-
-
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?