this ninth grade
or eighth grade?

The parabola opens up
when a is positive.

Just the same way, something
opens up when A, big A,

is positive here.

Then you have opening up.

When big A is negative,
then you have opening down.

So remember-- I'm going to make
smile here so you remember.

So I have it like that.

So I suspect that
it's going to look

like a surface of some
sort that maybe is not

surface of revolution.

You should tell me what it is.

You should think about this and
do the cross sections with z

constant and tell me
what surface that is.

But in any case, what do I care?

I care that I'm
looking at the origin.

And this is where
my special point is.

That's going to be
the value point.

How do I know?

Because A, which
is 6, is positive.

At this point, I know
what I'm left with.

I know that my surface is
going to look like a valley.

So how do I know again?

I'm not going to draw it.

But it's going to look
something like that.

At the origin, this
is going to be 7.

Are you guys with me?

And it's going to open up.

And so you should not attempt
intersecting with z equals 5

or z equals 1.

Because you're not
going to get anything.

But if you intersect,
for example, at z

equals 9, what are
you going to get?

If you intersect
at z equals 9, you

get 3x plus 2xy plus
y squared equals 2.

And what is that?

It's a rotated
form of an ellipse.

It's hard to see, because
it's missing [INAUDIBLE].

But this is exactly what
discriminant is saying.

So this is going to be an x.

Good, so I know what
I'm going to get.

What do you have to
say on the midterm

or on the final
about this problem?

STUDENT: The point is--

PROFESSOR: The
point is the origin.

I classified it.

I got delta positive.

I got A positive.

So it's a valley.

It's a relative minimum.

And that's it.

I have a relative min at the
point P of coordinates 0, 0,

and 7.

And that's the valley.

Yes, sir.

STUDENT: Why is it
A that determines

whether it's a relative
min or a relative max?

PROFESSOR: It's a whole story.

You can prove it.

I don't remember if we proved
this in the book or not.

But it can be
proved, so the fact

that it has to do with
concavity and convexity.

When you had a second
derivative, let's say,

what's the equivalent
of the Calculus I

notion that you know about?

In Calculus I, you had
functions of one variable,

and life was so easy like that.

And f prime positive meant
that the function increased.

And f prime negative meant
that the function decreased.

And f double prime was
just like your-- you sense

that the second partials
must have something

to do with it, especially the
first one with respect to x.

If you were in
plane, and you have

f double prime with respect
to x, when was this a valley?

When you had the smile.

When did you have a smile?

When f double prime was
positive, you have concave up.

When f double
prime was negative,

you have concave down.

Remember, guys?

So you have a smile or a frown.

This is how we know.

For the same reason that
would take about two pages

to write down the proof, you
have a smile for A positive.

And the smile means
actually in all directions

you have a smile locally
around the origin.

OK, look in the book.

I'm not sure how
much should we do.

Do we give a sketch of a proof,
or we give the entire proof?

But more likely, a sketch.

Yes.

STUDENT: I asked the
slightly wrong question,

but I answered it myself.

I wanted to ask, why is it
dependent on A and not on C?

PROFESSOR: Not on C.

STUDENT: But then I realized
that it is dependent on C

as well, because
if A is positive,

then C must be positive.

PROFESSOR: Yes, yes, it
is dependent on both.

STUDENT: OK, there we go.

That was my question.

PROFESSOR: So guys, remember.

Imagine what happens when
you had no B, B was 0.

Then the matrix
is diagonalizable.

And here you have
A and C. And Alex

says, why would A be
more important than C?

It's not.

But practically, if A is
positive and C is negative,

that means these are the
principal directions in which

one bends like a valley up and
one bends like a peak down.

So this is what happens in the
direction of x, f double prime

in the direction of x, kind of.

And this is in the
direction of y.

So this is f double prime
in the direction of y,

which we don't denote like that.

We call it f sub xx and f
sub yy, which is A and C.

So A positive, A being 1
and C being negative 2,

means a valley here, means
the valley meets the horse.

Look, I'm drawing the
tail of the horse.

He's a little bit
fat, this horse.

And that's his mane, his eye.

I'm just taking a break.

STUDENT: That's a
pretty good drawing.

PROFESSOR: It looks more
like a dog or a plush horse

or something.

So A equals 1, and
C equals minus 2.

But if it were diagonalizable,
and A would be 1

and C would be 7, both of
them positive in any case,

then you'll have valley and
valley, an x direction valley

and y direction valley.

So it has to be a
valley everywhere.

These are the principal
directions that I have 1 and 2.

But then the ultimate
case, what happens

when A is negative
and-- hmm, OK,

then either you have them both
one positive, one negative,

or you have plus,
plus and minus, minus.

And then you have this
as your surface, right?

Which one is the x direction?

That's the y direction.

That's the second one.

The x direction is that.

In the x direction,
you have a frown.

So f sub yy is negative.

In the y direction,
you also have a frown.

So both of them are negative.

So you have a relative max.

Yes, sir, Matthew, tell me.

STUDENT: So isn't it possible
to have both A and C positive,

but then yet still not be
more positive than B squared?

PROFESSOR: No, because
there's a theorem that--

STUDENT: I was just
wondering like numbers-wise.

PROFESSOR: You have this matrix.

And there is a theorem that
shows you that you can actually

diagonalize this matrix.

You'll learn your linear
algebra [INAUDIBLE].

STUDENT: It makes
sense, because when

you were saying A is this way,
and that way there's no way

you could have 2 come
up, and then yet,

not be a-- you know
what I'm saying?

Because then they'd
be less than 0.

PROFESSOR: You can if
you don't have or the 2.

That's an excellent question.

If I would have x
to the 4 y to the 4

added together, like Ax to
the 4 plus By to the 4 plus

something, then I have the
so-called monkey saddle.

That's so funny.

You can have something
that looks like that.

So in your direction,
you can have this.

Then I've reached
two equal peaks

in the x and the y direction.

But in the between,
I also went down.

So depending on a higher
degree symmetric polynomial,

you can have a monkey saddle.

And then it's not just
like you can predict what's

going to happen in between.

In between, if I go
up, if I go valley

in the x direction and
valley in the y direction,

I know that's going to be
a valley everywhere-- no.

If a polynomial
is high in order,

it can go down,
valley, and up again,

and monkey saddle it looks like.

Guys, you have dealt
with it when you

went to Luna Park or Joyland.

It's one of those things
that look like-- I'm trying.

I cannot draw.

STUDENT: It sounds
more like an octopus.

PROFESSOR: Like an octopus.

And one of those
things-- exactly--

that are shaped so that
they are undulated,

in some directions are
going up, in some directions

are going down.

STUDENT: Like an
egg carton, almost?

PROFESSOR: Yeah,
really undulated.

Imagine even a surface made
of metal that's undulated

and rotating at the same time.

They have some of
those in Disney World.

Have you been to Orlando?

STUDENT: I was
there last semester.

PROFESSOR: But you didn't take
me with you, which is bad.

Because that's one of
my favorite places.

STUDENT: I was just trying to
think of what you were talking

about so I could visualize it.

PROFESSOR: Maybe we
could make a proposal

to teach Calculus
III at Disney World

so that we could have examples
of motion and surfaces

all around and study the
motion of all sorts of gadgets,

velocity and trajectory.

Last night I couldn't sleep
until 1:00, and I was thinking,

I gave examples of
the winter sports

like bobsled and all
sorts of skiing and so on.

But I never thought
about a screw curve

with curvature and torsion that
is based on the roller coaster.

And the roller
coaster is actually

the best place to study the
[INAUDIBLE], the velocity,

the tangent unit, the
normal, the bi-normal.

And when you have in a
plane the roller coaster

goes like that, like this and
like that, like in a plane,

you have nothing but bending,
which means curvature.

But then when the roller coaster
goes away from the plane,

you have the torsion.

And that makes you sick
really to the stomach.

So we would have to
experience that to understand

Calculus III better.

So our next proposal is
we ask the administration

instead of study abroad
courses, the domestic study

at Disney World for Calc III.

It's Applied Calculus III.

OK, something else that
I want you to do-- I

had prepared an example.

This is an absolute extrema.

And you say, what the heck
are the absolute extrema?

Because she only talked to
us about relative maximum

and relative minimum.

And she never said anything
about absolute extrema.

And that will be the table.

And these will be the extrema.

I want to refresh your memory
first just a little bit.

This will be the last example.

Because it's actually
two examples in one.

And what if you have, let's say,
f of x equals e to the minus

x squared over the
interval minus 1, 1?

You are in Calc I.
You will build a time

machine from Disney World.

And we went back in time when
you actually took Calc I.

And you struggled
with this at first.

But then you loved
it so much that you

said, oh, that's my favorite
problem on the final.

They asked us for two things--
relative extrema, min or max,

min/max theory, and
they say absolute.

But for the absolute, your
teacher said, attention,

you have to know how
to get to the absolute.

You are constrained to be
on the segment minus 1, 1.

You see, the fact
that they introduced

this extra constraint
and they don't

let you move with x on the whole
real line is a big headache.

Why is that a big headache?

Your life would be much
easier if it were just e

to the negative x squared.

Because in that
case, you say, OK,

f prime of x equals minus
2xe to the minus x squared.

Piece of cake.

x0 is 0.

That's the only critical point.

And I want to study what kind
of critical point that is.

So I have to do f
double prime of x.

And if I don't know the
product rule, I'm in trouble.

And I go, let's
say, minus 2 times

e to the negative x
squared from prime of this

and this non-prime, plus
minus 2x un-prime times e

to the minus x squared
times minus 2x again.

So it's a headache.

I pull out an e to
the minus x squared.

And I have 4x squared--
4x squared-- minus 2.

But you say, but wait
a minute, Magdalena,

I'm not going to compute
the inflection points.

The inflection points
will be x equals

plus/minus 1 over root 2.

I only care about
the critical point.

And the only critical
point I have is at 0.

Compute f double prime of 0 to
see if it's a smile or a frown.

And you do it.

And you plug in, and you say,
e to the 0 is 1, 0 minus 2.

So you get a negative.

Do you care what it is?

No, but you care it's negative.

So at 0, so you
draw, and you know

at 0 you're going to
have some sort of a what?

Relative max.

Where?

At 0, and when you
plug 0 again, 1.

So you draw a table.

And you say,
relative max at 0, 1.

And then you're not done.

Because you say,
wait a minute, I

am to study my function in
Calc I at minus 1 and 1.

It's like you have a
continuous picture,

and you chop, take scissors,
and cut and cut at the extrema.

And there you can
get additional points

where you can get a relative
max or relative min.

Absolute max or min will be
the lowest of all the values

and the highest
of all the values.

OK, so I get at the point
minus 1-- how shall I put here?

x equals minus 1.

What do I get for y?

And for plus 1,
what do I get for y?

This is the question.

I plug it in, and I
get minus minus 1/e.

And then when I have
1, what do I get?

1/e again.

So do I have a
relative min here?

No, but I have an
absolute something.

And what do I have here?

Here I have an absolute max.

So how do we check the absolute
maxima and absolute minima?

We look for critical points.

We get many of them,
finitely many of them.

We compute all the
values of z for them,

all the function values.

And then we look
at the end points,

and we compare all three
of them, all the three

values in the end.

So in the end, you
compare 1/e to 1/e to 1.

And that's all you can get.

So the lowest in one will
be the highest in one.

Good.

In Calculus III, it's
more complicated.

But it's not much
more complicated.

Let's see what's
going to happen.

You can have a
critical point inside.

We are just praying we
don't have too many.

So how do I get to step one?

Critical point means f sub
x equals e to the x squared

minus y squared times 2x.

All righty, it looks good.

f sub y equals e to the x
squared minus y squared times

minus y.

I'm full of hope.

Because I only have one
critical point, thank god.

Origin is my only
critical point.

I don't know what that
is going to give me.

But it can give
me a relative max

or relative min or a saddle.

I don't know what
it's going to be.

Who tells me what
that is going to be?

Well, did I do this further?

I did it further and
a little bit lazy.

But I'm not asking the
nature of the point.

So for the time being, I only
want to see what happens at 0,

0.

So I have 1.

So in my table I will put, OK,
this is x, y, and this is z.

For 0, 0, I'm interested.

Because that's the critical
point inside the domain.

The domain will be the unities.

And inside the origin,
something interesting happens.

I get a 1.

And I hope that's going to
be my absolute something.

But I cannot be sure.

Why?

There may be other values
coming from the boundary.

And just like in
Calculus I, the only guys

that can give you other
absolute max or min,

they can come from the
boundary, nothing else.

Nowhere else in the interior
of the disc am I going to look.

I'm not interested.

I'm only interested in x
squared plus y squared equals 1.

This is where
something can happen,

nothing else interesting in the
inside, just like in Calc I.

So to take x squared plus y
squared equals 1 into account,

I pull y squared, who is
married to x-- the poor guy.

He's married to x.

He's dependent on x
completely, y squared

equals 1 minus x squared.

And I have to push him
back into the function.

So at the boundary, f becomes
a function of one variable.

He becomes f of x
only equals e to the x

squared minus 1 plus x squared.

Are you guys with me?

So f of x will become e
to the 2x squared minus 1

along the boundary, along
the circle, only here.

Now what else do I need to do?

I need to compute the critical
values for this function of one

variable, just the way
I did it in Calc I.

So f prime of x will give
me e to the 2x squared

minus 1 times-- what comes
down from the chain rule?

STUDENT: 4x.

PROFESSOR: 4x, so life is
hard but not that hard.

Because I can get what?

I can get only x
at 0 equals 0 here.

OK, so that's a critical point
that comes from the boundary.

But guys, you have
to pay attention.

When x is 0, how many
y's can I have for that 0

on the boundary?

This is on the boundary--
on the boundary.

STUDENT: Two.

PROFESSOR: Two of
them-- I can have 1,

or I can have negative 1.

There is one more tricky thing.

This is a function
of one variable only.

But this stinking
function is not

defined for arbitrary x real.

So I make a face again.

So I go, oh, headache.

Why?

x is constrained.

x is constrained, you see?

If you were inside the disc, x
must be between minus 1 and 1.

So I have to take into account
that x is not any real number,

but x is between minus 1 and 1.

Those are endpoints
for this function.

And in Calc I, I
learned, OK, I have

to also evaluate what
happens at those endpoints.

But thank god that will exhaust
my list, so I have a list.

Minus 1 for x and 1
for x-- thank god.

That will give you
what y on the boundary?

When x is 1 and x
is minus 1, you're

interested in what happens,
maximization or minimization,

for this function
at the endpoints.

But fortunately, since you are
on the boundary, y must be 0.

Because that's
how you got y out.

If x is plus/minus 1 on
the boundary, y must be 0.

So my list contains how
many interesting points?

One, two, three, four,
five-- for all of them,

we need to compute,
and we are done.

Of all of them, the lowest z
is called absolute minimum.

And the highest z is
the absolute maximum.

And we are done.

You guys need to help me,
because I'm running out of gas.

So x is 0.

Y is 1.

What is z?

STUDENT: [INAUDIBLE].

PROFESSOR: e to the
minus 1, you were

fast, 1/e, thank you, guys.

So when x is 0 and y is minus 1?

STUDENT: [INAUDIBLE].

PROFESSOR: Huh?

STUDENT: [INAUDIBLE].

PROFESSOR: No, no,
no, it's the same.

Because x is 0. y is minus 1.

I get e to the minus
1, which is 1/e.

So far, so good-- I'm
circling all the guys

that I want to compare after.

So for the final four points
there, what do I have?

My final candidates could
be x equals plus/minus 1

and y equals 0-- e and e.

Who's the biggest?

Who's the smallest?

STUDENT: e is the biggest.

PROFESSOR: e is the biggest,
and 1/e is the smallest.

So how do I write conclusion?

Conclusion-- we have
two absolute maxima

at minus 1, 0 and 1, 0.

And we have two absolute
minima at 0, minus 1 and 0, 1.

OK, now I have to-- now
that's like a saddle.

Can you see it with the
eyes of your imagination?

It's hard to see it.

This is the disc.

And the four
points, the cardinal

points-- OK, this is the disc.

We are looking at this
disc from perspective.

And the five points,
one is in the middle.

One is here.

Minus 1 is 0.

One is here, 1, 0.

One is here, 0, minus
1, and one here, 0, 1.

At minus 1, 0 and 1,
0 I get the maximum.

So the way it's going to be
shaped would be like that.

In this direction,
it will be like that.

And cut the cake here.

You see it's like that.

It's going to be like this.

OK, passing through the
origin, with my hands

I'm molding the surface made of
Play-Doh or something for you.

So I'm starting here,
and I'm going up.

And at this points, I'm here.

Are you with me?

The same height.

In the other direction,
I'm going from 0.

But I'm not so high.

I'm going only up
to-- what is 1/e?

About 1/3, meh,
something like that.

So I'm going to get here.

So the problem is that
one will be in between.

So if you really want to
see what it looks like,

we are here at 1.

We grow from 1,
altitude 1, you see?

We grow from 1 to about
2.71718283 for both of these.

And from 1, in this direction
I have to go down to 1/e.

So it looks like that.

I'll try to draw, OK?

Do you see the patch
around the origin?

So here's e.

And here's 1/e
above the sea level.

And this is 1.

And you have one just like that
in the back that is the-- it

still looks like a saddle.

It is a saddle.

It's symmetric.

But it's another kind of saddle.

There are all sorts
of saddles made

in Texas, different
ranches, different saddles.

So that was the harder one.

The ones that I actually
saw on the finals,

some of the last
three or four finals,

were much easier in the sense
that the table you had to draw

was much shorter than
this one-- in principle,

one critical value and
one max and one min point.

But you have to be prepared
more, rehearse more,

so when you see the
problem in the midterm,

you say, oh, well that is
easier than I'm used to.

That's the idea.

OK, go home.

Send me emails by WeBWorK.

We still have time to talk about
the homework if you get stuck.

And I'll see you Thursday.

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