TTU Math2450 Calculus3 chap. 10-11-12 review

Exam review

PROFESSOR: Questions so far?

STUDENT: [INAUDIBLE]

PROFESSOR: Yes, ma'am.

You are Megan?

STUDENT: Yes.

PROFESSOR: OK.

STUDENT: I was just wondering
if we get like a form

of [INAUDIBLE], note cards--

PROFESSOR: No, you
[INAUDIBLE] sheet whatsoever.

So I think it's better that
I review some of the formulas

today that you are
expected to know by heart,

because they are also-- they
require you know to expect you

to know the same formulas
by heart for the final

with no cheat sheet.

So the final will have
exactly the same policy,

at the end of [INAUDIBLE].

No calculator, no formula sheet,
no cheat sheet, no nothing,

but you know I'm telling
you guys this, except what

you remember.

Let me remind you
that you are expected

to know the equation
of the tangent plane.

I'm not going to give them
in chronological order,

but I think it's a good idea to
review for midterm and final,

some of the must-know formulas.

One, well, I discussed this
before but I didn't remind you,

differential of a function
of several variables.

In particular, two
variables most likely

are the examples we've worked
on a lot this semester.

Number two, the definition
and especially formula,

main formula for responding to
directional derivatives of F

at P of coordinates X 0, Y 0,
in the direction U 1 and U 2,

equals U.

Just to test you,
OK, well, I believe

you know the formula
of the differential.

But without me reminding you
what was that of two variables,

I expect you to say d F equals--

STUDENT: F [INAUDIBLE].

D X plus F Y, D Y.

PROFESSOR: How
about-- thank you--

how about the
directional derivative

of F at P in the
direction of the vector U?

You will need the formula.

Good, [INAUDIBLE].

Yeah, that's the easiest
way to remember it,

but that's not the first thing
I want you to say, right?

How did I write this?

[INAUDIBLE]

Of course, F of X-- thank
you-- and [INAUDIBLE] 0,

times Z 1 plus
derivative inspect Y,

and X 0, Y 0 times U 2.

What do we assume about it, F
C 1 on the domain [INAUDIBLE]?

Which means differential goal
with continuous derivatives.

This is what we assume
through chapter 11.

Number three, I think I
told you, but I'm not sure.

But I think I did.

Review the tangent
plane formula,

formulas-- how about both?

Well, only one is the one
I consider relative for us.

Which is Z equals F of X,
Y, will imply that at P

with on the surface,
even as a reference, we

have a tangent plane
of formula Z minus Z 0,

equals-- who does it?

OK, now you have
to remember this.

Of course, it's your midterm.

Review all of these things
by [INAUDIBLE] Thursday.

Variable S of X,
[INAUDIBLE] 0 at 0, times--

STUDENT: X minus X 0.

PROFESSOR: Thank you,
Roberto, X 0 minus X 0,

plus the same kind o thing in
a different color, because I

like to play [INAUDIBLE] orange,
S Y, X 0, Y 0, Y minus Y 0.

Don't come to the midterm-- you
better not come to the midterm,

and you get a 0 for not
knowing the formulas, right?

Now maybe you will see
on this midterm, maybe

not, maybe you'll see
it on the final-- what

happens when you don't have
the graph of a surface?

Maybe you'll have an implicit
equation, an implicit equation

where we write F of coordinates,
X, Y and Z, equals a constant.

Why is the tangent plane a P?

Tangent plane, tangent plane
in both cases should be Y.

Well, if you consider
the first formula

as a consequence
of the second one,

that would be simply
easy, because you

will have to write F of
X Y minus Z equals 0.

And there you are, the same
kind of formula in this.

So what do you write--
remember the surface,

the implicit formula.

Who gave you the normal to
the surface of a point P?

No?

The gradient of who?

Not the gradient of the left,
don't confuse-- the gradient

of the big F, right?

OK, at P. And the tangent
plane represents a what?

The tangent plane represents
exactly the perpendicular plane

that passes through the
point P, and is [INAUDIBLE]

to the normal.

So you're going to have
your surface, your normal,

and the tangent plane, which
is perpendicular to the normal.

Is this easy to remember,
maybe for your final?

I want to check if you know--
make a list, this list,

you have to post it in
the bathroom or somewhere,

on the wall or a closet.

Because you need to know
these things by the final.

S of X at point P becomes
S minus X 0 plus what?

The same kind of thing, right?

But it [INAUDIBLE] Y and Z.

So if you have the
curiosity to want

to prove that the first colorful
formula for the tangent plane,

using the red formula for the
tangent plane, it would come,

is an immediate [INAUDIBLE].

We've actually done that before.

We even did the implicit
function theorem.

There are some very
nice things you

can do when you have a
function of several variables.

And in particular, for a
function of two variables,

makes it really easy.

I'm gonna erase one,
two, three, and continue.

So I guess when
I leave the room,

I have to be careful not
to leave the actual midterm

in the room, although I
know that you wouldn't even

try to check my papers.

I did also something
in this for finding

a direction in which
the function increases

most rapidly.

I don't have to
write it down, but I

can remind you of the concept.

So it's just the concept now,
no formula to actually memorize.

But I'll still say number
four, problem number four,

because that's what I set
up on the actual exam.

So what is the direction of
highest ascend, deepest ascend?

STUDENT: [INAUDIBLE].

PROFESSOR: Is the
direction of the plane.

And what is the direction
of the steepest descent?

The opposite of the
direction of the grade.

So the direction of
steepest ascend and descend

is the direction of for the
graph Z equals F of X 1.

This is the function [INAUDIBLE]
that I'm talking about.

Five, the direction
of U that you

found at the previous problem,
I didn't ask if it's unique, OK?

Because that was one--
of course it's unique.

Because we [INAUDIBLE] sizes.

How do you say units
of sizing, [INAUDIBLE]?

By deriving with it,
[INAUDIBLE] a second,

you have a U and a -U.

STUDENT: So isn't the direction
that the actual [INAUDIBLE]

for is the gradient
of a normal vector?

PROFESSOR: So yeah,
so the way I-- OK, you

want me to read the problem?

I'm going to read the actual
function So find the direction

U, in which the function F of
X Y, blah, blah, blah, blah,

is here, it increases
most rapidly.

So what do you have to do?

So the direction
of that is, what is

the direction of that or this?

U equals [INAUDIBLE]
respectively minus U at P.

Five, this direction U that you
found at the previous problem,

could be perpendicular
to a certain line, which

of the following planes?

I may give you multiple choice.

Now what do you have
to do when you think

the direction of-- the way
it's actually formulated

is zero direction
is parallel to one

of the following [INAUDIBLE]
planes, which one?

Let me give you an example.

Z equals X [INAUDIBLE]
squared, at P coordinates 1, 1.

So that means X 0 is 1,
Y 0 is 1, and Z 0 is two.

Find the direction
of the gradient of F.

Let me put Z for alpha-- I'm
abusing my [INAUDIBLE]-- at P.

And state which of the following
lines is parallel for this

direction?

A, lines in plane.

X equals 2.

B, Y equals 3.

Or C, X plus 1 equals 0.

D, these are lines in
plane in the plane,

X Y. X plus Y. E,
none of the above.

So how are you going
to do that quickly?

Well, it's easy, right?

So what do I do?

I say gradient 2 X
2 Y, at the point

1, 1-- you don't have to
write down everything.

It's going to be the gradient
of F at P, will be 2, 2.

That means U will
be normalized 2, 2.

What do you get?

STUDENT: [INAUDIBLE]

PROFESSOR: Well, what
do people do normally

if they want to do
it by the definition?

They [INAUDIBLE]
the vector 2, 2,

by the square root [INAUDIBLE].

Well, you could be a little
bit smarter than that,

and say, F is the same
as the direction 1,

1 divided by the square
root of the sums,

of a sum of the squares.

It doesn't matter
which one you pick.

All the co-linear ones
will reveal the unique U.

And that's exactly what
I was trying to say,

was this thinking by in just two
or three moves ahead of that.

STUDENT: So that's the
same as 2, 2 over 4?

PROFESSOR: Yes, sir.

It's the same as 2, 2 over
the square root of 4 plus 4.

But it's easier, why it's
sort of faster to do it.

So why is that true
actually, Ryan is very right?

Why is that true?

Exactly because of that
uniqueness that I told you

about last time, when you
said, well, [INAUDIBLE],

what is that?

So you get 1 over square root of
2, and 1 over square root of 2,

is that you [INAUDIBLE]
vector direction.

Now without doing
further work, this

is just a simple
multiple question,

of [INAUDIBLE] question.

You are in front of your exam,
and you see lines in play.

You close your eyes and see
all of-- I will see my what?

You see all the lines in plane.

Of all these lines,
your favorite line

has to have the same direction
as the vector U. Is X equals 2?

No, that's nothing.

Y equals 3?

Those are horizontal, vertical.

That's the direction of
the first [INAUDIBLE].

So is this true of C?

No.

STUDENT: No, that's
for parallel lines.

PROFESSOR: D?

STUDENT: Yes.

PROFESSOR: Right?

So the incline X, Y--
the first bisector

is X equals Y [INAUDIBLE].

Number C is Y minus X, which
is called the second bisector.

You've seen that in college
algebra-- high school algebra,

more likely.

So we call this first
bisector, second bisector.

All right, so the
answer is D. Do you have

the same thing [INAUDIBLE]?

On the two multiple
choice things

you have, you see very
well, OK, I'm testing you.

I didn't say anything.

It was three feet away, OK?

We have just a quick
answer, and it's

going to be easy,
without algebra,

without computational stuff.

Just from the first glance,
you'll be able to answer.

Number six, what
is the maximum rate

of increase in the same
case as in problem five?

You say [INAUDIBLE], what
is the maximum, maximal rate

of increase of a [INAUDIBLE]?

And we all know what
I'm talking about,

although maybe not everybody.

But this is the gradient.

Who is giving you the
maximum rate of increase?

As I said last
time in the review,

that's actually the
directional derivative

in the direction
of the gradient.

But you are supposed to
know without proving again

that the directional
derivative and the direction

of the gradient will
give you that what?

Gradient of norm of--

STUDENT: [INAUDIBLE].

PROFESSOR: Exactly, the
magnitude of this F. So

what does that-- what
would that be in my case?

[INAUDIBLE] pay
attention, please.

Don't look at this,
if it's confusing you.

Look at that, right?

How much is that?

All right, [INAUDIBLE].

So you can put this
as [INAUDIBLE].

So your multiple choice-- how
many multiple choices do you

have?

Only two.

It may seem like what
is the maximum rate?

1, 0, 0, is telling-- that
means you have no increase.

You're not moving.

You're just lying
there on the plane.

OK.

What else?

2 root 2, 2 [INAUDIBLE]
to infinity.

I don't know.

I'm giving some
nonsensical choices.

But one of them
would be 2 root 2.

So you would see, it would
jump in front of your eyes.

Number seven, I think I'm
going-- I thought about this,

and I said one of
you guys asked me,

can you re-open any homework?

And I said, nope.

Why?

Because once the homework
closes, automatically

a few seconds later, all
the answers are gonna be up.

Do I have other
problems handy to create

a make-up for that individual
person who had the problem?

My cat almost died this
week, but she said,

but I have a treatment, and
hopefully she's going to live.

So in situations of [INAUDIBLE],
like an accident, a problem,

[INAUDIBLE] hospitalization,
and so on, I'm sorry,

I cannot re-open the homework.

The homework is already up
there with all the answers.

When I extend homework, it's
still doing that interval when

you cannot see the answers.

So I can extend it
by there [INAUDIBLE],

that was an exception.

So you have until the
fourth-- is the the fourth?

OK.

But once that closes,
I cannot re-open it.

However, I thought of giving
you a compensation midterm exam,

contains an extra
credit problem.

Because once you told
me that, I started

feeling bad for the two
people who have problems.

There were two or three people
who had very serious problems

this past weekend.

So in the midterm, you have
that extra credit problem,

that is meant to touch up
a little bit of let's say

if you missed a few
problems from the homework,

you had some bad day, whatever.

So you have ten
problems plus one.

Seven, you've seen that
before I told you about it.

It's an easy problem.

You have Z equals
F over X Y. And I'm

saying compute the volume of the
body that lies below the graph

and above the unit
[INAUDIBLE] D. Fine.

Eight, unfortunately
eight have [INAUDIBLE]

was disclosed because
Ryan was dreaming

of the problems in the midterm.

But it was something like
that, very good information.

So I would say a problem
like that, maybe a plane that

is cut in what?

The plane's coordinates form
something like a tetrahedron,

find the volume,
something like that.

Nine, again without giving
you the exact values,

you will have a function
F of X, and U of X,

I'd say positive over
a certain interval.

Set up the double integral,
set up a double integral

for the area of the domain
between F and G contained.

Compute that, and also reverse
the order of integration

to check your work.

Your answer, because here,
a multiple choice answer,

[INAUDIBLE] answer, no guesses.

It's going to not be
hard at all-- very

nice, friendly functions, very
nice, friendly [INAUDIBLE]

functions.

Just I have done this
before, but I'm not

going to repeat what it was.

I did it in the-- it's like
the one I did last week.

All right, so remember you
write the vertical strip thing,

integration with respect to Y
first, and then with respect

to X. You switch to
the horizontal strip

method of integration
with respect

to X first, and then with
respect to Y. Okie-doke?

And the actual algebra
here will be [INAUDIBLE]

expect to be done in one line.

So you will have something
extremely simple.

Ten-- it's another long exam.

So I have to try
to test everything

you know without you
spending more than one minute

per problem, just to conceive
the result. Formally,

hold on-- now nine, I split it.

Because I felt pity for you.

So I put [INAUDIBLE], I put
just set up the level integral,

and reverse the
order of integration.

So you have to write integral
integral equals integral

integral, nothing else,
no answer, no number.

And ten, actually compute
any of the two integrals

at number nine to find
the area of the domain.

Just like we did
last time, and you

don't have a calculator, OK?

Suppose your answer will
be-- what was last time?

One over six, I don't know.

If you give me decimals,
I will be very upset.

You have to give me the precise
answer for that problem,

because it's so easy to
compute that you would have

no need for using a calculator
or software, or any kind

of electronic device.

And finally number
11-- and number 11, I

shouldn't say what it is,
because it's extra credit.

But I'll still say what it is.

It's some simple
integral where you

are going to have to use
spherical coordinates.

And shut up,
[INAUDIBLE], because you

are talking too much.

So again, number 11 will
be a triple integral

that is easy to compute.

And when you're going-- well,
you don't have to use vehicle.

You can still do it with
cylindrical coordinates,

for example.

But it's a big pain doing
the cylindrical coordinates

for that kind of problem.

So imagine maybe I'm looking
at the domain to be a sphere.

The problems we worked
as a training in class,

are actually harder than
the ones I put on the exam.

I have a professor who's a grad
student, and he used to say,

the easy problems
are the professor

to work in classes examples.

The hard problems are the
students who have on the exam.

I think exactly the opposite.

Because when you will in
training for any kind of sports

or a skill or music, you
have to train yourself

above the level of
your competition.

Otherwise your
competition will be bored.

So what you're
doing with training

should not always
be how whether you

are an athlete or a
mathematician or a violinist

or whatever.

So you're not going to see
something like intersector

cylinders, passing one through
the other, one the cone,

ice cream cone will
be doing a parabola,

then the cone is full of ice
cream-- nothing like that.

Something simpler.

And you may guess what it is,
but keep in mind that force

of speed components, you
have to know the Jacobian,

don't hesitate.

That's assumed to be memorized.

Don't ask me in the
middle of the exam.

Why was the Jacobian
[INAUDIBLE] components?

You are supposed to
know that as being what?

What was that?

Roberto knows.

[INTERPOSING VOICES]

PROFESSOR: [INAUDIBLE]
assign by, and by

was what, for a friend of yours?

The latitude from Santa Claus,
measured down all the way

to [INAUDIBLE].

Theta is the longitude
from 0 to 2 pi.

You are going to have
some very nice domain.

All right.

That's it, guys.

That's what the exam will say.

I'm asking you for a few things.

First of all, you are already
prepared, I guarantee it.

Do not stay up late at night.

The biggest mistake
students make

is staying up the night before a
midterm or a final because they

want to study everything.

That's bad.

The next day you will be tired
and you won't perform as well.

Second of all, do not
be nervous at all.

You have no reason
to be nervous.

You have plenty of time.

You have plenty of
things to write down.

OK, about the way I grade.

If you leave the problem
completely blank,

yes, that's a zero.

But if you provide me with at
least a hint, a formula that

serves you-- not just
an arbitrary formula

that has nothing to do with it.

But a formula that's
in the regulation,

I give you partial
credit for everything.

So you have no
reason to freak out.

Even if you mess up, let's
say, one or two problems,

your algebra at the
end, you should still

gather together lots of credit.

I wrote the exam especially
because I, myself,

hate some medical
answers from web work.

I made the answers to
be fat and sassy, not

like the ones in the web works.

So something that you
can do even mentally,

not have to struggle
with the answer.

Several people have
asked me to go over

the last two problems of the
homework, and I'll go forward.

I'll give you an example
of a problem that bothered

a few bothered a few people.

And it's somewhat interesting
because, of course,

I make the algebra
easier than it is.

You have 2, 3, 7,
9, I don't know.

You have a function x and y that
are both functions of u and v.

And instead of asking you
for the determinant-- well,

one of them may have asked
you for the determinant,

the functions x, derivative
of x, y with [INAUDIBLE] u,

v as Jacobian.

But the other one was asking
you for just the opposite.

Well, several people
didn't see that.

And they kept asking me, so
I answered some 20 questions

during the weekend exactly
about problems like that.

Which doesn't bother me.

I think I would have had
the same problem when

I was like you.

The easy part on
the first Jacobian

is that you have 1, 1, 1
minus 1, whatever that is.

The definition is
x sub u, x sub v,

y sub u, y sub v. These
are the partial derivatives

and that's called Jacobian.

And what you have, you
have an easy answer.

In this case, you have
the answer negative 2.

And if you, however, are asked
by the author of the problem,

whoever created the
problem of this.

And you put negative
2, it's going

to say no, this is not correct.

And this is what happened
to several people.

Now, there are two
ways around it.

There are two ways
you can solve that.

STUDENT: On the work, it
says the reverse [INAUDIBLE].

That Jacobian times the
reverse Jacobian [INAUDIBLE].

PROFESSOR: I want
to say why that is.

For a student who doesn't know
why this Jacobian is exactly

j inverse, there are still
chances the student can say,

well, here's how smart I am.

I'm going to say u out, v
out in terms of x and y.

I inverse the
functions because they

are linear functions
[INAUDIBLE] linear system.

So I say x plus y.

This is elimination called--
when we were little,

this was called elimination 2u.

x minus y equals 2v.

So u is x plus y over 2.

That means 1/2 x, 1/2 y.

Right, guys?

I'm right?

STUDENT: Mm-hmm.

PROFESSOR: OK.

And 1/2 of x and minus 1/2 of y.

And then, what does
the student say?

I know what I'm going to do.

Just by the same definition,
I say the du, v dx,

y as I have an inverse function.

And I knew how to
invert the system.

I get 1/2.

Not matrix, Magdalena,
now, determinant.

1/2, 1/2 and 1/2, minus 1/2.

And guess what?

What do I get?

Exactly what it was saying,
but I did it the long way.

I got minus 1 over 4 minus 1
over 4, which is minus 1/2.

Which is--

STUDENT: Inverse.

PROFESSOR: The inverse of that.

And you are going to ask me,
OK, I don't understand why.

That's why I want to
tell you a story that I

think is beautiful.

The book doesn't start like
that, because the book doesn't

necessarily have enough
space to remind you

everything you learned in
Calc 1 when you are in Calc 3.

But if you think of what you
learned in Calc 1, in Calc 1

your professor-- I'm sure that
he or she showed you this.

If you have a function
y equals f of x,

assume this is a c1 function
and everything is nice.

And then you have that
f prime of x exists

and it's continuous everywhere.

That's what it means c1.

And you want to
invert this function.

You want to invert this
function around the point x0.

So you know that at
least for some interval

that f is one-to-one.

So it's invertible.

What is the derivative
of [INAUDIBLE]?

Somebody asks you, so what
is the-- the derivative

of the inverse
function is a function

of x with respect to x.

[INAUDIBLE]?

I don't know.

Remind yourself
how you did that.

Was this hard?

Anybody remembers the formula
for the inverse function?

STUDENT: [INAUDIBLE].

PROFESSOR: 1 over f prime of x.

So assume that you do
that at the next 0,

assume that f prime of
x0 is different from 0.

Now, how would you
prove that and how--

well, too much memorization.

This is what we are doing
in-- the derivative of e

to the x was what?

What was the derivative
of natural log of this?

STUDENT: [INAUDIBLE].

PROFESSOR: 1/x.

Now, when you have an
arbitrary function f

and you compose with inverse,
what is it by definition?

X equals x.

So this is the
identity function.

Chain rule tells
you, wait a minute.

Chain rule tell you how
to prime the whole thing.

So I prime-- what
is this animal?

F of f inverse of x is the
composition of functions,

right?

So apply chain rule to f of
f inverse of x, all prime,

with respect to x.

What is x prime
with respect to x?

x prime with respect to x is 1.

All right.

Chain rule.

What does the chain rule say?

Chain rule says f
prime of f inverse

of x times a function on
the outside prime first.

We go from the outside to the
inside one step at a time.

Derivative of the guy--
you cover f with your hand.

Derivative of the guy inside,
the core function inside,

that will simply be f inverse
of x prime with respect to x

equals 1.

That was x prime.

So the derivative of the
inverse function-- all right.

f inverse prime is 1 over
f prime of f inverse of x.

So if you think of
this being the y,

you have f inverse prime at
y equals 1 over-- well, yeah.

If you put it at x, it's
f prime of f inverse of x.

Because the f
inverse-- this is x.

This is f of x.

This is the y [INAUDIBLE].

When you have f inverse,
x is the image of y.

So f inverse has an input.

How is this called?

In the domain of f inverse.

That means, who is
in the domain of f?

f inverse of x.

So one is x, one is y.

So keep that in mind that when
you have to invert a function,

what do you do?

You say 1 over the
derivative of the initial--

so the derivative of
the inverse function

is 1 over derivative
of our initial function

at the corresponding point.

This is how you did the
derivative for the Calc 1

people.

All right.

So how do I apply that formula?

Well, I have two functions here.

One is e to the x and
one is natural log of x.

How do I know they are
inverse to one another?

Their graphs should be
symmetric with respect to the?

STUDENT: y equals x.

PROFESSOR: With respect
to the first [INAUDIBLE].

Assume that f of
x is e to the x.

OK.

f inverse of-- well,
let's say f inverse of y.

That would be natural
log of y, right?

So what if you put
here, what is f inverse?

Natural log.

Let's say a simple way to
write this, simple division.

According to that formula,
how would you do the math?

You go f inverse prime of x must
be 1 over the derivative of f

with respect of f inverse x.

So you go, wait a minute.

OK, who is f inverse of x?

Sorry, if f of x is e to the
x, who is f inverse of x?

You want me to change a letter?

I can put a y here.

But in any case, I want to
convince you that this is 1/x.

Why?

Because f prime is e to the x.

This is going to be e
to the f inverse of x,

which is e to the natural
log of x, which is x.

That's why I have x here.

So again, if f of x
equals e to the x, then f

inverse of x is the
national log of x.

By this formula,
you know that you

have to compute natural
log-- this is f inverse.

Natural log of x prime, right?

What is this by that formula?

1 over the derivative of f prime
computed at f inverse of x.

Now, who is f inverse of x?

f inverse of x is
natural log of x.

So again, let me write.

All this guy here in the orange
thing, this [INAUDIBLE] f

inverse of x is ln x.

Who is f prime?

f prime of x is e to the x.

So f prime of
natural log of x will

be e to the natural log of x.

Applied to natural
log of x, which is x.

So you got it.

All right?

So remember, this formula,
professors actually avoid.

They say, oh my god.

My students will never
understand this composition

thing, derivative.

So Magdalena, I don't care.

You are the
undergraduate director.

I'll never give it like that.

That's a mistake.

They should show it
to you like that.

f inverse prime of x equals
1 over f prime of what?

Of the inverse image of x.

Because you act on x.

So if x is acting on--
this is f of x, then

you have to invert by
acting on f of x, like this

and like that.

If x is in the domain of f
inverse, that means what?

That in the domain of f, you
have f inverse of x as input.

So instead of giving
you the formula,

they just make you
memorize the formulas

for the inverse functions,
like-- believe me,

you take e to the x
[INAUDIBLE] derivative.

You take natural log,
it's 1/x derivative.

Don't worry about the fact that
they are inverse to one another

and you an relate the
derivatives of two

inverse functions.

They try to stay out of trouble
because this is hard to follow.

You could see that you had
[INAUDIBLE] a little bit

and concentrate.

What is this woman saying?

This looks hard.

But it's the same process
that happens in the Jacobian.

So in the Jacobian of a
function of two variables.

Now, remember the signed
area that I told you about.

Signed area notion.

What did we say?

We said that dA is
dx dy, but it's not

the way they explain in
the book because it's

more like a wedge thing.

And that wedge thingy had
a meaning in the sense

that if you were to not take
the exterior derivative dx dy,

but take dy wedge dx,
it would change sign.

So we thought of
signed area before.

When we did dx wedge
dy, what did we

get in terms of Jacobian?

We get j d r
[INAUDIBLE] coordinates.

Do you remember what this j was?

STUDENT: r.

PROFESSOR: Very good, r.

In the case, in the simple
case of Cartesian versus polar,

Cartesian going to polar,
you have a function f.

Coming back it's called
the inverse function.

So I'm asking, this is the
Jacobian of which function?

This is the Jacobian of
the function that goes

from [INAUDIBLE] theta to x, y.

If I want the Jacobian
of the function that

goes from x, y into
[INAUDIBLE] theta,

I should write--
well, d r d theta

will be something times dx dy.

And now you understand
better what's going on.

1/j.

1/j.

So Matthew was
right, in the sense

that he said why are you so
clumsy and go ahead and compute

again u, v?

Express u, v in terms of x, y.

You waste your time
and get minus a 1/2.

What was that, guys?

Minus 1/2.

When I'm telling you that
for the inverse mapping,

the Jacobian you get is
the inverse of a Jacobian.

It's very simple.

It's a very simple relationship.

I could observe that.

And he was right.

So keep in mind that when you
have Jacobian of the map where

x, y are functions of u, v,
this is 1 over the Jacobian

where you have u, v
as functions of x, y.

So you have inverse mapping.

In Advanced Calculus, you
may learn a little bit more

about the inverse
mapping theorem.

This is what I'm talking about.

For the inverse
mapping theorem, you

go, well, if the derivative
of these two with respect

to these two are done as
j Jacobian, the derivative

of these two with
respect to these two

in a Jacobian [INAUDIBLE]
exactly j inverse, or 1/j.

j is a real number.

So for a real number, whether
I write 1/j or j inverse,

it's the same.

So as an application, do
you have to know all this?

No, you don't.

But as an application, let
me ask you the following.

Something harder than
[INAUDIBLE] in the book.

In the book, you have
simple transformations.

What is the Jacobian
of r theta-- theta, phi

or phi, theta.

It doesn't matter.

If I swap the two, I
still have the same thing.

If a determinant swaps
two rows or two columns,

do you guys know what happens?

You took linear algebra.

STUDENT: Swap.

PROFESSOR: You swap two
rows or two columns.

STUDENT: [INAUDIBLE].

PROFESSOR: It's going to pick
up a minus sign, very good.

But only three people in
this class figured it out.

How shall I denote?

Not j, but the notation
was [INAUDIBLE].

And this is j.

So [INAUDIBLE] phi theta
over [INAUDIBLE] x, y, z.

How do you compute them?

You say, no, I'm not
going to compute it

by hand because until tomorrow
I'm not going to finish it.

STUDENT: Does that
need a 3 by 3 matrix?

PROFESSOR: It's a determinant.

So when you were to
write this, you're

not going to do it because it's
a killer for somebody to work

like that in spherical
coordinates with only

those inverse functions.

Do you remember as a review
what spherical coordinates were?

x, y, z versus r, theta, phi.

We reviewed that.

Theta was the longitude.

Phi was the latitude
from the North Pole.

So x was-- who remembers that?

[INTERPOSING VOICES]

PROFESSOR: Cosine theta
r sine phi sine theta.

And z was the adjacent guy.

Remember, this was the thingy?

And this was the phi.

And to express x, the
phi was adjacent to it.

And that's why you
have cosine phi.

It's a killer if somebody wants
to pull out the r, phi, theta.

First of all, r will be easy.

But the other ones are a
little bit of a headache.

And with all those
big functions,

you would waste a lot of time
to compute the determinant.

What do you do?

You say, well, didn't
you say that if I

take the inverse mapping,
the Jacobian would be

1 over the original Jacobian?

Yes, I just said that.

So go ahead and remember what
the original Jacobian was

and leave us alone
you're going to say.

And you're right.

What was that I just said
the other Jacobian was?

STUDENT: [INAUDIBLE].

PROFESSOR: You told me.

I forgot it already.

r squared sine phi, right?

So if somebody's asking
you to solve this problem,

you don't need to
write out anything.

Just 1 over [INAUDIBLE].

I'm done.

But I'm not going to ask you.

Of course, I saw
this problem exactly.

Find the Jacobian of
the inverse mapping

for the spherical coordinates.

That was given at Princeton
in Advanced Calculus.

There were three variables, and
then there was a generalization

to [INAUDIBLE] variables.

But based on this
at Princeton, I'm

not going to give you
anything like that

to compute in the exam.

And I just expect that you
know your basics about how

to compute triple integrals.

Use the Jacobians and
be successful with it.

Let's do one last
problem about the review.

Although, it's not
in the midterm,

but I would like to-- I'd
like to see how you solve it.

A student from another
class, Calc 3, came to me.

And I was hesitant about even
helping him on the homework

because we're not supposed
to help our college students.

So I told him, did you go
to the tutoring center?

And he said yes, but they
couldn't help him much.

So I said, OK.

So let me see the problem.

He showed me the
problem and I wanted

to talk about this
problem with you.

This is not a hard problem, OK?

You just have to see
what this is about.

Understand what this is about.

So you have the z equals x
squared plus y squared, which

is the [INAUDIBLE].

Sorry about my typos.

We didn't write this
problem in the book.

So I suspect that his instructor
came up with this problem.

This is a cone.

We only look at
the upper halves.

Do these surfaces intersect?

Draw the body between
them if the case.

And compute the
volume of that body.

And what do you think
my reaction was?

Oh, this is a piece of cake.

And it is a piece of cake.

But you need to
learn Calc 3 first

in order to help other
people do Calc 3 problems.

Especially if they
are not in the book.

So one has to have a very good
understanding of the theory

and of geometry, analytic
geometry, and conics

before they move onto
triple integrals and so on.

Can you imagine these with
the eyes of your imagination?

Can we draw them?

Yeah.

We better draw them because
they are not nasty to draw.

Of course this looks
like the Tower of Pisa.

Let me do it again.

Better.

x, y, and z.

And then I'll take the cone.

Well, let me draw
the paraboloid first.

Kind of sort of.

And then the cone.

I hate myself when
I cannot draw.

If you were to cut, slice
up, it could be this.

And who asked me last
time, was it Alex, or Ryan,

or maybe somebody else, who
said maybe we could do that even

in Calc 2 by--

STUDENT: Yeah.

PROFESSOR: You asked me.

STUDENT: [INAUDIBLE].

PROFESSOR: If you
take a leaf like that

and you rotate it
around the body,

like in-- using one
of the two methods

that you learned in Calc 2.

Well, we can do that.

But you see we have in Calc 3.

So I would like to
write that in terms

of the volume of the body
faster with knowledge I have.

Do they intersect?

And where do they intersect?

And how do I find this out?

STUDENT: [INAUDIBLE].

PROFESSOR: Yes.

I have to make them equal and
solve for z, and then the rest.

How do I solve for z?

Well, z equals z0 gives
me two possibilities.

One is z equals 0
and 1 is z equals 1

because this is the same as
writing z times z minus 1

equals 0.

So where do they intersect?

They intersect
here at the origin

and they intersect
where z equals 1.

And where z equals 1, I'm
going to have what circle?

The unit circle.

I'll draw over--
I'll make it in red.

This is x squared plus y squared
equals 1 at the altitude 1,

z equals 1.

This is the plane z equals 1.

OK, so how many ways
to do this are there?

When we were in Chapter 12,
we said the triple integral

will give me the volume.

So the volume will
be triple integral

of a certain body-- of
1 over a certain body

dv, where the body is
the body of revolution

created by the motion
of-- what is this thing?

What shall we call it?

A wing.

[INAUDIBLE]

Domain D. No, domain D is
usually what's on [INAUDIBLE].

I don't know.

STUDENT: L for leaf?

PROFESSOR: L for leaf.

Wonderful.

I like that.

L. OK.

So I can write it up as
a triple integral how?

Is it easy to use it in
spherical coordinates?

No.

That's not a spherical
coordinate problem.

That's a cylindrical
coordinate problem.

Why is that?

I'm going to have to
think where I live.

I live above a beautiful disk,
which is the shadowy plane.

And that beautiful disk
has exactly radius 1.

So we are lucky.

That's the unit disk, x squared
plus y squared less than 1

and greater than 0.

So when I revolve, I'm
using polar coordinates.

And that means I'm using
cylindrical coordinates, which

is practically the same thing.

r will be between what and what?

STUDENT: [INAUDIBLE].

PROFESSOR: 0 to 1, very good.

Theta?

STUDENT: [INAUDIBLE].

PROFESSOR: 0 to 2 pi.

How about z?

z is the z from
cylindrical coordinates.

STUDENT: Square root x
squared plus y squared--

PROFESSOR: Who is on the bottom?

STUDENT: 0.

PROFESSOR: So the z is between--
let me write it in x first,

and then switch to polar.

Is that OK?

STUDENT: Yeah

PROFESSOR: All right.

So what do I write on
the left-hand side?

I need water.

STUDENT: [INAUDIBLE].

PROFESSOR: Who is smaller?

Who is smaller?

Square root of x
squared y squared or x

squared plus y squared?

STUDENT: Square root over.

PROFESSOR: This is smaller.

Why?

STUDENT: Because [INAUDIBLE].

PROFESSOR: So it's less than 1.

I mean, less than 1.

This is less than 1.

It's between 0 and 1.

So I was trying to explain
this to my son, but I couldn't.

But he's 10.

It's so hard.

So I said compare square
root of 0.04 with 0.04.

This is smaller, obviously.

This is 0.2.

He can understand.

So this is what we're doing.

We are saying that
this is x squared

plus y squared, the round
thing on the bottom.

And this is going to be on the
top, square root of x squared

plus y squared
from the cylinder,

from the cone-- sorry
guys, the upper half.

Because I only work
with the upper half.

Everything is about
the sea level.

Good, now let's write
out the whole thing.

So I have integral from
the polar coordinates,

from what to what?

STUDENT: r squared to r.

PROFESSOR: r squared to
r, 0 to 1, 0 to 2 pi.

So the order of integration
would be dz dr d theta.

And what's inside here?

STUDENT: [INAUDIBLE].

PROFESSOR: No.

STUDENT: r.

PROFESSOR: r,
excellent, r-- why r?

Because 1 was 1.

But dv is Jacobian times dr
d theta dz-- dz, dr, d theta.

So this is going to be the r
from the change of coordinates,

the Jacobian.

Is this hard?

Well, let's do it.

Come on, this shouldn't be hard.

We can even separate
the functions.

And I got you some tricks.

The first one we
have to work it out.

We have no other choice.

So I'm going to have
the integral from 0

to 2 pi, integral from 0 to 1.

And then I go what?

I go integral of what you see
with z, the z between r and r

squared times r dr d theta.

Who is going on my nerves?

Not you guys.

Here, there is a guy that
goes on my nerves-- the theta.

I can get rid of him, and I
say, I don't need the theta.

I've got things
to do with the r.

So I go 2 pi, which is the
integral from 0 to 2 pi of 1

d theta.

2 pi goes out.

Now 2 pi times integral
from 0 to 1 of what?

What's the simplest
way to write it?

STUDENT: [INAUDIBLE].

PROFESSOR: r squared in the end.

I mean, I do the whole
thing in the end.

I have r squared minus
r cubed, right guys?

Are you with me?

dr, so this is when I did it.

But I didn't do the
anti-derivative, not yet.

I did not apply the fundamental.

Now you apply the
fundamental [INAUDIBLE]

and tell me what you get.

What is this?

STUDENT: [INAUDIBLE] 1/12.

PROFESSOR: 1/12, that's very
good-- r cubed over 3 minus r

to the 4 over 4, 1/3 minus
1/4, 1/12, very good.

So you have 2 pi times
1/12 equals pi over 6.

Thank god, we got it.

Was it hard?

Would you have spent two
days without doing this?

I think you would have
gotten it by yourselves.

Am I right, with no problem?

Why is that?

Because I think you
worked enough problems

to master the material,
and you are prepared.

And this is not a
surprise for you

like it is for many
students in other classes.

Yes, sir.

STUDENT: Can you put that
one in spherical coordinates?

PROFESSOR: You can.

That is going to be a hassle.

I would do one more
problem that is not

quite appropriate
for spherical work,

but I want to do it [INAUDIBLE].

Because it looks like the
ones I gave you as a homework,

and several people
struggled with that.

And I want to see how it's done
since not everybody finished

it.

Given [INAUDIBLE] numbers,
you have a flat plane

z equals a at some
altitude a and a cone

exactly like the cone
I gave you before.

And of course this is
not just like you asked.

This is not very appropriate
for spherical coordinates.

It's appropriate
for cylindrical.

But they ask you
to do it in both.

Remember that problem, guys?

So you have the volume of, or
some function, or something.

And they say, put it in
both spherical coordinates

and cylindrical coordinates.

And let's assume that
you don't know what

function you are integrating.

I'm working too
much with volumes.

Let's suppose that
you are simply

integrating in function
F of x, y, z dV, which

is dx dy dx over the body of
the [INAUDIBLE], of the-- this

is the flat cone, the
flat ice cream cone.

Then somebody licked your
ice cream up to this point.

And you are left
with the ice cream

only under this at the level
of the rim of the waffle.

Let's break this into two-- they
don't ask you to compute it.

They ask you to set up
cylindrical coordinates

and set up the
spherical coordinates.

But thank you for the idea.

That was great.

So let's see, how hard is it?

I think it's very easy in
cylindrical coordinates.

What do you do in
cylindrical coordinates?

You say, well, wait a minute.

If z equals a has
to be intersected

with z squared
equals x squared, I

know the circle that
I'm going to get

is going to be a piece of cake.

x squared plus y squared
equals a squared.

So really my ice cream
cone has the radius a.

Are you guys with me?

Is it true?

Is it true that the radius
of this licked ice cream cone

is a?

STUDENT: Mhmm.

PROFESSOR: It is true.

Whatever that a was-- yours
was 43, 34, 37, god knows what,

doesn't matter.

I would foresee-- I'm not
a prophet or even a witch.

I am a witch.

But anyway, I would
not foresee somebody

giving you a hard
problem to solve

like that computationally.

But on the final, they can
make you set up the limits

and leave it like that.

So how do we do cylindrical?

Is this hard?

So r will be from 0 to a.

And god, that's easy.

0 to 2 pi is going
to be for the theta.

First I write dz.

Then I do dr and d theta.

Theta will be between 0 and
2 pi, r between 0 and a.

z-- you guys have to
tell me, because it's

between a bottom and a top.

And I was about to
take this to drink.

STUDENT: [INAUDIBLE].

PROFESSOR: r is the
one on the bottom,

and a is the one on the top.

And I think that's clear
to everybody, right?

Is there anything
missing obviously?

So what do I do when they
ask me on the final--

when I say this is a
mysterious function, what

do you put in here?

F of x, y, z, yes,
but yes and no.

Because you say F of x of r z
theta, y of r z theta, z of god

knows z, z.

z is the same, do
you understand?

So you indicate
to the poor people

that I'm not going to stay in
x, y, z, because I'm not stupid.

I'm going to transform
the whole thing

so it's going to be expressed
in terms of these letters-- r

theta and z.

Do you have to write all this?

If you were a professional
writing the math paper, yes,

you have to, or a math book
or whatever, you have to.

But you can also skip it
and put the F. I'm not

going to take off points.

I will understand.

Times r-- very good.

Never forget about
your nice Jacobian.

If you forget the r, this
is no good, 0 points,

even with all the setup you
tried to do going into it.

OK, finally let's see.

How you do this in
spherical is not-- yes, sir.

STUDENT: When you're
finding the volume,

isn't it with a triple integral,
don't you just put a 1?

PROFESSOR: Hm?

STUDENT: When you're
finding a volume?

PROFESSOR: No, I didn't
say-- I just said,

but you probably were
thinking of [INAUDIBLE].

I said, I gave you
too many volumes.

I just said, and
I'm tired of saying

volume of this, volume of that.

And in the actual
problem, they may

ask you to do triple
integral of any function,

differentiable function
or continuous function,

over a volume, over a body.

So this could be-- in
the next chapter we're

going to see some applications.

I maybe saw some in 12.6 like
mass moment, those things.

But in three coordinates,
you have other functions

that are these functions.

You'll have that included,
row z, x, y, z, and so on.

OK, good.

When you would integrate a
density function in that case,

you will have a mass.

Because you integrate this, d
over volume, you'd have a mass.

OK, in this case,
we have to be smart,

say F times r squared
sine phi is the Jacobian.

This is a function in r
phi and theta, right guys?

We don't care what it is.

We are going to have
the d something,

d something, d something.

The question is, which ones?

Because it's not obvious
at all, except for theta.

Theta is nice.

He's so nice.

And we say, OK theta,
we are grateful to you.

We put you at the end, because
it's a complete rotation.

And we know you are between 0
and 2 pi, very reliable guy.

Phi is not so reliable.

Well, phi is a nice guy.

But he puts us through
a little bit of work.

Do we like to work?

Well, not so much,
but we'll try.

So we need to know a little
bit more about this triangle.

We need to understand a little
bit more about this triangle.

STUDENT: Well, the angle
between the angle at the bottom

is 45 degrees.

PROFESSOR: How can you say?

STUDENT: Because the
slope of that line is 1.

PROFESSOR: Right, so say,
now I'm going to observe z

was a as well.

So that means it's a
right isosceles triangle.

If it's a right
isosceles triangle,

this is 45 degree angle.

So this is from d phi from
0 to pi over 4, excellent.

Finally, the only one that gives
us a little bit of a headache

but not too much of a
headache is the radius r.

Should I change the color?

No, I'll leave it
r dr. So we have

to think a little bit of
the meaning of our rays.

Drawing vertical strips
or horizontal strips

or whatever strips
is not a good idea.

When we are in
spherical coordinates,

what do we need to draw?

Rays, like rays of sun
coming from a source.

The source is here at the
origin in spherical coordinates.

These are like rays of
sun that are free to move.

But they bump.

They just bump against
the plane, the flat roof.

So they would reflect if
this were a physical problem.

So definitely all
your rays start at 0.

So you have to put 0 here.

But this is a question mark.

STUDENT: [INAUDIBLE].

STUDENT: a square root 2.

PROFESSOR: No, it's
not a fixed answer.

So you have z will be a fixed.

But who was z in
spherical coordinates?

That was the only
thing you can ask.

So z equals a is your
tradition that is the roof.

STUDENT: That would
be r [INAUDIBLE].

PROFESSOR: Very good,
r cosine of phi,

of the latitude
from the North Pole.

This is 45.

But I mean for a point like
this, phi will be this phi.

Do you guys understand?

Phi could be any point where the
point inside [INAUDIBLE], phi

will be the latitude
from the North Pole.

OK, so the way you do it
is r is between 0 and z

over cosine phi.

And that's the hard thing.

Since z at the
roof is a, you have

to put here a over-- a is
fixed, that 43 of yours,

whatever it was-- cosine phi.

So when you guys integrate
with respect to r,

assume this F will be 1,
just like you asked me, Alex.

That would make my life
easier and would be good.

When I integrate
with respect to r,

would it be hard
to solve a problem?

Oh, not so hard.

Why?

OK, integrate this
with respect to r.

We have r cubed.

Integrate r squared.

We have r cubed over 3, right?

Let's do this, solve the
same problem when F is 1.

Solve the same problem when
F would be 1, for F equals 1.

Then you get integral
from 0 to 2 pi,

integral from 0 to pi
over 4, integral from 0

to a over cosine phi, 1 r
squared sine phi dr d phi d

theta.

The guy that sits on my
nerves is again theta.

He's very nice.

He can be eliminated
from the game.

So 2 pi out, and I
will focus my attention

to the product of function.

Well, OK, I have to
integrate one at a time.

So I integrate with
respect to what?

STUDENT: [INAUDIBLE].

PROFESSOR: So I get r cubed
over 3, all right, the integral

from 0 to pi over 4.

STUDENT: [INAUDIBLE].

PROFESSOR: r cubed over 3
between-- it's a little bit

of a headache. r equals
a over cosine phi.

And I bet you my video
doesn't see anything,

so let me change the colors.

r equals a over cosine phi.

And r equals 0 down.

That's the easy part.

Inside I have r
cubed over 3, right?

All right, and sine
phi, and all I'm

left with is a phi
integration, is an integration

with respect to phi.

Let's see-- yes, sir.

STUDENT: [INAUDIBLE].

PROFESSOR: Well, I should be
able to manage with this guy.

STUDENT: [INAUDIBLE].

PROFESSOR: I'm writing
just as you said, OK?

Now, how much of a headache
do you think this is?

STUDENT: It's not much
of one, because it's

the same as a tangent
times the secant squared

with a constant pulled out.

STUDENT: So psi and
cosine don't [INAUDIBLE].

Tangents will give
you 1 over cosine--

PROFESSOR: What's the simplest
way to do it without thinking

of tangent and cotangent, huh?

STUDENT: [INAUDIBLE].

PROFESSOR: Instead, a
u substitution there?

What is the u substitution?

STUDENT: [INAUDIBLE].

PROFESSOR: Is this good?

STUDENT: No.

PROFESSOR: No?

STUDENT: [INAUDIBLE].

PROFESSOR: It's
u to the minus 3.

And that's OK.

So I have 2 pi a cubed
over 3 [INAUDIBLE]

because they are in my way
there making my life miserable,

integral.

And then I have u to the
minus 3 times-- for du

I get a minus that
that is sort of ugh.

I have to invent the minus, and
I have to invent the minus here

in front as well.

So they will compensate
for one another.

And I'll say du.

But these limit points, of
course I can do them by myself.

I don't need your help.

But I pretend that
I need your help.

What will be u when phi is 0?

STUDENT: 1.

PROFESSOR: 1.

What will be u when
phi is pi over 4?

STUDENT: [INAUDIBLE].

PROFESSOR: And from now on
you should be able to do this.

So I have minus 2 pi a cubed
over 3 times-- I integrate.

So I add the power, I add
the 1, and I add the 1.

So you have u to the
minus 2 over minus 2.

Are you guys with
me-- between u equals

1 and u equals root 2 over 2.

I promise you if
you have something

like that in the final
and you stop here,

I'm not going to be blaming you.

I'll say, very good, leave
it there, I don't care.

Because from this
point on, what follows

is just routine algebra.

So we have-- I hate this.

I'm not a calculator.

But it's better for me to write
1 over root 2, like you said.

Because in that case, the
square will be 1 over 2.

And when I invert 1
over 2, I get a 2.

So I have 2 over minus 2.

Are you guys with me again?

So I'm thinking the
same-- 1 over root 2.

Square it, you have 1 over 2.

Take it as a minus,
you have exactly 2.

And you have 2 over
minus-- is this a minus?

I'm so silly, look
at me, minus 2.

STUDENT: It's a cubed
over 3, not over 2.

PROFESSOR: It's going to be--

STUDENT: You've got an a
cubed over 2 right there.

And it was--

PROFESSOR: Huh?

STUDENT: You just wrote
2 pi a cubed over 2.

It's a cubed over 3.

PROFESSOR: Yes,
it's my silliness.

I looked, and I say
this instead of that.

Thank you so much.

What do I have here?

1 over minus 2.

In the end, what does this mean?

Let's see, what does this mean?

When I plug in, I subtract.

This is what?

This is minus 1 plus
1/2 is minus 1/2.

But that minus
should not scare me.

Because of course
a minus in a volume

would be completely wrong.

But I have a minus from before.

So it's plus 2 pi times a
cubed over 3, and times 1/2.

So in the end, the
answer, the total answer,

would be answered what?

STUDENT: [INAUDIBLE].

PROFESSOR: Pi a cubed
over 3, pi a cubed over 3.

It looks very-- huh?

It looks pretty.

Actually yes, it looks
pretty because-- now,

OK, I'm asking you a question.

Would we have done
that without calculus?

If somebody told you [INAUDIBLE]
it has a volume of some cone,

what's the volume of a cone?

Area of the base times
the height divided by 3.

So you could have very nicely
cheated on me on the exam

by saying, you
have this cone that

has pi is squared times a-- pi
is squared times a-- divided

by 3 equals pi cubed over 3.

When can you not
cheat on this problem?

STUDENT: When you say,
you've got to do it with a--

PROFESSOR: Exactly, when I
say, do it with a-- well,

I can say, OK, if we say, set up
the integral and write it down,

you set up the integral
and write it down.

If we say, set up the
integral and compute it,

you set up the integral,
you fake the computation,

and you come up with this.

If we say, set up the integral
and show all your work,

then you're in trouble.

But I'm going to try to advocate
that for a simple problem, that

is actually elementary.

One should not have
to show all the work.

All right, but keep in mind
when you have 2 minuses

like that-- that reminds me.

So there was a professor whose
sink didn't work anymore.

And he asked for a plumber
to come to his house.

He was a math professor.

So the plumber comes to his
house and fixes this, and says,

what else is wrong?

Fixes the toilet, fixes
everything in the house,

and then he shows the
professor the bill.

So the guy said, oh my god, this
is 1/3 of my monthly salary.

So the plumber said,
yeah, I mean, really?

You're a smart guy.

You're a professor.

You make that little money?

Yeah, really.

I'm so sorry for you.

Why don't you apply
to our company

and become a plumber if you're
interested, if you crave money?

No, of course, I need
money desperately.

I have five children
and a wife [INAUDIBLE].

OK, he applies.

And he says, pay attention.

Don't write that you are a
professor or you have a PhD.

Just say you just finished
high school or say,

I didn't finish high school.

So he writes, I didn't
finish high school.

I went to 10th grade.

They accept him.

They give him a job.

And they say, this
is your salary.

But there is something new.

Everybody has to
finish high school.

So they have to
take AP Calculus.

So he goes, oh my god.

They all go.

And there comes a TA from
the community college.

The class was full.

He tries to solve a
problem-- with calculus

compute the area inside
this disc of radius a.

So the TA-- OK, I did this.

I got minus pi a squared.

And the professor says,
OK, you cannot get that.

Let me explain to you.

He goes, I don't know
where he made a mistake.

Because I still get-- where
is minus pi a squared?

I don't see where
the mistake is.

And then the whole
class, 12, 15-- reverse

the integral limits.

Change the integral limits
and you'll get it right.

So we can all pretend that
we want to do something else

and we didn't finish high school
and we'll get a lot more money.

The person who came to
fix my air conditioner

said that he actually
makes about $100 an hour.

And I was thinking, wow.

Wow, I'll never get there.

But that's impressive.

Just changing some
things and fix,

press the button, $100 an hour.

STUDENT: But they don't
work full time. [INAUDIBLE].

PROFESSOR: Yeah, and I think
they are paid by the job.

But in any case, whether
it's a simple job

and they just-- there
is a contact that's

missing or something
trivial, they still

charge a lot of money.

STUDENT: [INAUDIBLE].

PROFESSOR: A professor?

STUDENT: [INAUDIBLE].

PROFESSOR: What?

[LAUGHING] No.

STUDENT: I know the professor.

I won't tell you who.

PROFESSOR: OK, I
don't want to know.

I don't want to know.

But anyway, it's interesting.

STUDENT: But he doesn't
do in the college.

He does outside the college
by just advising it.

PROFESSOR: Oh, you mean
like consulting or tutoring

or stuff like that?

STUDENT: Not tutoring,
consulting for the--

PROFESSOR: Consulting.

Actually, I bet that
if we did tutoring,

which we don't have time for,
we would make a lot of money.

But the nature of
my job, for example,

is that I work about
60 hours a week, 65,

and I will not have any time
left to do other things,

like consulting,
tutoring, and stuff.

STUDENT: [INAUDIBLE].

PROFESSOR: I don't need
normally that much.

I don't crave money that much.

STUDENT: [INAUDIBLE].

PROFESSOR: I have a
friend who got a masters.

She didn't get a PhD.

She got an offer from
this-- I told you about her.

She moved to California.

She was a single mom.

She earns a lot of
money working for Pixar.

And she helped with all
the animation things.

It was about 15 years
ago that she started.

And it was really hard.

We were all on Toy Story
and that kind of-- what

was that called?

There were two
rendering algorithms,

rendering algorithms.

Two masters students
were interested in that.

They got in immediately.

To be hired, I think
a post-doc with a PhD

was making about $40,000.

That was my offer.

My first offer was a post-doc
at Urbana-Champaign for $38,000

while she was at the hundred
and something thousand

dollars to start with
working at Disney.

Imagine-- with just a masters,
no aspiration for a PhD

whatever.

So in a way, if you're
thinking of doing this,

a masters in mathematics
is probably paying off.

Because it opens a
lot of doors for you.

And that's just in general.

I mean, masters in engineering
opens a lot of doors.

But in a way, you
pay a price after

if you want to start
even further, get a PhD,

stay in academia.

Then you pay a price.

And if you want to
augment your salary,

you really have to be very
good and accomplish some--

get [INAUDIBLE]
two or three times

and get higher up
each [INAUDIBLE].

But we all struggle
with these issues.

It's a lot of work.

But having a masters
in math is not so hard.

If you like math,
it's easy to get it.

It's a pleasure.

It's not a lot of hours.

I think in 36 hours in most
schools you can get a masters.

And it's doable.

All right, let's go back to
review Chapter 11 briefly here.

Is this on the midterm?

No, but it's going
to be on the final.

Assume you have a
x equals u plus v,

y equals u minus v. Write
the following derivative.

dx/dv where u of t equals
t squared and v equals t.

Do these both directly
and by writing a chain

rule for the values you have.

OK, how do we do this directly?

It's probably the simplest way.

Replace u by t squared, replace
v by t and see what you have.

So 1, directly.

X of t equals t squared plus t.

y of t equals t squared minus t.

Good.

So it's a piece of cake.
dx/dt equals 2t plus 1.

Unfortunately, this is just
the first part of the problem.

And it's actually
[INAUDIBLE] to show the chain

rule for the mappings we have.

And what mappings do we have?

We have a map from t
to u of t and v of t.

And then again, from u of t and
v of t to x of t and y of t.

And the transformation is
what? x equals u plus v,

y equals u minus v. And we have
another transformation here.

So how do you write dx/dt?

x is a function
of u and v, right?

So first you say
that dx/dv round,

which means we do it
with the first variable.

I'll write it for
you to see better,

that initially your x and y
were functions of u and v.

Times-- what is that?
dv/dt plus dx/du.

You can change the order.

If you didn't like
that I started with v,

I could have started with
the u, and the u, and the v,

and the v here.

It doesn't matter.

Guys, do you mind, really?

v, v. u, u.

Shooting cowboys?

Doesn't matter, remember just
that they're [INAUDIBLE].

D sorry, d.

Because where there
is no other variable,

we would put v. So dx/dt?

Lets see if we get
the same answer.

We should.

What is dx/dt?

1, from here.

What is dv/dt?

1 plus dx/du.

1, du/dt.

2t.

If we were to do
the same thing-- so

we got the same answer.

If you want to do
the same thing,

quickly with respect
to say dy/dt,

suppose that most finals
ask you to do both.

I have students
who didn't finish

because they didn't
have the time to finish,

but that was just my policy.

When I grade it,
I gave them 100%,

no matter if they
stopped here, because I

said you prove to me that
you know the chain rule.

Why would I punish you further?

So that's what I do.

But I want you to do it
now, without my help.

Both ways, dy/dt.

First you do it
with the chain rule.

First you write those
three [INAUDIBLE].

dy del y.

del u, du/dt, plus del y.

del v, dv/dt.

I'm not going to write it
down, you write it down.

What I'm going to
write down is what

you tell me the numbers are.

For everything.

STUDENT: Dy divided by d

PROFESSOR: Or just give me
the final answer in terms of

[INAUDIBLE].

What are the two [INAUDIBLE]?

Tell me.

Tell me, this times this,
plus this times that.

What?

So let's write down.

Let's write it down together.

dy/du, du/dt, plus dy/dv dv/dt.

Alright.

1 This is 1.

How much is dy/dt?

Or du/dt, I'm sorry.

I said dy, it's du/dt.

Plus minus 1, excellent.

Times 1.

Of course you would have
done the same thing,

by plugging in the
variables and saying well,

I have y, which is this
is t squared, this is t,

and I have t squared minus
t prime is 2t minus 1.

That's a simpler way
to verify [INAUDIBLE].

OK.

So remember to do that,
have this in mind,

because on the final you may
have something like that.

As we keep going in
the month of April,

I'm going to do as much review
as possible for the final.

Mark a star, or F, not the grade
F, but F around for the final,

put F and circle there to say
review this for the final.

And since we are still
in chapter 11 review,

we'll do another problem
of F, final review

that I didn't put on the midterm
but it may be on the final.

Let's say given the
constraint x squared

plus y squared plus z squared
equals 5, compute z sub x

and z sub y.

How do you do that?

What is this called,
actually, and why is it

so important for the final?

It's called implicit
differentiation

and it appears on almost every
final, at least once a year,

so there is always
a big possibility

that you are going to
see something like that.

I taught you how to think in
terms of implicit functions.

If you think of z as
a function of x and y.

That's a way of changing
your perspective.

So you say, OK, I
understand that z

has to be viewed as a
function of x and y.

I'm just changing
my perspective.

STUDENT: For that one,
wouldn't you just solve for z?

PROFESSOR: No.

Solving for z would make
your life a lot harder.

The point of
implicit functions is

that you don't separate them.

If you're going
to separate them,

you have to separately
integrate these.

And it's a headache.

It's easier-- actually
it's a good question.

It's easier to do z sub x,
z sub y without splitting it

into two cases.

Step two.

Differentiate this
with respect to x.

What do we have?

2x plus 0 plus the chain rule--
don't write the chain rule.

2 jumping down, it jumped down.

2z times-- cover the
2 with your hand.

z sub x, very good.

z prime with respect
to x equals zero.

Good.

So z sub x, step three.

And the last step.

See sub x will be what?

Pull it out.

Pull this 2 out.

Minus x over z.

The other one is symmetric.

Alex said let's be smart
and not do the whole thing

all over again.

Look at beautiful
symmetric polynomial.

You would have to be a
little bit careful with when

you have a 1 here and
y would have a 2 here.

It wouldn't be
symmetric in x and y.

But here, if you reverse
the roles of x and y,

it's not a big deal.

Are you guys with me?

Here we are. z sub y
equals minus y over z.

Am I right?

Keep this in mind
for-- I also saw,

when I was looking at
the [INAUDIBLE] library

files, [INAUDIBLE].

I also saw exams, and I was
looking at your reviews there.

I was looking at [INAUDIBLE].

The University of Houston has
a very beautiful online, free

library of calculus 1
and calculus 2 exams

that I found very useful.

Now, one of them--
listen to me so you

don't fall through this crack.

On the Cal 2 exam, they
wrote something like that.

You don't have to write 1 over
x squared, and then compute.

You just say, OK, if the natural
part of the of the argument

is 5, then the
argument is a constant.

And I don't care
what constant it

is, it it's something
that prime will give me 0,

it's the same problem.

Are you guys with me?

So in that case, I'm going
have just what kind of change?

This will be to the 5.

And I still have 0.

It's the same answer.

They just wanted to play
games, and you can play games.

For example, you can make this.

If you really have
a working mind,

and most mathematicians do,
give this to your students.

I mean, most people
freak out so bad

when they see that, the
won't even touch it.

It's just all in the head.

Remember that log in base
17 of a would be what?

STUDENT: If it's a
constant, it's to the 17th.

PROFESSOR: Who knows?

STUDENT: What do you
mean, you don't do that?

PROFESSOR: No, no, expressed
in terms of natural logs.

STUDENT: Natural log?

The natural log of a
over natural log of 17.

PROFESSOR: Very good.

So what does this matter?

In the end, you multiply 2, you
do the derivative, you still

get the same answer.

Some people are trying
to make things scarier

than they are, just to impress.

When you think of the
problem, it's a piece of cake.

So don't be afraid of it.

Oh, by the way, the final
exam-- so the midterm would

be 10 problems pus 1 extra one.

And did I tell
you how much time?

It's going to be approximately--
I say, in actual time.

Needed time.

For average student,
it'll be about 40 minutes.

Allowed time one
hour and 40 minutes.

So you have from 12:10 to 1:50.

On the final, just a guess,
about 15-16 problems.

Two hours and a half.

STUDENT: Is that allowed time?

PROFESSOR: Not allowed time.

If I manage to review very
well with you on these concepts

guys, I guarantee you're not
going to need more than 1.5.

This is the allowed time.

The allowed time for somebody
who hasn't practiced enough.

Let me ask you what you
think would be good.

I have a bunch of finals.

All the finals for Cal 3
look very similar in nature.

The same kind of topics
as the ones I review.

I would like to know
what you would prefer.

I would have two or
three finals to give you.

Would you prefer that you
try them yourselves first,

and then I give
you the solutions?

STUDENT: Yes.

PROFESSOR: Or I give you the
solutions from the beginning?

I'll give you the
solutions anyway, but--

STUDENT: Can it just
be on a separate sheet,

where we could go through--

PROFESSOR: No, no, they are
already on a separate sheet.

For example, I have Fall
2013, or Spring 2012.

They are from
different semesters.

They are all very similar.

So I'll give you-- I have
two files on this blog.

The exam itself
and the solutions.

I'll give you the exam, I'll
let you work if for two weeks,

and then I'll give
you the solutions.

How about that?

Put you'll work on it, you
don't cheat on me and any way.

Because working things
yourself, you're learning.

If you expect other people
to feed you the solutions,

you're not learning as much.

You are learning some, but
you're not learning as much.

OK, it's getting ready.

I have a few more
things to tell you.

Chapter 13, necessary reminders.

The gradient is very important.

Gradient of a function
f from r 2 to r.

We write that as z equals
f of x and y, usually.

And what was the gradient?

This is good review
for the midterm,

but that's the beginning
of section 13.1.

So I'm actually
doing two things,

I'm giving you the
beginning of section 13.1,

while doing review
for the final.

You have gradient of f of x,
y-- some people are ask me,

do you prefer that I
write on the exams,

on the midterm, on the
final a granular bracket?

Or do you prefer I write this in
this form in the standard base

i, j.

Standard [INAUDIBLE].

It doesn't make a difference.

In linear algebra,
you would have

to say what bases you are using.

But in calculus, we
assume that you are using

the bases which is 1, 0, 0, 1.

So you have space in a plane.

I'm indifferent.

This is OK, you can
use whatever you like.

If you have a function of
three variables, of course

you have a gradient.

But I prefer to write f sub x,
i plus f sub y, j plus f sub z,

the beginning of some ck.

Has anybody heard of
divergence before?

What is divergence?

Gradient is something
you've heard before.

But divergence, have you
ever heard of divergence?

Maybe in mechanical engineering,
have you heard of it before?

No?

OK.

Suppose that you
have a function,

and that is a vector
value function.

What does it mean?

A vector in itself will
have coordinates at x, y.

And it's assumed that
will be f1 of x, y-- no,

this is not a vector,
that's scalar.

Times i, plus f2 x, y, j.

And somebody, one of you
actually showed me-- of course

in mechanics-- you were
using divergence in that.

And I feel bad that I was
not the first maybe for some

of you, I was not the first to
tell you what divergence means.

Divergence f, assuming that f
would be missing one function.

What does this mean?

It means that it's differential,
but its derivatives

are continuous.

We did note that this
is the diff of f.

But in engineering, they denoted
most of the time like that.

There's not a lot of symbols,
but you saw the gradient

with a little dot after that.

If you don't put the dot,
it doesn't make sense

with what I'm saying.

So pay attention to the dot.

Alright.

What does this mean?

It means that you
have the derivative

of the first component
with respect to x.

Plus it's going to
be a value function,

the derivative of the second
component with respect to y.

How do you generalize
for higher powers?

What if you have a function--
assume you have a function

f that looks like that.

If x1, x2, x n variables,
i plus the last one will be

a [INAUDIBLE] of x1, x2
x n variables times--

eij doesn't make any sense.

So e1, e2, e n would
be the standard bases.

[INAUDIBLE] doesn't
make the [INAUDIBLE]

for a computer scientist,
an ordered set of components

and values.

And would be 7, 17, 29.

Some natural numbers.

So all these values are taken
in r, with every r x is in on.

What do you think that
the divergence of u

would be in that case?

If you were to generalize y.

First component, prime with
respect to the first variable.

Alright.

Only plus second component with
respect to the second variable,

and so on.

Last component with respect
to the last variable.

So that would be the
general definition.

And now I'm asking you, assume
x that somebody gives you

a function, f of x, y.

And with domain in the plane.

And f is c1.

[INAUDIBLE] with
continuous radius.

Actually no, I want more.

I want c2.

So twice differential bond,
with continuous variables.

Compute a.

Gradient double f.

b, divergence of gradient
of f, which you can also

write divergence like engineers
do, or gradient to our left.

Do you know what name that's
the last thing we need today,

the name for this operator.

Underlined here.

So what would be a good
name for this kind?

I'm curious if any of you
know if from engineering.

But we will see.

So we are in 13.

a will be the gradient of
f, that's a piece of cake.

She only wants the
definition, let

me give her the definition
of f sub xi, plus f sub y j.

And if we don't
know what those are,

this is the variable
with respect to x.

And then for dy, df/dy j.

Good.

So we know what a gradient is.

What will this divergence
with the gradient be?

That sounds really weird.

According to this
definition, we have

to see what big
F1 and big F2 are.

Or, big F1 and big F2.

I'm going to take
them in breaths.

Big F1, and big F2.

The components of the vector,
you apply divergence to it.

So now that I'm finishing,
what do I have to do?

Somebody tell me.

So yeah, I can write it
f sub x plus f sub y,

and that shows that you are
fast, and very [INAUDIBLE].

I can also write
it like this, which

is what I meant-- this is what
the book shows first course.

This is the same thing.

Now I really doubt that
somebody knows that,

but I want to give a
dollar to the person who

would know the name of this.

Let me see if I have a dollar.

Maybe I have $0.35 and a candy.

Does anybody know
the name of this?

Maybe I can help you a little
bit. $0.25 $0.85, $0.95.

Do you know what this is?

I'll give you a hint, because I
know in mechanical engineering,

I already introduced this.

And some physics classes and
we would try angle in front,

and we did all of this
triangle operators in the way.

And we can play a game.

It's a letter that
starts with L.

But $0.95 we have
two more minutes.

STUDENT: [INAUDIBLE]?

PROFESSOR: No.

You are getting close though,
because-- [INTERPOSING VOICES]

What kind of operator is this?

You're getting close, $0.95.

Tomorrow, I don't need this.

When I go to the
airports, I don't

like to have coins with me.

STUDENT: Laplace?

PROFESSOR: $0.95!

I wish I had a dollar.

Yes, this is the famous
Laplace operator.

Laplace was a mathematician.

And remember it.

If you take-- how
many of you-- you all

have to take differential
equations, right?

They will kill you with that.

You're going to see
this all the time.

This Laplace operator
is really famous.

I will tell you more
when I come back.

I'm going to see you on Tuesday.

We'll knock out the midterm.

For you, the people who feel
overly prepared for midterm

can go ahead and
read section 13.1

and see a little bit
about Laplace's operator.

[INTERPOSING VOICES]