TTU Math2450 Calculus3 Sec 11.5 and 11.6

Chain rules, Directrional derivatives and Gradient

MAGDALENA TODA: So what's
your general feeling

about Chapter 11?

STUDENT: It's OK.

MAGDALENA TODA: It's OK.

So functions of
two variables are

to be compared all the time with
the functions of one variable.

Every nothing you
have seen in Calc 1

has a corresponding
the motion in Calc 3.

So really no questions about
theory, concepts, Chapter 11

concepts, previous concepts?

Feel free to email
me this weekend.

Don't think it's the
weekend because we

are on a 24/7 availability.

People, we use WeBWork.

Not just me, but
everybody who uses

WeBWork is on a
24/7 availability,

answering questions
about WeBWork problems.

Saturday and Sunday is when
most of you do the homework.

It's convenient for us as well
because we are with the family,

but we don't have many
meetings to attend.

So I'll be happy to
answer your questions.

Last time, we discussed a
little bit about preparation

for The Chain Rule.

In Calc 3.

So the chain rule in Calc
3 was something really--

this is section 11.5.

The preparation
was done last time,

but I'm going to
review it a little bit.

Let's see what we discussed.

I'm going to split,
again, the board in two.

And I'll say, can we review
the notions of The Chain Rule.

When you start with a
variable-- let's say it's time.

Time going to f of t, which goes
into g of f of t by something

called composition.

We've done that since we
were kids in college algebra.

What?

You never took college algebra?

Except in high school, you
took high school algebra,

most of you.

So what did you do in
high school algebra?

We said g composed with l.

This is a composition
of two functions.

What I'm skipping here is the
theory that you learned then

that to a compose well, F of t
has to be in the domain of g.

So the image F of t, whatever
you get from this image,

has to be in the domain of g.

Otherwise, the composition
could not exist.

Now if you have
differentiability,

assuming that this
is g composed with F,

assuming to be c1-- c1
meaning differentiable

and derivatives are continuous--
assuming both of them are c1,

they compose well.

What am I going to do next?

I'm going to say the
d, dt g of F of t.

And we said last time,
we get The Chain Rule

from the last function
we applied, g prime.

And so you have dg,
[? d2 ?] at F of t.

I'm calling this guy u variable
just for my own enjoyment.

And then I go du, dt.

But du, dt would be
nothing but a prime of t,

so remember the cowboys
shooting at each other?

The du and du.

I will replace the u by prime of
t, just like you did in Calc 1.

Why?

Because I want to a mixture of
notations according to Calc 1

you took here.

The idea for Calc 3 is the
same with [INAUDIBLE] time,

assuming everything
composes well,

and has differentiability, and
the derivatives are continuous.

Just to make your life easier.

We have x of t, y of t.

Two nice functions
and a function

of these variables,
F of x and y.

So I'm going to have
to say, how about x

is a function of t and
y is a function of t?

So I should be able to go
ahead and differentiate

with respect to the t.

And how did it go?

Now that I prepared
you last time,

a little bit, for this kind
of new picture, new diagram,

you should be able to tell me,
without looking at the notes

from last time, how this goes.

So I'll take the function
F of x of t, y of t.

And when I view it like that,
I understand it's ultimately

a big function, F of t.

It's a real valued
function of t,

ultimately, as the composition.

This big F.

So does anybody
remember how this went?

Let's see.

The derivative,
with respect to t,

of this whole thing,
F of x of t, y of t?

Thoughts?

The partial derivative
of F with respect

to x, evaluated at
x of t and y to t.

So everything has to be
replaced in terms of t

because it's going to be y.

We assume that this derivative
exists and it's continuous.

Why?

Just to make your life
a little bit easier.

From the beginning,
we had dx, dt,

which was also
defined everywhere

and continuous, plus df, 2y at
the same point times dy, dt.

Notice what happens here with
these guys looking diagonally,

staring at each other.

Keep in mind the plus sign.

And of course, some of you
told me, well, is that OK?

You know favorite, right?

F of x at x of dy of t.

That's fine.

I saw that.

In engineering you use it.

Physics majors also use
a lot of this notation

as sub [INAUDIBLE] Fs of t.

We've seen that.

We've seen that.

It comes as no
surprise to us, but we

would like to see if there
are any other cases we

should worry about.

Now I don't want to
jump to the next example

until I give you
something that you

know very well from Calculus 1.

It's an example that you saw
before that was a melting ice

sphere.

It appears a lot in problems,
like final exam problems

and stuff.

What is the material
of this ball?

It's melting ice.

And if you remember, it
says that at the moment t0,

assume the radius was 5 inches.

We also know that the rate of
change of the radius in time

will be minus 5.

But let's suppose that we say
that inches per-- meaning,

it's really hot in the room.

Not this room, but
the hypothetic room

where the ice ball is melting.

So imagine, in 1
minute, the radius

will go down by 5 inches.

Yes, it must be really hot.

I want to know the derivative,
dv, dt at the time 0.

So you go, oh my god, I don't
remember doing this, actually.

It is a Calc 1 type of problem.

Why am I even
discussing it again?

Because I want to fool you a
little bit into remembering

the elementary formulas for
the volume of a sphere, volume

of a cone, volume of a cylinder.

That was a long time ago.

When you ask you teachers in
K12 if you should memorize them,

they said, by all
means, memorize them.

That was elementary geometry,
but some of you know them

by heart, some of you don't.

Do you remember
the volume formula

for a ball with radius r?

[INTERPOSING VOICES]

What?

[? STUDENT: High RQ. ?]

STUDENT: 4/3rds.

MAGDALENA TODA: Good.

I'm proud of you guys.

I've discovered lots of people
who are engineering majors

and they don't
know this formula.

So how are we going to
think of this problem?

We have to think, Chain Rule.

And Chain Rule means that you
view this radius as a shrinking

thing because
that's why you have

the grade of change negative.

The radius is shrinking,
it's decreasing,

so you view r as
a function of t.

And of course, you
made me cube it.

I had to cube it.

And then v will be a function
of t ultimately, but you see,

guys, t goes to r of t,
r of t goes to v of t.

What's the formula
for this function?

v equals 4 pi i cubed over 3.

So this is how the diagram goes.

You look at that composition
and you have dv, dt.

And I remember teaching as
a graduate student, that

was a long time ago,
in '97 or something,

with this kind of diagram with
compositions of functions.

And my students had told
me, nobody showed us

this kind of diagram before.

Well, I do.

I think they are very
useful for understanding

how a composition will go.

Now I would just going ahead and
say v prime because I'm lazy.

And I go v prime of t is 0.

Meaning, that this
is the dv, dt at t0.

And somebody has to help me
remember how we did The Chain

Rule in Calc 1.

It was ages ago.

4 pi over 3 constant times.

Who jumps down?

The 3 jumps down and he's
very happy to do that.

3, r squared.

But r squared is not an
independent variable.

He or she depends on t.

So I'll be very happy to
say 3 times that times.

And that's the essential part.

I'm not done.

STUDENT: It's dr over dt.

MAGDALENA TODA: dr, dt.

So I have finally
applied The Chain Rule.

And how do I plug
in the data in order

to get this as the final answer?

I just go 4 pi
over 3 times what?

3 times r-- who is
r at the time to 0,

where I want to view
the whole situation?

r squared at time
to 0 would be 25.

Are you guys with me?

dr, dt at time to
0 is negative 5.

All right.

I'm done.

So you are going to ask me,
if I'm taking the examine,

do I need this in
the exam like that?

Easy.

Oh, it depends on the exam.

If you have a multiple choice
where this is simplified,

obviously, it's not the right
thing to forget about it,

but I will accept
answers like that.

I don't care about the
numerical part very much.

If you want to do more, 4
times 25 is hundred times 5.

So I have minus what?

STUDENT: 500 pi.

MAGDALENA TODA: 500 pi.

How do we get the unit of that?

I'm wondering.

STUDENT: Cubic
inches per minute.

MAGDALENA TODA: Cubic
inches per minute.

Very good.

Cubic inches per minute.

Why don't I write it down?

Because I couldn't care less.

I'm a mathematician.

If I were a physicist, I would
definitely write it down.

And he was right.

Now you are going
to find this weird.

Why is she doing this review
of this kind of melting ice

problem from Calc 1?

Because today I'm
being sneaky and mean.

And I want to give
you a little challenge

for 1 point of extra credit.

You will have to compose
your own problem,

in Calculus 3,
that is like that.

So you have to compose a problem
about a solid cylinder made

of ice.

Say what, Magdalena?

OK.

So I'll write it down.

Solid cylinder made of ice
that's melting in time.

So compose your own problem.

Do you have to solve
your own problem?

Yes, I guess so.

Once you compose
your own problem,

solve your own problem
For extra credit, 1 point.

Compose, write, and solve--
you are the problem author.

Write and solve
your own problem,

so that the story includes--

STUDENT: A solid cylinder.

MAGDALENA TODA: Yes.

Includes-- instead of a
nice ball, a solid cylinder.

And necessarily, you cannot
write it just a story--

I once had an ice cylinder,
and it was melting,

and I went to watch a movie,
and by the time I came back,

it was all melted.

That's not what I want.

I want it so that the problem
is an example of applying

The Chain Rule in Calc 3.

And I won't say more.

So maybe somebody
can help with a hint.

Maybe I shouldn't
give too many hits,

but let's talk as if we
were chatting in a cafe,

without me writing
too much down.

Of course, you can take
notes of our discussion,

but I don't want
have it documented.

So we have a cylinder right.

There is the cylinder.

Forget about this.

So there's the cylinder.

It's made of ice
and it's melting.

And the volume should be a
function of two variables

because otherwise, you
don't have it in Calc 3.

So a function of two variables.

What other two variables
am I talking about?

STUDENT: The radius
and the height.

MAGDALENA TODA: The radius
would be one of them.

You don't have to say x and y.

This is r and h.

So h and r are in that formula.

I'm not going to
say which formula,

you guys should know of
the volume of the cylinder.

But both h and r, what do they
have in common in the story?

STUDENT: Time.

MAGDALENA TODA: They are
both functions of time.

They are melting in time.

STUDENT: Can I ask
a quick question?

MAGDALENA TODA: Yes, sir.

STUDENT: What if we solve
for-- what is the negative 500

[? path? ?]

MAGDALENA TODA: This is the
speed with which the volume is

shrinking at time to 0.

So the rate of change of
the volume at time to o.

And this is
something-- by the way,

that's how I would
like you to state it.

Find the rate of change
of the volume of the ice--

wasn't that a good cylinder?

At time to 0, if you
know that at time to 0

something happened.

Maybe r is given, h is given.

The derivatives are given.

You only have one
derivative given here,

which was our
prime of t minus 5.

Now I leave it to you.

I ask it to you, and I'll leave
it to you, and don't tell me.

When we have a
piece of ice-- well,

there was something in the
news, but I'm not going to say.

There was some nice, ice
sculpture in the news there.

So do the dimensions decrease
at the same rate, do you think?

I mean, I don't know.

It's all up to you.

Think of a case when the
radius and the height

would shrink at the same speed.

And think of a case when
the radius and the height

of the cylinder made
of ice would not

change at the same
rate for some reason.

I don't know, but
the simplest case

would be to assume that
all of the dimensions

shrink at the same speed,
at the same rate of change.

So you write your own problem,
you make up your own data.

Now you will appreciate
how much work people

put into that work book.

I mean, if there is a bug,
it's one in a thousand,

but for a programmer to be
able to write those problems,

he has to know calculus,
he has to know C++ or Java,

he has to be good-- that's
not a problem, right?

STUDENT: No.

That's fine.

MAGDALENA TODA: He or she has
to know how to write a problem,

so that you guys,
no matter how you

input your answer, as long as it
is correct, you'll get the OK.

Because you can put answers
in many equivalent forms

and all of them have to be--

STUDENT: The right answer.

MAGDALENA TODA: Yes.

To get the right answer.

So since I have new
people who just came--

And I understand you guys
come from different buildings

and I'm not mad for people who
are coming late because I know

you come from other
classes, I wanted

to say we started from a melting
ice sphere example in Calc 1

that was on many finals
in here, at Texas Tech.

And I want you to compose your
own problem based on that.

This time, involving
a cylinder made

of ice whose dimensions are
doing something special.

That shouldn't be hard.

I'm going to erase this
part because it's not

the relevant one.

I'm going to keep this
one a little bit more

for people who
want to take notes.

And I'm going to move on.

Another example we
give you in the book

is that one where x and y,
the variables the function f,

are not just
functions of time, t.

They, themselves, are functions
of other two variables.

Is that a lot more different
from what I gave you already?

No.

The idea is the same.

And you are imaginative.

You are able to come up
with your own answers.

I'm going to ask you to think
about what I'll have to write.

This is finished.

So assume that you have
function z equals F of x,y.

As we had it before,
this is example 2

where x is a function
of u and v itself.

And y is a function
of u and v itself.

And we assume that all
the partial derivatives

are defined and continuous.

And we make the
problem really nice.

And now we'll come
up with some example

you know from before where
x equals x of uv equals uv.

And y equals y of
uv equals u plus v.

So these functions are
the sum and the product

of other variables.

Can you tell me how I am going
to compute the derivative of 0,

or of f, with the script
of u at x of uv, y of uv?

Is this hard?

STUDENT: It is.

MAGDALENA TODA: I don't know.

You have to help me because--
why don't I put d here?

STUDENT: Because [INAUDIBLE].

MAGDALENA TODA:
Because you have 2.

So the composition
in itself will

be a function of two variables.

So of course, I
have [INAUDIBLE].

I'm going to go ahead and do
it as you say without rushing.

Of course, I know
you are watching.

What will happen?

STUDENT: 2x and 2y.

MAGDALENA TODA: No, in general.

Over here, I know you
want to do it right away,

but I would like you to give
me a general formula mimicking

the same thing you had before
when you had one parameter, t.

Now you have u and d separately.

You want it to do it straight.

So we have df, dx
at x of uv, y of uv.

Shut up, Magdalene.

Let people talk and help
you because you're tired.

It's a Thursday.

df, dx.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: dx.

Again, [INAUDIBLE] notation,
partial with respect

to u, plus df, dy.

So the second argument--
so I prime in respect

to the second argument,
computing everything

in the end, which means
in terms of u and v times,

again, the dy with respect to u.

You are saying that.

Now I'd like you
to see the pattern.

Of course, you see the
pattern here, smart people,

but I want to
emphasize the cowboys.

And green for the other cowboy.

I'm trying to match the
college beautifully.

And the independent
variable, Mr. u.

Not u, but Mr. u.

Yes, ma'am?

STUDENT: Is it the
partial of dx, du?

Or is it-- like you did
the partial for the--

MAGDALENA TODA: So did
I do anything wrong?

I don't think I
did anything wrong.

STUDENT: So it is the
partial for dx over du?

MAGDALENA TODA: So I go du with
respect to the first variable,

times that variable
with respect to u.

STUDENT: But is it the partial?

That's my question.

MAGDALENA TODA: But it has
to be a partial because x is

a function of u and
v, so I cannot put d.

And then the same plus
the same idea as before.

df with respect to
the second argument

times that second argument
with respect to the u.

You see, Mr. u is
replacing Mr. t.

He is independent.

He's the guy who is moving.

We don't care
about anybody else,

but he replaces the time
in this kind of problem.

Now the other one.

I will let you speak.

Df, dv.

The same idea, but somebody
else is going to talk.

STUDENT: It would
be del f, del y.

MAGDALENA TODA: Del f, del x?

Well, let's try
to start in order.

And I tried to be
organized and write neatly

because I looked at-- so these
videos are new and in progress.

And I'm trying to see what
I did well and I didn't.

And at times, I wrote neatly.

At times, I wrote not so neatly.

I'm just learning about myself.

It's one thing, what you think
about yourself from the inside

and to you see yourself
the way other people

see from the outside.

It's not fun.

STUDENT: Can you say it again?

MAGDALENA TODA: This is
v. So I'll say it again.

We all have a certain
impression about ourselves,

but when you see a
movie of yourself,

you see the way
other people see you.

And it's not fun.

STUDENT: So what--

MAGDALENA TODA: So
let's see the cowboys.

Ryan is looking at the
[? man. ?] He is all [? man. ?]

And y is here, right?

And who is the time
variable, kind of, this time?

This time, which
one is the time?

v. And v is the only ultimate
variable that we care about.

So everything you did
before with respect to t,

you do now with
respect to u, you

do now with respect
to v. It shouldn't

be hard to understand.

I want to work the
example, of course.

With your help, I will work it.

Now remember how my students
cheated on this one?

So I told my colleague, he did
not say, five or six years ago,

by first writing The Chain Rule
for functions of two variables,

express all the df, du, df,
dv, but he said by any method.

Of course, what they
did-- they were sneaky.

They took something
like x equals uv

and they plugged it in here.

They took the function
[? u and v, ?]

they plugged it in here.

They computed everything
in terms of u and v

and took the partials.

STUDENT: Why don't
you [INAUDIBLE]?

MAGDALENA TODA: It depends
how the problem is formulated.

STUDENT: So if you
make it [INAUDIBLE],

then it's [INAUDIBLE].

MAGDALENA TODA: So when they
give you the precise functions,

you're right.

But if they don't give
you those functions,

if they keep them a
secret, then you still

have to write the
general formula.

If they don't give you
the functions, all of them

explicitly.

So let's see what
to do in this case.

df, du at x of u, vy
of uv will be what?

Now people, help me, please.

I want to teach you how
engineers and physicists very,

very often express
those at x and y.

And many of you know
because we talked

about that in office hours.

2x, I might write,
but evaluated at--

and this is a very
frequent notation

image in the engineering
and physicist world.

So 2x evaluated at where?

At the point where x is
uv and y is u plus v.

So I say x of uv, y of uv.

And I'll replace later
because I'm not in a hurry.

dx, du.

Who is dx, du?

The derivative of x
or with respect to u?

Are you guys awake?

STUDENT: Yes.

So it's v.

MAGDALENA TODA: v. Very good.
v plus-- the next term, who's

going to tell me what we have?

STUDENT: 2y evaluated at--

MAGDALENA TODA: 2y evaluated
at-- look how lazy I am.

Times the derivative
of y with respect to u.

So you were right because of 2y.

Attention, right?

So it's dy, du is 1.

It's very easy to
make a mistake.

I've had mistakes who
made mistakes in the final

from just miscalculating
because when

you are close to
some formula, you

don't see the whole picture.

What do you do?

At the end of your
exams, go back and rather

than quickly turning in
a paper, never do that,

go back and check
all your problems.

It's a good habit.

2 times x, which is uv, I plug
it as a function of u and v,

right?

Times a v plus-- who is 2y?

That's the last of the Mohicans.

One is out.

STUDENT: 2.

MAGDALENA TODA: 2y 2 times
replace y in terms of u and v.

And you're done.

So do you like it?

I don't.

And how would you write it?

Not much better than that,
but at least let's try.

2uv squared plus 2u plus 2v.

You can do a little
bit more than that,

but if you want to list it
in the order of the degrees

of the polynomials, that's OK.

Now next one.

df, dv, x of uv, y of uv.

Such examples are in the book.

Many things are in the
book and out of the book.

I mean, on the white board.

I don't know why it gives you so
many combinations of this type,

u plus v, u minus-- 2u
plus 2v, 2u you minus 2v.

Well, I know why.

Because that's a
rotation and rescaling.

So there is a
reason behind that,

but I thought of something
different for df, dv.

Now what do I do?

df, dx.

STUDENT: You [? have to find
something symmetrical to that.

?]

MAGDALENA TODA:
Again, the same thing.

2x evaluated at whoever times--

STUDENT: u.

MAGDALENA TODA: Because you
have dx with respect to v,

so you have u plus--

STUDENT: df, dy.

MAGDALENA TODA: df, dy,
which is 2y, evaluated

at the same kind of guy.

So all you have to do is
replace with respect to u and v.

And finally, multiplied by-

STUDENT: dy.

MAGDALENA TODA: dy, dv.

dy, dv is 1 again.

Just pay attention
when you plug in

because you realize you
can know these very well

and understand it as a process,
but if you make an algebra

and everything is out.

And then you send me
an email that says,

I've tried this
problem 15 times.

And I don't even hold
you responsible for that

because I can make
algebra mistakes anytime.

So 2uv times u plus 2 times u
plus v. So what did I do here?

I simply replaced the given
functions in terms of u and v.

And I'm done.

Do I like it?

No, but I'd like you to notice
something as soon as I'm done.

2u squared v plus 2u plus 2v.

Could I have expected that?

Look at the beauty
of the functions.

Z is a symmetric function.

x and y have some of
the symmetry as well.

If you swap u and v, these
are symmetric polynomials

of order 2 and 1.

[INAUDIBLE]

Swap the variables, you
still get the same thing.

Swap the variables u and
v, you get the same thing.

So how could I have
imagined that I'm

going to get-- if I were smart,
without doing all the work,

I could figure out
this by just swapping

the u and v, the rows of u
and v. I would have said,

2vu squared, dv plus 2u and
it's the same thing I got here.

But not always are you so
lucky to be given nice data.

Well, in real life, it's a mess.

If you are, let's say, working
with geophysics real data,

you two parameters and for
each parameter, x and y,

you have other parameters.

You will never have
anything that nice.

You may have nasty
truncations of polynomials

with many, many
terms that you work

with approximating polynomials
all the time. [INAUDIBLE]

or something like that.

So don't expect these miracles
to happen with real data,

but the process is the same.

And, of course,
there are programs

that incorporate all of
the Calculus 3 notions

that we went over.

There were people
who already wrote

lots of programs that enable
you to compute derivatives

of function of
several variables.

Now let me take your
temperature again.

Is this hard?

No.

It's sort of logical you just
have to pay attention to what?

Pay attention to not making too
many algebra mistakes, right?

That's kind of the idea.

More things that I
wanted to-- there

are many more things I wanted
to share with you today,

but I'm glad we reached
some consensus in the sense

that you feel there is logic and
order in this type of problem.

I tried to give you a little
bit of an introduction to why

the gradient is so
important last time.

And I'm going to
come back to that

again, so I'm not going
to leave you in the air.

But before then,
I would like to do

the directional
derivative, which

is a very important section.

So I'm going to start again.

And I'll also do, at the same
time, some review of 11.5.

So I will combine them.

And I want to
introduce the notion

of directional
derivatives because it's

right there for us to grab it.

And you say, well,
that sounds familiar.

It sounds like I dealt
with direction before,

but I didn't what that was.

That's exactly true.

You dealt with it before, you
just didn't know what it was.

And I'll give you the
general definition,

but then I would like you to
think about if you have ever

seen that before.

I'm going to say I have the
derivative of a function, f,

in the direction, u.

And I'm going put u bar
as if you were free,

not a married man.

But u as a direction as
always a unit vector.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: I
told you last time,

just to prepare you, direction,
u, is always a unit vector.

Always.

Computed at x0y0.

But x0y0 is a given view point.

And I'm going to say
what that's going to be.

I have a limit.

I'm going to use the h.

And you say, why in the
world is she using h?

You will see in a second--
h goes to 0-- because we

haven't used h in awhile.

h is like a small displacement
that shrinks to 0.

And I put here, f of x0
plus hu1, y0 plus hu2,

close, minus f of x0y0.

So you say, wait a minute,
Magdalena, oh my god, I've

got a headache.

I'm not here.

Z0 is easy to understand
for everybody, right?

That's going to be
altitude at the point x0y0.

It shouldn't be hard.

On the other hand,
what am I doing?

I have to look at a real
graph, in the real world.

And that's going to be a
patch of a smooth surface.

And I say, OK, this
is my favorite point.

I have x0y0 on the ground.

And the corresponding
point in three dimensions,

would be x0y0 and z0,
which is the f of x0y0.

And you say, wait a
minute, what do you mean

I can't move in a direction?

Is it like when took
a sleigh and we went

to have fun on the hill?

Yes, but I said that
would be the last time

we talked about the hilly
area with snow on it.

It was a good preparation
for today in the sense that--

Remember, we went somewhere when
I picked your direction north,

east?

i plus j?

And in the direction
of i plus j,

which is not quite
the direction and I'll

ask you why in a second, I was
going down along a meridian.

Remember last time?

And then that was the direction
of the steepest descent.

I was sliding down.

If I wanted the direction
of the steepest ascent,

that would have been
minus i minus j.

So I had plus i plus
j, minus i minus j.

And I told you last time, why
are those not quite directions?

STUDENT: Because
they are not unitary.

MAGDALENA TODA: They
are not unitary.

So to make them like
this u, I should

have said, in the
direction i plus

j, that was one minus
x squared minus y

squared, the parabola way,
that was the hill full of snow.

So in the direction
i plus j, I go down

the fastest possible way.

In the direction i plus
j over square root of 2,

I would be fine
with a unit vector.

In the opposite direction, I
go up the fastest way possible,

but you don't want to because
it's-- can you imagine hiking

the steepest possible
direction in the steepest way?

Now with my direction.

My direction in plane
should be the i vector.

And that magic vector should
have length 1 from here

to here.

And when you measure this
guy, he has to have length 1.

And if you decompose, you have
to decompose him along the--

what is this?

The x direction and
the y direction, right?

How do you split a vector
in such a decomposition?

Well, Mr. u will be u1i plus 1i.

It sounds funny.

Plus u2j.

So you have u1
from here to here.

I don't well you can draw.

I think some of you can
draw really well, especially

better than me because you
took technical drawing.

How many of you took technical
drawing in this glass?

STUDENT: Only in this class?

MAGDALENA TODA: In anything.

STUDENT: In high school.

MAGDALENA TODA: High
school or college.

STUDENT: I went to
it in middle school.

So it gives you so
that [INAUDIBLE]

and you'd have to
draw it. [INAUDIBLE].

MAGDALENA TODA: It's
really helping you

with the perspective view,
3D view, from an angle.

So now you're looking
at this u direction

as being u1i plus u2j.

And you say, OK, I think
I know what's going on.

You have a displacement
in the direction of the x

coordinate by 1 times h.

So it's a small displacement
that you're talking about.

And-- yes?

STUDENT: Why 1 [INAUDIBLE]?

MAGDALENA TODA: Which one?

STUDENT: You said 1 times H.

MAGDALENA TODA: u1.

You will see in a second.

That's the way you define it.

This is adjusted information.

I would like you to tell me
what the whole animal is, if I

want to represent it later.

And if you can give
me some examples.

And if I go in a y direction
with a small displacement,

from y0, I have to leave and go.

So I am here at x0y0.

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

And when I displace a little
bit, I displace with the green.

I displace in this direction.

I will have to displace
and see what happens here.

And then in this direction--
I'm not going to write it yet.

So I'm displacing
in this direction

and in that direction.

Why am I keeping it h?

Well, because I have the
coordinates x0y0 plus--

how do you give me a collinear
vector to u, but a small one?

You say, wait a minute,
I know what you mean.

I start from the point x0, this
is p, plus a small multiple

of the direction you give me.

So here, you had it
before in Calc 2.

You had t times uru2,
which is my vector, u.

So give me a very small
displacement vector

in the direction u, which
is u1u2, u2 as a vector.

You like angular graphics.

I don't, but it doesn't matter.

STUDENT: So basically, h.

MAGDALENA TODA:
So basically, this

is x0 plus-- you want t or h?

t or h, it doesn't matter.

hu1, ui0 plus hu2.

Why not t?

Why did I take h?

It is like time parameter
that I'm doing with h,

but h is a very
small time parameter.

It's an infinitesimally
small time.

It's just a fraction of
a second after I start.

That's why I use little
h and not little t.

H, in general, indicates a
very small time displacement.

So tried to say,
where am I here?

I'm here, just one step further
with a small displacement.

And that's going to p
at this whole thing.

Let's call this F of--
the blue one is F of x0y0.

And the green altitude, or the
altitude of the green point,

will be what?

Well, this is
something, something,

and the altitude would be F
of x0 plus hu1, y0 plus hu2.

And I measure how far
away the altitudes are.

They are very close.

The blue altitude and
the green altitude

varies the displacement.

And how can I draw that?

Here.

You see this one?

This is the delta z.

So this thing is like
a delta z kind of guy.

Any questions?

It's a little bit hard, but
you will see in a second.

Yes, sir?

STUDENT: Is it like
a small displacement

that has to be perpendicular
to the [INAUDIBLE]?

MAGDALENA TODA: No.

STUDENT: It's a
result of [INAUDIBLE]?

MAGDALENA TODA: It
is in the direction.

STUDENT: In the direction?

MAGDALENA TODA: So
let's model it better.

I don't have a three
dimensional-- they sent me

an email this morning
from the library saying,

do you want your three
dimensional print--

do you want to support the idea
of Texas Tech having a three

dimensional printer available
for educational purposes?

STUDENT: Did you
say, of course, yes?

MAGDALENA TODA: Of
course, I would.

But I don't have a three
dimensional printer,

but you have
imagination and imagine

we have a surface that,
again, looks like a hill.

That's my hand.

And this engagement ring
that I have is actually p0,

which is x0y0zz.

And I'm going in a
direction of somebody.

It doesn't have to be u.

No, [INAUDIBLE].

So I'm going in the
direction of u-- yu2,

is that horizontal thing.

I'm going in that direction.

So this is the
direction I'm going in

and I say, OK, where do I go?

We'll do a small displacement,
an infinitesimally small

displacement in
that direction here.

So the two points are
related to one another.

And you say, but there's such
a small difference in altitudes

because you have an
infinitesimally small

displacement in that direction.

Yes, I know.

But when you make the ratio
between that small delta

z and the small h, the ratio
could be 65 or 120 minus 32.

You don't know what you get.

So just like in general limit
of the difference quotient

being the derivative, you'll get
the ratio between some things

that are very small.

But in the end, you can
get something unexpected.

Finite or anything.

Now what do you think
this guy-- according

to your previous Chain
Rule preparation.

I taught you about Chain Rule.

What will this be
if we compute them?

There is a proof for this.

It would be like a
page or a 2 page proof

for what I'm claiming to have.

Or how do you think
I'm going to get

to this without doing the
limit of a difference quotient?

Because if I give
you functions and you

do the limit of the
difference quotients

for some nasty functions,
you'll never finish.

So what do you think
we ought to do?

This is going to be some
sort of derivative, right?

And it's going to be
a derivative of what?

Yes, sir.

STUDENT: Well, it's going to
be like a partial derivative,

except the plane you're
using to cut the surface

is not going to be in the x
direction or the y direction.

It's going to be
along the [? uz. ?]

MAGDALENA TODA: Right.

So that is a very
good observation.

And it would be like I would the
partial not in this direction,

not in that direction,
but in this direction.

Let me tell you what this is.

So according to a
theorem, this would

be df, dx, exactly
like The Chain Rule,

at my favorite point
here, x0y0 [INAUDIBLE]

p times-- now you say, oh,
Magdalena, I understand.

You're doing some
sort of derivation.

The derivative of that with
respect to h would be u1.

Yes.

It's a Chain Rule.

So then I go times u1 plus
df, dy at the point times u2.

And you say, OK, but
can I prove that?

Yes, you could, but
to prove that you

would need to play a game.

The proof will involve that
you multiply up and down

by an additional expression.

And then you take
limit of a product.

If you take product,
the product of limits,

and you study them separately
until you get to this Actually,

this is an application
of The Chain.

But I want to come back to
what Alexander just notice.

I can explain this
much better if we only

think of derivative in the
direction of i and derivative

in the direction of j.

What the heck are those?

What are they going to be?

The direction of deritivie--
if I have i instead of u, that

will make you understand the
whole notion much better.

So what would be the
directional derivative

of in the direction of i only?

Well, i for an i.

It goes this way.

This is a hard lesson.

And it's advanced calculus
rather than Calc 3,

but you're going to get it.

So if I go in the
direction of i,

I should have the df, dx, right?

That should be it.

Do I?

STUDENT: Yes, but [INAUDIBLE] 0.

MAGDALENA TODA: Exactly.

Was I able to
invent something so

when I come back to what I
already know, I recreate df, dx

and nothing else?

Precisely because for i as
being u, what will be u1 and u2?

STUDENT: [INAUDIBLE].

MAGDALENA TODA: u1 is 1.

u2 is 0.

Right?

Because when we write i
as a function of i and j,

that's 1 times i plus 0 times j.

So u1 is 1, u2 is zero.

Thank god.

According to the anything,
this difference quotient

or the simpler way to define
it from the theorem would

be simply the second goes away.

It vanishes.

u1 would be 1 and what
I'm left with is df, dx.

And that's exactly
what Alex noticed.

So the directional
derivative is defined,

as a combination of vectors,
such that you recreate

the directional derivative
in the direction of i

being the partial, df, dx.

Exactly like you
learned before in 11.3.

And what do I have
if I try to recreate

the directional derivative
in the direct of j?

x0y0.

We don't explain this
much in the book.

I think on this one, I'm doing
a better job than the book.

So what is df in
the direction of j?

j is this way.

Well, [INAUDIBLE]
is that 1j-- you

let me write it
down-- is 0i plus 1j.

0 is u1.

1 is u2.

So by this formula,
I simply should

get the directional
deritive-- I mean,

directional derivative is the
partial deritive-- with respect

to y at my point times a 1
that I'm not going to write.

So it's a concoction, so that
in the directions of i and j,

you actually get the
partial deritives.

And everything else
is linear algebra.

So if you have a problem
understanding the composition

of vectors, the sum
of vectors, this

is because-- u1 and
u2 are [INAUDIBLE],

I'm sorry-- this is
because you haven't taken

the linear algebra yet, which
teaches you a lot about how

a vector decomposes in
two different directions

or along the standard
canonical bases.

Let's see some
problems of the type

that I've always put in the
midterm and the same kind

of problems like we
have seen in the final.

For example 3, is it, guys?

I don't know.

Example 3, 4, or
something like that?

STUDENT: 3.

MAGDALENA TODA: Given
z equals F of xy--

what do you like best,
the value or the hill?

This appeared in
most of my exams.

x squared plus y squared,
circular [INAUDIBLE]

was one of my favorite examples.

1 minus x squared
minus y squared

was the circular
parabola upside down.

Which one do you prefer?

I don't care.

Which one?

STUDENT: [INAUDIBLE].

MAGDALENA TODA: The [INAUDIBLE]?

The first one.

It's easier.

And a typical problem.

Compute the directional
derivative of z

equals F of x and y at the
point p of coordinates 1, 1, 2

in the following
directions-- A, i.

B, j.

C, i plus j.

D, the opposite, minus
i, minus j over square 2.

And E--

STUDENT: That's a square root 3.

MAGDALENA TODA: What?

STUDENT: You wrote
a square root 3.

MAGDALENA TODA: I
wrote square root of 3.

Thank you guys.

Thanks for being vigilant.

So always keep an eye on me
because I'm full of surprises,

good and bad.

No, just kidding.

So let's see.

What do I want to put here?

Something.

How about this?

3 over root 5, pi
plus [? y ?] over 5j.

Is this a unit vector or not?

STUDENT: No.

STUDENT: Yes, it is.

So you're going to
drag the [INAUDIBLE].

MAGDALENA TODA: Why
is that a unit vector?

STUDENT: It's
missing-- no, it's not.

MAGDALENA TODA: Then how
do I make it a unit vector?

STUDENT: [INAUDIBLE].

STUDENT: [INAUDIBLE].

STUDENT: I have to take down--
there's a 3 that has to be 1.

[INAUDIBLE]

And the second one has
to be 1, on the top,

to make it a unit vector.

MAGDALENA TODA: Give
me a unit vector.

Another one then
these easy ones.

STUDENT: 3 over 5 by 4 or 5.

MAGDALENA TODA: What?

STUDENT: 3 over 5 by 4 over 5j.

MAGDALENA TODA: 3
over-- I cannot hear.

STUDENT: 3 over 5--

MAGDALENA TODA: 3 over 5.

STUDENT: And 4 over 5j.

MAGDALENA TODA: And 4 over 5j.

And why is that a unit vector?

STUDENT: Because 3
squared is [INAUDIBLE].

MAGDALENA TODA: And what
do we call these numbers?

You say, what is that?

And interview?

Yes, it is an interview.

Pythagorean numbers.

3, 4, and 5 are
Pythagorean numbers.

So let me think a little
bit where I should write.

Is this seen by
the-- yes, it's seen

by the-- I'll just leave
what's important for me

to solve this problem.

A. So what do we do?

The same thing. i is 1.i plus
u, or 1 times i plus u times j.

So simply, you can write
the formula or you can say,

the heck with the formula.

You know that df is df, dx.

The derivative of
this at the point p.

So what you want to do is say,
2x-- are you guys with me?

STUDENT: Yes.

MAGDALENA TODA: At the
value 1, 1, 2, which is 2.

And at the end of
this exercise, I'm

going to ask you if there's
any connection between--

or maybe I will
ask you next time.

Oh, we have time.

What is d in the direction of j?

The partial derivative
with respect to y.

Nothing else, but
our old friend.

And our old friend
says, I have 2y

computed for the
point p, 1, 1, 2.

What does it mean?

Y is 1, so just plug
this 1 into the thingy.

It's 2.

Now do I see some--
I'm a scientist.

I have to find
interpretations when

I get results that coincide.

It's a pattern.

Why do I get the same answer?

STUDENT: Because your
functions are symmetric.

MAGDALENA TODA: Right.

And more than that, because
the function is symmetric,

it's a quadric that I love,
it's just a circular problem.

It's rotation is symmetric.

So I just take one parabola,
one branch of a parabola,

and I rotate it by 360 degrees.

So the slope will be the same
in both directions, i and j,

at the point that I have.

Well, it depends on the point.

If the point is,
itself, symmetric

like that, x and y are
the same, one in one,

I did it on purpose-- if
you didn't have one and one,

you had an x variable and
y variable to plug in.

But your magic point is where?

Oh my god.

I don't know how to
explain with my hands.

Here I am, the frame.

I am the frame. x, y, and z.

1, 1.

Go up.

Where do you meet the vase?

At c equals 2.

So it's really symmetric
and really beautiful.

Next I say, oh, in
the direction i plus

j, which is exactly the
direction of this meridian

that I was talking about, i
plus j over square root 2.

Now I've had students-- that's
where I was broken hearted.

Really, I didn't
know what to do,

how much partial credit to give.

The definition of direction
derivative is very strict.

It says you cannot take
whatever 1 and 2 that you want.

You cannot multiply
them by proportionality.

You have to have u
to be a unit vector.

And then the directional
derivative will be unique.

If I take 1 and 1 for u1 and
u2, then I can take 2 and 2,

and 7 and 7, and 9 and 9.

And that's going to
be a mess because

the directional derivative
wouldn't be unique anymore.

And that's why whoever
gave this definition,

I think Euler-- I tried
to see in the history who

was the first
mathematician who gave

the definition of the
directional derivative.

And some people
said it was Gateaux

because that's a french
mathematician who first talked

about the Gateaux
derivative, which

is like the
directional derivative,

but other people said,
no, look at Euler's work.

He was a genius.

He's the guy who discovered
the transcendental number

e and many other things.

And the exponential
e to the x is also

from Euler and everything.

He was one of the
fathers of calculus.

Apparently, he knew the first
32 decimals of the number e.

And how he got to
them is by hand.

Do you guys know of them?

2.71828-- and that's all I know.

The first five decimals.

Well, he knew 32 of them
and he got to them by hand.

And they are non-repeating,
infinitely remaining decimals.

It's a transcendental number.

STUDENT: And his 32 are correct?

MAGDALENA TODA: What?

STUDENT: His 32 are correct?

MAGDALENA TODA: His first
32 decimals were correct.

I don't know what--
I mean, the guy

was something like-- he
was working at night.

And he would fill out,
in one night, hundreds

of pages, computations, both
by hand formulas and numerical.

So imagine-- of course, he would
never make a WeBWork mistake.

I mean, if we built
a time machine,

and we bring Euler back,
and he's at Texas Tech,

and we make him solve
our WeBWork problems,

I think he would take
a thousand problems

and solve them in one night.

He need to know
how to type, so we

have to teach him how to type.

But he would be able to
compute what you guys have,

all those numerical
answers, in his head.

He was a scary fellow.

So u has to be [INAUDIBLE]
in some way, made unique.

u1 and u2.

I have students-- that's
where the story started--

who were very good, very smart,
both honors and non-honors, who

took u1 to be 1, u2 to be 2
because they thought direction

1 and 1, which is not made
unique as a direction, unitary.

And they plugged in here
1, they plugged in here 1,

they got these correctly, what
was I supposed to give them, as

a [? friend? ?]

STUDENT: [INAUDIBLE].

MAGDALENA TODA: What?

STUDENT: [INAUDIBLE].

MAGDALENA TODA: I gave them.

How much do you think?

You should know me.

STUDENT: [INAUDIBLE].

STUDENT: Full.

MAGDALENA TODA: 60%.

No.

Some people don't
give any credit,

so pay attention to this.

In this case, this has
to be 1 over square root

of 2 times the derivative of f
at x, which is computed before

at the point, plus 1 over square
root of 2 times the derivative

of the function.

Again, compute it
at the same place.

Which is, oh my god, square
root of 2 plus square root of 2,

which is 2 square root of 2.

And finally, the derivative
of F at the same point-- I

should have put at the point.

Like a physicist
would say, at p.

That would make you
familiar with this notation.

And then measured at what?

The opposite direction,
minus i minus j.

And now I'm getting lazy
and I'm going to ask you

what the answer will be.

STUDENT: 2 minus
square root of 2.

MAGDALENA TODA: So you see,
there is another pattern.

In the opposite
direction, the direction

of the derivative in this case
would just be the negative one.

What if we took this directional
derivative in absolute value?

Because you see,
in this direction,

there's a positive
directional derivaty.

In the other direction, it's
like it's because-- I know why.

I'm a vase.

So in the direction i plus
j over square root of 2,

the directional derivative
will be positive.

It goes up.

But in the direction
minus i minus

j, which is the opposite, over
square root of 2, it goes down.

So the slope is negative.

So that's why we have negative.

Everything you get
in life or in math,

you have to find
an interpretation.

Sometimes in life and
mathematics, things are subtle.

People will say one thing
and they mean another thing.

You have to try to see
beyond their words.

That's sad.

And in mathematics, you have to
try to see beyond the numbers.

You see a pattern.

So being in opposite
directions, I

got opposite signs of the
directional derivative

because I have opposite slopes.

What else do I want to
learn in this example?

One last thing.

STUDENT: E.

MAGDALENA TODA: E. So
I have the same thing.

So it's not going to
matter, the direction

is the only thing that changes.

These guys are the same.

The partials are the
same at the same point.

I'm not going to
worry about them.

So I get 2 or both.

What changes is the blue guys.

They are going to be
3 over 5 and 4 over 5.

And what do I get?

I get-- right?

Now I want to tell
you something--

I already anticipated
something last time.

And let me tell you
what I said last time.

Maybe I should not
erase-- well, I

have to erase this
whether I like it or not.

And now I'll review
what this was.

What was this? d equals
x squared plus y squared?

Yes or no?

STUDENT: Yes.

MAGDALENA TODA: So what
did I say last time?

We have no result. We
noticed it last time.

We did not prove it.

We did not prove it, only
found it experimentally

using our physical common sense.

When you have a function
z equals F of xy,

we studied the
maximum rate of change

at the point x0y0 in the domain,
assuming this is a c1 function.

I don't know.

Maximum rate of change
was a magic thing.

And you probably thought,
what in the world is that?

And we also said, this
maximum for the rate of change

is always attained in the
direction of the gradient.

So you realize that it's
the steepest ascent,

the way it's called in
many, many other fields,

but mathematics.

Or the steepest descent.

Now if it's an ascent, then it's
in the direction gradient of F.

But if it's a
descent, it's going

to be in the opposite
direction, minus gradient of F.

But then I [INAUDIBLE]
first of all,

it's not the same direction,
if you have opposites.

Well, direction is sort
of given by one line.

Whether you take this or the
opposite, it's the same thing.

What this means is
that we say direction

and we didn't
[? unitarize ?] it.

So we could say,
or gradient of F

over length of gradient of
F. Or minus gradient of F

over length of gradient of F.
Can this theorem be proved?

Yes, it can be proved.

We are going to discuss a little
bit more next time about it,

but I want to tell you
a big disclosure today.

This maximum rate of change
is the directional derivative.

This maximum rate
of change is exactly

the directional derivative
in the direction

of the gradient, which is also
the magnitude of the gradient.

And you'll say,
wait a minute, what?

What did you say?

Let's first verify my claim.

I'm not even sure
my claim is true.

We will see next time.

Can I verify my
claim on one example?

Well, OK.

Maximum rate of change
would be exactly

as the directional
derivative and the direction

of the gradient?

I don't know about that.

That all sounds crazy.

So what do I have to compute?

I have to compute that
directional derivative

of, let's say, my function F in
the direction of the gradient--

what is the gradient?

We have to figure it out.

We did it last time,
but you forgot.

So for this guy, nabla F,
what will be the gradient?

Where is my function?

Nabla F will be 2x, 2y, right?

Which means 2xi plus 2yj, right?

But if I'm at the point
p, what does it mean?

At the point p, it means that I
have 2 times i plus 2 times j,

right?

And what is the magnitude
of the gradient?

Yes.

The magnitude of the gradient is
somebody I know, which is what?

Which is square root of
2 squared plus 2 squared.

I cannot do that now.

What's the square root of 8?

STUDENT: 2 root 2.

MAGDALENA TODA: 2 root 2.

This is a pattern.

2 root 2.

I've seen this 2 root
2 again somewhere.

Where the heck have I seen it?

STUDENT: That was the
directional derivative.

MAGDALENA TODA: The
directional derivative.

So the claim may be right.

It says it is the directional
derivative in the direction

of the gradient.

But is this really the
direction of the gradient?

Yes.

Because when you compote the
direction for the gradient, 2y

plus 2j, you don't mean 2i
plus 2j as a twice i plus j,

you mean the unit vector
correspondent to that.

So what is the
direction corresponding

to the gradient 2i plus 2j?

STUDENT: i plus j [? over 2. ?]

MAGDALENA TODA: Exactly.

U equals i plus j
divided by square 2.

So this is the
directional derivative

in the direction of the gradient
at the point p, which is 2 root

2.

And it's the same thing-- for
some reason that's mysterious

and we will see next time.

For some mysterious reason
you get exactly the same

as the length of
the gradient vector.

We will see about this
mystery next time.

I have you enough to
torment you until Tuesday.

What have you promised me
besides doing the homework?

STUDENT: To read the book.

MAGDALENA TODA:
To read the book.

You're very smart.

Please, read the book.

All the examples in the book.

They are short.

Thank you so much.

Have a wonderful
weekend and I'll

talk to you on Tuesday about
anything you have trouble with.

When is the homework due?

STUDENT: Saturday.

MAGDALENA TODA: On Saturday.

I was mean.

I should have given it you
until Sunday night, but--

STUDENT: Yes.

MAGDALENA TODA: Do you want me
to make it until Sunday night?

STUDENT: Yes.

MAGDALENA TODA: At midnight?

STUDENT: Yes.

MAGDALENA TODA: I'll do that.

I will extend it.

STUDENT: She asked, I said yes.

STUDENT: Why did
you do that, dude?

Come on, my life is ruined
now because I have more time

to work on my homework.

MAGDALENA TODA: And
I've ruined your Sunday.

STUDENT: Yes.

No.

MAGDALENA TODA: No.

Actually, I know why I did that.

I thought that the
28th of February

is the last day of the month,
but it's a short month.

So if we [? try it, ?] we
have to extend the months

a little bit by pulling
it by one more day.

STUDENT: We did?

MAGDALENA TODA: The first
of March is Sunday, right?

STUDENT: Yes.

[INTERPOSING VOICES]

STUDENT: You're going
to miss the speech.

STUDENT: Oh, we're doing that?

STUDENT: You're in English?

STUDENT: [INAUDIBLE].

STUDENT: You don't know English?

Why are you talking English?

That's what my
father used to say.

You don't know your own tongue?