TTU Math2450 Calculus3 Sec 13.3

Conservative vector fields

MAGDALENA TODA: Welcome
to our review of 13.1.

How many of you didn't
get your exams back?

I have your exam, and yours.

And you have to wait.

I don't have it with me.

I have it in my office.

If you have questions
about the score,

why don't you go ahead and
email me right after class.

Chapter 13 is a very
physical chapter.

It has a lot to do with
mechanical engineering,

with mechanics,
physics, electricity.

You're going to see things,
weird things like work.

You've already seen work.

Do you remember the definition?

So we define the work
as a path integral

along the regular curve.

And by regular curve-- I'm
sorry if I'm repeating myself,

but this is part of the
deal-- R is the position

vector in R3 that is class C1.

That means differentiable and
derivatives are continuous.

Plus you are not
allowed to stop.

So no matter how drunk,
the bug has to keep flying,

and not even for a
fraction of a second is he

or she allowed to
have velocity 0.

At no point I want
to have velocity 0.

And that's the position vector.

And then you have some force
field acting on you-- no,

acting on the particle
at every moment.

So you have an F that is
acting at location xy.

Maybe if you are in space,
let's talk about the xyz,

where x is a function of
t, y is a functional of t,

z is a function of t,
which is the same as saying

that R of t, which is the given
position vector, is x of t

y of t.

Let me put angular bracket,
although I hate them,

because you like angular
brackets for vectors.

F is also a nice function.

How nice?

We discussed a
little bit last time.

It really doesn't
have to be continuous.

The book assumes it continues.

It has to be
integrable, so maybe it

could be piecewise continuous.

So I had nice enough, was
it continues piecewise.

And we define the work as
being the path integral over c.

I keep repeating, because
that's going to be on the final

as well.

So all the notions
that are important

should be given enough
attention in this class.

Hi.

So do you guys remember
how we denoted F?

F was, in general, three
components in our F1, F2, F3.

They are functions of
the position vector,

or the position xyz.

And the position is
a function of time.

So all in all, after
you do all the work,

keep in mind that when you
multiply with a dot product,

the integral will give you what?

A time integral?

From a time T0 to a time
T1, you are here at time T0

and you are here at time T1.

Maybe your curve is
piecewise, differentiable,

you don't know what it is.

But let's assume just a
very nice, smooth arc here.

Of what?

Of F1 times what is that?
x prime of t plus F2 times

y prime of t, plus F3
times z prime of t dt.

So keep in mind that
Mr. dR is your friend.

And he was-- what was he?

Was defined as the
velocity vector multiplied

by the infinitesimal element dt.

Say again, the
velocity vector prime

was a vector in F3 quantified
by the infinitesimal element dt.

So we reduce this Calc
3 notion path integral

to a Calc 1 notion, which was a
simple integral from t0 to t1.

And we've done a
lot of applications.

What else have we done?

We've done some
integral of this type

over another curve, script c.

I'm repeating mostly for Alex.

You're caught in the process.

And there are two or three
people who need an update.

Maybe I have another
function of g and ds.

And this is an integral that
in the end will depend on s.

But s itself depends on t.

So if I were to re-express
this in terms of d,

how would I re-express
the whole thing?

g of s, of t, whatever that
is, then Mr. ds was what?

STUDENT: s prime of t.

MAGDALENA TODA: Right.

So this was the-- that s
prime of t was the speed.

The speed of the arc of a curve.

So you have an R of
t and R3, a vector.

And the speed was,
by definition,

arc length element was
by definition integral

from 2t0 to t.

Of the speed R prime
magnitude d tau.

I'll have you put tau
because I'm Greek,

and it's all Greek to me.

So the tau, some people call
the tau the dummy variable.

I don't like to call it dumb.

It's a very smart variable.

It goes from t0 to t, so what
you have is a function of t.

This guy is speed.

So when you do that
here, ds becomes speed,

R prime of t times dt.

This was your old friend ds.

And let me put it on top
of this guy with speed.

Because he was so
important to you,

you cannot forget about him.

So that was review of--
reviewing of 13.1 and 13.2

There were some things in
13.3 that I pointed out

to you are important.

13.3 was independence of path.

Everybody write, magic-- no.

Magic section.

No, have to be serious.

So that's independence of path
of certain type of integrals,

of some integrals.

And an integral like
that, a path integral

is independent of path.

When would such an animal--
look at this pink animal,

inside-- when would
this not depend

on the path you are taking
between two given points?

So I can move on another
arc and another arc

and another regular arc, and
all sorts of regular arcs.

It doesn't matter
which path I'm taking--

STUDENT: If that
force is conservative.

MAGDALENA TODA: If the
force is conservative.

Excellent, Alex.

And what did it mean for a
force to be conservative?

How many of you
know-- it's no shame.

Just raise hands.

If you forgot what it is,
don't raise your hand.

But if you remember what it
means for a force F force

field-- may the force be
with you-- be conservative,

then what do you do?

Say F is conservative
by definition.

When, if and only, F
there is a so-called--

STUDENT: Scalar.

MAGDALENA TODA: --potential.

Scalar potential, thank you.

I'll fix that.

A scalar potential function f.

Instead of there is, I
didn't want to put this.

Because a few
people told me they

got scared about
the symbolistics.

This means "there exists."

OK, smooth potential
such that-- at least

is differential [INAUDIBLE]
1 such that the nabla of f--

what the heck is that?

The gradient of this little
f will be the given F.

And we saw all sorts of wizards
here, like, Harry Potter,

[INAUDIBLE] well,
there are many,

Alex, Erin, many,
many-- Matthew.

So what did they do?

They guessed the
scalar potential.

I had to stop because
there are 10 of them.

It's a whole school
of Harry Potter.

How do they find the little f?

Through witchcraft.

No.

Normally you should--

STUDENT: I've actually
done it through witchcraft.

Tell you that?

MAGDALENA TODA: You did.

I think you can do it
through witchcraft.

But practically everybody
has the ability to guess.

Why do we have the ability
to guess and check?

Because our brain does
the integration for you.

Whether you tell your
brain to stop or not,

when your brain, for example,
sees is kind of function--

and now I'm gonna
test your magic skills

on a little harder one.

I didn't want to do an
R2 value vector function.

Let me go to R3.

But I know that you have
your witchcraft handy.

So let's say somebody
gave you a force field

that is yz i plus xzj plus xyk.

And you're going to jump and
say this is a piece of cake.

I can see the scalar potential
and just wave my magic wand,

and I get it.

STUDENT: [INAUDIBLE]

MAGDALENA TODA: Oh my god, yes.

Guys, you saw it fast.

OK, I should be proud of you.

And I am proud of you.

I've had made classes
where the students couldn't

see any of the scalar
potentials that I gave them,

that I asked them to guess.

How did you deal with it?

You integrate this
with respect to F?

In the back of
your mind you did.

And then you guessed
one, and then you

said, OK so should be xyz.

Does it verify my
other two conditions?

And you say, oh yeah, it does.

Because of I prime with respect
to y, I have exactly xz.

If I prime with respect to c I
have exactly xy, so I got it.

And even if somebody
said xyz plus 7,

they would still be right.

In the end you can have
any xyz plus a constant.

In general it's not
so easy to guess.

But there are lots of examples
of conservative forces where

you simply cannot see the scalar
potential or cannot deduce it

like in a few seconds.

Expect something easy,
though, like that,

something that you can see.

Let's see an example.

Assume this is your force field
acting on a particle that's

moving on a curving space.

And it's stubborn and it
decides to move on a helix,

because it's a-- I don't
know what kind of particle

would move on a
helix, but suppose

a lot of particles, just a
little train or a drunken bug

or something.

And you were moving
on another helix.

Now suppose that helix will
be R of t equals cosine t

sine t and t where you
have t as 0 to start with.

What do I have at 0?

The point 1, 0, 0.

That's the point,
let's call it A.

And let's call this B. I
don't know what I want to do.

I'll just do a
complete rotation,

just to make my life easier.

And this is B. And that
will be A at t equals 0

and B equals 2 pi.

So what will this be at B?

STUDENT: 1, 0, 2 pi.

MAGDALENA TODA: 1, 0, and 2 pi.

So you perform a complete
rotation and come back.

Now, if your force is
conservative, you are lucky.

Because you know the theorem
that says in that case

the work integral will
be independent of path.

And due to the theorem in-- what
section was that again-- 13.3,

independence of path,
you know that this

is going to be-- let me rewrite
it one more time with gradient

of f instead of big F.

And this will become what,
f of the q-- not the q.

In the book it's f of q minus
f of q. f of B minus f of A,

right?

What does this mean?

You have to measure
the-- to evaluate

the coordinates of
this function xyz

where t equals 2 pi minus
xyz where t equals what?

0.

And now I have to be
careful, because I

have to evaluate them.

So when t is 0 I have x
is 1, y is 0, and t is 0.

In the end it doesn't matter.

I can get 0-- I
can get 0 for this

and get 0 for that as well.

So when this is 2 pi I get
x equals 1, y equals 0,

and t equals 2 pi.

So in the end, both products
are 0 and I got a 0.

So although the [INAUDIBLE]
works very hard-- I mean,

works hard in our perception
to get from a point

to another-- the work is 0.

Why?

Because it's a vector
value thing inside.

And there are some
annihilations going on.

So that reminds me
of another example.

So we are done
with this example.

Let's go back to our washer.

I was just doing
laundry last night

and I was thinking of
the washer example.

And I thought of a small
variation of the washer

example, just assuming that
I would give you a pop quiz.

And I'm not giving you
a pop quiz right now.

But if I gave you
a pop quiz now,

I would ask you example
two, the washer.

It is performing
a circular motion,

and I want to know
the work performed

by the centrifugal force
between various points.

So have the circular motion,
the centrifugal force.

This is the
centrifugal, I'm sorry.

I'll take the centrifugal force.

And that was last
time we discussed

that, that was extending
the radius of the initial--

the vector value position.

So you have that in
every point, xi plus yj.

And you want F to
be able xi plus yj.

But it points outside
from the point

on the circular trajectory.

And I asked you, find
out what you performed

by F in one full rotation.

We gave the equation of motion,
being cosine t y sine t,

if you remember from last time.

And then W2, let's
say, is performed by F

from t equals 0 to t equals pi.

I want that as well.

And W2 performed by F from
t-- that makes t0 to t

equals pi-- t equals 0
to t equals pi over 4.

These are all very
easy questions,

and you should be able to
answer them in no time.

Now, let me tell you something.

We are in plane, not in space.

But it doesn't matter.

It's like the third quadrant
would be 0, piece of cake.

Your eye should be so
well-trained that when

you look at the force
field like that,

and people talk about what
you should ask yourself,

is it conservative?

And it is conservative.

And that means little f is what?

Nitish said that yesterday.

Why did you go there?

You want to sleep today?

I'm just teasing you.

I got so comfortable with
you sitting in the front row.

STUDENT: I took his spot.

STUDENT: She doesn't like
you sitting over here.

MAGDALENA TODA: It's OK.

It's fine.

I still give him credit
for what he said last time.

So do you guys remember,
he gave us this answer?

x squared plus y squared over
2, and he found the scalar

potential through witchcraft
in about a second and a half?

OK.

We are gonna conclude something.

Do you remember that I found the
answer by find the explanation?

I got W to be 0.

But if I were to find another
explanation why the work would

be 0 in this case, it
would have been 0 anyway

for any force field.

Even if I took the F
to be something else.

Assume that F would be
G. Really wild, crazy,

but still differentiable
vector value function.

G differential.

Would the work that we
want be the same for G?

STUDENT: Yeah.

MAGDALENA TODA: Why?

STUDENT: Because of
displacement scenario.

MAGDALENA TODA: Since
it's conservative,

you have a closed loop.

So the closed loop
will say, thick F

at that terminal point minus
thick F at the initial point.

But if a loop motion,
your terminal point

is the initial point.

Duh.

So you have the
same point, the P

equals qe if it's
a closed curve.

So for a closed curve--
we also call that a loop.

With a basketball, it
would have been too easy

and you would have gotten a
dollar for free like that.

So any closed curve
is called a loop.

If your force field is
conservative-- attention,

you might have examples
like that in the exams--

then it doesn't matter
who little f is,

if p equals q you get 0 anyway.

But the reason why I
said you would get 0

on the example of last time
was a slightly different one.

What does the engineer
say to himself?

STUDENT: Force is perpendicular.

MAGDALENA TODA: Yeah.

Very good.

Whenever the force
is perpendicular

to the trajectory, I'm going
to get 0 for the force.

Because at every
moment the dot product

between the force and the
displacement direction,

which would be like dR, the
tangent to the displacement,

would be [INAUDIBLE].

And cosine of [INAUDIBLE] is 0.

Duh.

So that's another reason.

Reason of last time
was F perpendicular

to the R prime
direction, R prime

being the velocity-- look,
when I'm moving in a circle,

this is the force.

And I'm moving.

This is my velocity, is
the tangent to the circle.

And the velocity and the normal
are always perpendicular,

at every point.

That's why I have 0.

So note that even if I
didn't take a close look,

why would the answer
be from 0 to pi?

Still?

0 because of that.

0.

How about from 0 to pi over 4?

Still 0.

And of course if somebody
would not believe them,

if somebody would not
understand the theory,

they would do the work and
they would get to the answer

and say, oh my
god, yeah, I got 0.

All right?

OK.

Now, what if somebody--
and I want to spray this.

Can I go ahead and
erase the board

and move onto example
three or whatever?

Yes?

OK.

All right.

STUDENT: Could you say
non-conservative force?

MAGDALENA TODA: Yeah,
that's what I-- exactly.

You are a mind reader.

You are gonna guess my mind.

And I'm going to
pick a nasty one.

And since I'm doing
review anyway,

you may have one like that.

And you may have both one that
involves a conservative force

field and one that does not
involve a conservative force

field.

And we can ask you, find us the
work belong to different path.

And I've done this
type of example before.

Let's take F of
x and y in plane.

In our two I take xyi
plus x squared y of j.

And the problem would
involve my favorite picture,

y equals x squared and y
equals x, our two paths.

One is the straight path.

One is the [INAUDIBLE] path.

They go from 0,
0 to 1, 1 anyway.

And I'm asking you to
find W1 along path one

and W2 along path two.

And of course,
example three, if this

were conservative
you would say, oh,

it doesn't matter
what path I'm taking,

I'm still getting
the same answer.

But is this conservative?

STUDENT: No.

Because you said it wasn't.

MAGDALENA TODA: Very good.

So how do you know?

That's one test
when you are in two.

There is the magic test that
says-- let's say this is M,

and let's say this is N. You
would have to check if M sub--

STUDENT: y.

MAGDALENA TODA: y.

Very good.

I'm proud of you.

You're ready for
3350, by the way.

Is equal to N sub x.

M sub y is x.

N sub x is 2xy.

They are not equal.

So that's me crying that I have
to do the work twice and get--

probably I'll get two
different examples.

If you read the book--
I'm afraid to ask

how many of you opened the
book at section 13.2, 13.3.

But did you read
it, any of them?

STUDENT: Nitish read it.

MAGDALENA TODA: Oh, good.

There is another criteria for
a force to be conservative.

If you are, it's piece of cake.

You do that, right?

MAGDALENA TODA: Yes, sir?

STUDENT: Curl has frequency 0.

MAGDALENA TODA: The curl
criteria, excellent.

The curl has to be zero.

So if F in R 3 is
conservative, then you'll

get different order curve.

Curl F is 0.

Now let's check what
the heck was curl.

You see, mathematics
is not a bunch

of these joint discussions
like other sciences.

In mathematics, if you don't
know a section or you skipped

it, you are sick, you
have a date that day,

you didn't study, then it's
all over because you cannot

understand how to work out the
problems and materials if you

skip the section.

Curl was the one
where we learned

that we used the determinant.

That's the easiest story.

It came with a t-shirt,
but that t-shirt really

doesn't help because
it's easier to,

instead of memorizing
the formula,

you set out the determinant.

So you have the operator
derivative with respect

to x, y z followed by what?

F1, F2, F3.

Now in your case, I'm
asking you if you did it

for this F, what is
the third component?

STUDENT: The 0.

MAGDALENA TODA: The
0, so this guy is 0.

This guy is X squared
Y, and this guy is xy.

And it should be
a piece of cake,

but I want to do
it one more time.

I times the minor derivative
of 0 with respect to y

is 0 minus derivative
of x squared

y respect to 0, all
right, plus j minus

j because I'm alternating.

You've known enough
in your algebra

to know why I'm expanding
along the first row.

I have a minus, all
right, then the x

of 0, 0 derivative of xy
respect to the 0 plus k times

the minor corresponding
to k derivative 2xy.

Oh, and the derivative--

STUDENT: Yeah, this
is the n equals 0.

MAGDALENA TODA: Oh,
yeah, that's the one

where it's not a because
that's not conservative.

So what do you get.

You get 2xy minus x, right?

But I don't know how to
write it better than that.

Well, it doesn't matter.

Leave it like that.

So this would be 0 if it only
if x would be 0, but otherwise y

was 1/2.

But in general, it
is not a 0, good.

So F is not
conservative, and then we

can say goodbye
to the whole thing

here and move on to
computing the works.

What is the only
way we can do that?

By parameterizing
the first path,

but I didn't say which
one is the first path.

This is the first path, so
x of t equals t, and y of t

equals t is your
parameterization.

The simplest one, and then
W1 will be integral of-- I'm

too lazy to write down x of t,
y of t, but this is what it is.

Times x prime of
t plus x squared

y times y prime of t dt where--

STUDENT: Isn't that just
xy dx y-- never mind.

MAGDALENA TODA: This is F2.

And this is x prime,
and this is y prime

because this thing is
just-- I have no idea.

STUDENT: Right,
but what I'm asking

is that not the same as just
F1 dx because we're going

to do a chain rule anyway.

MAGDALENA TODA: If I put
the x, I cannot put this.

OK, this times that is dx.

This guy times this guy is dx.

STUDENT: But then
you can't use your

MAGDALENA TODA: Then I
cannot use the t's then.

STUDENT: All right, there we go.

MAGDALENA TODA: All right, so
I have integral from 0 to 1 t,

t times 1 t squared.

If I make a mistake, that would
be a silly algebra mistake

[INAUDIBLE].

All right, class.

t cubed times 1dt,
how much is this?

t cubed over 3 plus t
to the fourth over 4.

STUDENT: It's just 2-- oh, no.

MAGDALENA TODA: Very good.

Do not expect that we kill you
with computations on the exams,

but that's not what we want.

We want to test if you have
the basic understanding of what

this is all about, not to
kill you with, OK, that.

I'm not going to say that
in front of the cameras,

but everybody knows that.

There are professors
who would like

to kill you with computations.

Now, we're living in
a different world.

If I gave you a long
polynomial sausage here

and I ask you to
work with it, that

doesn't mean that I'm smart
because MATLAB can do it.

Mathematica, you get some
very nice simplifications

over there, so I'm
just trying to see

if rather than being able
to compute with no error,

you are having the basic
understanding of the concept.

And the rest can been done
by the mathematical software,

which, nowadays, most
mathematicians are using.

If you asked me 15
years ago, I think

I knew colleagues at all the
ranks in academia who would not

touch Mathematica or
MATLAB or Maple say

that's like tool from
evil or something,

but now everybody uses.

Engineers use mostly
MATLAB as I told you.

Mathematicians use both
MATLAB and Mathematica.

Some of them use Maple,
especially the ones who

have demos for K-12
level teachers,

but MATLAB is a wonderful
tool, very pretty powerful

in many ways.

If you are doing any kind
of linear algebra project--

I noticed three or four of you
are taking linear algebra-- you

can always rely on MATLAB being
the best of all of the above.

OK, W2.

For W2, I have a parabola, and
it's, again, a piece of cake.

X prime will be 1,
y prime will be 2t.

When I write down
the whole thing,

I have to pay a little
bit of attention

when I substitute
especially when I'm

taking an exam under pressure.

x squared is t
squared, y is t squared

times y prime, which is 2t.

So now this is x prime.

This is y prime.

Let me change colors.

All politicians change colors.

But I'm not a
politician, but I'm

thinking it's useful for you
to see who everybody was.

This is the F1 in terms of t.

That's the idea of what that
is, and this is F2 in terms of t

as well.

Oh, my God, another answer?

Absolutely, I'm going to
get an another answer.

Is it obviously to everybody
I'm going to get another answer?

STUDENT: Yeah.

MAGDALENA TODA: So I don't
have to put the t's here,

but I thought it was
sort of neat to see

that t goes from 0 to 1.

And what do I get?

This whole lot of them is t
cubed plus 2 t to the fifth.

So when I do the integration,
I get t to the 4 over 4 plus--

shut up, Magdalena, get people--

STUDENT: [INAUDIBLE].

MAGDALENA TODA: Very good.

Yeah, he's done
the simplification.

STUDENT: You get
the same values.

Plug in 1, you get 7/12 again.

MAGDALENA TODA: So I'm
asking you-- OK, what was it?

Solve 0, 1-- so I'm
asking why do you

think we get the same value?

Because the force
is not conservative,

and I went on another path.

I went on one path, and
I went on another path.

And look, obviously my
expression was different.

It's like one of those
math games or UIL games.

And look at the algebra.

The polynomials are different.

What was my luck here?

I took 1.

STUDENT: Yeah.

MAGDALENA TODA: I
could have taken 2.

So if instead of 1, I would
have taken another number,

then the higher the power,
the bigger the number

would have been.

I could have taken 2--

STUDENT: You could
have taken negative 1,

and you still wouldn't
have got the same answer.

MAGDALENA TODA: Yeah, there
are many reasons why that is.

But anyway, know that when you
take 1, 1 to every power is 1.

And yeah, you were lucky.

But in general, keep in
mind that if the force is

conservative, in
general, in most examples

you're not going to get the
same answer for the work

because it does depend on
the path you want to take.

I think I have reviewed quite
everything that I wanted.

So I should be ready
to move forward.

So I'm saying we are done
with sections 13.1, 13.2,

and 13.3, which was my
favorite because it's not

just the integral of
the path that I like,

but it's the so-called
fundamental theorem of calculus

3, which says, fundamental
theorem of the path integral

saying that you have f of the
endpoint minus f of the origin,

where little f is
that scalar potential

as the linear function
was concerned.

I'm going to call it the
fundamental theorem of path

integral.

Last time I told you the
fundamental theorem of calculus

is Federal Trade Commission.

We refer to that in Calc 1.

But this one is the fundamental
theorem of path integral.

Remember it because at
least one problem out of 15

or something on the
final, and there are not

going to be very many.

It's going to ask you to
know that result. This is

an important theorem.

And another important theorem
that is starting right now

is Green's theorem.

Green's theorem is
a magic result. I

have a t-shirt with it.

I didn't bring it today.

Maybe I'm going to bring
it next time First,

I want you to see
the result, and then

I'll bring the t-shirt
to the exam, so OK.

Assume that you have a
soup called Jordan curve.

You see, mathematicians don't
follow mathematical objects

by their names.

We are crazy people, but
we don't have a big ego.

We would not say a theorem
of myself or whatever.

We never give our names to that.

But all through calculus you
saw all sorts of results.

Like you see the Jordan
curve is a terminology,

but then you see
everywhere the Linus rule.

Did Linus get to
call it his own rule?

No, but Euler's
number, these are

things that were
discovered, and in honor

of that particular
mathematician,

we call them names.

We call them the name
of the mathematician.

Out of curiosity for
0.5 extra credit points,

find out who Jordan was.

Jordan curve is a closed
curve that, in general,

could be piecewise continuous.

So you have a closed
loop over here.

So in general, I could
have something like that

that does not enclose.

That encloses a
domain without holes.

Holes are functions
of the same thing.

STUDENT: So doesn't it
need to be continuous?

MAGDALENA TODA:
No, I said it is.

STUDENT: You said, piecewise.

MAGDALENA TODA: Ah, piecewise.

This is piecewise.

STUDENT: Oh, so it's piecewise.

OK.

MAGDALENA TODA: So you
have a bunch of arcs.

Finitely many, let's
say, in your case.

Finitely many arcs,
they have corners,

but you can see define the
integral along such a path.

Oh, and also for another
0.5 extra credit,

find out who Mr.
Green was because he

has several theorems that
are through mathematics

and free mechanics and
variation calculus.

There are several identities
that are called Greens.

There is this famous
Green's theorem,

but there are Green's
first identity,

Green's second identity,
and all sorts of things.

And find out who Mr.
Green was, and as a total,

you have 1 point extra credit.

And you can turn in a regular
essay like a two-page thing.

You want biography of these
mathematicians if you want,

just a few paragraphs.

So what does Green's theorem do?

Green's theorem is
a remarkable result

which links the path integral
to the double integral.

It's a remarkable
and famous result.

And that links the path
integral on the closed

curve to a double integral
over the domain enclosed.

I can see the domain
inside, but you

have to understand it's
enclosed by the curve.

All right, and assume
that you have--

M and N are C1 functions of
x and y, what does it mean?

M is a function of xy.

N is a function of xy in plane.

Both of them are differentiable
with continuous derivative.

They are differentiable.

You can take the
partial derivatives,

and all the partial
derivatives are continuous.

That's what we mean
by being C1 functions.

And there the magic
happens, so let me show you

where the magic happens.

This in the box,
the path integral

over c of M dx plus Ndy is
equal to the double integral

over the domain enclosed.

OK, this is the c.

On the boundary you
go counterclockwise

like any respectable
mathematician

would go in a trigonometric
sense, just counterclockwise.

And this is the domain
being closed by c.

And you put here the
integral, which is magic.

This is easy to
remember for you.

This is not easy to
remember unless I

take the t-shirt to
the exam, and you

cheat by looking at my t-shirt.

No, by the time of the
exam, I promised you

you are going to have at least
one week, seven days or more,

10-day period in which
we will study samples,

various samples of old finals.

I'm going to go ahead and
send you some by email.

Do you mind?

In the next week
after this week, we

are going to start reviewing.

And by dA I mean dxdy, the
usual area limit in Cartesian

coordinates the way you
are used to it the most.

And then, Alex is looking
at it and said, well,

then I tell her that
the most elegant way

is to put it with dxdy.

This is what we call a
one form in mathematics.

What is a one form.

It is a linear combination of
this infinitesimal elements

dxdy in plane with some
scalar functions of x

and y in front of her.

OK, so what do we do?

We integrate the one form.

The book doesn't talk about one
forms because the is actually

written for the average
student, the average freshman

or the average
sophomore, but I think

we have an exposure to
the notion of one form,

so I can get a little bit
more elegant and more rigorous

in my speech.

If you are a graduate
student, you most likely

would know this is a one form.

That's actually the
definition of a one form.

And you'll say, what is this?

This is actually two
form, but you are

going to say, wait a minute.

I have a scalar
function, whatever

that is, from the integration
in front of the dxdy

you want but you never said
that dxdy is a two form.

Actually, I did, and I
didn't call it a two form.

Do you remember that
I introduced to you

some magic wedge product?

And we said, this is
a tiny displacement.

Dx infinitesimal is small.

Imagine how much the
video we'll there

is an infinitesimal
displacement dx

and an infinitesimal
displacement dy,

and you have some
sort of a sign area.

So we said, we don't
just take dxdy,

but we take a product
between dxdy with a wedge,

meaning that if I
change the order,

I'm going to have minus dy here.

This is typical exterior
derivative theory-- exterior

derivative theory.

And it's a theory that
starts more or less

at the graduate level.

And many people get their
master's degree in math

and never get to see it, and
I pity them, but this life.

On the other hand, when
you have dx, which dx--

the area between dx and dx is 0.

So we're all very happy
I get rid of those.

When I have the sign
between the displacement,

dy and itself is 0.

So these are the
basic properties

that we started
about the sign area.

I want to show you what happens.

I'm going to-- yeah,
I'm going to erase here.

I'm going to show
you later I'm going

to prove this theorem to you
later using these tricks that I

just showed you here.

I will provide proof
to this formula, OK?

And let's take a look at
that, and we say, well,

can I memorize that by
the time of the final?

Yes, you can.

What is beautiful about
this, it can actually

help you solve problems that you
didn't think would be possible.

For example, example
1, and I say,

that would be one of
the most basic ones.

Find the geometric meaning of
the integral over a c where

c is a closed loop.

OK, c is a loop.

Piecewise define Jordan
curve-- Jordan curve.

And I integrate out
of something weird.

And you say, oh, my God.

Look at her.

She picked some weird function
where the path from the dx

is M, and the path in front of
dy is N, the M and N functions.

Why would pick like that?

You wouldn't know yet, but
if you apply Green's theorem,

assuming you believe
it's true, you

have double integral over the
domain enclosed by this loop.

The loop is enclosing
this domain of what?

Now, I'm trying to shut up,
and I'm want you to talk.

What am I going to
write over here?

STUDENT: 1 plus 1.

MAGDALENA TODA: 1 plus
1, how fun is that?

Y minus 1, 1 plus 1 equals
2 last time I checked,

and this is dA.

And what do you think
this animal would be?

The cast of 2 always can escape.

So if we don't want
it, just kick it out.

What is the remaining
double integral for d of DA?

You have seen this guy all
through the Calculus 3 course.

You're tired of it.

You said, I cannot wait for
this semester to be over

because this is the double
integral of 1dA over d.

What in the world is that?

That is the--

STUDENT: --area.

MAGDALENA TODA: Area, very good.

This is the area of the
domain d inside the curve.

The shaded area is this.

So you have discovered
something beautiful

that the area of the domain
enclosed by a Jordan curve

equals 1/2 because you pull
the two out in front here,

it's going to be 1/2 of the path
integrals over the boundary.

This is called
boundary of a domain.

c is the boundary of the domain.

Some mathematicians--
I don't know

how far you want to go with your
education, but in a few years

you might become
graduate students.

And even some engineers use this
notation boundary of d, del d.

That means the boundaries,
the frontier of a domain.

The fence of a ranch.

That is the del d, but
don't tell the rancher

because he will take his gun
out and shoot you thinking

that you are off the hook or
you are after something weird.

So that's the boundary
of the domain.

And then you have
minus ydx plus xdy.

MAGDALENA TODA: We discover
something beautiful.

Something important.

And now I'm asking,
with this exercise--

one which I could even--
I could even call a lemma.

Lemma is not quite a
theorem, because it's based--

could be based on a theorem.

It's a little result that can
be proved in just a few lines

without being something
sophisticated based

on something you
knew from before.

So this is called a lemma.

When you have a sophisticated
area to compute--

or even can you prove-- if you
believe in Green's theorem,

can you prove that the
area inside the circle

is pi r squared?

Can you prove that the
area inside of an ellipse

is-- I don't know what.

Do you know the area
inside of an ellipse?

Nobody taught me in school.

I don't know why
it's so beautiful.

I learned what an ellipse
was in eleventh grade

in high school and again
a review as a freshman

analytic geometry.

So we've seen conics again--

STUDENT: I think we did
conics in 10th grade.

We might have seen it.

MAGDALENA TODA:
But nobody told me

like-- I give you an ellipse.

Compute the area inside.

I had no idea.

And I didn't know
the formula until I

became an assistant professor.

I was already in my thirties.

That's a shame to see that
thing for the first time OK.

So let's see if we believe
this lemma, and the Green's

theorem of course.

But let's apply the
lemma, primarily

from the Green's theorem.

Can we actually prove
that the area of the disk

is pi r squared and the
area of the ellipse--

inside the ellipse
will be god knows what.

And we will discover
that by ourselves.

I think that's the
beauty of mathematics.

Because every now and then
even if you discover things

that people have known
for hundreds of years,

it still gives you
the satisfaction

that you did something by
yourself-- all on yourself.

Like, when you feel
build a helicopter or you

build a table.

There are many more
beautiful tables

that were built before
you, but still it's

a lot of satisfaction that
you do all by yourself.

It's the same with mathematics.

So let's see what we can
do now for the first time.

Not for the first time.

We do it in other ways.

Can you prove using the lemma
or Green's theorem-- which

is the same thing-- either
one-- that the area of the disk

of radius r-- this is the r.

so this the radius
r is pi r squared.

I hope so.

And the answer is, I hope so.

And that's all.

Area of the disk of radius r.

Oh my god.

So you go, well.

If I knew the parameterization
of that boundary C,

it would be a piece of cake.

Because I would just-- I know
how to do a path integral now.

I've learned in the
previous sections,

so this should be easy.

Can we do that?

So let's see.

Without computing
the double integral,

because I can always do
that with polar coordinates.

And we are going to do that.

Let's do that as
well, as practice.

Because so you
review for the exam.

But another way to
do it would be what?

1/2 integral over the circle.

And how do I parametrize a
circle of fixed radius r?

Who tells me?

x of t will be--
that was Chapter 10.

Everything is a
circle in mathematics.

STUDENT: r cosine t.

MAGDALENA TODA: r cosine t.

Excellent.

y of t is?

STUDENT: r sine t.

MAGDALENA TODA: r sine t.

So, finally I'm going to
go ahead and use this one.

And I'm going to say, well,
minus y to be plugged in.

This is minus y.

Multiply by dx.

Well, you say, wait a
minute. dx with respect.

What is dx?

dx is just x prime dt.

Dy is just y prime dt.

And t goes out.

It's banished.

No, he's the most important guy.

So t goes from something
to something else.

We will see that later.

What is x prime dt?

X prime is minus r sine
theta-- sine t, Magdalena.

Minus r sine t.

That was x prime.

Change the color.

Give people some
variation in their life.

Plus r cosine t,
because this x--

STUDENT: [INAUDIBLE].

MAGDALENA TODA: --times
the y, which is r cosine t.

So it suddenly became beautiful.

It looks-- first it looks ugly,
but now it became beautiful.

Why?

How come it became beautiful?

STUDENT: Because you got sine
squared plus cosine square.

MAGDALENA TODA:
Because I got a plus.

If you pay attention, plus sine
squared plus cosine squared.

So I have, what is sine
squared plus cosine squared?

I heard that our
students in trig--

Poly told me-- who still don't
know that this is the most

important thing you
learn in trigonometry--

is Pythagorean theorem.

Right?

So you have 1/2 integral

STUDENT: r squared--

MAGDALENA TODA:
r-- no, I'm lazy.

I'm going slow-- r.

dt.

T from what to what?

From 0 times 0.

I'm starting whatever
I want, actually.

I go counterclockwise
I'm into pi.

STUDENT: Why is
that not r squared?

It should be r squared.

MAGDALENA TODA: I'm sorry, guys.

I'm sorry.

I don't know what
I am-- r squared.

1/2 r squared times 2 pi.

So we have pi r squared.

And if you did not
tell me it's r squared,

we wouldn't have
gotten the answer.

That's good.

What's the other way to do it?

If a problem on
the final would ask

you prove in two different
ways that the rubber

disk is pi r squared using
Calc 3, or whatever--

STUDENT: Would require--

MAGDALENA TODA: The
double integral, right?

Right?

STUDENT: Could have done
Cartesian coordinates as well.

If that counts as a second way.

MAGDALENA TODA: Yeah.

You can-- OK.

What could this be?

Oh my god.

This would be minus 1
to 1 minus square root

1 minus x squared to square
root 1 minus x squared.

Am i right guys?

STUDENT: Yep.

MAGDALENA TODA: 1 dy dx.

Of course it's a pain.

STUDENT: You could double that
and set the bottoms both equal

to 0.

MAGDALENA TODA: Right.

So we can do by symmetry--

STUDENT: Yeah.

MAGDALENA TODA: I'm--
shall I erase or leave it.

Are you understand
what Alex is saying?

This is 2i is the integral
that you will get.

STUDENT: Just write
it next to it--

MAGDALENA TODA: I tell
you four times, you

see, Alex, because you have--

STUDENT: Oh, yeah.

MAGDALENA TODA: --symmetry
with respect to the x-axis,

and symmetry with
respect to y-axis.

And you can take 0
to 1 and 0 to that.

And you have x from 0 to 1.

You have y from 0 to stop.

Square root of 1 minus x square.

Like the strips.

And you have 4
times that A1, which

would be the area of
the first quadratic.

You can do that, too.

It's easier.

But the best way to do that is
not in Cartesian coordinates.

The best way is to do
it in polar coordinates.

Always remember
your Jacobian is r.

So if you have
Jacobian r-- erase.

Let's put r here again.

And then dr d theta.

But now you say, wait
a minute, Magdalena.

You said r is fixed.

Yes.

And that's why I
need to learn Greek,

because it's all Greek to me.

Instead of r I put
rho as a variable.

And I say, rho is
between 0 and r.

r is fixed.

That's my [INAUDIBLE].

Big r is not usually written
as a variable from 0 to some.

I cannot use that.

So I have to us a Greek letter,
whether I like it or not.

And theta is from 0 to 2 pi.

And I still get the same thing.

I get r-- rho squared
over 2 between 0 and r.

And I have 2 pi.

And in the end that means
pi r squared, and I'm back.

And you say, wait,
this is Example 4.

Whatever example.

Is it Example 4, 5?

You say, this is
a piece of cake.

I have two methods showing
me that area of the disk

is so pi r squared.

It's so trivial.

Yeah, then let's move
on and do the ellipse.

Or we could have been
smart and done the ellipse

from the beginning.

And then the circular
disk would have

been just a trivial, particular
example of the ellipse.

But let's do the ellipse
with this magic formula

that I just taught you.

In the finals-- I'm going to
send you a bunch of finals.

You're going to be
amused, because you're

going to look at
them and you say,

regardless of the year and
semester when the final was

given for Calc 3,
there was always

one of the problems at the
end using direct application

of Green's theorem.

So Green's theorem
is an obsession,

and not only at Tech.

I was looking UT Austin,
A&M, other schools--

California Berkley-- all the
Calc 3 courses on the final

have at least one application--
direct application

applying principal.

OK.

So what did I say?

I said that we have
to draw an ellipse.

How do we draw an ellipse
without making it up?

That's the question.

STUDENT: Draw a circle.

MAGDALENA TODA: Draw a circle.

Good answer.

OK.

All right.

And guys this
started really bad.

So I'm doing what I can.

I should have tried
more coffee today,

because I'm getting
insecure and very shaky.

OK.

So I have the ellipse
in standard form

of center O, x squared over
x squared plus y squared

over B squared equals 1.

And now you are going to
me who is A and who is B?

What are they called?

Semi--

STUDENT: Semiotics.

MAGDALENA TODA: Semiotics.

A and B. Good.

Find the area.

I don't like-- OK.

Let's put B inside, and let's
put C outside the boundary.

So area of the ellipse domain
D will be-- by the lemma-- 1/2

integral over C.

This is C. Is not f.

Don't confuse it.

It is my beautiful
script C. I've

tried to use it many times.

Going to be minus y dx plus xdy.

Again, why was that?

Because we said this
is M and this is N,

and Green's theorem will give
you double integral of N sub x

minus M sub y.

So you have 1 minus
minus 1, which is 2.

And 2 knocked that out.

OK.

That's how we prove it.

OK.

Problem is that I do not the
parametrization of the ellipse.

And if somebody doesn't help me,
I'm going to be in big trouble.

And I'll start
cursing and I'm not

allowed to curse in
front of the classroom.

But you can help me on
that, because this reminds

you of a famous Greek identity.

The fundamental trig identity.

If this would be cosine
squared of theta,

and this would be sine squared
of theta, as two animals,

their sum would be 1.

And whenever you have sums
of sum squared thingies,

then you have to think trig.

So, what would be
good as a parameter?

OK.

What would be good
as a parametrization

to make this come true?

STUDENT: You have the cosine
of theta would equal x over x.

MAGDALENA TODA: Uh-huh.

So then x would be A times--

STUDENT: The cosine of theta.

MAGDALENA TODA:
Do you like theta?

You don't, because
you're not Greek.

That's the problem.

If you were Greek,
you would like it.

We had a colleague who
is not here anymore.

Greek from Cypress.

And he could claim that the
most important-- most important

alphabet is the
Greek one, and that's

why the mathematicians
adopted it.

OK?

B sine t.

How do you check?

You always think, OK.

This over that is cosine.

This over this is sine.

I square them.

I get exactly that
and I get a 1.

Good.

I'm in good shape.

I know that this
implicit equation--

this is an implicit
equation-- happens if and only

if I have this system of
the parametrization with t

between-- anything I want,
including the basic 0 to 2

pi interval.

And then if I were to move
all around for time real t

I would wind around that the
circle infinitely many times.

Between time equals
minus infinity--

that nobody remembers-- and
time equals plus infinity--

that nobody will
ever get to know.

So those are the values of it.

All the real values, actually.

I only needed from 0 to 2
pi to wind one time around.

And this is the idea.

I wind one time around.

Now people-- you're going
to see mathematicians

are not the greatest people.

I've seen engineers and
physicists use a lot this sign.

Do you know what this means?

STUDENT: It means
one full revolution.

MAGDALENA TODA: It
means a full revolution.

You're going to have
a loop-- loops, that's

whatever you want.

Here and goes counterclockwise.

And they put this
little sign showing

I'm going counterclockwise on
a closed curved, or a loop.

All right.

Don't think they are crazy.

This was used in lots
of scientific papers

in math, physics, and
engineering, and so on.

OK.

Let's do it then.

Can we do it by ourselves?

I think so.

That's see.

1/2 is 1.

And I don't like
the pink marker.

Integral log.

Time from 0 to 2 pi
should be measured.

y minus B sine t.

dx-- what tells me that?

STUDENT: B minus--

MAGDALENA TODA: Very good.

Minus A sine t.

How hard is that?

It's a piece of cake Plus x--

STUDENT: A cosine.

MAGDALENA TODA: Very good.

A cosine t.

TImes--

STUDENT: B cosine t.

MAGDALENA TODA: --B cosine t.

And dt.

And this thing-- look at it.

It's huge.

It looks huge, but it's
so beautiful, because--

STUDENT: AB.

MAGDALENA TODA: AB.

Why is it AB?

It's AB because sine squared
plus cosine squared inside

becomes 1.

And I have plus AB,
plus AB, AB out.

Kick out the AB.

Kick out the A and
the B and you get

something beautiful-- sine
squared t plus cosine squared

t is your old friend.

And he says, I'm 1.

Look how beautiful
life is for you.

Finally, we proved it.

What did we prove?

We are almost there.

We got a 1/2.

A constant value kick out, AB.

STUDENT: Times 2 pi.

MAGDALENA TODA: Times 2 pi.

Good.

2 goes away.

And we got a magic thing that
nobody taught us in school,

because they were mean.

They really didn't want
us to learn too much.

That's the thingy.

AB pi.

AB pi is what we were
hoping for, because, look.

I mean it's almost
too good to be true.

Well, it's a disk of radius
r, A and B are equal.

And they are the
radius of the disk.

And that's why we
have pi r squared

as a particular
example of the disk

of the area of this ellipse.

When I saw it the first
time, I was like, well,

I'm glad that I lived to be
30 or something to learn this.

Because nobody had shown it
to me in K-12 or in college.

And I was a completing-- I was
a PhD and I didn't know it.

And then I said, oh,
that's why-- pi AB.

Yes, OK.

All right.

So it's so easy to
understand once you-- well.

Once you learn the section.

If you don't learn
the section you

will not be able to understand.

OK.

All right.

I'm going to go
ahead and erase this.

And I'll show you
an example that

was popping up like an obsession
with the numbers changed

in most of the final exams
that happen in the last three

years, regardless of
who wrote the exam.

Because this problem really
matches the learning outcomes,

oh, just about any university--
any good university

around the world.

So you'll say, wow.

It's so easy.

I could not believe it
that-- how easy it is.

But once you see it, you
will-- you'll say, wow.

It's easy.

OK.

[CHATTER]

Let's try this one.

You have a circle.

and the circle will be
a circle radius r given

and origin 0 of 4, 9, 0, and 0.

And I'm going to
write-- I'm going

to give you-- first I'm going
to give you a very simple one.

Compute in the
simplest possible way.

If you don't want to
parametrize the circle--

you can always
parametrize the circle.

Right?

But you don't want to.

You want to do it the
fastest possible way

without parameterizing
the circle.

Without writing down
what I'm writing down.

You are in a hurry.

You have 20-- 15 minutes
left of your final.

And you're looking
at me and say, I

hope I get an A in this final.

So what do you have to
remember when you look at that?

M and M. M and M.
No, M and N. OK.

And you have to remember
that you are over a circle

so you have a closed loop.

And that's a Jordan curve.

That's enclosing a disk.

So you have a relationship
between the path

integral along the C and the
area along the D-- over D.

Which is of what?

Is N sub x minus M sub y.

So let me write it
in this form, which

is the same thing my students
mostly prefer to write it as.

N sub x minus M sub y.

The t-shirt I have
has it written

like that, because it was
bought from nerdytshirt.com

And it was especially
created to impress nerds.

And of course if you
look at the del notation

that gives you that kind
of snobbish attitude

that you aren't a scientist.

OK.

So what is this
going to be then?

Double integral over d.

And sub x is up
here so it gave 5.

And sub y is a piece of cake.

3 dx dy equals 2 out times
the area of the disk, which

is something you know.

And I'm not going to ask you
to prove that all over again.

So you have to say 2.

I know the area of the
disk-- pi r squared.

And that's the answer.

And you leave the room.

And that's it.

It's almost too
easy to believe it,

but it was always there in
the simplest possible way.

And now I'm wondering, if I
were to give you something hard,

because-- you know my theory
that when you practice

at something in
the classroom you

have to be working on harder
things in the classroom

to do better in the exam.

So let me cook up
something ugly for you.

The same kind of disk.

And I'm changing the functions.

And I'll make it
more complicated.

Let's see how you
perform on this one.

We avoided that one,
probably, on finals

because I think the
majority of students

wouldn't have understood what
theorem they needed to apply.

It looks a little bit scary.

But let's say that I've
given you the hint,

apply Greens theorem
on the same path

integral, which is a circle
of origin 0 and radius r.

I now draw counterclockwise.

You apply Green's theorem and
you say, I know how to do this,

because now I know the theorem.

This is M. This is N. And I--
my t-shirt did not say M and N.

It said P and Q. Do you
want to put P and Q?

I put P and Q.

So I can-- I can have this
like it is on my t-shirt.

So this is going
to be P sub x-- no.

Q sub x.

Sorry.

M and N. So the second
one with respect to x.

The one that sticks to the y
is prime root respect to x.

The one that sticks to dx is
prime root with respect to y.

And I think one
time-- the one time

when that my friend and
colleague wrote that,

he did it differently.

He wrote something
like, just-- I'll

put-- I don't remember what.

He put this one.

Then the student
was used to dx/dy

and got completely confused.

So pay attention to
what you are saying.

Most of us write it
in x and y first.

And we can see that the
derivative with respect

to x of q, because that is
the one next to be the y.

When he gave it to me
like that, he messed up

everybody's notations.

No.

Good students steal data.

So you guys have to
put it in standard form

and pay attention to
what you are doing.

All right.

So that one form can
be swapped by people

who try to play games.

Now in this one-- So you
have q sub x minus b sub y.

You have 3x squared minus
minus, or just plus, 3y squared.

Good.

Wonderful.

Am I happy, do you
think I'm happy?

Why would I be so happy?

Why is this a happy thing?

I could have had
something more wild.

I don't.

I'm happy I don't.

Why am I so happy?

Let's see.

3 out over the disk.

Is this ringing a bell?

Yeah.

It's r squared if I
do this in former.

So if I do this in former,
its going to be rdr, d theta.

So life is not as
hard as you believe.

It can look like
a harder problem,

but in reality, it's not really.

So I have 3 times-- now, I
have r squared, I have r cubed.

r cubed dr d theta, r between.

r was between 0 and big
R. Theta will always

be between 0 and 2 pi.

So, I want you, without
me to compute the answer

and tell me what you got.

STUDENT: Just say it?

MAGDALENA TODA: Yep.

STUDENT: 3/2, pi
r to the fourth.

MAGDALENA TODA: So
how did you do that?

You said, r to the
4 over 4, coming

from integration times the 2 pi,
coming from integration times

3.

Are you guys with me?

Is everybody with me on this?

OK so, we will simplify
the answer, we'll do that.

What regard is the
radius of the disk?

STUDENT: How did he
solve that integral

without switching the poles?

MAGDALENA TODA: It would
have been a killer.

Let me write it out.

[LAUGHTER]

Because you want to
write it out, of course.

OK, 3 integral, integral
x squared plus y

squared, dy/dx, just to make
my life a little bit funnier,

and then y between minus
square root-- you're

looking for trouble, huh?

Y squared minus x squared to
r squared minus s squared.

And again, you could
do what you just

said, split into four integrals
over four different domains,

or two up and down.

And minus r and are
you guys with me?

And then, when you go
and integrate that,

you integrate with respect
to y-- [INAUDIBLE].

Well he's right,
so you can get x

squared y plus y cubed over 3.

Between those points, minus 12.

And from that moment,
that would just

leave it and go for a walk.

I will not have the
patience to do this.

Just a second, Matthew.

For this kind of
stuff, of course

I could put this in Maple.

You know Maple has these
little interactive fields,

like little squares?

And you go inside there
and add your endpoints.

And even if it looks very ugly,
Maple will spit you the answer.

If you know your
syntax and do it right,

even if you don't
switch to polar

coordinates or put
it in Cartesian.

Give it the right data, and
it's going to spit the answer.

Yes, Matthew?

STUDENT: I was
out of the room, I

was wondering why
it's now y cubed.

MAGDALENA TODA: Because if
you integrate with respect

to y first--

STUDENT: Because when I
walked out, it was negative y.

MAGDALENA TODA: If
I didn't put minus.

STUDENT: It's a new problem.

That's what he's confused about.

He walked out of the room
during the previous problem

and came back after this one.

And now he's confused.

MAGDALENA TODA: You don't
care about what I just asked?

STUDENT: Oh.

No.

I like the polar coordinates.

MAGDALENA TODA: Let
me ask you a question

before I talk any further.

I was about to put a plus here.

What would have been the problem
if I had put a plus here?

If I worked this out,
I would have gotten

x squared minus y squared.

Would that have been
the end of the world?

No.

But it would have complicated
my life a little bit more.

Let's do that one as well.

STUDENT: I was
just curious of how

you do any of these problems
when you can't switch to polar.

MAGDALENA TODA: Right, let's see
what-- because Actually, even

in this case, life is not so
hard, not as hard as you think.

The persistence in that matters.

You never give up on a
problem that freaks you out.

That's the definition
of a mathematician.

3x squared minus 3y
squared over dx/dy.

Do it slowly because
I'm not in a hurry.

We are almost done with 13.4.

This is OK, right?

Just the minus sign again?

STUDENT: Well not
the minus sign.

I was just wondering because
in the previous problem

you were doing the ellipse, you
started out with the equation

with the negative y--

MAGDALENA TODA:
For this one that's

just the limit that says that
this is the go double integral

of the area of the domain.

It's just a consequence--
or correlate if you want.

It's a consequence
of Green's theorem.

When you forget that consequence
of Green's theorem and we say

goodbye to that.

But while you were out,
this is Green's theorem.

The real Green's theorem,
the one that was a teacher.

There are several
Greens I can give you.

The famous Green
theorem is the one

I said when you have--
this is what we apply here.

The integral of M dx plus M dy.

You have a double integral of
M sub x minus M sub y over c.

So I'm assuming we would have
had this case of maybe me not

paying attention, or
being mean and not wanting

to give you a simple problem.

And what do you
do in such a case?

It's not obvious to
everybody, but you will see.

It's so pretty at some
point, if you know

how to get out of the mess.

I was already thinking, but
I'm using polar coordinates.

So that's arc of sine, so I
have to go back to the basics.

If I go back to the
basics, ideas come to me.

Right?

So, OK.

r-- let's put dr d theta,
just to get rid of it,

because it's on my nerves.

This is 0 to 2 pi,
this is 0 to r.

And now, you say,
OK, in our mind,

because we are lazy
people, plug in those

and pull out what you can.

One 3 out equals for what?

Inside, you have r squared.

Do you agree?

And times your favorite
expression, which is cosine

squared theta, minus
i squared theta.

And you're going to ask me why.

You shouldn't ask me why.

You just square these
and subtract them,

and see what in the world
you're going to get.

Because you get r squared
times cosine squared,

minus i squared.

I'm too lazy to write
down the argument.

But you know we
have trigonometry.

Yes, you see why it's
important for you

to learn trigonometry
when you are little.

You may be 50 or
60, in high school,

or you may be freshman year.

I don't care when, but you
have to learn that this is

the cosine of the double angle.

How many of you remember that?

Maybe you learned that?

Remember that?

OK.

I don't blame you at all
when you don't remember,

because since I've been
the main checker of finals

for the past five years--
it's a lot of finals.

Yeah, the i is there.

That's exactly what
I wanted to tell you,

that's why I left some room.

This data would be t.

The double angle formula did
not appear on many finals.

And I was thinking
it's a period.

When I ask the
instructors, generally they

say students have trouble
remembering or understanding

this later on, by
avoiding the issue,

you sort of bound to it for
the first time in Cal 2,

because there are any
geometric formulas.

And then, you bump again
inside it in Cal 3.

And it never leaves you.

So this, just knowing this
will help you so much.

Let me put the r nicely here.

And now finally, we know
how to solve it, because I'm

going to go ahead and erase.

So why it is good for us is
that-- as Matthew observed

a few moments ago,
whenever you have

a product of a function, you
not only in a function in theta

only, your life becomes easier
because you can separate them

between the rhos.

In two different products.

So that's would be this theorem.

And you have 3 times-- the
part that depends only on r,

and the part that depends
only on theta, let's

put them separate.

We need theta, and
dr. And what do you

integrate when you integrate?

r cubed.

Attention, do not do rr.

From 0 to r.

OK?

STUDENT: And cosine theta?

MAGDALENA TODA: And then you
have a 0 to 2 pi, cosine 2.

now, let me give
you-- Let me tell you

what it is, because when
I was young, I was naive

and I always started with that.

You should always start with the
part, the trig part in theta.

Because that becomes 0.

So no matter how
ugly this is, I've

had professors who are
playing games with us,

and they were giving us
some extremely ugly thing

that would take you forever
for you to integrate.

Or sometimes, it would have
been impossible to integrate.

But then, the whole
thing would have been 0

because when you
integrate cosine 2 theta,

it goes to sine theta.

Sine 2 theta at 2 pi and 0
are the same things, 0 minus 0

equals z.

So the answer is z.

I cannot tell you how many
professors I've had who will

play this game with us.

They give us something
that discouraged us.

No, it's not a piece of cake,
compared to what I have.

Some integral value
will go over two lines,

with a huge polynomial
or something.

But in the end, the integral
was 0 for such a result. Yes?

STUDENT: So I have a question.

Could we take that force and
prove that it was conservative?

MAGDALENA TODA: So now
that I'm questioning this,

I'm not questioning
you, but I-- is

the force, that is with you--
what is the original force

that Alex is talking about?

If I take y cubed i plus x cubed
j-- and you have to be careful.

Is this conservative?

STUDENT: Yeah.

MAGDALENA TODA: Really?

Why would we pick
a conservative?

STUDENT: Y squared plus
x squared over 2 is--

MAGDALENA TODA: Why is
it not conservative?

IT doesn't pass the hole test.

So p sub y is not
equal to q sub x.

If you primed this with respect
to y, you get that dy squared.

Prime this with this respect
to x, you get 3x squared.

So it's not concerned with him.

And still, I'm
getting-- it's a loop,

and I'm getting a 0, sort
of like I would expect it

I had any dependence of that.

What is the secret here?

STUDENT: That is conservative,
given a condition.

MAGDALENA TODA: Yes,
given a condition

that your x and y are moving
on the serpent's circle.

And that happens, because this
is a symmetric expression,

and x and y are
moving on a circle,

and one is the cosine theta
and one is sine theta.

So in the end, it
simplifies out.

But in general, if I would
have this kind of problem--

if somebody asked me is this
conservative, the answer is no.

Let me give you a
few more examples.

One example that maybe will look
hard to most people is here.

The vector value function
given by f of x, y incline,

are two values.

No, I mean define two
values of [INAUDIBLE].

A typical exam problem.

And I saw it at
Texas A&M, as well.

So maybe some people like this
kind of a, b, c, d problem.

Is f conservative?

STUDENT: Yep

MAGDALENA TODA:
You already did it?

Good for you guys.

So if I gave you one that
has three components what

did you have to do?

Compute the curl.

You can, of course, compute
the curl also on this one

and have 0 for the
third component.

But the simplest thing
is to do f1 and f2.

f1 prime with respect to y
equals f2 prime with respect

to x.

So I'm going to
make a smile here.

And you realize that the authors
of such a problem, whether they

are at Tech or at Texas
A&M. They do that on purpose

so that you can use this
result to the next level.

And they're saying compute
the happy u over the curve

x cubed and y cubed equals 8 on
the path that connects points

2, 1 and 1, 2 in [INAUDIBLE].

Does this integral depend on f?

State why.

And you see, they don't tell
you find the scalar potential.

Which is bad, and
many of you will

be able to see it
because you have

good mathematical intuition,
and a computer process

planning in the background
over all the other processes.

We are very visual people.

If you realize that every time
just there with each other

through the classroom, there
are hundreds of distractions.

There's the screen,
there is somebody

who's next to you
who's sneezing,

all sorts of distractions.

Still, your computer
unit can still

function, trying to
integrate and find the scalar

potential, which is a miracle.

I don't know how we managed
to do that after all.

If you don't manage to do that,
what do you have to set up?

You have to say, find is
there-- well, you know there is.

So you're not going to question
the existence of the scalar

potential You know it exists,
but you don't know what it is.

What is f such that f sub
x would be 6xy plus 1,

and m sub y will be 3x squared?

And normally, you would
have to integrate backwards.

Now, I'll give you 10 seconds.

If in 10 seconds, you don't
find me a scalar potential,

I'm going to make you
integrate backwards.

So this is finding the scalar
potential by integration.

The way you should, if
you weren't very smart.

But I think you're
smart enough to smell

the potential-- Very good.

But what if you don't?

OK I'm asking.

So we had one or two
student who figured it out.

What if you don't?

If you don't, you can still do
perfectly fine on this problem.

Let's see how we do it
without seeing or guessing.

His brain was running
in the background.

He came up with the answer.

He's happy.

He can move on to
the next level.

STUDENT: Integrate both
sides with respect to r.

MAGDALENA TODA: Right, and
then mix and match them.

Make them in work.

So try to integrate
with respect to x.

6y-- or plus 1, I'm sorry guys.

And once you get it,
you're going to get--

STUDENT: 3x squared y plus x.

MAGDALENA TODA: And
plus a c of what?

And then take this fellow and
prime it with respect to y.

And you're going to
get-- it's not hard.

You're going to get dx
squared plus nothing,

plus c from the y, and it's
good because I gave you

a simple one.

So sometimes you can
have something here,

but in this case, it was just 0.

So c is kappa as a constant.

So instead of why we teach
found with a plus kappa here,

and it still does it.

So on such a problem,
I don't know,

but I think I would give equal
weights to it, B and C. Compute

the path integral
over the curve.

This is horrible, as
an increasing curve.

But I know that
there is a path that

connects the points 2, 1 and 1.

What I have to pay
attention to in my mind

is that these points
actually are on the curve.

And they are, because I
have 8 times 1 equals 8,

1 times 8 equals 8.

So while I was writing it,
I had to think a little bit

on the problem.

If you were to
draw-- well that's

for you have to find
out when you go home.

What do you think
this is going to be?

Actually, we have to
do it now, because it's

a lot simpler than
you think it is.

x and y will be positive,
I can also restrict that.

It looks horrible, but
it's actually much easier

than you think.

So how do I compute that path
integral that makes the points?

I'm going to have
fundamental there.

Which has f of x at q
minus f, with p, which

says that little f is here.

3x squared y plus
x at 2, 1 minus 3x

squared y plus x at 1, 2.

So all I have to do is
go ahead and-- do you

see what I'm actually doing?

It's funny.

Which one is the origin, and
which one is the endpoint?

The problem doesn't tell you.

It tells you only you are
connecting the two points.

But which one is the alpha,
and which one is the omega?

Where do you start?

You start here or
you start here?

OK.

Sort of arbitrary.

How do you handle this problem?

Depending on the direction--
pick one direction you move on

along the r, it's up to you.

And then you get an answer, and
if you change the direction,

what's going to happen
to the integral?

It's just change the
sign and that's all.

3 times 4, times 1, plus 2--
guys, keep an eye on my algebra

please, because I
don't want to mess up.

Am I right, here?

STUDENT: Yes.

MAGDALENA TODA: So how much?

14, is it?

STUDENT: It's 7.

MAGDALENA TODA: Minus 7.

Good.

Wonderful.

So we know what to get,
and we know this does not

depend on the fact.

How much blah, blah,
blah does the instructor

expect for you to get full
credit on the problem?

STUDENT: Just enough to explain.

MAGDALENA TODA: Just
enough to explain.

About 2 lines or 1 line saying
you can say anything really.

You can say this is the theorem
that either shows independence

of that integral.

If the force F vector value
function is conservative,

then this is what
you have to write.

This doesn't depend
on the path c.

And you apply the
fundamental theorem

of path integrals for
the scalar potential.

And that scalar potential
depends on the endpoints

that you're taking.

And the value of
the work depends--

the work depends only on the
scalar potential and the two

points.

That's enough.

That's more than enough.

What if somebody's
not good with wording?

I'm not going to write
her all that explanation.

I'm just going to say whatever.

I'm going to give
her the theorem

in mathematical compressed way.

And I don't care if she
understands it or not.

Even if you write this
formula with not much wording,

I still give you credit.

But I would prefer
that you give me

some sort of-- some
sort of explanation.

Yes, sir.

STUDENT: You said answer was 0.

Then it would have
been path independent?

MAGDALENA TODA: No, the
answer would not be for sure 0

if it was a longer loop.

If it were a longer
closed curve,

that way where it
starts, it ends.

Even if I take a weekly
road between the two points,

I still get 7, right?

That's the whole idea.

Am I clear about that?

Are we clear about that?

Let me ask you though,
how do you find out?

Because I don't know how
many of you figured out

what kind of curve that is.

And it looks like an enemy
to you, but there is a catch.

It's an old friend of
yours and you don't see it.

So what is the curve?

What is the curve?

And what is this arc of a
curve between 2, 1 and 1, 2?

Can we find out what that is?

Of course, or cubic.

It's a fake cubic.

It's a fake cubic--

STUDENT: To function together?

MAGDALENA TODA: Let's
see what this is.

xy cubed minus 2 cubed equals 0.

We were in fourth grade--
well, our teachers--

I think our teachers teach us
when we were little that this,

if you divided by a
minus- I wasn't little.

I was in high school.

Well, 14-year-old.

STUDENT: A cubed.

STUDENT: A squared.

MAGDALENA TODA: A squared.

STUDENT: Minus 2AB.

Plus 2AB.

MAGDALENA TODA: Very good.

Plus AB, not 2AB.

STUDENT: Oh, darn.

MAGDALENA TODA: Plus B squared.

Suppose you don't believe.

That proves this.

Let's multiply.

A cubed plus A squared
B plus AB squared.

I'm done with the
first multiplication.

Minus BA squared minus
AB squared minus B cubed.

Do they cancel out?

Yes.

Good.

Cancel out.

And cancel out.

Out, poof.

We've proved it, why?

Because maybe some of you--
nobody gave it to proof before.

So as an application,
what is this?

There.

Who is A and who is B?

A is xy, B is 2.

So you have xy minus 2 times
all this fluffy guy, xy

squared plus 2xy plus--

STUDENT: 4.

MAGDALENA TODA: 4.

And I also said, because
I was sneaky, that's why.

To make your life easier
or harder. xy is positive.

When I said xy was positive,
what was I intending?

I was intending for you to see
that this cannot be 0 ever.

So the only possible
for you to have 0 here

is when xy equals 2.

And xy equals 2 is a
much simpler curve.

And I want to know
if you realize

that this will have the points
2,1 and 1, 2 staring at you.

Have a nice day today.

Take care.

And good luck.

What is it?

STUDENT: [INAUDIBLE].

MAGDALENA TODA:
Some sort of animal.

It's a curve, a linear curve.

It's not a line.

What is it?

Talking about conics because
I was talking a little bit

with Casey about conics.

Is this a conic?

Yeah.

What is a conic?

A conic is any kind of
curve that looks like this.

In general form--
oh my god, ABCD.

Now I got my ABC
plus f equals 0.

This is a conic in plane.

My conic is missing
everything else.

And B is 0.

And there is a way where
you-- I showed you how you

know what kind of conic it is.

A, A, B, B, C. A is
positive is-- no, A is 0,

B is-- it should be 2 here.

So you split this in half.

1/2, 1/2, and 0.

The determinant of this is
negative, the discriminant.

That's why we call it
discriminant about the conic.

So it cannot be an ellipse.

So what the heck is it?

STUDENT: [INAUDIBLE].

MAGDALENA TODA: Well, I'm silly.

I should have pulled out for y.

And I knew that it
goes down like 1/x.

But I'm asking you, why in
the world is that a conic?

Because you say, wait.

Wait a minute.

I know this curve since I was
five year old in kindergarten.

And this is the point 2, 1.

It's on it.

And there is a symmetric
point for your pleasure here.

1, 2.

And between the
two points, there

is just one arc of a curve.

And this is the path that
you are dragging some object

with force f.

You are computing
the work of a-- maybe

you're computing the work of
a neutron between those two

locations.

It's a--

STUDENT: Hyperbola?

MAGDALENA TODA: Hyperbola.

Why Nitish?

Yes, sir.

STUDENT: I was just
wondering, couldn't we

have gone to xy equals 2 plane?

STUDENT: Yeah, way quicker.

STUDENT: x cubed, y
cubed equals 2 cubed.

Then you'd just do both sides--

MAGDALENA TODA:
That's what I did.

STUDENT: The cubed root.

MAGDALENA TODA:
Didn't I do that?

No, because in
general, it's not--

you cannot say if and only
if xy equals 2 in general.

You have to write to
decompose the polynomial.

You were lucky
this was positive.

STUDENT: Well, because
we divided by x cubed.

We could have just
divided everything

by x cubed, and then taken
the cube root of both sides.

MAGDALENA TODA: He's
saying the same thing.

But in mathematics, we don't--
let me show you something.

STUDENT: It would
work for this case,

but not necessarily
for all cases?

MAGDALENA TODA: Yeah.

Let me show you some other
example where you just-- how

do you solve this equation?

By the way, a math
field test is coming.

No, only if you're a math major.

Sorry, junior or senior.

In one math field test,
you don't have to take it.

But some people who
go to graduate school,

if they take the math field
test, that replaces the GRE,

if the school agrees.

So there was this questions,
how many roots does it have

and what kind?

Two are imaginary
and one is real.

But everybody said
it only had one root.

How can it have one root
if it's a cubic equation?

So one root.

x1 is 1.

The other two are imaginary.

This is the case in this also.

You have some imaginary roots.

So those roots
are funny, but you

would have to
solve this equation

because this is x minus 1
times x squared plus x plus 1.

So the roots are minus
1, plus minus square root

of b squared minus 4ac
over 2, which are minus 1

plus minus square
root of 3i over 2.

Do you guys know
how they are called?

You know them because in
some countries we learn them.

But do you know the notations?

STUDENT: What they call them?

MAGDALENA TODA: Yeah.

There is a Greek letter.

STUDENT: Iota.

MAGDALENA TODA: In
India, probably.

In my country, it was omega.

But I don't think--

STUDENT: In India, iota.

MAGDALENA TODA: But we call
them omega and omega squared.

Because one is the
square of the other.

They are, of course,
both imaginary.

And we call this the
cubic roots of unity.

You say Magdalena, why would
you talk about imaginary numbers

when everything is real?

OK.

It's real for the time being
while you are still with me.

The moment you're going
to say goodbye to me

and you know in 3350 your
life is going to change.

In that course,
they will ask you

to solve this equation just like
we asked all our 3350 students.

To our surprise,
the students don't

know what imaginary roots are.

Many, you know.

You will refresh your memory.

But the majority of
the students didn't

know how to get to
those imaginary numbers.

You're going to need to not
only use them, but also express

these in terms of trigonometry.

So just out of curiosity, since
I am already talking to you,

and since I've preparing you a
little bit for the differential

equations class where you
have lots of electric circuits

and applications
of trigonometry,

these imaginary numbers
can also be put-- they

are in general of the form
a plus ib. a plus minus ib.

And we agree that in
3350 you have to do that.

Out of curiosity,
is there anybody

who knows the trigonometric
form of these complex numbers?

STUDENT: Isn't it r e to the j--

MAGDALENA TODA: So you would
have exactly what he says here.

This number will
be-- if it's plus.

r e to the i theta.

He knows a little bit
more than most students.

And that is cosine
theta plus i sine theta.

Can you find me the
angle theta if I

want to write cosine theta
plus i sine theta or cosine

theta minus i sine theta?

Can you find me
the angle of theta?

Is it hard?

Is it easy?

What in the world is it?

Think like this.

We are done with this
example, but I'm just

saying some things that
will help you in 3350.

If you want cosine
theta to be minus 1/2

and you want sine theta to be
root 3 over 2, which quadrant?

Which quadrant are you in?

STUDENT: Second.

MAGDALENA TODA: The
second quadrant.

Very good.

All right.

So think cosine.

If cosine would be a half and
sine would be root 3 over 2,

it would be in first quadrant.

And what angle would that be?

STUDENT: 60.

STUDENT: That's 60--

MAGDALENA TODA: 60 degrees,
which is pi over 3, right?

But pi over 3 is your
friend, so he's happy.

Well, he is there somewhere.

STUDENT: 120.

MAGDALENA TODA: Where you
are here, you are at what?

How much is 120-- very good.

How much is 120 pi?

STUDENT: 4 pi?

MAGDALENA TODA: No.

STUDENT: 2 pi over 3.

MAGDALENA TODA: 2 pi over 3.

Excellent.

So 2 pi over 3.

This would be if you
were to think about it--

this is in radians.

Let me write radians.

In degrees, that's 120 degrees.

So to conclude my detour
to introduction to 3350.

When they will ask you to solve
this equation, x cubed minus 1,

you have to tell them like that.

They will ask you to put
it in trigonometric form.

x1 is 1, x2 is cosine of 2 pi
over 3 plus i sine 2 pi over 3.

And the other one
is x3 equals cosine

of 2 pi over 3 minus
i sine of 2 pi over 3.

The last thing.

Because I should let you go.

There was no break.

I squeezed your brains
really bad today.

We still have like 150 minutes.

I stole from you--
no, I stole really big

because we would have-- yeah,
we still have 15 minutes.

But the break was 10 minutes,
so I didn't give you a break.

What would this be if
you wanted to express it

in terms of another angle?

That's the last thing
I'm asking of you.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: Not minus.

Like cosine of an angle
plus i sine of an angle.

You would need to go to
another quadrant, right?

And which quadrant?

STUDENT: 4.

MAGDALENA TODA:
You've said it before.

That would be 4 pi over 3.

And 4 pi over 3.

Keep in mind these things
with imaginary numbers because

in 3350, they will rely on
you knowing these things.

STUDENT: Then you apply
Euler's formula up there.

MAGDALENA TODA: Oh, yeah.

By the way, this is
called Euler's formula.

STUDENT: In middle
school, they teach you,

and they tell you when
discriminant is small,

there's no solutions.

MAGDALENA TODA: Yeah.

STUDENT: And you
go to [INAUDIBLE].

MAGDALENA TODA: When the
discriminant is less than 0,

there are no real solutions.

But you have in pairs
imaginary solutions.

They always come in pairs.

Do you want me to
show you probably

the most important problem
in 3350 in 2 minutes,

and then I'll let you go?

STUDENT: Sure.

MAGDALENA TODA: So somebody
gives you the equation

of the harmonic oscillator.

And you say, what
the heck is that?

You have a little spring
and you pull that spring.

And it's going to come back.

You displace it, it comes back.

It oscillates back and forth,
oscillates back and forth.

If you were to write the
solutions of the harmonic

oscillator in electric
circuits, there

would be oscillating functions.

So it has to do with sine and
cosine, so they must be trig.

If somebody gives
you this equation,

let's say ax squared-- y
double prime of x minus b.

Plus.

Equals to 0.

And here is a y equals 0.

Why would that
show up like that?

Well, Hooke's law tells
you that there is a force.

And there is a
force and the force

is mass times acceleration.

And acceleration is like this
type of second derivative

of the displacement.

And F and the displacement
are proportional,

when you write F
equals displacement,

let's call it y of x.

When you have y of x, x is time.

That's the displacement.

That's the force.

That's the k.

So you have a certain
Hooke's constant.

Hooke's law constant.

So when you write
this, Hooke's law

is going to become like that.

Mass times y double prime of
x equals-- this is the force.

k times y of x.

But it depends because
you can have plus minus.

So you can have plus or minus.

And these are
positive functions.

You have two equations
in that case.

One equation is the form y
double prime plus-- give me

a number.

Cy equals 0.

And the other one would be y
double prime minus cy equals 0.

All right.

Now, how hard is to
guess your solutions?

Can you guess the
solutions with naked eyes?

STUDENT: e to the x--

MAGDALENA TODA: So if you have--
you have e to the something.

If you didn't have a c, it
would make your life easier.

Forget about the c.

The c will act the
same in the end.

So here, what are the
possible solutions?

STUDENT: e to the--

MAGDALENA TODA: e to the t
is one of them. e to the x

is one of them, right?

So in the end, to
solve such a problem

they teach you the method.

You take the equation.

And for that, you associate
the so-called characteristic

equation.

For power 2, you put r squared.

Then you minus n for-- this
is how many times is it prime?

No times.

0 times.

So you put a 1.

If it's prime one times,
y prime is missing.

It's prime 1 time,
you would put minus r.

Equals 0.

And then you look at
the two roots of that.

And what are they?

Plus minus 1.

So r1 is 1, r2 is 2.

And there is a
theorem that says--

STUDENT: r2 is minus 1.

MAGDALENA TODA: r2 is minus 1.

Excuse me.

OK, there's a theorem that
says all the solutions

of this equation come as
linear combinations of e

to the r1t and e to the r2t.

So linear combination
means you can

take any number a and any
number b, or c1 and c2, anything

like that.

So all the solutions of
this will look like e

to the t with an a in
front plus e to the minus

t with a b in front.

Could you have seen
that with naked eye?

Well, yeah.

I mean, you are smart
and you guessed one.

An you said e to
the t satisfied.

Because if you put e to the
p and prime it as many times

as you want, you
still get e to the t.

So you get 0.

But nobody thought of-- or maybe
some people thought about e

to the minus t.

STUDENT: Yeah.

I was about to go
through that one.

MAGDALENA TODA: You were about.

STUDENT: That's for a selection.

MAGDALENA TODA: So even if
you take e to the minus t,

you get the same answer.

And you get this thing.

All right, all the combinations
will satisfy the same equation

as well.

This is a superposition
principle.

With this, it was easy.

But this is the so-called
harmonic oscillator equation.

So either you have it simplified
y double prime plus y equals 0,

or you have some constant c.

Well, what do you
do in that case?

Let's assume you have 1.

Who can guess the solutions?

STUDENT: 0 and cosine--

MAGDALENA TODA: No, 0
is the trivial solution

and it's not going to count.

You can get it from the
combination of the--

STUDENT: y equals sine t.

MAGDALENA TODA:
Sine t is a solution

because sine t prime is cosine.

When you prime it
again, it's minus sine.

When you add sine and
minus sine, you get 0.

So you just guessed
1 and you're right.

Make a face.

Do you see another one?

STUDENT: Cosine t.

MAGDALENA TODA: Cosine.

They are independent,
linear independent.

And so the multitude
of solutions

for that-- I taught you
a whole chapter in 3350.

Now you don't have
to take it anymore--

is going to be a equals sine t--

STUDENT: How about e to the i t?

MAGDALENA TODA: Plus b sine t.

I tell you in a second.

All right, we have to
do an e to the i t.

OK.

So you guessed that all the
solutions will be combinations

like-- on the monitor when you
have cosine and sine, if you

add them up-- multiply
and add them up,

you get something like the
monitor thing at the hospital.

So any kind of
oscillation like that

is a combination of this kind.

Maybe with some different
phases and amplitudes.

You have cosine of 70 or
cosine of 5t or something.

But let me show
you what they are

going to show you [INAUDIBLE]
for the harmonic oscillator

equation how the method goes.

You solve for the
characteristic equation.

So you have r squared
plus 1 equals 0.

Now, here's where most of
the students in 3350 fail.

They understand that.

And some of them say, OK,
this has no solutions.

Some of them even say this
has solutions plus minus 1.

I mean, crazy stuff.

Now, what are the
solutions of that?

Because the theory
in this case says

if your solutions are
imaginary, then y1

would be e to the ax cosine bx.

And y2 will be e
to the ax sine bx

where your imaginary
solutions are a plus minus ib.

It has a lot to do with
Euler's formula in a way.

So if you knew the theory in
3350 and not be just very smart

and get these by yourselves
by guessing them,

how are you supposed
to know that?

Well, r squared
equals minus 1, right?

The square root of minus 1 is i.

STUDENT: Or negative.

MAGDALENA TODA: Or negative i.

So r1 is 0 plus minus i.

So who is a?

a is 0.

Who is b?

b is 1.

So the solutions are e to
the 0x equal cosine 1x and e

to the 0x sine 1x, which
is cosine x, sine x.

Now you know why you can
do everything formalized

and you get all these
solutions from a method.

This method is an
entire chapter.

It's so much easier than in 350.

So much easier than Calculus 3.

You will say this is easy.

It's a pleasure.

You spend about one
fourth of the semester

just on this method.

So now you don't have
to take it anymore.

You can learn it all by
yourself and you're going

to be ready for the next thing.

So I'm just giving you courage.

If you do really, really well in
Calc 3, 3350 will be a breeze.

You can breeze through that.

You only have the probability
in stats for most engineers

to take.

Math is not so complicated.