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.
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: 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?