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.