what we discussed last time in 11.6
which was-- do you remember the topics we discussed?
We discussed the valuation of a derivative.
What else have we discussed about them?
I'm gonna split the fields, although they are related.
Gradient and the steepest ascent and descent.
So in which direction will z equals
f of x and y, differential cofunction,
and with a derivative that is continuous?
So will this have the maximum rate of change?
This is just review.
Why am I doing review?
Well, after talking to you on a personal basis,
like one-on-one basis by email and a little bit in person,
I realized that you like very much when I review.
When I briefly review some of the notions.
I will give you the essentials for the section, that
was 11.5 is embedded in this.
Embedded.
So 11.5 is embedded in 11.6, and in one shot
we can talk of them.
In 11.7 you're gonna see some extrema of functions
of two variables, like max and min,
and then we have 11.8, which is Lagrange multipliers.
What have we done about this differentiable function
with continuous partial derivatives?
Let's assume that it's smooth, OK?
And in that case, we define the partial--
the directional derivative in the direction of u at u,
v with unit vector at the point x, y, but let's fix it
x0, y0, using a limit of a difference quotient,
just like any derivative should be introduced.
But I'm not gonna repeat that definition.
Why is that?
Because I want you to give me an alternative definition,
which is something that is simpler
to use in applications, which is what?
Who remembers?
The partial derivative of f with respect to x
measured at the point x0, y0.
I'm gonna use this color, and then I'll change the color.
And I'll say times u1.
Plus-- change the color again-- f sub y at x0, y0, times u2.
And this is just a times-- multiplication
between real values.
Because u1 and u2 will be nobody but the components--
the real value components of the unit vector direction
that we have.
So, guys, remember, direction in this books means unit vector.
Every time I say direction, I should
say that's a unit vector.
Is there any way, any other way, to express this?
Maybe with a vector multiplication of some sort--
multiplication between vectors.
See, what if I had a pink vector with pink components,
and a blue vector with blue components?
I'm getting somewhere.
And I'm sneaky.
And you feel-- you know where I'm
getting because you had chapter nine fresh in your mind
and a certain product between vectors.
So what is this?
It's a dot product, excellent.
Rachel, right?
So it's a dot product.
And I'm gonna run it, and it's going
to be just the gradient of f at the point p0.
But I'm going to write it again, x0, y0,
although it drives me crazy to have
to write that all the time.
Dot product or scalar product with what vector u?
I'm not gonna put a bar on that because the-- that would
be like an oxymoron.
The gradient is a bar thing.
It's always a vector, so I'm not gonna write a bar.
But I write a bar on u, reminding you
that u is a free vector.
All right, do you like this?
Yes.
It's a compactified form of the fluffy expression
we had before.
It's much easier to remember than the limit definition.
Of course, it's equivalent to it.
And in applications, I-- well, we ask about that all the time.
What was the gradient?
So see, in mathematics everything is related.
And talking about-- speaking about the devil--
I mean, not you, but we weren't talking about you--
just this gradient.
Gradient f at the point p will simply
be the vector whose components in the direction of i and j
are the partials.
The partial derivative with respect
to x, plus the partial derivative with respect to y.
OK, last time we dealt even with gradients of-- functions
of more variables.
If I have n variables, so x1, x2,
x3, xn, then this vector will have n components-- f sub x1,
f sub x2, f sub x3, f sub x4.
Somebody stop me.
F sub xn.
So I have n of them, and it's an n [INAUDIBLE].
It's a sum ordered [INAUDIBLE].
OK.
A set of n elements in this order.
It's an order set, [INAUDIBLE] n.
All right.
So the order matters.
Now, steepest ascent and descent.
In which direction do I have the maximum rate of change?
And the answer is-- this is Q and this
is A-- the maximum rate of change
happens in the direction of the gradient at every point--
every point of the domain where I
have smoothness or [INAUDIBLE] or [INAUDIBLE] whatever.
All right.
And what else?
I claimed last time, but I didn't
prove, that in that direction that I claimed c
from [INAUDIBLE].
The directional derivative is maximized exactly
in the direction of the gradient.
Can we prove that?
Now we can prove it.
Before I couldn't prove it because you couldn't see it,
because I didn't look at it as a dot product.
And we were all blind, like guiding each other in the dark.
Me blind, you blind.
We couldn't see.
Now we can see.
So now we can see how to prove my claim
that the directional derivative, which
is measuring the maximum-- measuring
the instantaneous rate of change in a direction-- compass
direction, like, what is that, east?
East, northeast, southwest, whatever.
Those are the compass directions like i, j, i
plus j over square root 2 and so on,
those are called compass directions
because you hold the compass in your hand as horizontal
as you can, and you refer to the floor.
Even if you are on a slope.
Maybe you imagine me on a slope, hiking or whatever.
Going down, going up.
But the compass should always be kept horizontal.
Do you hike, Alex?
STUDENT: Yeah.
PROFESSOR: I'm sorry, I've put you on the spot.
So whatever you do when you think of a path up or down
is measured in the direction of the horizontal plane,
the compass direction.
And that is the gradient I was talking about.
You see, it's a function.
It's a vector in plane.
That means a geographic compass direction.
Prove it.
Let's prove the claim.
Let's prove the claim, because this is Tuesday,
and it's almost weekend, and we have to prove something
this week, right?
Now, do you like what you see?
Well, I have no idea.
What if I want to measure this.
This could be a negative number, but it doesn't matter.
Assume that I take the dot product,
and I think, well, it's a scalar product, it must be positive.
What do I get?
This guy-- if I'm a physicist, I'm going to say this guy
is the length of the first vector at p0.
Gradient at p0.
That's the length of u.
Duh, that is 1.
Last time I checked, u was unitary, so that's silly of me,
but I'll write it anyway.
Times the cosine of the angle between-- oh, cosine
of the parenthesis angle between nabla f and u.
OK.
When is this maximized?
STUDENT: When the angle between nabla f and u is 0.
PROFESSOR: Exactly.
When this is pi, so this quantity is maximized-- gosh,
I hate writing a lot.
I had to submit homework in the past two days,
and one was about biological research and the other one
about stress management.
The stress management class stresses me out the most.
I shouldn't make it public.
Really, because we have to write these essays of seven or eight
pages every Tuesday, every end of the week.
Twice a week.
, So I realized how much I hate writing down a lot,
and what a blessing it is to be a mathematician.
You abbreviate everything.
You compress everything.
I love formulas.
So what we have here is maximized when the cosine is 1.
And if you have become 0 to 2 pi open,
then theta 0 is your only option.
Well, if you take an absolute value,
you could also have it in the other direction,
cosine pi, but in that case you change the sign.
So what you get-- you get a maximum.
Let's say you hike, right?
I'm just hiking in my brain.
The maximum rate of change in this direction,
climbing towards the peak.
And then the steepest descent is the exactly minus gradient
direction.
So I could have 0 or pi for the angle.
That's the philosophical meaning of that.
All right.
So the directional derivative, which is this guy,
when does it become maximum?
When the angle is 0.
So I'm done.
QED.
What does it mean?
That I know this happens when-- when direction u is
the direction of the gradient.
Can I write u equals gradient of f?
Not quite.
I should say divided by norm or magnitude.
Why is that?
And you say, Magdalena, didn't you say like 10 times
that u is a unit vector?
You want u to be a unit vector direction.
So the direction should be the direction of the gradient
in order to maximize this directional derivative.
But then you have to take the gradient
and divide it by its magnitude.
Let's compute it.
Let's see what we get.
Now, I'm sorry about my beautiful handwriting,
but I'm-- well, I'm gonna have to-- I have room here.
And actually I can use this formula.
So in the direction of the gradient,
when u is the gradient.
Let's take u to be-- what did I say over there?
Gradient of f over the magnitude of gradient of f.
I'll take this guy and drag him here.
[INAUDIBLE]
What will the value of the directional derivative be?
We've done last time that-- the same thing on an example.
We've done it on a function, beautiful f of x, y
equals x squared plus y squared.
This is the type of function I like,
because they are the fastest to deal with.
But anyway, we'll have all sorts of other functions.
What am I going to write here?
I have to write.
Well, by definition, now, by my new way
to look at the definition, I'm gonna
have a gradient at the point, the vector,
dot-- who is u again?
The gradient this time, that special value,
divided by absolute-- by magnitude of the gradient.
But what in the world is that?
This animal is-- what if you take
a vector multiplied by itself?
Dot product.
No, not dot product.
What you get is the magnitude squared.
All right.
So although the pinkie guy and the pinkie guy
are magnitude of f squared divided by magnitude of f.
And Alex said, but wait, that's just-- I know.
I didn't want to just jump ahead too fast.
We get gradient of f in magnitude.
So, beautiful.
So we know who that maximum is.
The maximum of the rate of change
will be for-- this equals f of x, y.
The max of the rate of change is-- what is that again?
Magnitude of lambda f, in the direction nabla-- nabla f
divided by its magnitude.
This is what we discovered.
And now I'm going to ask you, what
is the minimum rate of change at the same point?
STUDENT: [INAUDIBLE].
PROFESSOR: Just parallel opposite.
So it's gonna-- I'm gonna have the so-called highest--
steepest, not highest, highest means maximum.
The lowest value.
So I'm going to have the lowest value, which
indicates the steepest descent.
Me going down on-- in the snow, I'm dreaming, on a sleigh,
or on a plastic bag.
That would give me the steepest descent,
and the steepest descent will correspond to what?
I'm going to make an NB.
Nota bene.
In Latin.
Note the minimum will be minus magnitude
of f, [INAUDIBLE] nabla f, in the opposite direction.
Shall I write it in words?
Let me write it in as O-P-P from opposite-- no,
from opposite-- opposite direction.
What do I mean opposite direction?
Opposite direction to the gradient.
Which is the same direction, if you think about,
because it's the same line.
So it's going to be minus nabla f over magnitude of nabla f.
It's like when we were in [INAUDIBLE], which
was 1 minus x squared minus y squared, whatever it was--
we had i plus j for the descent, and minus i minus
j after the ascent-- the steepest descent
and the steepest ascent.
We started with examples because it's easier
to understand mathematics-- actually
it's easier to understand anything on an example.
And then-- if the example is good.
If the example is bad, it's confusing.
But if the example is good, you understand just
about any concept, and then you move on to the theory,
and this is the theory.
And it looks very abstract.
When somebody steps in this classroom
and they haven't taken more than calc 2 they will get scared,
and they will never want to take calc 3.
Well, that's why-- I didn't want to scare you off yet.
OK, so this is what you have to remember from section 11.6
with 11.5 embedded in it.
Now, one thing that I would like to see
would be more examples and connections to other topics.
So one example that I picked-- and I think it's a nice one.
I just copied from the book.
I usually don't bring cheat sheets.
I don't like professors who bring books to the class
and start reading out of the book.
I think that's ridiculous.
I mean, as if you guys couldn't read your own book at home.
And I try to make up examples that are easier than the ones
from the book to start with.
But here's example 4, which is not so easy to deal with,
but it's not hard either.
And I picked it because I saw this browsing
through the previous finals, I saw it
as a pattern coming every now and then.
Find the direction in which f increases or decreases
most rapidly.
I have to write down beautifully.
Find the directions in which f increases or decreases most
rapidly at p0 coordinates 2, 1.
And, what is the maximum-- that's
a question-- what is the maximum rate of change or of increase?
This type of problem is also covered in the Khan Academy
videos and also the MIT library, but I don't
feel they do a very good job.
They cover it just lightly, as if they
were afraid to speak too much about-- give too many examples
and talk too much about the subject.
So what do you guys want to do with this one?
Help me solve the problem.
That was kind of the idea.
So we start with computing the what?
The animal called-- what's the animal?
STUDENT: Gradient.
PROFESSOR: Gradient.
Thank you very much.
So we do that and we start differentiating.
With respect to x, we get a product [INAUDIBLE].
So your product [INAUDIBLE] 1-- it's differentiated-- times e
to the 2y minus x [INAUDIBLE] plus x undifferentiated.
The second guy, prime.
Copy and paste, Magdalena.
Times-- don't forget the minus 1,
because-- I'm talking to myself.
Because if you do, you get a 0 on this in the final.
I'm talking to myself.
OK?
All right.
Plus, parenthesis-- the same procedure with respect to y.
When I do it with respect to y, this
is good review of the whole chapter.
Yes, sir.
STUDENT: [INAUDIBLE].
PROFESSOR: This primed with respect to x.
Am I doing something wrong?
No.
Are you with me?
So this guy primed with respect to x, I'm
going to write it es copied.
And take that out, you maintain and differentiate with respect
to x.
So I did it right.
OK, the second one, x is a constant for me right now.
Who is the variable y?
So I copy and paste e to the whole argument
times-- I cover everything else with my hand,
and I differentiate the argument with respect to y.
And I get prime sub 2 and a j, and I say, thank
god, this was a little bit long.
You realize that if you make the slightest algebra mistake,
it's all over for you.
In that case, I ask my colleagues,
what do you guys do when guy missed that or missed this?
0, 0, no points-- OK, maybe a little bit,
maybe a tiny bit of extra credit.
But pay attention to your math.
So you know what you need to do.
Now I'm going to go on and say, but I am at the point 0.
By the way, I really don't like what we did in the book.
OK, I should not say that out loud, but it's too late.
The book denotes that sometimes, well, we
try not to do that too often, but not by F sub 0.
Because some other books use that.
I don't like that.
So every time-- you should never do that.
Because it gives the feeling that
you're differentiating a constant or something.
OK, so I always try to say, gradient [? up ?]
is-- which means, I have a fixed value.
But I don't fix the value before I took the gradient.
This is too confusing as a notation.
Don't do it.
Close your eyes when you get to it
when you're reading the book.
OK, now I have to plug in instead of x sub 2--
I tried to remember that-- y equals 1.
So I go 1 times e to the 2 times 1 minus 2
plus 2 times e to the 2 times 1 minus--
I wish I had Data with me, I mean the guy from Star Trek.
Because he could do this in just a fraction of a second
without me having to bother with this whole thing.
STUDENT: But then if Data existed,
why would we be math majors?
PROFESSOR: Exactly, so we do this
so that we can program people, I mean androids,
and eventually learn how to clone ourselves.
So let's see what we have.
e to the 0 is 1.
e to the 0 is 1, 1 minus 2-- are you guys with me?
I'm going too fast?
STUDENT: No, the [? board ?] says
the y component is equal to 1, x component is equal to 2.
PROFESSOR: x component equals to 2.
x0 is 2, and y0 is 1.
And I plug in x0 equals [INAUDIBLE].
So 2 times e to the 2 times 1 minus 2-- you think I like it?
I don't like it.
But anyway, it's my life.
I have to go on.
So we have 1 minus 2 plus minus 1, right?
Minus 1i-- minus 1i sounds scary.
OK, plus 4j-- minus i plus 4j.
Is it lovely?
No, I hate it.
The magnitude is going to be square root of 17.
But that's life.
I mean, you as engineering majors
see that all the time, and even worse than that.
So I'm going to say the gradient of F at P0 in magnitude
will be square root of 17.
What is that?
That is the maximum rate of change, right guys?
But in which direction does that happen?
My beloved book says, in the direction of minus i plus 4j.
That's our answer, so in the direction of--
STUDENT: Over root 17.
PROFESSOR: No, let me tell you what.
We fought about that as the authors when we wrote the book.
So I said, if you're saying, in the direction of,
then you have to say, over root 17.
And my coauthor said, no, actually, Magdalena,
as a matter of English-- since you're not a native,
you don't understand.
In the direction of a certain vector
means the direction of a certain vector,
the direction could be that.
But I say equivalently, in the direction double dot
minus i plus 4j over square root of 17,
meaning that this is the direction.
It's a matter of interpretation.
I don't understand it, but it's your language.
Yes, sir.
STUDENT: [INAUDIBLE].
PROFESSOR: I'm saying--
STUDENT: Which one do you like?
PROFESSOR: Which one do I like?
This one.
STUDENT: [INAUDIBLE].
PROFESSOR: Thank you.
So if we define direction to be a unit vector,
let's be consistent and not say, of me, of you, of my cousin,
of whatever.
All right, was this hard?
No.
Do I have a caveat about this kind of problem?
Yeah, that was kind of the idea.
If I put that in the midterm, guys,
please do it two or three times, make
sure you didn't make any algebra mistakes.
Because if you know the theory, I will still
give you like 30% or something.
But if you mess up with the algebra,
I have no choice but giving up the 70%, whatever that is.
I try to be fair and give you something for everything
you do and know.
But try not to mess it up too badly,
because it's very easy to mess it up.
Yes, sir.
STUDENT: So which way is increasing?
PROFESSOR: OK, find the direction
in which r increases or decreases most rapidly.
OK, the direction in which I could draw it,
this is increasing in the direction of that.
At what rate?
At the rate of square root of 17.
Good question.
How about the other one?
In the direction of plus i minus 4j over square root of 17,
I get the rate of change minus square root of 17.
And that's it.
Do we have to say that?
Ehh, I give you extra credit if you say that.
But at this point, I'm saying I'm happy with what I just
wrote on the board.
STUDENT: [INAUDIBLE].
PROFESSOR: Yeah, so it's like finding
on which path are you going to get to the top
the fastest when you climb a mountain.
It's the same kind of question.
Because it's all about z being an altitude.
z equals F of x.
I will go ahead and erase the whole thing.
And I wish you good luck now.
I wish you good luck because I was asked by [INAUDIBLE]
to solve a problem like the one you gave me in the web work,
but I don't remember it.
But it's OK.
it was a review of what we did last time.
And we said, instead of that, where the tangent plane--
we know what that is.
At P0 was-- guys, by the final I want this memorized.
There is no question.
So z minus z0 is like Taylor's formula
in the linear approximation.
You truncate.
You throw away the second order, and so on.
So what is that?
f sub x at delta x.
Oh, x equals 0.
I'm lazy.
Times x minus x0-- this is the delta x.
This is the delta z.
And this is not the round surface, curvy and everything.
This is the plane approximation, the planar approximation--
A-P-P-R. I'm not done-- plus f sub y y minus y0.
So this is the equation of the plane that's tangent pi at 0.
And let me draw the surface pink.
Because I'm a girl, and because I want to draw this in pink.
Let's call it some S, for Surface.
Can we paint an S here?
OK, but if I'm giving the same picture
for a different equation-- so I have F of x, y, z
equals constant, that's the implicit form.
I make this face.
Why do I make this face?
Because I've got three confessions.
I'm more like a priestess in mathematics.
People don't like implicit differentiation.
We will do a little bit of that today,
because you told me your stories from implicit differentiation,
and I got scared.
Those were horror stories.
And I don't want those to repeat in the final or the midterm.
So I'm going to do something with implicit differentiation
as well.
What did I want to say?
If we apply this, we'd be wrong.
But we have to remember what the normal would be.
And the normal was the gradient of big F.
So those will be F sub xi plus F sub yj plus big F sub zk.
Then force, such a surface, even implicitly,
the tangent plane looks a little bit differently.
But it's the same story.
And I prove that.
You may not believe it, or may not remember.
But I proved that, that it's one and the same thing.
So I get F sub x.
I get 0i0.
This was the A dot, times x minus x plus.
What's next?
From the coefficients coming from the gradient--
that was the gradient.
So this is F of x, y, z equals C.
And the gradient was-- this is the normal, F sub x, F sub y,
F sub z angular bracket equals gradient of F, which
is more or less than normal.
The unit normal will be just gradient of F over length of F.
So let's continue-- F sub y at x0y0 times delta y.
This is B. I had the problem with memorizing these,
especially since when I was 18 I did not understand them
whatsoever first year of college.
I had to use markers, put them in markers
and glue them to my closet.
Because when I was 18, of course,
I was looking in the mirror all the time.
So whenever I got into the closet,
opened the door to the mirror, next to the mirror
there was this formula.
So whether I liked it or not-- I didn't-- I memorized it just
by seeing it every day when I opened the door.
Yes, sir.
STUDENT: That green one, should that be x minus 0 or z minus 0?
PROFESSOR: z minus 0.
That is my mistake.
Thank you.
So again, you have delta x, delta y, delta z.
Thank you.
Is it hard to memorize?
No, not if you put it in markers.
I think you will do just fine.
What have we done as time?
Let me review it really quickly.
We said, wait a minute.
How come they are one and the same?
You said, oh I'm getting a headache.
I don't understand why they are one and the same.
And we said, yes, but you see, this guy is nothing but z
minus F of x, y.
So from this form, you can make it implicit it by pulling out F
to the left and creating this big F of x, y, and z equals 0.
And what does it mean?
It means that F sub x will be oh my god, this has nothing
to do with us minus F sub x.
F sub y will be minus F sub y.
Am I right?
And F sub z, big F sub z, is simply-- there's no z here-- 1.
So coming back to the guide's A, B, C, forget about A,
B, C. I'll take the A, and I'll replace it with minus F
sub x, which doesn't write.
And it's time for him to go away-- minus F sub x.
And this is F sub y with the minus.
And finally, Mr. C, who is happy, he is 1.
So he says, I'm happy.
You're going to separate me.
We are going to separate him.
So this equation is nothing but what?
Let's write it from the left to the right.
Let's keep the green guy in the left hand side.
And everybody else goes for a moving sale.
The blue and the pink go away.
And when they go to the other side,
they have a minus, pick up a minus sign.
But with the minus, the minus here is a plus sign.
Are you guys with me?
So the blue guy has moved and changed sign.
Where's the pink?
The pink guy will also move.
And he picked up this.
So practically, this and that formula are one and the same.
They're both used for the same tangent plane.
It depends how you introduce the tangent plane.
And [INAUDIBLE], before class started, 20 or 25 minutes
before class, when I was in a hurry, I answered you briefly.
You made some algebra mistake in the-- you got it?
I would like to make up one like those.
But I forgot what your surface was.
Was it an ellipsoid?
STUDENT: Ellipsoid.
PROFESSOR: OK, I'll make up an ellipsoid.
STUDENT: [INAUDIBLE].
PROFESSOR: Yeah, mhmm, if you have it with you.
If you don't have it with you, that's fine.
So I'm going to go ahead and keep
just the implicit equation.
Because the ellipsoid is given by the implicit equation.
And everything else I will erase.
And that was problem 24.
How many problems?
Well, you still have a lot, up to 49.
STUDENT: 42.
PROFESSOR: 42, OK, I reduced it.
Now, when I sent you an email on Sunday,
I said I was giving you an extension.
STUDENT: Till March.
PROFESSOR: Till a lot of March.
Because I thought March the 2nd, and I gave a few more days.
So you have one more week.
STUDENT: When's spring break?
PROFESSOR: Spring break is the 14th,
but this is due on the 9th right?
STUDENT: The 10th.
PROFESSOR: The 10th-- maybe I don't remember.
So what was your problem?
What's your problem?
What?
Can I take it?
So you have an ellipsoid, which comes
from the ellipse 3x squared [INAUDIBLE].
Then you have a 2z squared equals 9.
Then it says, you look at the point P
of coordinates minus 1-- I haven't even
checked if it's correct, but it should be.
So 3 plus-- I did not program the problem.
3 plus 4, 7, plus 2, 9, so he did a good job.
We want the tangent plane.
I'll put it here.
We want the tangent plane.
How do we compute the tangent plane?
You say, this is F of x, y, z, right?
So F sub x equals 6x.
F sub y equals 2y.
F sub z equals 4z.
Computing it at P0, what do we have?
x is minus 1, y is minus 2, z is minus 1.
I should get negative 6, negative 4, and minus 4.
And then I should plug in and get minus 6 times
x minus minus 1.
I have to pay attention myself.
It's not easy to get the algebra right-- minus 4 times y
plus 2 minus 4 times z plus 1 equals 0.
And I hope I get what you got-- minus 6x minus 4y minus 4z,
so many of those.
You've got to divide by 2.
I'm not getting that.
And then minus 6-- I'm going to write it down.
So it's even, right?
The whole thing minus 18, divide by 2 should be-- divide by 2,
and did you change the signs, [INAUDIBLE]?
What's your password?
No.
[LAUGHING] Check if I'm getting the same thing you got.
So I get 3x plus 2y plus 2z.
I divide by minus 2, right, plus 9?
Did you both get the same thing?
STUDENT: [INAUDIBLE].
PROFESSOR: You didn't?
Well, I'm trying to simplify all my answers
with this simplification.
So I guess of course if you enter it
like that, it's going to work.
Now, I need you guys to help me on this one.
Find the parametric form-- it's so easy,
but this is a problem session-- of the line passing
through the same point that's perpendicular
to the tangent plane.
Express your answer in the parametric form of the type
the one that you know that I don't like very much,
but I will write it down-- a2t [INAUDIBLE] b2 a3t plus b3.
This is how he wants me to write it, which I don't like.
I would've even preferred it to be in symmetric form.
It's the same.
I'll put the t, and I'm fine.
So x minus x0 over l, y minus y0 over m,
z minus z0 over n, that was the symmetric form.
I make it parametric by saying equal to t.
So what were the parametric equations?
x equals lt plus x0.
That's the normal.
y equals mt plus y0.
z equals nt plus z0.
So finally, my answer-- I'll check with [INAUDIBLE] answer
in a second-- should be take the normal from the tangent plane,
3, 2, 2, right?
2t plus whatever, this is 2t plus z
equals-- first it's 3, 3, 2, 2, 3, 2, 2, plus x0 y0 z0.
So erase the pluses and minuses.
And those should be the equations.
And I should write them down.
It's OK, you have the same.
Now you just have to take them, this, this, and that,
and put it in this form.
This is a combination problem.
Why do I say combination problem?
It's combining Chapter 11 with Chapter 9.
This was the review from Chapter 9.
Did you have trouble understanding
the radiant or the tangent planes
or anything like that, implicit form, explicit form?
Let me do an application, since I'm doing review anyway.
I'm done with the Section 11.6.
But before I want to go further, I
want to do some review of Chapter
11 sections 11.1 through 11.6.
I said something about implicit differentiation
being a headache for many of you.
One person asked me, how do you compute z sub x and/or z
sub y based on the equation x squared plus y squared plus z
squared equals 5?
And of course this is implicit differentiation.
Why implicit?
OK, because this is an implicit equation of the type F of x, y,
z equals constant.
When do we call it explicit?
When one of the variables, x or y or z,
is given explicitly in terms of the other two.
So if this would be-- well, here it's hard to pull it out.
But whether it be upper part then lower hemisphere,
z would be plus or minus.
So you have two caps, two hemispheres,
plus/minus square root 5 minus x squared minus y squared.
Well, that's two functions.
We don't like that.
We want to be able to do everything in one shot
without splitting it into two different graphs.
So how do we view z to be a function of x,
you're going to ask yourself.
You imagine inside this thing that x and y
are independent variables-- independent variables.
They can take whatever they want.
One is temperature.
One is time.
They run like crazies.
But z depends on both temperature and time, like us
unfortunately.
It's so cold outside.
I hate it.
OK, you promised me, and it came true.
Who promised me?
Matthew, I give you a brownie point for that.
Because you said last week it's going to be 80 degrees.
And it was.
So the prophecy came true.
On the other hand, it came back too bad.
And of course it's not Matthew's fault.
He didn't say what's going to happen this week.
All right, in this case, implicit differentiation
is just a philosophical thing.
It's a very important philosophical step
that you're taking-- think.
Think of z being a function of x and y.
And two, differentiate z with respect to x.
So what do you mean, differentiate
with respect to x?
By differentiating the entire equation,
both sides of an equation with respect to x.
So for you, x is the wanted variable.
y is like a constant.
z is a function of x is not hard at all.
So what is going to happen actually if you were to do it?
Theoretically, you would go like that.
If I'm going to differentiate this guy with respect to x,
what is the philosophy?
The chain rule tells me, differentiate F
with respect to the first variable, and then times dx/dx.
And you say, god, now that was silly, right?
Differentiate with respect to x.
That's the chain rule.
Plus differentiate F with respect
to the second place, second variable, and then
say, dy with respect to x.
But are dx and dy married?
Do they depend on one another?
Do they file an income tax return together?
They don't want to have anything to do with one another.
Thank god, so x and y are independent variables.
If you're taking statistics or researching
any other kind of physics, chemistry,
you know that these are called independent variables,
and this is called the dependent variable.
And then you have what's called the constraint.
In physics and engineering and mechanics, F of some variable
equals C. It's called constraint usually.
OK, so this guy is all silly.
These guys don't want to have to do anything with one another.
And then you get plus.
Finally, dF with respect to the third place,
and then that third place, z, is occupied by a function that's
a function of x.
So you go, dz/dx.
Why del and not d?
Because poor z is a function of two variables, x and y.
So you cannot say, dz/dx.
You have to say del z dx, equals 0.
Thank god, I got to the end where I wanted to get.
Now, if I want to see what's going on, it's a piece of cake.
That's 1.
And I get that Mr. z sub x, which other people write dz/dx,
but I don't, because I don't like it-- I keep mixing x.
Equals-- how do I pull this guy out?
How do I substitute for that?
I get Mr. First Fellow here to the other side.
He's going to pick up a minus at whatever d
I have divided by-- so this guy divided by this guy.
STUDENT: What happened to dF/dy, dy/dx?
PROFESSOR: So again, that is a very good thing.
So dF/dy was behaving.
He was nice.
But when we got to dy with respect to dx, y said,
I'm not married to dx.
I have nothing to do with dx.
I'm independent from this.
So dy/dx is 0.
And so this guy disappears. dx/dx is 1.
Duh, that's a piece of cake.
So I'm done.
This is actually a formula that looks sort of easy.
But there is a lot hidden behind it.
This is the implicit function theorem.
where you of course assume that these partial derivatives
exist, are continuous, everything is nice.
It's a beautiful result. People actually get
to learn it only when they are big, I mean big mathematically,
mature, in graduate school, first or second year
of graduate school.
We call that intermediate analysis
or advanced-- very advanced-- calculus.
Because this calculus is advanced enough,
but I'm talking about graduate level calculus.
And this is the so-called implicit function theorem.
So if you will ever be even not necessarily a graduate student
in mathematics, but a graduate student in physics or something
related to pure science, remember this result. So let me
see what's going to happen in practice.
In practice, do we have to learn this?
No, in practice we can build everything from scratch, again,
just the way we did it with the formula.
So for the example I gave you, it
should be a piece of cake to do the differentiation.
But I'm going to step by step.
Step one, think.
You have to think.
If you don't think, you cannot do math.
So you have x squared plus y squared,
the independent guys, and z, who is married to both of them.
Or maybe z is the baby.
These are two spouses that are independent from one another.
And z is their baby.
Because he depends on both of them.
So you thought you had a different approach
to the problem, different vision of what's going on.
Now finally, step two, differentiate with respect
to one only, x only.
You could of course do the same process with respect to y.
And in some of the final exam problems,
we are asking, compute z sub x and z sub y.
The secret is that-- maybe I shouldn't
talk too much again-- when I grade those finals, if you
do z sub x, I give you 100%.
Because z sub y is the same.
So I really don't care.
Sometimes there are so many things
to do that all I care is, did he or she cover
the essential work?
So with respect to x, x squared differentiated with respect
to x-- 2x.
Good first step, now, y squared differentiated with respect
to x.
0-- am I going to write 0?
Yes, because I'm silly.
But I don't have to.
2 times z of x, y.
2 jumps down, z of x, y-- I'm not done with the chain rule.
STUDENT: z sub x.
PROFESSOR: It's z sub x, very good.
This is dz/dx.
I'm not going to hide it completely like that.
That is the same thing.
y prime is 0, thank god.
So you say, if I were to keep in mind that that's
the derivative of big F with respect to x,
I could plug in everything in here.
I could plug in the formula.
But why memorize the formula and plug it
in when you can do everything from scratch all over again?
Math is not about memorization.
If you are good, for example, some people here-- I'm
not going to name them-- are in sciences that involve
a lot of memorization.
More power to them.
I was not very good at that.
So I'm going to go ahead and write z sub x pulled down
minus 2x divided by 2z.
I'm too lazy to remind you that z is the baby,
and he depends on his parents x and y.
I'm not going to write that.
And that's the answer.
So you have minus x/z.
So for example, if somebody says, compute z sub x
at the point on the sphere, that is 0, root 5, and 0,
what do you have to do?
You have to say, z sub x equals--
and now I'm asking you something that is minus 0/0.
Assuming that the expressions, the derivatives, are defined
and the denominator one is different from 0-- so
whenever you do the implicit function theorem,
you can apply with the condition that you are away
from points where derivative of F with respect to z are 0.
So this is a problem that's not well posed.
So to give you a well-posed problem, what
do I need to do on the final?
I have to say the same-- 2, 1, and 0.
STUDENT: z can't be 0.
PROFESSOR: No, I know.
So I go, z is 0 is too easy.
Let's have y to be 0.
STUDENT: 2, 0, 1.
PROFESSOR: Very good, x equals 2, z equals 1, excellent.
So z sub x at the point 2, 0, 1 will
be by the implicit function theorem minus 2/1
equals negative.
You see, that's a slope in a certain direction
if you were to look at z with respect to x in the plane x, z.
OK, what else?
Nothing-- that was review of chain rule and stuff.
And you have to review chain rule.
Make yourself a note.
Before the midterm, I have to memorize the chain rule.
Yes, sir.
STUDENT: [INAUDIBLE].
PROFESSOR: I will do that either in the review session today
or in the review for the midterm, OK?
And I'm thinking about that.
In March, I want to dedicate at least 10 days
for the review for the midterm.
Yes, sir.
STUDENT: When is the midterm?
PROFESSOR: The midterm is on the 2nd of April.
Several people asked me-- OK, I forgot about that.
I have to tell you guys.
Several people asked me questions
by email about the midterm.
So the midterm-- write down for yourselves-- will
be over the following chapters.
Chapter 10, no Chapter 9.
Chapter 9 is [INAUDIBLE].
Chapter 11 all, Chapter 10 only what
we have required-- 10.1, 10.2, and 10.3-- and Chapter 12,
all but Section 12.6.
Because I see that some of you study ahead of time.
More power to you.
You know what to read.
Skip Section 12.6.
And I'm planning to not give you anything after Chapter 13
on the midterm.
But of course, Chapter 13 will be on the final emphasized
in at least six problems out of the 15 problems
you'll have on the final, all right?
We still have plenty of time.
Chapter 9, guys, you were concerned about it.
It's some sort of embedded, you see?
Wherever you go, wherever you turn,
you bump into some parametric equations of a line
or bump into a tangent line.
That's the dot product that you dealt with, delta F dot N.
So it's like an obsession, repetitive review of Chapter 9
at ever step.
Vector spaces are very important.
Vectors in general are very important.
I'm going to move onto 11.7 right now.
We'll take a break.
Why don't we take a short break now, five minutes.
And then we have to go on until 2:50.
So practically we have one more hour.
Take a break, eat, drink something.
I don't want a big break.
Because then a big break we'll just fall asleep.
I'm tired as well.
So we have to keep going.
[BACKGROUND CHATTER]
PROFESSOR: All right.
I will start with a little bit of a review
of some friend of yours.
And since we are in Texas, of course
this counts as an obsession.
This is going to be extrema of functions of several variables.
Do I draw better lately?
I think I do.
That's why I stopped drinking coffee.
I'm drinking white tea.
It's good for you.
All right.
White tea.
For some reason, the black tea was giving me
the shaking and all that.
Too much black tea.
I don't know, maybe it has less caffeine.
Jasmine is good, green, or white.
STUDENT: I think green has less [INAUDIBLE].
PROFESSOR: OK.
So above this saddle is a function
of two variables-- you know a lot already,
but I'm asking you to compute the partial derivatives
and the gradient.
And you're going to jump on it and say
I'm doing [INAUDIBLE] anyway.
So I've got 2x, and this is minus 2y.
If I want to ask you the differential
on the final or midterm, you will say
that f sub xdx plus x of ygy.
Everybody knows that.
Don't break my heart.
Don't say 2x minus y, because I'll never recover.
Every time I see that, I die 100 deaths.
So don't forget about the x and the y,
which are the important guys of infinitesimal elements.
This is a 1 form.
In mathematics, any combination of a dx and dy
in a linear combination in the 1 form.
It's a consecrated terminology.
But I'm not asking you about the differential.
I'm asking you about the gradient.
All righty, and that is a f sub xi plus f sub yk, which
is exactly 2xi minus 2yj.
And you say all right, but I want to take a look,
I always have started with examples.
Hopefully they are good.
Let's look at the tangent vectors to the surface.
We discussed about the notion of tangent vector
before, remember, when we had r sub u and r sub
v form the parametrization.
Now look at the tangent vectors for this graph
along the x direction going this way,
and along the y direction going this way.
We see that both of them are horizontal at the origin.
And that's a beautiful thing.
And so this origin is a so-called critical point.
Critical point for a differentiable function.
Z equals f of xy is a point in plane x0i0
where the partial derivatives vanish.
And according to the book, and many books, all don't exist.
Well I don't like that.
Even our book says if you have a function in calc 1--
let's say b equal g of u, critical point.
Do you remember what a critical point was?
U0, in calc 1 we said either a point where g prime of u was 0,
or g prime of u 0 doesn't exist.
Although u is 0 is in the domain.
I don't like that.
You say wait a minute, why don't you like that?
I don't like that for many reasons practically.
If you have the absolute value function,
you'll say yeah, yeah, but look, I considered the corner
to be a point of non-differentiability,
but it's still an extreme value, a critical point.
According to our book in Calculus 1, yeah.
We extended this definition to ugly points,
points where you don't have a [? pick ?] or a value
or an inflection, but you have something ugly like
a [? cusp, ?] a corner, the ugliness.
I don't like that kind of ugliness,
because I want to have more information there.
I maybe even have a point with a bigger problem than that.
First of all, when I say critical point,
I have to assume the point is in the domain of the function.
But then what kind of ugliness I can have there?
I don't even want to think about it.
So in the context of my class-- in context
of my class-- calc 3 honors, I will denote a critical point.
Is the x0y0 such that f sub x at x0y0 is 0.
One slope is 0, the other slope is 0.
f sub y is x0y0, of course.
And no other are the points.
What am I going to call the [INAUDIBLE] points where
derivatives don't exist?
I simply say I have a singularity.
I have a singularity.
What type of singularity we can discuss in an advanced calculus
setting.
If you're math majors, you're going to have the chance
to discuss that later on.
So remember that I would prefer both in the context
of calculus 1 and calculus 3 to say critical value
is where the derivative becomes zero.
Not undefined, plus, minus, infinity, or something
really crazy, one on the left, one on the right.
So I don't want to have any kind of complications.
Now you may say, but I thought that since you
have those slopes both zero, that
means that the tangent plane at the point is horizontal.
And that's exactly what it is.
I agree with you.
If somebody would draw the tangent plane to the surface,
S-- S is for surface, but it's funny,
S is also coming from saddles.
So that's a saddle point, saddle surface.
Origin is so-called saddle point.
We don't know yet why.
The tangent plane at 0, at the origin,
will be 0.0 in this case.
Why?
Well, it's easy to see.
z minus 0 equals f sub x, x minus x0 plus f sub y,
y minus y0.
But this is 0 and that's 0, so z equals zero.
So thank you very much.
Poor horse.
I can take a horizontal plane, imaginary plane
and make it be tangent to the saddle in all directions
at the point in the middle.
All right.
STUDENT: So you're saying [? the critical ?] point
is where both--
PROFESSOR: Where both partial derivatives vanish.
They have to both vanish.
In case of calculus 1, of course there
is only one derivative that vanishes at that point.
What if I were in-- now, you see,
the more you ask me questions, the more I think
And it's a dangerous thing.
What if I had z equals f of x1, x2, x3, xn?
Critical point would be where all the partial derivatives
will be zero.
And then the situation becomes more complicated,
but it's doable.
The other is the classification of special points.
Classification of critical points
based on second partial derivatives.
The objects you want to study in this case are several.
One of the most important ones is the so-called discriminant.
What is the discriminant?
You haven't talked about discriminants since a long time
ago.
And there is a relationship between discriminant
in high school algebra and discriminant in calculus 3.
The discriminant the way we define
it is D, or delta-- some people denote it
like this, some people by delta--
and that is the following.
This is the determinant.
f sub xx, f sub xy, f sub yx, f sub yy,
computed at the point p0, which is critical.
So p0 first has to satisfy those two equations,
and then I'm going to have to compute the [INAUDIBLE]
at that point.
But you say wait a minute, Magdalena,
what the heck is this?
Well this is the second partial with respect
of x, one after the other, second partial with respect
to y, one after the other.
These guys are equal.
Remember that there was a German mathematician
whose name was Schwartz, the black cavalier, the black man.
Schwartz means black in German.
And he came up with this theorem that it
doesn't matter in which order you differentiate, f sub xy
or f sub yx is the same thing as long as the function is smooth.
So I'm very happy about that.
Now there are these other guys, A, B,
C. It's very easy to remember, it's from the song
that you all learned in kindergarten.
Once you know your ABC, you come back to the discriminant.
So f sub xx at the point p0, f sub xy at the point p0,
and f sub yy at the point p0.
Second partial with respect to x, second partial with respect
to x and y, mixed one, mixed derivative, and second
partial with respect to y.
You have to plug in the values for the p0 will be x0, y0.
The critical point you got from what?
From solving this system.
So you got x0y0 by solving that system.
Come back, plug in, compute those, get ABC as numbers.
And who is D going to be based on ABC?
According to the diagram that I drew, it's easy for you guys
to see that A and B and C are what?
Related to D. So D will simply be A, B, B, and C,
computed at the point p0.
So it's going to be now-- now that
will remind you of something.
AC minus B-squared.
OK?
When we had the quadratic formula in school--
I'm not going to write it.
I'm going to write it here.
So what was the quadratic formula?
ax-squared plus bx plus c equals 0.
That was algebra.
Baby algebra.
What do we call that?
High school algebra?
x12 plus minus b plus minus square root of b-squared minus
4ac divided by 2a.
Now don't don;t know what kind of professors you had.
But I had a teacher when I was in high school.
Every time she taught me something and I did not
absorb it, she was all over me.
She was preparing me for some math competitions,
and she taught me a trick.
She said look, Magdalena, pay attention.
If b would be an even number-- take b to be 2b prime,
2-- give me another letter.
2 big B. Right?
Then, the quadratic formula would be easier to use.
Because in that case, you get x1 2 equals minus-- b is 2b.
So you have just 2b like that.
Plus minus square root 4b squared
minus 4ac divided by 2a.
She explained this to me once and then
she expected me to remember it for the rest of my life.
And then she said minus big B plus minus square root of bb
squared minus ac.
Do you see why?
It's because you pull out the factor of 4, square root of 4
is 2.
2, 2, and 2 simplify.
And then she gave me to solve problems.
STUDENT: What about the a?
STUDENT: How about the a?
STUDENT: Because you divide it by [INAUDIBLE].
PROFESSOR: Divided by.
I forgot to write it down.
Because I didn't have space.
I said, I'm not going to bend and doodle.
So when you have x-squared plus 2x-- let's say minus 3.
And she gave me that.
And I said OK, let me do it.
Let me do it.
x1 2 minus 2 plus minus square root b-squared minus 4ac,
which is 12, divided by 2.
And she started screaming.
And she started screaming big time.
Do you know why?
She said, I just told you the half formula.
By half formula, I mean she meant this one.
So when-- and I said OK, OK, the half formula.
But then for maybe another seven years,
I did this with the formula-- with the formula
that everybody knows.
And at the end, I would remember I
could have done the half formula,
but I didn't do it because I'm in a routine.
So the way she wanted me to do this was what?
Who is the half of 2?
1.
So put minus 1 plus minus square root of big B
is 1-squared minus a times c, which
is plus 3, divided by 1 divided by nobody.
This way you don't have to simplify it further,
and you do it faster.
So you get minus 1 plus minus root 4, which is minus 5 and 3.
But of course, you could have done this by factoring.
So you could have said wait a minute.
Two numbers that multiply-- um--
STUDENT: [INAUDIBLE] square root.
PROFESSOR: I didn't do right.
So it's 4--
STUDENT: [INAUDIBLE].
PROFESSOR: Yeah.
So you get x plus 5 times--
STUDENT: It's x minus 1--
[INTERPOSING VOICES]
PROFESSOR: Oh, I think I--
STUDENT: Square root.
PROFESSOR: Square root.
I'm sorry, guys.
OK.
Thank you for that.
1 and minus 3.
So x plus 3 times x minus 1, which
is the same-- the exact same as x-squared plus 2x minus 3
equals.
All right?
So just the way she insisted that I learn the half formula.
I'm not insisting that you learn the half formula, god forbid.
But see here there is some more symmetry.
The four doesn't appear anymore.
b-squared minus 4ac appeared here, but here it doesn't.
Here you're going to have b-squared minus ac.
There is a reason.
This comes from a discriminant just like that.
And this is why I told you the whole secret
about the half quadratic formula.
Not because I wanted you to know about it,
but because I wanted you to see that there is a pattern here.
You have-- for the half formula, you
have plus minus square root of a new type of discriminant.
People even call this discriminant b-squared
minus 4ac.
b-squared minus ac.
So for us, it is ac minus b-squared.
It's just the opposite of that discriminant you have.
Now depending on the sign of this discriminant,
you can go ahead and classify the critical values you have.
So classification is the following.
Classification of special critical points.
If delta at p0 is negative, then p0 is a saddle point.
If delta at p0 is 0, nothing can be said yet
about the nature of the point.
So I make a face, a sad face.
If delta at p0 is greater than 0, then I have to ramify again.
And I get if a is positive, it's going to look like a smile.
Forget about this side.
It's going to look like a smile.
So it's going to be a valley point, what do we call that?
Relative minimum, or valley point.
Don't say valley point on the exam, OK?
Relative minimum.
If a is less than 0 at the point,
then locally the surface will look
like I have a peak-- a relative maximum Peaks and valleys.
Just the way you remember them in Calc 1.
Now it's a little bit more complicated
because the functions have two variables.
But some of the patterns can be recognized.
Let's go back to our original example
and say wait a minute, Magdalena.
You just gave us a saddle, but we didn't
do the whole classification.
Yes, we didn't, because I didn't go over the next steps.
z equals x-squared minus y-squared.
Again, we computed the gradient.
We computed the partial derivatives.
And then what was that one in finding the critical points?
So f sub x equals 0, f sub y equals 0.
Solve for x and y.
And that's good, because that's going
to give me a lot of information, a lot that
will give me exactly where the critical points may be.
So that is if and only if I need to solve 2x equals 0 minus 2y
equals 0.
Is this system hard to solve?
No.
That's exactly why I picked it.
Because it's easy to solve.
The only solution is x0 equals y0 equals 0.
So the origin-- that's exactly where
you put your butt on the saddle when you ride the horse.
That is the only critical point you have.
The only one.
Now if we want to classify that, what kind of-- is it a valley?
No.
It looks like a valley in the direction
of the axis of the horse, Because the saddle's
going to look like that.
This is the horse.
That's the head of the horse I'm petting.
And this is the tail of the horse.
So in this direction, the saddle will be shaped like a parabola,
like a valley.
But in the perpendicular direction,
it's going to be shaped going down,
like a parabola going down.
So it's neither a valley nor a peak.
It's a valley in one direction, and a peak
in another direction.
And that's the saddle point.
So say it again.
What is that?
It looks like a valley in one principle direction
and the peak in the other principle direction.
And then that's going to be a saddle point.
Indeed, how do we figure this out by the method I provided?
Well, who is A?
A is f sub xx at the point.
2x goes primed one time.
f sub x was 2x.
f sub y was 2y.
f sub xx is 2.
f sub xB is f sub xy.
What is that?
0.
Good, that makes my life easier.
C equals f sub yy.
What is that?
2.
OK, this is looking beautiful.
Because I don't have to plug in any values.
The D is there for me to see it.
And it's going to consist of the determinant having the elements
2, 0, 0, 2-- minus 2, minus 2.
I'm sorry, guys, I missed here the minus.
And it cost me my life-- 2x and minus 2y, and here minus 2.
STUDENT: It didn't cost you your life,
because you caught it before you were done with the problem.
PROFESSOR: I caught it up there.
I'm taking the final exam.
I still get 100%, because I caught it up
at the last minute.
So 2, 0, 0, minus 2-- I knew that I
had to get something negative.
So I said, for god's sake, I need to get a saddle point.
That's why it's the horse in the saddle.
So I knew I should get minus 4, negative.
All right, so the only thing I have
to say as a final answer is the only
critical point of this surface that I'm too lazy to write
about-- don't write that.
So the only critical point on the surface z
equals x squared minus y squared will
be at the origin O of corner 0, 0, 0
where the discriminant being negative
indicates it's going to be a saddle point.
And that's it-- nothing else.
You don't need more.
But there are more examples.
Because life is hard.
And I'm going to give you another example.
Well, OK, this one.
Suppose we have the surface-- that's
still going to be very easy.
But I want to make the first examples easy.
I have a reason why.
This is a function of two variables, right?
It's still a polynomial in two variables of order 2.
And how do I solve for the classification of the extrema?
I'm looking for local extrema, not absolute-- local extrema.
I'm not constrained.
I'm saying, what do you mean, no constraint?
Constrained would have been, let's say that x and y
are in the unit disc.
Or let's say x and y are on the circle x squared
plus y squared equals 1.
That would be a constraint.
But they're not constrained about anything.
x and y are real numbers.
They can take the whole plane as a domain.
So I get f sub x equals 0, f sub y equals 0, solve for x and y,
get the critical values.
I get very nice 2x.
I have to pay attention.
Because now this is not so easy anymore--
plus prime with respect to x, 2y, prime with respect to x, 0,
prime with respect to x, plus 3, prime of this, OK, equals 0.
f sub y-- 0 plus prime with respect
to y, 2x, plus prime with respect to y, 2y,
plus nothing, prime with respect to y equals 0.
And now you have to be very smart.
Well, you have to be perceptive and tell me what I got.
What is this that we mean?
Look at this system.
It looks like crazy.
STUDENT: [INAUDIBLE] the origin or-- because can't you
just subtract it down?
PROFESSOR: Is this possible?
And what does this mean?
What do we call such a system?
Inconsistent system-- we call it inconsistent.
How can I make this problem to be possible,
to have some critical points?
STUDENT: If you add 3x.
PROFESSOR: How about that, just remove the 3x
and see what's going to happen.
Oh, in that case, I have something
that's over-determined, right?
I have something that tells me the same thing.
So I'm priming with respect to x.
I get that.
I'm priming with respect to y.
I get this.
I get 0.
So I don't even need the second equation.
And that means the critical point
is any point of the form-- shall I
put a Greek letter alpha minus alpha or lambda minus lambda?
What shall I do?
So any point that is situated on the second bisector,
I mean the x, y plane.
And this is the x, and this is the y.
And I say, what does it mean, x plus y equals 0?
Not this line-- don't draw it.
That is x equals y.
The other one, called the second bisector-- y
equals negative x, so not this one, the diagonal,
but the diagonal that's on the corridor, this one.
All right, so any point of the form alpha minus alpha,
here's the critical point.
The question is, how am I going to get to the classification
for such points?
Can anybody help me?
So step two--
STUDENT: Solve the equation.
STUDENT: Solve alpha for one of the two variables first.
PROFESSOR: Take alpha minus alpha-- could be anything.
And then I'll say, f sub-- this is f sub x.
And this is f sub y.
What is f sub x?
f sub xx, I'm sorry.
STUDENT: [INAUDIBLE].
PROFESSOR: Huh?
2.
OK, are you with me?
So you know what it is.
f sub xy equals?
STUDENT: 2.
PROFESSOR: 2.
f sub yy equals?
STUDENT: 2
PROFESSOR: 2-- that's the mystery man.
The book doesn't give this example,
and it drives me crazy.
And I wanted to give you some bad example where
the classification doesn't work.
Because we always cook up nice examples for you
and claim everything is beautiful.
Life is not always beautiful.
So you get 0.
In that case, nothing can be said with this classification.
I make a face, sad face.
So what do I hope?
To get to Maple or MATLAB and be able to draw that, or a TI-92
if my mother would give me $200 and some.
I told her.
She asked me what to buy for my birthday.
I have a TI-83 or something.
And it was cheap.
I bought it on eBay, and then I stopped using it.
And then I saw this TI-92 that can draw surfaces
in three dimensions.
And I said, this is like MATLAB.
You just carry it in your pocket.
It's only a little bit too expensive.
All right, how about another kind?
Look at this one.
You cannot tell with the naked eye.
But you can go ahead and do this step one
looking for critical values.
So the system, f sub x will be 6x plus 2y equals 0.
f sub y will be-- who's going to tell me?
2x plus 2y equals 0.
Now, by elimination or by substitution or by anything
I want, I subtract the second from the first.
What do I get?
I get 4x equals 0.
And that gives me the only possibility is x0 equals 0.
And then I say, OK, if my only one is 0, then y is 0.
0 is 0.
So I only have one critical point, which is the origin.
Now, do I know, what am I going to get?
Not unless I'm a genius and I can
see two steps ahead of time.
I would need to do ABC quickly in my head.
Some of you are able, thank god.
But some of you, like me, are not.
So I have to take a few seconds to see what's going on.
A-- f sub xx at the point is 0.
B-- f sub xy.
C-- f sub yy.
What do we do?
We get 6.
Are we happy about it?
We don't know yet, to be happy or not.
f sub xy or f sub yx, you see, Mr. Schwarz
is now happy that he proved to you
that it doesn't matter which order you're
taking for a polynomial that's a smooth function.
You always have the same.
And finally, C is 2.
And you are ready to do the D. And I could smell that D,
but I didn't want to say anything.
6, 2, 2, and 2-- is that a nice thing?
Yeah, we haven't encountered this example yet.
Because according to the classification,
this is greater than 0.
Does it really matter what value it is?
No, it only matters that it is positive.
And if it's positive, that means I can move on with my life
and look at the classification.
From this point where delta or v is positive,
I'm going to get a ramification into separate cases.
And who is going to tell me next what to do?
Look at A. Oh, by the way, talking
about the quadratic formula from school,
from kindergarten, when we computed
the-- I'll use the general one, minus b plus minus square root
of b squared minus 4ac over 2a.
We were afraid of some special cases
when we were looking at that.
Especially when delta was negative,
that was really imaginary and so on.
But one thing we remember from ninth grade--
was this ninth grade or eighth grade?
The parabola opens up when a is positive.
Just the same way, something opens up when A, big A,
is positive here.
Then you have opening up.
When big A is negative, then you have opening down.
So remember-- I'm going to make smile here so you remember.
So I have it like that.
So I suspect that it's going to look
like a surface of some sort that maybe is not
surface of revolution.
You should tell me what it is.
You should think about this and do the cross sections with z
constant and tell me what surface that is.
But in any case, what do I care?
I care that I'm looking at the origin.
And this is where my special point is.
That's going to be the value point.
How do I know?
Because A, which is 6, is positive.
At this point, I know what I'm left with.
I know that my surface is going to look like a valley.
So how do I know again?
I'm not going to draw it.
But it's going to look something like that.
At the origin, this is going to be 7.
Are you guys with me?
And it's going to open up.
And so you should not attempt intersecting with z equals 5
or z equals 1.
Because you're not going to get anything.
But if you intersect, for example, at z
equals 9, what are you going to get?
If you intersect at z equals 9, you
get 3x plus 2xy plus y squared equals 2.
And what is that?
It's a rotated form of an ellipse.
It's hard to see, because it's missing [INAUDIBLE].
But this is exactly what discriminant is saying.
So this is going to be an x.
Good, so I know what I'm going to get.
What do you have to say on the midterm
or on the final about this problem?
STUDENT: The point is--
PROFESSOR: The point is the origin.
I classified it.
I got delta positive.
I got A positive.
So it's a valley.
It's a relative minimum.
And that's it.
I have a relative min at the point P of coordinates 0, 0,
and 7.
And that's the valley.
Yes, sir.
STUDENT: Why is it A that determines
whether it's a relative min or a relative max?
PROFESSOR: It's a whole story.
You can prove it.
I don't remember if we proved this in the book or not.
But it can be proved, so the fact
that it has to do with concavity and convexity.
When you had a second derivative, let's say,
what's the equivalent of the Calculus I
notion that you know about?
In Calculus I, you had functions of one variable,
and life was so easy like that.
And f prime positive meant that the function increased.
And f prime negative meant that the function decreased.
And f double prime was just like your-- you sense
that the second partials must have something
to do with it, especially the first one with respect to x.
If you were in plane, and you have
f double prime with respect to x, when was this a valley?
When you had the smile.
When did you have a smile?
When f double prime was positive, you have concave up.
When f double prime was negative,
you have concave down.
Remember, guys?
So you have a smile or a frown.
This is how we know.
For the same reason that would take about two pages
to write down the proof, you have a smile for A positive.
And the smile means actually in all directions
you have a smile locally around the origin.
OK, look in the book.
I'm not sure how much should we do.
Do we give a sketch of a proof, or we give the entire proof?
But more likely, a sketch.
Yes.
STUDENT: I asked the slightly wrong question,
but I answered it myself.
I wanted to ask, why is it dependent on A and not on C?
PROFESSOR: Not on C.
STUDENT: But then I realized that it is dependent on C
as well, because if A is positive,
then C must be positive.
PROFESSOR: Yes, yes, it is dependent on both.
STUDENT: OK, there we go.
That was my question.
PROFESSOR: So guys, remember.
Imagine what happens when you had no B, B was 0.
Then the matrix is diagonalizable.
And here you have A and C. And Alex
says, why would A be more important than C?
It's not.
But practically, if A is positive and C is negative,
that means these are the principal directions in which
one bends like a valley up and one bends like a peak down.
So this is what happens in the direction of x, f double prime
in the direction of x, kind of.
And this is in the direction of y.
So this is f double prime in the direction of y,
which we don't denote like that.
We call it f sub xx and f sub yy, which is A and C.
So A positive, A being 1 and C being negative 2,
means a valley here, means the valley meets the horse.
Look, I'm drawing the tail of the horse.
He's a little bit fat, this horse.
And that's his mane, his eye.
I'm just taking a break.
STUDENT: That's a pretty good drawing.
PROFESSOR: It looks more like a dog or a plush horse
or something.
So A equals 1, and C equals minus 2.
But if it were diagonalizable, and A would be 1
and C would be 7, both of them positive in any case,
then you'll have valley and valley, an x direction valley
and y direction valley.
So it has to be a valley everywhere.
These are the principal directions that I have 1 and 2.
But then the ultimate case, what happens
when A is negative and-- hmm, OK,
then either you have them both one positive, one negative,
or you have plus, plus and minus, minus.
And then you have this as your surface, right?
Which one is the x direction?
That's the y direction.
That's the second one.
The x direction is that.
In the x direction, you have a frown.
So f sub yy is negative.
In the y direction, you also have a frown.
So both of them are negative.
So you have a relative max.
Yes, sir, Matthew, tell me.
STUDENT: So isn't it possible to have both A and C positive,
but then yet still not be more positive than B squared?
PROFESSOR: No, because there's a theorem that--
STUDENT: I was just wondering like numbers-wise.
PROFESSOR: You have this matrix.
And there is a theorem that shows you that you can actually
diagonalize this matrix.
You'll learn your linear algebra [INAUDIBLE].
STUDENT: It makes sense, because when
you were saying A is this way, and that way there's no way
you could have 2 come up, and then yet,
not be a-- you know what I'm saying?
Because then they'd be less than 0.
PROFESSOR: You can if you don't have or the 2.
That's an excellent question.
If I would have x to the 4 y to the 4
added together, like Ax to the 4 plus By to the 4 plus
something, then I have the so-called monkey saddle.
That's so funny.
You can have something that looks like that.
So in your direction, you can have this.
Then I've reached two equal peaks
in the x and the y direction.
But in the between, I also went down.
So depending on a higher degree symmetric polynomial,
you can have a monkey saddle.
And then it's not just like you can predict what's
going to happen in between.
In between, if I go up, if I go valley
in the x direction and valley in the y direction,
I know that's going to be a valley everywhere-- no.
If a polynomial is high in order,
it can go down, valley, and up again,
and monkey saddle it looks like.
Guys, you have dealt with it when you
went to Luna Park or Joyland.
It's one of those things that look like-- I'm trying.
I cannot draw.
STUDENT: It sounds more like an octopus.
PROFESSOR: Like an octopus.
And one of those things-- exactly--
that are shaped so that they are undulated,
in some directions are going up, in some directions
are going down.
STUDENT: Like an egg carton, almost?
PROFESSOR: Yeah, really undulated.
Imagine even a surface made of metal that's undulated
and rotating at the same time.
They have some of those in Disney World.
Have you been to Orlando?
STUDENT: I was there last semester.
PROFESSOR: But you didn't take me with you, which is bad.
Because that's one of my favorite places.
STUDENT: I was just trying to think of what you were talking
about so I could visualize it.
PROFESSOR: Maybe we could make a proposal
to teach Calculus III at Disney World
so that we could have examples of motion and surfaces
all around and study the motion of all sorts of gadgets,
velocity and trajectory.
Last night I couldn't sleep until 1:00, and I was thinking,
I gave examples of the winter sports
like bobsled and all sorts of skiing and so on.
But I never thought about a screw curve
with curvature and torsion that is based on the roller coaster.
And the roller coaster is actually
the best place to study the [INAUDIBLE], the velocity,
the tangent unit, the normal, the bi-normal.
And when you have in a plane the roller coaster
goes like that, like this and like that, like in a plane,
you have nothing but bending, which means curvature.
But then when the roller coaster goes away from the plane,
you have the torsion.
And that makes you sick really to the stomach.
So we would have to experience that to understand
Calculus III better.
So our next proposal is we ask the administration
instead of study abroad courses, the domestic study
at Disney World for Calc III.
It's Applied Calculus III.
OK, something else that I want you to do-- I
had prepared an example.
This is an absolute extrema.
And you say, what the heck are the absolute extrema?
Because she only talked to us about relative maximum
and relative minimum.
And she never said anything about absolute extrema.
And that will be the table.
And these will be the extrema.
I want to refresh your memory first just a little bit.
This will be the last example.
Because it's actually two examples in one.
And what if you have, let's say, f of x equals e to the minus
x squared over the interval minus 1, 1?
You are in Calc I. You will build a time
machine from Disney World.
And we went back in time when you actually took Calc I.
And you struggled with this at first.
But then you loved it so much that you
said, oh, that's my favorite problem on the final.
They asked us for two things-- relative extrema, min or max,
min/max theory, and they say absolute.
But for the absolute, your teacher said, attention,
you have to know how to get to the absolute.
You are constrained to be on the segment minus 1, 1.
You see, the fact that they introduced
this extra constraint and they don't
let you move with x on the whole real line is a big headache.
Why is that a big headache?
Your life would be much easier if it were just e
to the negative x squared.
Because in that case, you say, OK,
f prime of x equals minus 2xe to the minus x squared.
Piece of cake.
x0 is 0.
That's the only critical point.
And I want to study what kind of critical point that is.
So I have to do f double prime of x.
And if I don't know the product rule, I'm in trouble.
And I go, let's say, minus 2 times
e to the negative x squared from prime of this
and this non-prime, plus minus 2x un-prime times e
to the minus x squared times minus 2x again.
So it's a headache.
I pull out an e to the minus x squared.
And I have 4x squared-- 4x squared-- minus 2.
But you say, but wait a minute, Magdalena,
I'm not going to compute the inflection points.
The inflection points will be x equals
plus/minus 1 over root 2.
I only care about the critical point.
And the only critical point I have is at 0.
Compute f double prime of 0 to see if it's a smile or a frown.
And you do it.
And you plug in, and you say, e to the 0 is 1, 0 minus 2.
So you get a negative.
Do you care what it is?
No, but you care it's negative.
So at 0, so you draw, and you know
at 0 you're going to have some sort of a what?
Relative max.
Where?
At 0, and when you plug 0 again, 1.
So you draw a table.
And you say, relative max at 0, 1.
And then you're not done.
Because you say, wait a minute, I
am to study my function in Calc I at minus 1 and 1.
It's like you have a continuous picture,
and you chop, take scissors, and cut and cut at the extrema.
And there you can get additional points
where you can get a relative max or relative min.
Absolute max or min will be the lowest of all the values
and the highest of all the values.
OK, so I get at the point minus 1-- how shall I put here?
x equals minus 1.
What do I get for y?
And for plus 1, what do I get for y?
This is the question.
I plug it in, and I get minus minus 1/e.
And then when I have 1, what do I get?
1/e again.
So do I have a relative min here?
No, but I have an absolute something.
And what do I have here?
Here I have an absolute max.
So how do we check the absolute maxima and absolute minima?
We look for critical points.
We get many of them, finitely many of them.
We compute all the values of z for them,
all the function values.
And then we look at the end points,
and we compare all three of them, all the three
values in the end.
So in the end, you compare 1/e to 1/e to 1.
And that's all you can get.
So the lowest in one will be the highest in one.
Good.
In Calculus III, it's more complicated.
But it's not much more complicated.
Let's see what's going to happen.
You can have a critical point inside.
We are just praying we don't have too many.
So how do I get to step one?
Critical point means f sub x equals e to the x squared
minus y squared times 2x.
All righty, it looks good.
f sub y equals e to the x squared minus y squared times
minus y.
I'm full of hope.
Because I only have one critical point, thank god.
Origin is my only critical point.
I don't know what that is going to give me.
But it can give me a relative max
or relative min or a saddle.
I don't know what it's going to be.
Who tells me what that is going to be?
Well, did I do this further?
I did it further and a little bit lazy.
But I'm not asking the nature of the point.
So for the time being, I only want to see what happens at 0,
0.
So I have 1.
So in my table I will put, OK, this is x, y, and this is z.
For 0, 0, I'm interested.
Because that's the critical point inside the domain.
The domain will be the unities.
And inside the origin, something interesting happens.
I get a 1.
And I hope that's going to be my absolute something.
But I cannot be sure.
Why?
There may be other values coming from the boundary.
And just like in Calculus I, the only guys
that can give you other absolute max or min,
they can come from the boundary, nothing else.
Nowhere else in the interior of the disc am I going to look.
I'm not interested.
I'm only interested in x squared plus y squared equals 1.
This is where something can happen,
nothing else interesting in the inside, just like in Calc I.
So to take x squared plus y squared equals 1 into account,
I pull y squared, who is married to x-- the poor guy.
He's married to x.
He's dependent on x completely, y squared
equals 1 minus x squared.
And I have to push him back into the function.
So at the boundary, f becomes a function of one variable.
He becomes f of x only equals e to the x
squared minus 1 plus x squared.
Are you guys with me?
So f of x will become e to the 2x squared minus 1
along the boundary, along the circle, only here.
Now what else do I need to do?
I need to compute the critical values for this function of one
variable, just the way I did it in Calc I.
So f prime of x will give me e to the 2x squared
minus 1 times-- what comes down from the chain rule?
STUDENT: 4x.
PROFESSOR: 4x, so life is hard but not that hard.
Because I can get what?
I can get only x at 0 equals 0 here.
OK, so that's a critical point that comes from the boundary.
But guys, you have to pay attention.
When x is 0, how many y's can I have for that 0
on the boundary?
This is on the boundary-- on the boundary.
STUDENT: Two.
PROFESSOR: Two of them-- I can have 1,
or I can have negative 1.
There is one more tricky thing.
This is a function of one variable only.
But this stinking function is not
defined for arbitrary x real.
So I make a face again.
So I go, oh, headache.
Why?
x is constrained.
x is constrained, you see?
If you were inside the disc, x must be between minus 1 and 1.
So I have to take into account that x is not any real number,
but x is between minus 1 and 1.
Those are endpoints for this function.
And in Calc I, I learned, OK, I have
to also evaluate what happens at those endpoints.
But thank god that will exhaust my list, so I have a list.
Minus 1 for x and 1 for x-- thank god.
That will give you what y on the boundary?
When x is 1 and x is minus 1, you're
interested in what happens, maximization or minimization,
for this function at the endpoints.
But fortunately, since you are on the boundary, y must be 0.
Because that's how you got y out.
If x is plus/minus 1 on the boundary, y must be 0.
So my list contains how many interesting points?
One, two, three, four, five-- for all of them,
we need to compute, and we are done.
Of all of them, the lowest z is called absolute minimum.
And the highest z is the absolute maximum.
And we are done.
You guys need to help me, because I'm running out of gas.
So x is 0.
Y is 1.
What is z?
STUDENT: [INAUDIBLE].
PROFESSOR: e to the minus 1, you were
fast, 1/e, thank you, guys.
So when x is 0 and y is minus 1?
STUDENT: [INAUDIBLE].
PROFESSOR: Huh?
STUDENT: [INAUDIBLE].
PROFESSOR: No, no, no, it's the same.
Because x is 0. y is minus 1.
I get e to the minus 1, which is 1/e.
So far, so good-- I'm circling all the guys
that I want to compare after.
So for the final four points there, what do I have?
My final candidates could be x equals plus/minus 1
and y equals 0-- e and e.
Who's the biggest?
Who's the smallest?
STUDENT: e is the biggest.
PROFESSOR: e is the biggest, and 1/e is the smallest.
So how do I write conclusion?
Conclusion-- we have two absolute maxima
at minus 1, 0 and 1, 0.
And we have two absolute minima at 0, minus 1 and 0, 1.
OK, now I have to-- now that's like a saddle.
Can you see it with the eyes of your imagination?
It's hard to see it.
This is the disc.
And the four points, the cardinal
points-- OK, this is the disc.
We are looking at this disc from perspective.
And the five points, one is in the middle.
One is here.
Minus 1 is 0.
One is here, 1, 0.
One is here, 0, minus 1, and one here, 0, 1.
At minus 1, 0 and 1, 0 I get the maximum.
So the way it's going to be shaped would be like that.
In this direction, it will be like that.
And cut the cake here.
You see it's like that.
It's going to be like this.
OK, passing through the origin, with my hands
I'm molding the surface made of Play-Doh or something for you.
So I'm starting here, and I'm going up.
And at this points, I'm here.
Are you with me?
The same height.
In the other direction, I'm going from 0.
But I'm not so high.
I'm going only up to-- what is 1/e?
About 1/3, meh, something like that.
So I'm going to get here.
So the problem is that one will be in between.
So if you really want to see what it looks like,
we are here at 1.
We grow from 1, altitude 1, you see?
We grow from 1 to about 2.71718283 for both of these.
And from 1, in this direction I have to go down to 1/e.
So it looks like that.
I'll try to draw, OK?
Do you see the patch around the origin?
So here's e.
And here's 1/e above the sea level.
And this is 1.
And you have one just like that in the back that is the-- it
still looks like a saddle.
It is a saddle.
It's symmetric.
But it's another kind of saddle.
There are all sorts of saddles made
in Texas, different ranches, different saddles.
So that was the harder one.
The ones that I actually saw on the finals,
some of the last three or four finals,
were much easier in the sense that the table you had to draw
was much shorter than this one-- in principle,
one critical value and one max and one min point.
But you have to be prepared more, rehearse more,
so when you see the problem in the midterm,
you say, oh, well that is easier than I'm used to.
That's the idea.
OK, go home.
Send me emails by WeBWorK.
We still have time to talk about the homework if you get stuck.
And I'll see you Thursday.
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