TTU Math2450 Calculus3 Sec 11.2 and 11.3

Functions of several variables: continuity and partial derivatives

People asked me if I'm
going to go over homework.

Of course I will.

Let me explain.

Out of the four
hours you have, three

should be more or
less lecture time.

And the fourth hour, which
is the instructor's latitude,

where they put it-- it's
applications, problems,

homework like problems, all
sorts of practice for exams

and so on.

It's not a recitation.

It's some sort of workshop that
the instructor conducts himself

personally.

All right.

If you don't have
questions, I'm just

going to go ahead and
review a little bit of what

we discussed last time.

Something new and exciting
was chapter 11, section 11.1.

And we did 11.2.

And what was that about?

That was about functions
of several variables.

And we discussed
several examples,

but then we focused
our attention mainly

to explicit functions, which
means z equals f of x, y,

of two variables.

And we call this a graph
because it is a graph.

In 3D, it's a surface whose
domain is on the floor.

And the altitude is z, and
that is the-- this is the-- OK.

How many of you are
non-math majors?

Can you raise hands?

Oh, OK.

So you know a little
bit about research

from your own classes,
science classes

or from science
fairs from school.

These are the independent
variables, x, y.

And z is the dependent variable.

We don't use this kind of
terminology in this class.

But so that you know-- we
discussed domain last time.

This was about what?

Domain, range.

After range, what did we do?

We talked about level curves.

What is the level curve?

Level curves are curves x,
y in the plane corresponding

to f of x, y equals constant.

These are called
level curves in plane,

in the plane called x, y plane.

What else have we discussed?

We went straight into 11.2.

In 11.2, we were very
happy to remember

a little bit of Calculus
1, which was practically

a review of limits from Calc 1.

And what did we do?

We did epsilon delta, which
was not covered in Calculus 1.

And where is Aaron?

OK.

Thank you, Aaron.

And today, I was thinking,
I want to show you actually

an example that is quite
easy of how you use epsilon

delta for continuity, to show
if the function is continuous,

but for a function
of true variables.

And that's not hard.

You may think, oh, my god.

That must be hard.

That's not hard at all.

I'm going to move on to the
second part of 11.2, which

is continuity.

11.2, second part.

The first part was what?

It was limits of
functions, right, guys?

We discussed
properties of limits,

algebraic properties of
adding sums and taking a limit

of a sum, taking a limit
of a product of functions,

taking the limit of a quotient
of function, when it exists,

when it doesn't.

Now the second part of
11.2 is called continuity.

Continuity of what?

Well, I'm too lazy
to right down,

but it's continuity of functions
of two variables, right?

Now in Calc 1-- you
reminded me last time.

I tried to remind you.

You tried to remind me.

Let's remind each other.

This is like a discussion.

What was the meaning of f of x
being a continuous function x0,

which is part of the domain?

x0 has to be in the domain.

This is if and only if what?

Well, what kind of
function is that?

A one variable
function, real value.

It takes values on, let's say,
an interval on the real line.

What was the group
of properties that

have to be
simultaneously satisfied,

satisfied at the same time?

And you told me it has
to be at the same time.

And I was very happy because
if one of the three conditions

is missing, then
goodbye, continuity.

One?

STUDENT: It's defined
at that point.

MAGDALENA TODA: Yes,
sir. f of x0 is defined.

Actually, I said that
here in the domain.

I'll remove it because
now I said it better.

Two?

STUDENT: The limit exists.

MAGDALENA TODA: Very good.

The limit, as I approach
x0 with any kind of value

closer and closer,
exists and is finite.

Let's give it a name.

Let's call it L.

STUDENT:
[? The following value ?]

equals the limit.

MAGDALENA TODA: Yes, sir.

That's the last thing.

And I'm glad I didn't
have to pull the truth out

of your mouth.

So the limit will-- the limit
of f of x when x goes to x0

equals f of x0.

No examples.

You should know
Calc 1, and you do.

I'm just going to
move on to Calc 3.

And let's see what the
definition of continuity

would mean for us in Calc 3.

Can anybody mimic the properties
that-- well, f of x, y

is said to be
continuous at x0, y0

if and only if the following
conditions are-- my arm hurts.

Are simultaneously satisfied.

I don't like professors who
use PDF files or slides.

Shh.

OK.

I don't want anything premade.

The class is a
construction, is working,

is something like
a work in progress.

We are building things together.

This is teamwork.

If I come up with
some slides that were

made at home or a PDF file.

First of all, it means I'm lazy.

Second of all, it
means that I'm not

willing to take it
one step at a time

and show you how
the idea's revealed.

One.

Who is telling me?

I'm not going to say it.

It's a work in progress.

STUDENT: [INAUDIBLE]

MAGDALENA TODA: f of--

STUDENT: [INAUDIBLE]

MAGDALENA TODA: Of
x0, y0 is defined.

And why not?

Well, just to have
a silly [? pun ?].

Two.

Limit as the pair x, y
approaches x0, x0-- and guys,

when you close your eyes--
no you close your eyes--

and you imagine
x, y going to x0,

y0 by any possible paths
in any possible way,

it's not that you have a
predetermined path to x0, y0,

because you may be trapped.

You may have-- as you've
seen last time, you may have,

coming from this direction,
the limit will exist,

will be this one.

Coming from that direction,
the limit will exist,

would be another one.

And then you don't
have overall limits.

So the limit-- when I call that,
that means the overall limit

exists, exists and
equals L. It's finite.

That's what I mean.

And three, the value
of the function at x0,

y0 must be equal to the limit
of the function that value

as you approach it, x0, y0.

And equals L, of course.

So great.

So it's so obvious
that we are following

exactly the same
type of definition,

the same type of pattern.

I'm going to ask you
to help me, to help

me solve a harder problem
that involves continuity.

And I'm asking you, if I
have the following function--

I'm going to erase the
definition of continuity

from Calc 1.

I'm going to ask you, what if
I have this funny function?

You've seen it before, and
I gave you a little bit

of a warning about it.

Limit as x, y goes
to 0, 0 of x squared

plus y squared times sine of 1
over x squared plus y squared.

Does that exist?

And also--

STUDENT: It's actually--
so the limit is actually

approaching a plane rather
than a set of [INAUDIBLE].

MAGDALENA TODA: So
well, actually, it's

not approaching a plane.

Let's see what's
happening when--

STUDENT: Sorry, sorry.

Not a plane, a [? line. ?]

MAGDALENA TODA: Yes.

STUDENT: And is the z-axis--
the entire z-axis is 0, 0?

MAGDALENA TODA: So
this is the z-axis.

And that means exactly that
x and y-- it will be 0.

Now I am just looking
at what happens

in the plane, in the
floor plane x, y.

The pairs x, y are wiggly.

They are like
little wormy worms.

And they float on the
water on the floor.

And these squiggly
things approach

x, y from any possible path.

They go like this.

They go like that.

They go in every possible way.

Let's see what happens.

Continuity-- is this continuous?

Well, you say,
Magdalena, come on.

You cannot have this
continuous at 0, 0,

because it's undefined at 0, 0.

Yes.

But maybe I can extend
it by continuity.

So let me introduce-- this
is my favorite, f of x, y.

But I'll say, f of x, y
is not defined at 0, 0.

But how about g of x, y as
being my f of x, y for any x,

y different from 0, 0.

And at the origin, at the
very origin, I will say,

I want to have--
when x, y equals 0,

0, I want to have a value.

Which value do you
think might extend

this function by continuity?

STUDENT: The limit.

MAGDALENA TODA: The
limit if it exists

and if-- well, you know already,
I think, what the limit is

because some of you
thought about this at home

for extra credit.

So it's not fair, right?

No, I'm just kidding.

So I claim that maybe--
if I put a 0 here,

will this be continuous?

Will g be continuous?

So prove, prove either way,
prove, justify your answer

by a proof, a complete
proof with epsilon delta.

Proof.

OK.

OK.

So now is a worried face.

Like, oh, my god.

This guy is worried
because, oh, my god.

Epsilon delta.

Oh, my god.

But the principle--
the intuition

tells us that we should look
first at some sort of a graph,

just like Ryan pointed out.

One should close their eyes and
imagine a graph of a function

with-- it's hard to visualize in
3D the graph of a function that

is a surface.

This is a surface. z
equals the whole shebang.

But when I'm going to look
at the one dimensional case

from last time, we
remember the sine of 1/x

was a crazy function.

We called it the harmonica,
well, 20-something years ago

when I was in high school.

I was in an advanced
calculus class.

And our teacher was
not funny at all.

He was also not teaching much,
gave us a lot of homework,

very challenging.

So in order to make our
life a little bit easier,

we always worked in
groups, which was allowed.

So we called it a harmonica
because it was oscillating

like that to the point
that-- you've seen

the harmonica-- the accordion.

When you bring it back to
the-- harmonica came to my mind

from the harmonic function.

So the accordion is--
when you actually

squeeze it, all that oscillation
things, the cusps are

closer and closer to a line.

So what you have here is
this kind of oscillation,

very, very rapid
oscillation for sine of 1/x.

When we want to multiply by
an x, what's going to happen?

Well, this has not limit at 0
because it takes all the values

infinitesimally close to 0.

It keeps going through all the
values between minus 1 and 1,

closer and closer.

So that was no good.

But if we take this guy--
that's going to go to-- well,

I cannot do better.

MATLAB can do better than me.

Mathematica can do better.

You can do that.

In most engineering
classes, if you are--

who is an electrical
engineering major?

But even if you are
not, you are going

to see this type
of function a lot.

And you're going to see it
again in differential equations.

How can I imagine-- this
graph is hard to draw.

Don't ask me to draw that.

But ask me if I can use epsilon
delta to prove continuity.

So what would it mean,
proving continuity?

I have a feeling--

STUDENT: Well, actually, if this
is-- going back to that graph,

doesn't that graph look like--

MAGDALENA TODA: This goes to 0.

The limit exists for x
sine of 1/x, and it is 0.

Why?

Ryan?

RYAN: Wouldn't the graph
with the x squared plus

y squared times that
side-- wouldn't that

just look like a ripple
in a circle going out

from the center?

MAGDALENA TODA: Yeah,
it will be ripples.

STUDENT: Just like a
[INAUDIBLE] from an epicenter

going outwards [INAUDIBLE].

MAGDALENA TODA: And I think--
yes, we managed to-- you

have a concentric image, right?

STUDENT: Yeah.

MAGDALENA TODA: Like those
ripples, exactly like--

STUDENT: So that's
what that looks like?

MAGDALENA TODA: --when you
throw a stone into the water,

this kind of wave.

But it's infinitesimally close.

It's like acting weird.

But then it sort
of shrinks here.

And that-- it
imposes the limit 0.

How come this goes
to 0, you say?

Well, Magdalena, this
guy is crazy, right?

Sine of 1/x goes
between minus 1 and 1

infinitely many times
as I go close, close,

closer and closer, more rapidly,
more and more rapidly close

to 0.

This will oscillate
more rapidly,

more rapidly, and more rapidly.

This is crazy, right?

How does this guy, x-- how
is this guy taming this guy?

STUDENT: Because
as 0 [INAUDIBLE].

Something really small
times something [INAUDIBLE].

MAGDALENA TODA:
Something very small

that shrinks to 0 times
something bounded.

Ryan brought the main idea.

If something goes strongly to
0, and that multiplies something

that's bounded, bounded
by a finite number,

the whole problem will go to 0.

Actually, you can prove
that as a theorem.

And some of you did.

In most honors
classes unfortunately,

epsilon delta was not covered.

So let's see how we prove
this with epsilon delta.

And, oh, my god.

Many of you read from the book
and may be able to help me.

So what am I supposed to
show with epsilon delta?

The limit of x squared plus
y squared sine of 1 over x

squared plus y squared is
0 as I approach the origin

with my pair, couple, x, y,
which can go any one path that

approaches 0.

So you say, oh, well, Magdalena,
the Ryan principle-- this

is the Ryan theorem.

It's the same because
this guy will be

bounded between minus 1 and 1.

I multiplied with a guy
that very determinedly goes

to 0 very strongly.

And he knows where he's going.

x squared plus y squared
says, I know what I'm doing.

I'm not going to change my mind.

This is like the guy who changes
his major too many times.

And this guy knows
what he's doing.

He's going there, and he's
a polynomial, goes to 0,

0 very rapidly.

Now it's clear what
happens intuitively.

But I'm a mathematician.

And if I don't publish
my proof, my article

will be very nicely rejected
by all the serious journals

on the market.

This is how it goes
in mathematics.

Even before journals
existed, mathematicians

had to show a rigorous
proof of their work,

of their conjecture.

OK.

So I go, for every epsilon
positive, no matter how small,

there must exist a
delta positive, which

depends on epsilon-- that
depends on epsilon-- such that

as soon as-- how did
we write the distance?

I'll write the distance
again because I'm lazy.

The distance between the
point x, y and the origin

is less than delta.

It follows that the
absolute value--

these are all real numbers--
of f of x, y or g of x,

y-- g of x, y is the extension.

f of x, y minus 0, which
I claim to be the limit,

will be less than epsilon.

So you go, oh, my god.

What is this woman doing?

It's not hard.

I need your help though.

I need your help to do that.

So it's hard to see how you
should-- you take any epsilon.

You pick your favorite
epsilon, infinitesimally small,

any small number, but
then you go, but then I

have to show this delta exists.

You have to grab that delta
and say, you are my delta.

You cannot escape me.

I tell you who you are.

And that's the
hardest part in here,

figuring out who that delta must
be as a function of epsilon.

Is that hard?

How do you build
such a construction?

First of all,
understand what proof.

"Choose any positive epsilon."

Then forget about him,
because he's your friend,

and he's going to do whatever
you want to do with him.

Delta, chasing
after delta is going

to be a little bit harder.

"Chasing after delta
with that property."

Dot, dot, dot, dot, dot.

What is this distance?

You guys have
helped me last time,

you cannot let me down now.

So as soon as this distance,
your gradient distance

is less than delta,
you must have

that f of x, y [INAUDIBLE].

Could you tell me
what that would be?

It was Euclidean, right?

So I had squared
root of-- did I?

Square root of x minus 0
squared plus y minus 0 squared.

You say, but that's
silly, Magdalena.

So you have to write
it down like that?

STUDENT: It's the [INAUDIBLE].

MAGDALENA TODA: Huh?

Yeah.

So square root of this
plus square root of that

plus then delta,
that means what?

If and only if x squared plus
y squared is less than delta

squared.

And what do I want to do,
what do I want to build?

So we are thinking how
to set up all this thing.

How to choose the delta.

How to choose the delta.

OK, so what do I--
what am I after?

"I am after having" double dot.

F of x, y must be Mr. Ugly.

This one.

So absolute value of x squared
plus y squared, sine of 1

over x squared plus
y squared minus 0.

Duh.

I'm not going to write it.

We all know what that means.

Less than epsilon.

This is what must
follow as a conclusion.

This is what must
follow, must happen.

Must happen.

Now I'm getting excited.

Why?

Because I am thinking.

I started thinking.

Once I started thinking,
I'm dangerous, man.

So here sine of 1 over x squared
plus y squared is your friend.

Why is that your friend?

Sine of 1 over x squared
plus y squared, this

is always an absolute value.

The absolute value of that
is always less than 1.

OK?

STUDENT: Can't it be 4?

MAGDALENA TODA:
So-- so-- so what

shall I take in terms of
delta-- this is my question.

What shall I take
in terms of delta?

"Delta equals 1 as a
function of epsilon

in order to have the
conclusion satisfied."

You say, OK.

It's enough to choose delta
like that function of epsilon,

and I'm done, because then
everything will be fine.

So you chose your own epsilon,
positive, small, or God

gave you an epsilon.

You don't care how
you got the epsilon.

The epsilon is arbitrary.

You pick positive and small.

Now, it's up to
you to find delta.

So what delta would
satisfy everything?

What delta would
be good enough--

you don't care
for all the good--

it's like when you get married.

Do you care for all the
people who'd match you?

Hopefully not, because
then you would probably

have too large of a pool,
and it's hard to choose.

You only need one that satisfies
that assumption, that satisfies

all the conditions you have.

So what is the delta that
satisfies all the conditions

that I have?

[INTERPOSING VOICES]

MAGDALENA TODA: [INAUDIBLE].

Who?

[INTERPOSING VOICES]

MAGDALENA TODA: For example,
delta equals epsilon.

Would that satisfy?

Well, let's see.

If I take delta to
be epsilon, then x

squared plus y squared would
be less than epsilon squared.

Now the question is is epsilon
squared less than epsilon?

Not always.

If epsilon is between 0
and 1, then epsilon squared

is less then epsilon.

But if I choose epsilon
to be greater than 1,

then oh, my God.

Then if it's greater than
1, then epsilon squared

is greater than 1--
greater than it.

So what if I choose
delta to be what?

STUDENT: 0?

MAGDALENA TODA: No, no, no.

Delta cannot be 0.

So delta-- look, there exists
delta strictly bigger than 0,

that depends on epsilon.

Maybe if epsilon is very small,
in a way Alexander was right.

But the delta [INAUDIBLE],
we don't go with epsilon

greater than 1.

Come on.

Be serious.

Epsilon is always
between 0 and 1.

I mean, it's a lot
smaller than that.

It's infinitesimal small.

So in the end, yes, in
that case epsilon squared

would be less than epsilon,
which would be OK for us

and that would be fine.

OK?

So that would be a
possibility to say, hey,

since epsilon-- Alexander,
if you write that as a proof

I'll be OK.

You say, I took my epsilon
to be a very small number,

so anyway it's going
to be less than 1.

So epsilon squared
is less than epsilon.

So when I take
delta to be epsilon,

for sure this guy will be less
than epsilon squared, which

is less than epsilon,
so I'm satisfied.

I'll give you a 100%.

I'm happy.

Is that the only way?

STUDENT: But what
about the sine?

What about [INAUDIBLE].

STUDENT: Yeah.

MAGDALENA TODA: So
this doesn't matter.

Let me write it down.

So note that x squared
plus y squared sine of 1

over x squared plus
y square would always

be less than absolute
value of x squared

plus y, which is positive.

Why is that?

Is this true?

Yeah.

Why is that?

STUDENT: Because the sine can
only be one of these negatives.

MAGDALENA TODA: So
in absolute value,

sine of 1 over x squared plus y
squared is always less than 1.

STUDENT: Can't it equal 1?

MAGDALENA TODA: Well,
when does it equal 1?

STUDENT: Wouldn't it be x
squared plus y squared equals 1

[INAUDIBLE]?

MAGDALENA TODA: Less
than or equal to.

For some values it will.

STUDENT: Yeah.

OK.

MAGDALENA TODA: Now, will
that be a problem with us?

No.

Let's put it here.

Less than or equal to x
squared plus y squared, which

has to be less than epsilon
if and only if-- well,

if delta is what?

So, again, Alexander said,
well, but if I take delta

to be epsilon, I'm done.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: How
about square root?

Can I take delta to be
square root of epsilon.

STUDENT: That's what I said.

MAGDALENA TODA: No.

You said epsilon.

STUDENT: I said square
root of epsilon.

MAGDALENA TODA: OK.

If delta is square
root of epsilon,

then everything will be perfect
and it will be a perfect match.

In what case?

STUDENT: If epsilon
is in between 0 and 1

and if delta is equal
to bigger than epsilon.

MAGDALENA TODA: So that's
exactly the same assumption.

Epsilon should be
made in less than.

STUDENT: But I thought
delta was supposed

to be less than
epsilon in every case.

So if epsilon is between 0 and
1, the square root of epsilon

is going to be [INAUDIBLE].

MAGDALENA TODA: So when
both of them are small,

delta squared will be-- if
I take delta-- so take delta

to be square root of epsilon.

STUDENT: Then anything less
than 1 and greater than 0,

epsilon would be great
than [INAUDIBLE].

MAGDALENA TODA: "Delta to
be square root of epsilon,

then x squared plus y squared
less than delta squared equals

epsilon."

Then x squared plus
y squared sine of 1

over x squared plus
y squared less than

or equal to x squared
plus y squared.

I dont' need the absolute value.

I can [INAUDIBLE].

Less than epsilon [INAUDIBLE].

Qed.

STUDENT: Well, but
you told us delta

has to be less than epsilon.

Well, if--

MAGDALENA TODA: No,
I didn't say that.

I didn't say that delta has
to be less than epsilon.

Absolutely--

STUDENT: Yeah.

You said for all the values
of epsilon greater than 0,

there's a value of delta that is
greater than 0 that [INAUDIBLE]

such that as soon as the
distance between is less than

delta-- I don't remember what--

MAGDALENA TODA: OK, so, again--

STUDENT: Such that the
distance is less than--

MAGDALENA TODA: So again,
for epsilon positive,

there is a delta
positive, very small.

Very small means very small, OK?

I'm not threatened by-- what?

For epsilon greater
than 0, very small,

there is a delta greater
than 0, very small,

which depends on epsilon-- I
didn't say it cannot be equal

to epsilon-- that depends on
epsilon such that whenever x,

y is within delta
distance from origin,

[INAUDIBLE] that f of x, y
is within epsilon of from l.

All right?

And now I will actually give
you another example where

maybe delta will be epsilon.

And let me challenge you
with another problem that's

not hard.

OK?

So let me give
you the function g

of x, y equals x sine
of 1 over y as x, y.

y is equal [? to delta 0. ?]
And let's say 0 for the rest.

Can you show-- can you check
if g is continuous at 0, 0?

This is one of the
problems in your book.

So how do you check
that with epsilon delta?

Again, we recite the poetry.

We have to say that.

"For every epsilon
positive, small, very small,

there is a delta
positive that depends

on epsilon, such that as soon
as--" how is the distance?

Square root of x squared plus
y squared is less than delta.

This is the distance
between point and origin.

"It follows that absolute value
of x sine of 1 over y minus--"

so practically x, y no 0.

x, y different from 0.

OK?

I"m careful here, because
if y is 0, then I blow up.

And I don't want to blow up.

So x sine of 1 over y minus who?

Minus 0 is less than epsilon.

So now you're thinking,
OK, you want me

to prove there is such a delta?

Yes.

That depends on epsilon?

Yes.

And what would that delta be?

The simplest choice you
can have in this case.

So you go, oh, my God.

How do I do that?

You have to always
think backwards.

So "we need to satisfy
absolute value of x sine of 1

over y less than epsilon."

Is this hard?

What is your advantage here?

Do you have any advantage?

Remark absolute value
of x sine of 1 over y

is smaller than who?

Smaller than the product
of absolute values.

Say it again?

Yes?

STUDENT: But, like, for
example, the only condition

for that equation is that
y must not be equal to 0.

What if you used
another point for x?

Would the answer for
delta be different?

MAGDALENA TODA:
Well, x is-- you can

choose-- you were right here.

You can say, OK, can you be
more restrictive, Magdelena,

and say, for every point
of the type x equals 0

and y not 0, it's still OK?

Yes.

So you could be a
professional mathematician.

So practically all I care
about is x, y in the disk.

What disk?

What is this disk?

Disk of radius 0 when--
what is the radius?

Delta-- such that your
y should not be 0.

So a more rigorous
point would be

like take all the
couples that are

in this small disk
of radius delta,

except for those where y is 0.

So what do you actually remove?

You remove this stinking line.

But everybody else in this
disk, every couple in this disk

should be happy,
should be analyzed

as part of this thread.

Right?

OK.

x sine of 1 over y less
than-- is that true?

Is that less than the
absolute value of x?

STUDENT: Yeah.

MAGDALENA TODA: Right.

So it should be-- less
than should be made

should be less than epsilon.

When is this happening
on that occasion?

If I take delta-- meh?

STUDENT: When delta's epsilon.

MAGDALENA TODA: So if
I take-- very good.

So Alex saw that, hey,
Magdelena, your proof is over.

And I mean it's over.

Take delta, which is delta
of epsilon, to be epsilon.

You're done.

Why?

Let me explain what
Alex wants, because he

doesn't want to explain
much, but it's not his job.

He's not your teacher.

Right?

So why is this working?

Because in this case,
note that if I take delta

to be exactly epsilon,
what's going to happen?

x, Mr. x, could be
positive or negative.

See, x could be
positive or negative.

Let's take this guy and
protect him in absolute value.

He's always less than square
root of x square plus y

squared.

Why is that, guys?

STUDENT: Because y can't be 0.

MAGDALENA TODA: So this
is-- square it in your mind.

You got x squared less than
x squared plus y squared.

So this is always true.

Always satisfied.

But we chose this to
be less than delta,

and if we choose delta to be
epsilon, that's our choice.

So God gave us the epsilon,
but delta is our choice,

because you have to prove
you can do something

with your life.

Right?

So delta equals epsilon.

If you take delta
equals epsilon,

then you're done, because
in that case absolute value

of x is less than epsilon, and
your conclusion, which is this,

was satisfied.

Now, if a student
is really smart--

one time I had a student,
I gave him this proof.

That was several
years ago in honors,

because we don't do epsilon
delta in non-honors.

And we very rarely do
it in honors as well.

His proof consisted of this.

Considering the fact that
absolute value of sine

is less than 1, if I
take delta to be epsilon,

that is sufficient.

I'm done.

And of course I gave
him 100%, because this

is the essence of the proof.

He didn't show any details.

And I thought, this is the
kind of guy who is great.

He's very smart, but he's not
going to make a good teacher.

So he's probably going to
be the next researcher,

the next astronaut, the next
something else, but not--

And then, years later, he
took advanced calculus.

He graduated with
a graduate degree

in three years sponsored
by the Air Force.

And he works right
now for the Air Force.

He came out dressed
as a captain.

He came and gave a talk this
year at Tech in a conference--

he was rushed.

I mean, if I talk
like that, my student

wouldn't be able to follow me.

But he was the same brilliant
student that I remember.

So he's working on some very
important top secret projects.

Very intelligent guy.

And every now and than going
to give talks at conferences.

Like, research talks
about what he's doing.

In his class-- he took
advanced calculus with me,

which was actually graduate
level [INAUDIBLE]--

I explained epsilon delta, and
he had it very well understood.

And after I left the classroom
he explained it to his peers,

to his classmates.

And he explained
it better than me.

And I was there listening,
and I remember being jealous,

because although
he was very rushed,

he had a very clear
understanding of how

you take an epsilon, no
matter how small, and then

you take a little ball
here, radius delta.

So the image of that little
ball will fit in that ball

that you take here.

So even if you
shrink on the image,

you can take this
ball even smaller

so the image will
still fit inside.

And I was going, gosh,
this is the essence,

but I wish I could convey
it, because no book

will say it just-- or show you
how to do it with your hands.

STUDENT: [INAUDIBLE]

MAGDALENA TODA: Right.

So he was rushed, but he
had a very clear picture

of what is going on.

OK.

11.3 is a completely new start.

And you are gonna read
and be happy about that

because that's
partial derivatives.

And you say, Magdalena,
finally, this is piece of cake.

You see, I know these things.

I can do them in
my-- in my sleep.

So f of x and y
is still a graph.

And then you say,
how do we introduce

the partial derivative with
respect to one variable only.

You think, I draw the graph.

OK.

On this graph, I
pick a point x0, y0.

And if I were to take x to
be 0, what is-- what is the z

equals f of x0, y?

So I'll try to draw it.

It's not easy.

This is x and y and z, and you
want your x0 to be a constant.

STUDENT: [INAUDIBLE]

MAGDALENA TODA: It's a
so-called coordinate curve.

Very good.

It's a curve, but I want to
be good enough to draw it.

So you guys have
to wish me luck,

because I don't-- didn't have
enough coffee and I don't feel

like I can draw very well.

x0 is here.

So x is there, so you
cut with this board-- are

you guys with me?

You cut with this board
at the level x0 over here.

You cut.

When you cut with
this board-- you

cut your surface
with this board--

you get a curve like that.

And we call that a
curve f of x0, y.

Some people who are a little bit
in a hurry and smarter than me,

they say x equals x0.

That's called coordinate curve.

So, the thing is, this--
it's a curve in plane.

This is the blue plane.

I don't know how to call it.

Pi.

You know I love to call it pi.

Since I'm in plane with
a point in a curve--

a plane curve-- this curve
has a slope at x0, y0.

Can I draw that slope?

I'll try.

The slope of the
blue line, though.

Let me make it red.

The slope of the red line--
now, if you don't have colors

you can make it a dotted line.

The slope of the dotted line
is-- who the heck is that?

The derivative of f with respect
to y, because x0 is a constant.

So how do we write that?

Because x0 is sort of in
our way, driving us crazy.

Although he was fixed.

We keep him fixed by
keeping him in this plane.

x0 is fixed.

We have to write
another notation.

We cannot say f prime.

Because f depends
on two variables.

f prime were for when we
were babies in calculus 1.

We cannot use f prime anymore.

We have two variables.

Life became too complicated.

So we have to say--

STUDENT: Professor?

MAGDALENA TODA: --instead
of df dy-- yes, sir.

May you use a subscript?

MAGDALENA TODA: You use--
yeah, you can do that as well.

That's what I do.

Let me do both.

f sub y at-- who
was fixed? x0 and y.

But this is my
favorite notation.

I'm going to make a
face because I love it.

This is what engineers love.

This is what we physicists love.

Mathematicians, though,
are crazy people.

They are.

All of them.

And they invented
another notation.

Do you remember
that Mr. Leibniz,

because he had nothing better to
do, when he invented calculus,

he did df dy, or df dx?

What is that?

That was the limit of
delta f, delta y, right?

That's what Leibniz did.

He introduced this
delta notation,

and then he said if you have
delta space over delta time,

then shrink both, and you
make a ratio in the limit,

you should read-- you
should write it df dy.

And that's the so-called
Leibniz notation, right?

That was in calc 1.

But I erased it because
that was calc 1.

Now, mathematicians, to
imitate the Leibniz notation,

they said, I cannot use df dy.

So what the heck shall I use?

After they thought
for about a year,

and I was reading through
the history about how

they invented this,
they said, let's take

the Greek-- the Greek d.

Which is the del.

That's partial.

The del f, del y, at x0, y.

When I was 20--
no, I was 18 when

I saw this the first time--
I had the hardest time making

this sign.

It's all in the wrist.

It's very-- OK.

Now.

df dy.

If you don't like it,
then what do you do?

You can adopt this notation.

And what is the meaning
of this by definition?

You say, you haven't even
defined it, Magdalena.

It has to be limit of
a difference quotient,

just like here.

But we have to be happy
and think of that.

What is the delta f
versus the delta y?

It has to be like that.

f of Mr. x0 is fixed.

x0, comma, y.

We have an increment in y.

y plus delta y. y plus
delta y minus-- that's

the difference quotient.

f of what-- the original
point was, well--

STUDENT: x0, y0.

MAGDALENA TODA:
x0-- let me put y0

because our original
point was x0, y0.

x0, y0 over-- over delta y.

But if I am at x0, y0, I better
put x0, y0 fixed point here.

And I would like you to
photograph or put this thing--

STUDENT: So is that a delta
that's in front of the f?

MAGDALENA TODA: Let me
review the whole thing

because it's very important.

Where shall I start,
here, or here?

It doesn't matter.

So the limit--

STUDENT: [INAUDIBLE] start at m.

MAGDALENA TODA: At m?

At m.

OK, I'll start at m.

The slopes of this line at
x0, y0, right at my point,

will be, my favorite
notation is f sub y at x0,

y0, which means partial
derivative of f with respect

to y at the point--
fixed point x0, y0.

Or, for most mathematicians,
df-- of del-- del f,

del y at x0, y0.

Which is by definition the limit
of this difference quotient.

So x0 is held fixed
in both cases.

y0 is allowed to
deviate a little bit.

So y0 is fixed, but you
displace it by a little delta,

or by a little-- how did we
denote that in calc 1, h?

Little h?

STUDENT: Yeah.

MAGDALENA TODA: So
delta y, sometimes it

was called little h.

And this is the
same as little h.

Over that h.

Now you, without my
help, because you

have all the knowledge
and you're smart,

you should tell me how I
define f sub x at x0, y0,

and shut up, Magdalena,
let people talk.

This is hard.

[INTERPOSING VOICES]

MAGDALENA TODA: No.

I hope not.

As a limit of a
difference quotient,

so it's gonna be an
instantaneous rate of change.

That's the limit of a
difference quotient.

Limit of what?

Shut up.

I will zip my lips.

STUDENT: Delta x

MAGDALENA TODA:
Delta x, excellent.

Delta x going to 0.

So you shrink-- you displace
by a small displacement

only in the direction of x.

STUDENT: So f.

MAGDALENA TODA: f.

STUDENT: [INAUDIBLE] this
time, x is changing, so--

[INTERPOSING VOICES]

MAGDALENA TODA: X0 plus
delta x, y0 is still fixed,

minus f of x0, y0.

Thank God this is always fixed.

I love this guy.

STUDENT: Delta--

MAGDALENA TODA:
Delta x, which is

like the h we were
talking about.

Now in reality,
you never do that.

You would die if for every
exercise, derivation exercise,

you would have to compute a
limit of a difference quotient.

You will go bananas.

What we do?

We do exactly the same thing.

How can I draw?

Can anybody help me draw?

For y0, I would need to take
this other plane through y0.

Where is y0?

Here.

Is my drawing good enough?

I hope so.

So it's something like
I have this plane with,

oh, do you see that, guys?

OK.

So what is that, the other
curve, coordinate curve, look

like?

Oh my God.

Looks like that.

Through the same point,
and then the slope

of the line will be a
blue slope and the slope

will be f sub-- well OK.

So here I have in the red
one, which was the blue one,

this is f sub y, and for
this one, this is f sub x.

Right?

So guys, don't look
at the picture.

The picture's confusing.

This is x coming
towards me, right?

And y going there
and z is going up.

This is the graph.

When I do the
derivative with respect

to what is this, y, the
derivative with respect to y,

with respect to y, y
is my only variable,

so the curve will be like that.

And the slope will be for a
curve that depends on y only.

When I do derivative
with respect to x,

it's like I'm on top of a hill
and I decide to go skiing.

And I'm-- and I point
my skis like that,

and the slope is going down,
and that's the x direction.

OK?

And what I'm going to
describe as a skier

will be a plane curve going
down in this direction.

Zzzzsssshh, like that.

And the slope at every
point, the slope of the line,

of y trajectory, will
be the derivative.

So I have a curve like
that, and a curve like this.

And they're called
coordinate curves.

Now this is hard.

You'll see how
beautiful and easy

it is when you actually
compute the partial derivatives

of functions by hand.

Examples?

Let's take f of x, y to be
x squared plus y squared.

I'm asking you, who
is f sub x at x, y?

Who is f sub x at 1
minus 1, 1, 0, OK.

Who is f sub y at x, y?

And who is f sub y at 3 and 2.

Since I make up my
example-- I don't

want to copy the
examples from the book,

because you are supposedly
going to read the book.

This is-- should be another
example, just for you.

So who's gonna help me-- I'm
pausing a little bit-- who's

gonna help me here?

What's the answer here?

So how do I think?

I think I got-- when I
prime with respect to x, y

is like a held constant.

He's held prisoner.

Poor guy cannot leave his cell.

That's awful.

So you prime with respect to x.

Because x is the only variable.

And he is--

STUDENT: So then it's 2x plus y?

MAGDALENA TODA: 2x plus 0.

Plus 0.

Because y is a constant and
when you prime a constant,

you get 0.

STUDENT: So when you
take partial derivatives,

you-- when you're
taking it with respect

to the first derivative, the
first variable [INAUDIBLE]

MAGDALENA TODA: You
don't completely

know because it
might be multiplied.

But you view it as a constant.

So for you-- very good, Ryan.

So for you, it's like,
as if y would be 7.

Imagine that y would be 7.

And then you have x squared plus
7 squared prime is u, right?

STUDENT: So then that means
f of 1-- or f x of 1,0

is [INAUDIBLE]

MAGDALENA TODA: Very good.

STUDENT: OK.

And in this case, f sub y,
what do you think it is?

STUDENT: 2y.

MAGDALENA TODA: 2y.

And what is f y of 3, 2?

STUDENT: 4.

MAGDALENA TODA: It's 4.

And you say, OK, that
makes sense, that was easy.

Let's try something hard.

I'm going to build them
on so many examples

that you say, stop,
Magdalena, because I became

an expert in partial
differentiation

and I-- now everything is so
trivial that you have to stop.

So example A, example B. A was f
of x, y [INAUDIBLE] x, y plus y

sine x.

And you say, wait,
wait, wait, you're

giving me a little
bit of trouble.

No, I don't mean to.

It's very easy.

Believe me guys,
very, very easy.

We just have to
think how we do this.

f sub x at 1 and 2, f
sub y at x, y in general,

f sub y at 1 and
2, for God's sake.

OK.

All right.

And now, while you're
staring at that,

I take out my beautiful
colors that I paid $6 for.

The department told me that
they don't buy different colors,

just two or three basic ones.

All right?

So what do we do?

STUDENT: First
one will be the y.

MAGDALENA TODA: It's like y
would be a constant 7, right,

but you have to keep in
mind it's mister called y.

Which for you is a constant.

So you go, I'm priming this
with respect to x only--

STUDENT: Then you get y.

MAGDALENA TODA: Very good.

Plus--

STUDENT: y cosine x.

MAGDALENA TODA: y cosine x.

Excellent.

And stop.

And stop.

Because that's all I have.

You see, it's not hard.

Let me put here a y.

OK.

And then, I plug
a different color.

I'm a girl, of course I
like different colors.

So 1, 2. x is 1, and y is 2.

2 plus 2 cosine 1.

And you say, oh, wait a minute,
what is that cosine of 1?

Never mind.

Don't worry about it.

It's like cosine
of 1, [INAUDIBLE]

plug it in the
calculator, nobody cares.

Well, in the final, you
don't have a calculator,

so you leave it like that.

Who cares?

It's just the perfect--
I would actually hate it

that you gave me--
because all you

could give me would be an
approximation, a truncation,

with two decimals.

I prefer you give me the
precise answer, which

is an exact answer like that.

f sub y.

Now, Mr. x is held prisoner.

He is a constant.

He cannot move.

Mr. y can move.

He has all the freedom.

So prime with respect
to y, what do you have?

STUDENT: x--

[INTERPOSING VOICES]

MAGDALENA TODA: x plus
sine x is a constant.

So for God's sake,
I'll write it.

So then I get 1,
plug in x equals 1. y

doesn't appear in the picture.

I don't care.

1 plus sine 1.

And now comes-- don't erase.

Now comes the-- I mean,
you cannot erase it.

I can erase it.

Comes this mean professor
who says, wait a minute,

I want more.

Mathematicians always want more.

He goes, I want the
second derivative.

f sub x x of x, y.

And you say, what in
the world is that?

Even some mathematicians,
they denote it

as del 2 f dx 2, which
is d of-- d with respect

to x sub d u with respect to x.

What does it mean?

You take the first derivative
and you derive it again.

And don't drink and derive
because you'll be in trouble.

Right?

So you have d of dx primed
again, with-- differentiated

again with respect to x.

Is that hard?

Uh-uh.

What you do?

In the-- don't do it here.

You do it in general, right?

With respect to x as a variable,
y is again held as a prisoner,

constant.

So when you prime
that y goes away.

You're gonna get 0.

I'll write 0 like a silly
because we are just starters.

And what else?

STUDENT: Negative y sine of x.

MAGDALENA TODA:
Minus y sine of x.

And I know you've gonna
love this process.

You are becoming
experts in that.

And in a way I'm a little
bit sorry it's so easy,

but I guess not
everybody gets it.

There are students who
don't get it the first time.

So what do we get here?

Minus--

STUDENT: 0.

MAGDALENA TODA: Please
tell me-- sine 1, 0.

Good.

I could do the same
thing for f y y.

I could do this thing--
what is f sub x y?

By definition f sub x y--

STUDENT: Is that taking the
derivative of the derivative

with respect-- is that
taking the second derivative

with respect to y after
you take the derivative

of the-- first derivative
with respect to x?

MAGDALENA TODA: Right.

So when I write like that,
because that's a little bit

confusing, when students
ask me, which one is first?

First you do f sub
x, and then you do y.

And then f sub y x would be the
derivative with respect to y

primed again with respect to x.

Now, let me tell
you the good news.

They-- the book doesn't call
it any name, because we don't

like to call anybody names.

I'm just kidding.

It's called the
Schwartz principle,

or the theorem of Schwartz.

When I told my co-authors,
they said, who cares?

Well I care, because I was a
student when my professors told

me that this German
mathematician made

this discovery, which
is so beautiful.

If f is twice differentiable
with respect to x and y,

and the partial derivatives--
the second partial

derivatives-- are continuous,
then, now in English

it would say it doesn't
matter in which order

you differentiate.

The mixed ones are
always the same.

Say what?

f sub x y equals f sub
y x for every point.

For every-- do you remember
what I taught you for every x, y

in the domain.

Or for every x, y
where this happens.

So what does this mean?

That means that whether
you differentiate

first with respect to x and then
with respect to, y, or first

with respect to y and
then with respect to x,

it doesn't matter.

The mixed partial
derivatives are the same.

Which is wonderful.

I mean, this is one
of the best things

that ever happened to us.

Let's see if this
is true in our case.

I mean, of course it's true
because it's a theorem,

if it weren't true
I wouldn't teach it,

but let's verify it on a baby.

Not on a real baby,
on a baby example.

Right?

So, f sub x is y plus y
equals sine x primed again

with respect to y.

And what do we get out of it?

Cosine of x.

Are you guys with me?

So f sub x was y
plus y equals sine x.

Take this guy
again, put it here,

squeeze them up a little
bit, divide by-- no.

Time with respect to y, x is
a constant, what do you think?

Cosine of x, am I right?

STUDENT: 1 plus [INAUDIBLE].

MAGDALENA TODA: That's
what it starts with.

Plus [INAUDIBLE].

So cosine of x, [INAUDIBLE]
a constant, plus 1.

Another way to have done
it is, like, wait a minute,

at this point I go, constant
out-- are you with me?--

constant out, prime with respect
to y, equals sine x plus 1.

Thank you.

All right.

F sub yx is going to be f sub y.

x plus sine x, but I have
to take it from here,

and I prime again with respect
to x, and I get the same thing.

I don't know,
maybe I'm dyslexic,

I go from the right to the
left, what's the matter with me.

Instead of saying 1 plus,
I go cosine of x plus 1.

So it's the same thing.

Yes, sir.

STUDENT:I'm looking at
the f of xy from the--

MAGDALENA TODA: Which
one are you looking at?

Show me.

STUDENT: It's in the purple.

MAGDALENA TODA: It
is in the purple.

STUDENT: It's that
one right there.

So--

MAGDALENA TODA: This one?

STUDENT: Mmhm.

So, I'm looking at
the y plus y cosine x.

You got that from f of x.

MAGDALENA TODA: I
got this from f of x,

and I prime it again,
with respect to y.

The whole thing.

STUDENT: OK, so you're not
writing that as a derivative?

You're just substituting
that in for f of x?

MAGDALENA TODA: So,
let me write it better,

because I was a little bit
rushed, and I don't know,

silly or something.

When I prime this
with respect to y--

STUDENT: Then you get
the cosine of x plus 1.

MAGDALENA TODA: Yeah.

I could say, I can take
out all the constants.

STUDENT: OK.

MAGDALENA TODA: And that
constant is this plus 1.

And that's all I'm left with.

Right?

It's the same thing
as 1 plus cosine x,

which is a constant times y.

Prime this with respect
to y, I get the constant.

It's the same principal as when
you have bdy of 7y equals 7.

Right?

OK.

Is this too easy?

I'll give you a nicer function.

I'm imitating the one
in WeBWorK [INAUDIBLE]

To make it harder for you.

Nothing I can make at this
point is hard for you,

because you're becoming experts
in partial differentiation,

and I cannot
challenge you on that.

I'm just trying to
make it harder for you.

And I'm trying to
look up something.

OK, how about that?

This is harder than the
ones you have in WeBWorK.

But that was kind of
the idea-- that when

you go home, and open
those WeBWorK problem sets,

that's a piece of cake.

What we did in class was harder.

When I was a graduate
student, one professor said,

the easy examples are the
ones that the professor's

supposed to write in
class, on the board.

The hard examples
are the ones that

are left for the
students' homework.

I disagree.

I think it should be
the other way around.

So f sub x.

That means bfdx for
the pair xy, any xy.

I'm not specifying an x and a y.

I'm not making them a constant.

What am I going to
have in this case?

Chain -- if I catch you
not knowing the chain rule,

you fail the final.

Not really, but, OK,
you get some penalty.

You know it.

Just pay attention
to what you do.

I make my own
mistakes sometimes.

So 1 over.

What do you do here
when you differentiate

with respect to x?

You think, OK, from the outside
to the inside, one at a time.

1 over the variable
squared plus 1, right?

Whatever that variable,
it's like you call variable

of the argument xy, right?

STUDENT: [INAUDIBLE]

MAGDALENA TODA: Square plus 1.

Times-- cover it with your
hand-- prime with respect to x.

y, right?

Good!

And you're done.

You see how easy it was.

Just don't forget something
because it can cost you points.

Are you guys with me?

So, once we are done with
saying, 1 over argument

squared plus 1, I cover
this with my hand,

xy prime with
respect to 2x is y.

And I'm done.

And I'm done.

And here, pause.

What's the easiest
way to do that?

You look at it like,
she wants me to get

caught in the quotient rule.

She wants to catch me
not knowing this rule,

while I can do better.

One way to do it would
be numerator prime plus

denominator, minus
numerator [INAUDIBLE] What's

the easier way to do it?

STUDENT: x squared plus
y squared, all of it

to the negative one.

MAGDALENA TODA: Right.

So you say, hey,
you cannot catch me,

I'm the gingerbread man.

Good!

That was a good idea.

Chain rule, and
minus 1/2, times--

who tells me what's next?

I'm not going to say a word.

STUDENT: 2x plus y squared.

No, it's 2x.

x squared plus y squared.

MAGDALENA TODA: From the
outside to the inside.

From the outside-- to the what?

STUDENT: [INAUDIBLE]

MAGDALENA TODA: Good.

And now I'm done.

I don't see that anymore.

I focus to the core.

2x.

Times 2x.

And that is plenty.

OK, now, let me
ask you a question.

What if you would
ask a smart kid,

I don't know, somebody
who knows that,

can you pose the f sub y of xy
without doing the whole thing

all over again?

Can you sort of figure
out what it would be?

The beautiful
thing about x and y

is that these are
symmetric polynomials.

What does it mean,
symmetric polynomials?

That means, if you swap x
and y, and you swap x and y,

it's the same thing.

Just think of that--
swapping x and y.

Swapping the roles of x and y.

So what do you think
you're going to get?

OK, one student said,
this is for smart people,

not for people like me.

And I said, well, OK,
what's the matter with you?

I'm a hard worker.

I'm the kind of guy who takes
the whole thing again, and does

the derivation from scratch.

And thinking back in high
school, I think, even

for symmetric polynomials,

I'm sure that being
smart and being

able to guess the
whole thing-- but I

did the computation
many times mechanically,

just in the same way,
because I was a hard worker.

So what do you
have in that case?

1/xy squared plus 1 times x
plus-- the same kind of thing.

Attention, this is the symmetric
polynomial, and I go to that.

And then times 2y.

So, see-- that kind
of easy, fast thing.

Why is this a good
observation when

you have symmetric polynomials?

If you are on the final and
you don't have that much time,

or on any kind of exam when
you are in a time-crunch.

Now, we want those
exams so you are not

going to be in a time-crunch.

If there is something I hate,
I hate a final of 2 hours

and a half with 25
serious problems,

and you know nobody can do that.

So, it happens a lot.

I see that-- one of my jobs
is also to look at the finals

after people wrote
them, and I still

do that every semester-- I see
too many people making finals.

The finals are not
supposed to be long.

The finals are supposed
to be comprehensive, cover

everything, but not extensive.

So maybe you'll have 15
problems that cover practically

the material entirely.

Why?

Because every little problem
can have two short questions.

You were done with
a section, you

shot half of a chapter
only one question.

This is one example just--
not involving [INAUDIBLE]

of an expression like that, no.

That's too time-consuming.

But maybe just tangent of
x-squared plus y-squared,

find the partial derivatives.

That's a good exam
question, and that's enough

when it comes to
testing partials.

By the way, how
much-- what is that?

And I'm going to let
you go right now.

Use the bathroom.

And when you come back from the
bathroom, we'll fill in this.

You know I am horrible in the
sense that I want-- I'm greedy.

I need extra time.

I want to use more time.

I will do your
problems from now on,

and you can use the bathroom,
eat something, wash your hands.

I'll start in
about five minutes.

Don't worry.

Alexander?

Are you here?

Come get this.

I apologize.

This is long due back to you.

STUDENT: Oh.

Thank you.

STUDENT: Is there an
attendance sheet today?

MAGDALENA TODA: I will--
I'm making up one.

There is already on
one side attendance.

Let's use the other side.

Put today's date.

[INAUDIBLE]

[SIDE CONVERSATIONS]

MAGDALENA TODA: They
are spoiling me.

They give me new
sprays every week.

[INAUDIBLE] take care of this.

[SIDE CONVERSATIONS]

MAGDALENA TODA: So I'm
going to ask you something.

And you respond honestly.

Which chapter-- we already
browsed through three chapters.

I mean, Chapter 9
was vector spaces,

and it was all review
from-- from what?

From Calc 2.

Chapter 10 was curves in
[INAUDIBLE] and curves

in space, practically.

And Chapter 11 is functions
of several variables.

Now you have a flavor
of all of them,

which one was hardest for you?

STUDENT: 9 and 10, both.

MAGDALENA TODA: 9 and 10 both.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: This is so
much better than the other.

No, I think you guys
actually-- it looks better,

because you've seen a lot more
vectors and vector functions.

STUDENT: I didn't
understand any of 9 or 10.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: Yes, ma'am.

STUDENT: Could you go over
parametrization [INAUDIBLE]?

MAGDALENA TODA: I will
go over that again.

And I will go over some
other parametrizations today.

And I promised that at the
end, in those 20 minutes,

I will do that problem that
gave a few of you trouble.

Yes, sir?

STUDENT: Do we take
the same final exam

as all the other [INAUDIBLE]
classes? [INAUDIBLE]?

MAGDALENA TODA: Well, that's
what I was asked yesterday.

So practically, it's at the
latitude of the instructor who

teaches honors if they
write their own final,

and in general make
it harder, or they

take the general final
like everybody else.

For your formative
purposes, and as a study,

I would like you to
take the general final,

because I want to see
where you stand compared

to the rest of the population.

So you are my sample, and
they are the entire student

population of Calc
3, I want to make

the statistical analysis of your
performance compared to them.

STUDENT: So we'll
take the regular one?

MAGDALENA TODA: Yeah.

For this one, I just
have to make sure

that they also have that
extra credit added in.

Because if I have too much
extra credit in there,

well they also count that.

So that's what that means.

So we can [INAUDIBLE].

All right.

Let me finish this exercise.

And then [? stop ?]
[INAUDIBLE] and go

over some homework problems and
some parametrization problems.

And I will see what else.

So tangent of [INAUDIBLE].

Is this hard?

No, it's [INAUDIBLE].

But you have to
remind me, because I

pretend that I
forgot-- let me pretend

that I forgot what the
derivative [INAUDIBLE] notation

of tangent of t was.

STUDENT: Secant squared.

MAGDALENA TODA: You guys love
that secant squared thingy.

Why do you like secant squared?

I, as a student, I didn't
like expressing it like that.

I liked [INAUDIBLE].

Of course, it's the same thing.

But I always like it like
1 over cosine [INAUDIBLE].

And of course, I have
to ask you something,

because I'm curious to
see what you remember.

And you say yeah,
curiosity killed the cat.

But where did the
derivative exist?

Because maybe was
that tangent of T--

STUDENT: Wasn't
it a quotient rule

of sine and [? cosine x? ?]

MAGDALENA TODA: Good.

I'm proud of you.

That is the answer.

So [? my ?] [? have ?] this
blowing up, this blows up--

blows up where cosine
T was zero, right?

So where did that blow up?

[INAUDIBLE] blow up of
cosine and zero [INAUDIBLE].

The cosine was the
shadow on the x-axis.

So here you blow up here, you
blow up here, you blow up here,

you blow up here.

So [? what does ?] [INAUDIBLE].

It should not be what?

STUDENT: Pi over 2.

MAGDALENA TODA: Yeah.

And can we express
that OK, among 0pi,

let's say you go in
between 0 and 2pi only.

I get rid of pi over
2 and 3pi over 2.

But if I express that in
general for [INAUDIBLE] T

not restricted to 0
to T, what do I say?

STUDENT: It's k.

STUDENT: So it can
[? never be ?] pi over 2

plus pi?

MAGDALENA TODA: 2k plus 1.

2k plus 1.

Odd number over--

STUDENT: Pi over 2.

MAGDALENA TODA: Pi over 2.

Odd number, pi over 2.

And all the odd
numbers are 2k plus 1.

Right?

All right.

So you have a not
existence and-- OK.

Coming back.

I'm just playing, because
we are still in the break.

Now we are ready.

What is dfdx, del f, del x, xy.

And what is del f, del y?

I'm not going to ask you for
the second partial derivative.

We've had enough of that.

We also agreed that we have
important results in that.

What is the final answer here?

STUDENT: [INAUDIBLE] plus
x-squared [INAUDIBLE].

MAGDALENA TODA: 1
over [INAUDIBLE].

I love this one, OK?

Don't tell me what I
want to [INAUDIBLE].

I'm just kidding.

[INAUDIBLE] squared times--

STUDENT: 2x.

MAGDALENA TODA: 2x, good.

How about the other one?

The same thing.

Times 2y.

OK.

I want to tell you something
that I will repeat.

But you will see it
all through the course.

There is a certain
notion that Alexander,

who is not talking--
I'm just kidding,

you can talk-- he
reminded me of gradient.

We don't talk about gradient
until a few sections from now.

But I'd like to
anticipate a little bit.

So the gradient of
a function, wherever

the partial derivatives exist,
with the partial derivative--

that is, f sub x
and f sub y exist--

I'm going to have that
delta f-- nabla f.

nabla is a [INAUDIBLE].

Nable f at xy represents what?

The vector.

And I know you love vectors.

And that's why I'm going back
to the vector notation f sub x

at xy times i, i being
the standard vector i

unit along the x axis,
f sub y at xy times j.

STUDENT: So it's just like
the notation of [INAUDIBLE]?

MAGDALENA TODA: Just
the vector notation.

How else could I write it?

Angular bracket, f sub x x
at xy, comma, f sub y at xy.

And you know-- people who
saw my videos, colleagues

who teach Calc 3
at the same time

said I have a tendency of not
going by the book notations

all the time, and just give you
the [? round ?] parentheses.

It's OK.

I mean, different books,
different notations.

But what I mean is to represent
the vector in the standard way

[INAUDIBLE].

All right.

OK.

Can you have this
notion for something

like a function of
three variables?

Absolutely.

Now I'll give you an easy one.

Suppose that you have
x-squared plus y-squared

plus z-squared equals 1.

And that is called-- let's
call it names-- f of x, y, z.

Compute the gradient nabla f
at any point x, y, z for f.

Find the meaning of that
gradient-- of that-- find

the geometric meaning of it.

For this case, not in
general, for this case.

So you say, wait,
wait, Magdalena.

A-dah-dah, you're confusing me.

This is the gradient.

Hmm.

Depends on how many
variables you have.

So you have to show a vector
whose coordinates represent

the partial derivatives with
respect to all the variables.

If I have n variables, I have
f sub x1 comma f sub x2 comma

f sub x3 comma f
sub xn, and stop.

Yes, sir.

STUDENT: If the formula
was just f of xy,

wouldn't that be implicit?

MAGDALENA TODA:
That is implicit.

That's exactly what I meant.

What's the geometric
meaning of this animal?

Forget about the left hand side.

I'm going to clean it quickly.

What is that animal?

That is a hippopotamus.

What is that?

STUDENT: It's a sphere.

MAGDALENA TODA: It's a sphere.

But what kind of sphere?

Center 0, 0, 0 with radius 1.

What do we call that?

Unit sphere.

Do you know what notation
that mathematicians

use for that object?

You don't know but I'll
tell you. s1 is the sphere.

We have s2, I'm
sorry, the sphere

of dimension 2, which
means the surface.

s1 is the circle.

s1 is a circle.

s2 is a sphere.

So what is this number
here for a mathematician?

That's the dimension of
that kind of manifold.

So if I have just a
circle, we call it s1

because there is only a one
independent variable, which

is time, and we parameterize.

Why go clockwise?

Shame on me.

Go counterclockwise.

All right.

That's s1.

For s2, I have two
degrees of freedom.

It's a surface.

On earth, what are those
two degrees of freedom?

It's a riddle.

No extra credit.

STUDENT: The latitude
and longitude?

MAGDALENA TODA: Who said it?

Who said it first?

STUDENT: [INAUDIBLE].

MAGDALENA TODA: How many of
you said it at the same time?

Alexander said it.

STUDENT: I know there
was one other person.

I wasn't the only one.

STUDENT: I didn't.

STUDENT: [INAUDIBLE], sorry.

[INTERPOSING VOICES]

MAGDALENA TODA: I
don't have enough.

STUDENT: I'll take
the credit for it.

MAGDALENA TODA:
[INAUDIBLE] extra credit.

OK, you choose.

These are good.

They are Valentine's hearts,
chocolate [INAUDIBLE].

Wilson.

I heard you saying Wilson.

I have more.

I have more.

These are cough drops,
so I'm [INAUDIBLE].

You set it right
next time, Alexander.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: OK.

Anybody else?

Anybody needing cough drops?

OK.

I'll leave them here.

Just let me see.

Do I have more chocolate?

Eh, next time.

I'm going to get some
before-- we have-- we

need before Valentine's, right?

So it's Thursday.

I'm going to bring
you a lot more.

So in that case, what
is the gradient of f?

An x, y, z.

Aha.

I have three variables.

What's the gradient?

I can write it as a
bracket, angular notation.

Am I right?

Or I can write it 2xi
plus 2ij plus 2zk.

Can anybody tell me why?

What in the world are
these, 2x, 2y, 2z?

STUDENT: Those are the
partial derivatives.

MAGDALENA TODA: They are
exactly the partial derivatives

with respect to x, with respect
to y, with respect to z.

Does this have a
geometric meaning?

I don't know.

I have to draw.

And maybe when I
draw, I get an idea.

Is this a unit vector?

Uh-uh.

It's not.

Nabla s, right.

In a way it is.

It's not a unit vector.

But if I were to
[? uniterize ?] it--

and you know very well what it
means to [? uniterize it ?].

It means to--

STUDENT: Divide it by--

MAGDALENA TODA: Divide
it by its magnitude

and make it a unit vector
that would have a meaning.

This is the sphere.

What if I make like this?

n equals nabla f over
a magnitude of f.

And what is the meaning
of that going to be?

Can you tell me what
I'm going to get here?

In your head,
compute the magnitude

and divide by the magnitude,
and you have exactly 15 seconds

to tell me what it is.

STUDENT: [INAUDIBLE].

MAGDALENA TODA:
[? Ryan, ?] [? Ryan, ?]

you are in a Twilight Zone.

But I'm sure once I tell you,
once I tell you, [INAUDIBLE].

STUDENT: 1 divided by
the square root of 2

for the [? i controller. ?]

STUDENT: [INAUDIBLE].

MAGDALENA TODA: Well, OK.

Say it again, somebody.

STUDENT: x plus y plus z.

MAGDALENA TODA: xi plus yj
plus zk, not x plus x, y,

z because that
would be a mistake.

It would be a scalar function.
[INAUDIBLE] has to be a vector.

If I am to draw this vector,
how am I going to draw it?

Well, this is the
position vector.

Say it again.

This is the position vector.

When I have a point on this
stinking earth, whatever

it is, x, y, z, the
position vector is x, y, z.

It's xi plus yj plus zk.

I have this identification
between the point

and the vector.

This is our vector.

So I'm going to draw these
needles, all these needles,

all these vectors whose tips
are exactly on the sphere.

So why?

You say, OK.

I understand that is
the position vector,

but why did you put an n here?

And anybody who answers
that gets a cough drops.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: Because that is?

STUDENT: The normal
to the surface.

MAGDALENA TODA: You get a--

STUDENT: Yeah, cough drop.

MAGDALENA TODA: Two of them.

STUDENT: Aw, yeah.

MAGDALENA TODA: All right.

So that's the normal
to the surface, which

would be a continuation
of the position vector.

You see, guys?

So imagine you take
your position vector.

This is the sphere.

It's like an egg.

And these tips
are on the sphere.

If you continue from
sitting on the sphere,

another radius vector
colinear to that,

that would be the
normal to the sphere.

So in topology, we
have a name for that.

We call that the hairy ball.

The hairy ball in
mathematics, I'm not kidding,

it's a concentrated notations.

You see it in graduate
courses, if you're

going to become a graduate
student in mathematics,

or you want to do a
dual degree or whatever,

you're going to see the hairy
ball, all those normal vectors

of length 1.

It's also called
the normal field.

So if you ask Dr.
Ibragimov, because he

is in this kind of field
theory, [INAUDIBLE] normal field

to a surface.

But for the topologists
or geometers,

they say, oh, that's
the hairy ball.

So if you ask him what the
hairy ball is, he will say,

why are you talking
nonsense to me?

Right.

Exactly.

So here's where we stopped
our intrusion in chapter 11.

It's going to be as
fun as it was today

with these partial derivatives.

You're going to love them.

You have a lot of computations
like the ones we did today.

Let's go back to
something you hated,

which is the parameterizations.

So one of you--
no, three of you--

asked me to redo one
problem like the one

with the parameterization
of a circle.

But now I have to pay
attention to the data

that I come up with.

So write the parameterization
of a circle of radius.

Do you want specific
data or you want letters?

STUDENT: [INAUDIBLE].

MAGDALENA TODA: OK.

Let's do it [INAUDIBLE] r,
and then I'll give an example.

And center x0, y0 in plane
where-- what is the point?

Where is the particle
moving for time t equals 0?

Where is it located?

All right.

So review.

We had frame that we always
picked at the origin.

That was bad because we could
pick x0, y0 as a center,

and that has a separate radius.

And now, they want me to write
a parameterization of a circle.

How do you achieve it?

You say the circle is x minus x0
squared plus y minus y0 squared

equals r squared.

And one of you
asked me by email--

and that was a good
question-- you said, come on.

Look, it was [INAUDIBLE].

So you said, I was
quite good in math.

I was smart.

Why didn't I know the equations,
the parametric equations,

or even this?

I'll tell you why.

This used to be
covered in high school.

It's something called
college algebra.

We had a chapter,
either trigonometry

or college algebra.

We had a chapter called
analytic geometry.

This is analytic geometry.

It's the same chapter
in which you guys

covered conics, [INAUDIBLE],
ellipse, [INAUDIBLE], parabola.

It's no longer covered
in most high schools.

I asked around.

The teachers told
me that we reduced

the geometric
applications a lot,

according to the general
standards that are imposed.

That's a pity, because you
really need this in college.

All right.

So how do you come up
with a parameterization?

You say, I would like to
parameterize in such way

that this would be
easy to understand

this for Pythagorean theorem.

Oh, OK.

So what is the Pythagorean
theorem telling me?

It's telling you that if you are
in a unit circle practically,

then this is cosine and
theta and this is sine theta,

and the sum of
cosine theta squared

plus sine theta squared is 1.

This is 1, so that is
the Pythagorean theorem

[INAUDIBLE].

So xy plus x0 should be cosine
of theta times an R. Why an R?

Because I want, when I square,
I want the R squared up.

And here, this guy inside
will be our sine [? thing. ?]

Am I going to be in good shape?

Yes, because when I
square this fellow squared

plus this fellow squared will
give me exactly R squared.

And here is my
[INAUDIBLE] smiley face.

So I want to understand
what I'm doing.

x minus x0 must
be R cosine theta.

y minus y0 is R sine theta.

Theta in general is an
angular velocity, [INAUDIBLE].

But it's also time, right?

It has the meaning
of time parameter.

So when we wrote those--
and some of you are bored,

but I think it's not
going to harm anybody

that I do this again.

R cosine of t plus x0 y is R
sine t plus x0, or plus y0.

Now note, all those
examples in web work,

they were not very imaginative.

They didn't mean for
you to try other things.

Like if one would put here
cosine of 5t or sine of 5t,

that person would move five
times faster on the circle.

And instead of being back
at 2 pi, in time 2 pi,

they would be there
in time 2 pi over 5.

All the examples-- and each of
you, it was randomized somehow.

Each of you has a
different data set.

Different R,
different x0 with 0,

and a different place where
the particle is moving.

But no matter what
they gave you,

it's a response to
the same problem.

And at time t equals
0, you have M. Do

you want me to call it M0?

Yes, from my initial-- M0.

For t equals 0, you're
going to have R plus x0.

And for t equals 0, you have y0.

So for example, Ryan had-- Ryan,
I don't remember what you had.

You had some where theta R was--

STUDENT: 4 and 8.

MAGDALENA TODA: 7.

You, what did you have?

STUDENT: No, R was 7
and x was 3, y was 1.

MAGDALENA TODA: R
was 7 and x0 was--

STUDENT: 3, 1.

MAGDALENA TODA: 3, 1 was x0, y0
so in that case, the point they

gave here was 7 plus 3.

Am I right, Ryan?

You can always check.

I remember.

It was 10 and God
knows, and 10 and 1.

So all of the data that
you had in that problem

was created so that you
have these equations.

And at time 0, you were exactly
at the time t equals 0 replaced

the t.

All right.

OK.

STUDENT: What's the M0?

What is--

MAGDALENA TODA: M0
is Magdalena times 0.

I don't know.

I mean, it's the
point where you are.

I couldn't come up
with a better name.

So I'm going to
erase here and I'll

get to another problem, which
gave you guys a big headache.

And it's not so hard, but this
is the computational problem,

very pretty in itself.

[INAUDIBLE] cosine t i plus
e to the 3t sine t j plus e

to the 3tk.

And I think this was more or
less in everybody's homework

the same.

There's a position vector
given as parameterized form.

So since you love
parameterization so much,

I'm going to remind you what
that means for x and y and zr.

And what did they want from you?

I forget what number of
the problem that was.

They wanted the length of
the arc of a curve from t

equals-- I don't know.

STUDENT: 2 to 5.

MAGDALENA TODA: 2 to 5.

Thank you.

[INAUDIBLE] t equals 5.

So this is the beginning and the
end of the curve, the beginning

and the end of a curve.

So what is that going
to be [INAUDIBLE]?

How does [INAUDIBLE],
which we have

to write down 2 to 5
magnitude of r prime at t, dt.

And I don't know.

But I want to review this
because-- so what in the world?

Maybe I put this
on the midterm or I

make it a little bit easier,
but the same what I don't like,

it's time consuming.

But I can give you
something a lot easier

that tests the
concept, the idea, not

the computational power.

So r prime of t here with
a little bit of attention,

of course, most of you
computing this correctly.

You are just a little bit scared
of what happened after that,

and you should not be scared
because now I'll tell you

why you shouldn't be scared.

Chain rule, product rule.

So I have first prime--

STUDENT: 3.

MAGDALENA TODA: 3 into the
3e second and [? time ?]

cosine t plus-- I'm
going to do that later.

I know what you're thinking.

STUDENT: e 3t.

MAGDALENA TODA: e to
the 3t minus sine.

I'm not worried
about this minus now.

I'll take care of that later.

Times i.

Now with your
permission-- when you

say, why is she not writing the
whole thing in continuation?

Because I don't want to.

No.

Because I want to help
you see what's going on.

You do the same kind of stuff
for this individual one.

I want to put it
right underneath.

If I put it right underneath,
it's going to [? agree ?].

Otherwise it's not
going to [? agree ?].

E to the 3t times sine t
plus e to the 3t cosine t.

You didn't have a
problem because you

know how to differentiate.

You started having the
problem from this point on.

3 into the 3tk.

The problem came when
you were supposed

to identify the coordinates and
square them and squeeze them

under the same square root.

And that drove you crazy
when you have enough.

Let me put the minus here to
make it more obvious what's

going to happen.

When you're going
to have problems

like that in
differential equations,

you better have the eye
for it, [INAUDIBLE].

You should be able to recognize
this is like a pattern.

Have you seen the
movie A Beautiful Mind?

STUDENT: Yeah.

MAGDALENA TODA:
OK, so Nash, when

he was writing with the finger
on everything, on the walls

at Princeton, on the window,
he was thinking of patterns.

He's actually
trying to-- and it's

hard to visualize
without drawing,

but this is what most of us
recognize all the time when

a mathematician writes
down some computations

in a different way.

All we hope for is to get a
few steps behind that board

and see a pattern.

And when you do that,
you see the pattern.

This is an a minus b
and that's an a plus b.

And then you say, OK, if
I'm going to square them,

what's going to happen?

When you square an a minus
b and you square an a plus b

and you have this giggly
guy there-- leave him there.

He's having too much fun.

You actually develop these
guys and you put them one

under the other
and say wow, what

a beautiful simplification.

When I'm going to
add these guys,

this thing in the middle
will simply will cancel out,

but the a squared will double
and the b squared will double.

And that's the beauty
of seeing pattern.

You see how there is
something symmetric and magic

in mathematics that make
the answer simplified.

And that allows you to compress
your equations that originally

seemed to be a mess
into something that's

more easily expressed.

So when you're going
to compute this r

prime of t magic absolute
value of the magnitude, that's

going to be square root of--
instead of writing all the

[INAUDIBLE], I hate writing
and rewriting the whole thing

squared plus the whole thing
squared plus this squared.

If I love to write so much,
I'd be in humanities and not

in mathematics.

So as a mathematician, how
am I going to write that?

As a mathematician, I'm going
to use some sort of-- like the U

substitution.

So I say, I call this Mr.
A, and I call this Mr. B.

And that's A minus B,
and that's A plus B.

And that's somebody else.

So when I square
the first guy, and I

square the second component, and
I square the third component,

and I add them together,
I'm going to get what?

Square root of 2A
squared plus 2B squared.

Because I know that
these are the first two.

This guy squared
plus this guy squared

is going to be
exactly 2A squared

plus 2B squared,
nothing in the middle.

These guys cancel out.

STUDENT: A and B
are not the same.

MAGDALENA TODA: Well,
yeah, you're right.

Let me call-- you're
right, this is the same,

but these are different.

So let me call them
A prime plus B prime.

No, that's derivative.

Let me call them C
and D-- very good,

thank you-- C squared
plus 2CD plus D squared.

But the principle is the same.

So I'm going to have A
squared plus C squared.

This goes away.

Why?

Because this times that is
the same as this times that.

Say it again.

If we look in the
middle, the middle term

will have 3e to the 3t cosine
t times e to the 3t sine t.

Middle term here is 3e to the
3t e to the 3t sine and cosine.

So they will cancel
out, this and that.

So here I have the
sum of the square of A

plus the square
of C. And here I'm

going to have the square
of B plus the square of D.

OK, now when I square this
and that, what do I get?

The beauty of that-- let me
write it down then explicitly.

9e to the 3t cosine squared
t remains from this guy.

Plus from the square
of that, we'll

have 9e to the 3t-- no, just 3,
9 to the 6t, 9 to the 6t sine

squared.

So I take this guy.

I square it.

I take this guy.

I square it.

The middle terms will
disappear, thank god.

Then I have this guy, I square
it, that guy, I square it,

good.

Plus another parenthesis-- e
to the 6t sine squared t plus e

to the 6t cosine squared t.

So even if they don't
double because they're not

the same thing, what
is the principle

that will make my life easier?

The same pattern
of simplification.

What is that same pattern
of simplification?

Look at the beauty
of this guy and look

at the beauty of this guy.

And then there is
something missing,

the happy guy that was quiet
because I told him to be quiet.

That's 9e to the 6t.

He was there in the corner.

And you had to square this
guy and square this guy

and square this guy and
add them on top together.

Now what is the pattern?

The pattern is 9e to the 6t
with 9e to the 6t, same guy.

The orange guys-- that's
why I love the colors.

Cosine squared cosine
squared will be 1.

Another pattern like that, I
have e to the 6t, to the 6t,

and the same happy guys sine
squared t, sine squared t,

add them together is 1.

So all in all, this mess
is not a mess anymore.

So it becomes 9e to the 6t plus
e to the 6t plus 9e to the 6t.

Are you guys with me?

All right, now how many
e to the 6t's do we have?

9 plus 9 plus 1, 19, square
root of 19 e to the 6t.

So when we integrate,
we go integral

from 2 to 5 square root of 19.

Kick him out of your life.

He's just making
your life harder.

And then you have square root
of e to the 6t e to the 3t.

So after you kick
the guy out, you

have e to the 3t divided
by 3 between t equals 2

and t equals 5.

Actually, I took it right off
the WeBWorK problem you had.

So if you type this
in your WeBWorK--

you probably already did-- you
should get exactly the answer

as being correct.

On the exam, do not
expect anything that long.

The idea of simplifying
these patterns

by finding the sine cosine, sine
squared plus cosine squared is

1, is still going to be there.

But don't expect
anything that long.

Also, don't expect-- once
you get to this state,

I don't want an answer.

This is the answer.

That's the precise answer.

I don't want any approximation
or anything like that.

A few of you did this
with a calculator.

Well, you will not have
calculators in the final.

You are going to
have easy problems.

If you did that
with a calculator,

and you truncated
your answer later,

and if you were within
0.01 of the correct answer,

you were fine.

But some people
approximated too much.

And that's always a problem.

So it's always a good
idea to enter something

like that in WeBWorK.

I said I wouldn't do it
except in the last 20 minutes.

But I wanted to do
something like that.

I want to give you another
example, because you love

parametrization so much it just
occurred to me that it would

be very, very helpful--
maybe, I don't

know-- to give you another
problem similar to this one.

It's not in the book,
but it was cooked up

by one of my colleagues
for his homework.

So I'd like to show it to you.

e to the t i is
a parametrization

of a [INAUDIBLE] space.

Plus e to the minus t j
plus square root of 2 tk.

And how do I know?

Well, one of his
students came to me

and asked for help
with homework.

Well, we don't give help when
it comes from another colleague.

So in the end, the student
went to the tutoring center.

And the tutoring center
helped only in parts.

She came back to me.

So what was the deal here?

Find f prime of t in
the most simplified form

and find the absolute
value r prime of t

in the most simplified form.

And find the length of the
arc of this curve between t

equals 0 and t equals 1.

If this were given
by a physicist,

how would that physicist
reformulate the problem?

He would say-- he or she--
what is the distance travelled

by the particle between
0 seconds and 1 second?

So how do you write that?

Integral from 0 to 1 of
r prime of t [INAUDIBLE].

And you have to do the rest.

So arguably, this is
the Chapter 10 review.

It's very useful for
the midterm exam.

So although we are
just doing this review,

you should not erase
it from your memory.

Because I don't like to
put surprise problems

on the midterm.

But if you worked a
certain type of problem,

you may expect
something like that.

Maybe it's different
but in the same spirit.

r prime of t, who's going to
help me with r prime of t?

This fellow-- e to the t.

And how about that?

Negative e to the negative t.

STUDENT: I thought the arc
length was the square root of 1

plus f prime of t squared.

MAGDALENA TODA:
For a plane curve.

OK, let me remind you.

If you have a plane
curve y equals

f of x, then this thing
would become integral from A

to B square root of 1
plus f prime of x dx.

And that, did you do that
with your Calc II instructor?

How many of you
had Dr. Williams?

That was a wonderful
class, wasn't it?

And he taught that.

And of course he
was not supposed

to tell you that was the
speed of a parametric curve.

If you were to
parametrize here, x of t

was t and y of t
would be f of t.

He could have told you.

Maybe he told you.

Maybe you don't remember.

OK, let's forget about it.

That was Calc II.

Now, coming back here,
I have to list what?

Square root of 2 times
t prime is one k.

Who's going to help
me compute the speed

and put it in a nice formula?

Well, my god--

STUDENT: [INAUDIBLE]

MAGDALENA TODA: Ahh,
you are too smart.

Today you had some what is
that called with caffeine

and vitamins and--

STUDENT: You're
thinking of Red Bull.

MAGDALENA TODA: I know.

That was very nice.

I try to stay away.

What is that called
with the energy booster?

STUDENT: I wouldn't know.

STUDENT: 5-Hour Energy.

MAGDALENA TODA: 5-Hour, OK.

I used to have that.

When I had that, I could
anticipate two steps computing.

Just a joke, Alex,
don't take it up.

Very good observation.

So Alex saw.

He has a premonition.

He can see two steps in advance.

He said, OK, square that.

You have e to the 2t.

Square this.

The minus doesn't matter.

Plus e to the minus
2t, and square that.

Then he saw patterns.

Because he is the
wizard 101 today.

So what is the
witchcraft he performed?

Do you see?

Does anybody else
see the pattern?

[? Nateesh ?] sees the pattern.

Anybody illuminated?

I didn't see it from the start.

You guys saw it faster than me.

It took me about a
minute and a half

when I saw this
for the first time.

Is this a perfect square?

Of who?

e to the t plus e to
the minus 2 squared

is-- anybody else sees the
pattern I don't have candy.

Next time-- Alex,
[INAUDIBLE], anybody else?

Do you now see the
pattern, e to the 2t plus

e to the minus 2t plus
twice the product?

And that's where the student
was having the problem.

Where do you see the product?

The product is 1.

The product is 1 doubled.

So you get 2.

So it's indeed exactly
the perfect square.

So once-- it was a she.

Once she saw the perfect
square, she was so happy.

Because you get square
root of the square.

You get e to the t
plus e to the minus t.

And that's a trivial thing
to integrate that you

have no problem integrating.

It's a positive
function, very beautiful.

The professor who gave this was
Dr. [INAUDIBLE] from Denmark.

He's one of the best
teachers we have.

But he makes up his
homework as far as I know.

I think in the sixth
edition, this edition,

we actually stole his idea,
and we made a problem like that

in the book somewhere.

We doubled the number of
problems more or less.

So if you are to compute
0 to 1 of the speed,

what is the speed?

The speed is this
beautiful thing.

Because you were able
to see the pattern.

If you're not able
to see that, do you

realize it's
impossible, practically,

for you to integrate by hand?

You have to go to a
calculator, Matlab, whatever.

So this is easy.

Why is that easy? e to the t
minus e to the minus t at 1

and at 0-- you compare them.

You get at 1 e minus
e to the minus 1

minus the fundamental theorem
of calc e to the 0 minus

e to the 0.

Well, that's silly.

Why is that silly?

Because I'm going to give it up.

So the answer was
e to the minus 1/e.

And she knew what
the answer would be.

But she didn't know why.

So she came back to me.

I don't know how the tutoring
center helped her figure

out the answer.

But she did not
understand the solution.

So I said, I'm not going to
take anymore people coming

from Professor [INAUDIBLE].

I was also told it's not OK.

So don't go to another
professor with homework coming

for me or the other way around.

Because it's not OK.

But you can go to the tutoring
center asking them for hints.

They're open starting 9:00
AM and until around when?

Do you know?

They used to have until 4:00.

But now they're going to
work on an extended schedule

until 8:00 PM.

It's going to be
something crazy.

Now, the thing is, we want
the students to be better,

to do better, to not give
up, to be successful,

top one, two, three.

I'm a little bit
concerned, but maybe I

shouldn't be, about those hours.

So I don't know if they managed
to put a security camera

or not.

But having extended
hours may be a problem.

Take advantage of
those afternoon hours,

especially if you are busy.

Those late hours will
be a big help for you.

Do you know where it is?

Room 106 over there.

Any other questions related
to this type of problem

or related to anything
else in the material

that maybe I can
give you hints on,

at least the hint I'm
going to give you?

Sometimes I cannot stop, and
I just give the problem away.

I'm not supposed to do that.

Look at your WeBWorK, see what
kind of help I can give you.

You still have a
little bit of time.

STUDENT: [INAUDIBLE]

MAGDALENA TODA: That's
the maximum of what?

It was--

STUDENT: [INAUDIBLE]

MAGDALENA TODA: Was
this the problem?

STUDENT: e to the 2x
or something like that.

MAGDALENA TODA:
Something like that?

I erased it.

STUDENT: You erased
that? [INAUDIBLE].

I found an answer.

MAGDALENA TODA: It's
very computational I saw.

But before that, I
saw that seven of you

guys-- you two also did it.

So I wrote-- you have a
brownie waiting for that.

But then I erased it.

STUDENT: You erased the previous
one too in the homework one.

MAGDALENA TODA: Because
that had a bug in it.

That one, the one in the
homework one, had a bug in it.

It only worked for some data.

And for other data
it didn't work.

So every time you find
a bug, you tell me,

and I will tell the programmer
of those problems, who's

really careful.

But one in 1,000 you
are bound to find a bug.

And I'm going to
give you a chocolate

or something for every bug.

And any other questions?

STUDENT: So are you
saying this is too long?

MAGDALENA TODA: Actually,
it's very beautiful.

If you have a calculator,
it's easier to solve it.

You can do it by hand,
write it by hand, also.

But it's a long--

STUDENT: [INAUDIBLE]

MAGDALENA TODA: Right,
so let's do it now

for anybody who wants to stay.

You don't have to stay.

So practicing what you do--

[SIDE CONVERSATIONS]