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