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.
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]