PROFESSOR: I'll go over the exam.
It's good review for the final, and it's
a good feedback for you in case you have questions.
I do not change grades, I do not curve your exam.
I do not make adjustments after I give you the grade.
Therefore, it's very important for me
to explain why you got what you got.
Not everybody did well on this exam.
Most people did pretty good, and I'm
quite happy with what I see as a average for the class.
However, there are many open questions
from many people, things they didn't quite understand,
and I would like to discuss those.
First of all, the midterm exam was 11 questions.
10 were mandatory, so the maximum possible
percentage-wise was 110%.
So for somebody who did perfectly fine,
they would have 110%.
There is one person only who got the high.
I didn't disclose his name, but I would
like to say congratulations.
And I'm going to go ahead and solve each problem with you,
for you.
So you have the function f of x, y,
to be x squared minus y squared.
And the differential was f sub x dx plus f sub y dy,
whihc is 2x dx, minus 2y dy.
That was something very easy.
It was not supposed to give you any headache, and most of you
did a fine job on this one.
What created some problems to most students
was the second problem, though.
And I sorry to hear that, sorry to see that.
Find the directional derivative of a function,
the same function as before.
So I have taken advantage of the previous problem,
in order to make your do time shorter.
At the point p of coordinates x equals 0, y equals 1,
you will have a direction given by the vector.
What does it mean direction given?
Analyze that direction given by the vector means what?
Not the vector i minus j is y, because it's not a unit vector.
What is the corresponding direction given by it?
A corresponding direction given by it
is 1 over square root of 2i minus 1 over square root of 2j.
So it's a collinear vector-- that is, unit varies.
Say it again, [INAUDIBLE]?
The direction u represents a collinear vector.
So pointing in the same direction as v,
but it has to be unitary.
Why?
Because the definition of the directional derivative
is a function along the direction u at the point p,
was given by the formula partial derivative at p
and at 1, plus partial derivative at p times 2.
Did I expect to write all this down?
Yes, I did, as I showed you last time.
So you have 2x evaluated at-- what is x?
0.
1 times u1.
This is u1 minus 2y, evaluated at 0, 1 times
minus 1 over root 2, which is u2.
Well that means the first term goes away,
because this is going to be 0.
And after the second term, you have a plus.
y is 1, thank god, that's easy.
2 over square root of 2, the answer is root 2.
So any other answer would normally receiving a 0.
The answer was b, square root 2.
Now, on number three, the function given is different.
f of x, y equals e to the xy.
And they say, the gradient of this function
is at an arbitrary point.
Say is again?
The gradient of this function is at an arbitrary point.
That was only part of the problem.
A little bit of credit for just writing the gradient.
This is actually easy, a piece of cake.
y is that.
You have f sub x i plus f sub y j.
It equals y to the xy i plus x to the xyj.
That's very good.
Alright, OK?
Then, which direction-- it's just the gradient right?
The direction corresponds to the gradient.
They don't ask you for the u.
Actually, you don't need the u.
You just need the tangent plane in this case.
And if you know the equation of the tangent plane,
as I told you to remember that, that would be very helpful.
Write your answer in the space provided.
So what did I expect you to do?
First this, and then write the equation z minus z0
equals f sub x times x minus x0, plus f sub y, y minus y0.
Here at the point p, evaluate it at the point p.
But attention, what is the point p?
Well, p is the origin, because we say at the origin.
Oh, so that makes things easier.
I'm not done.
Half of the problem is still coming.
If you did until this point, I can only give you 5 out of 10
or something like that.
Many people made a mistake at z0.
Attention guys, you plus in that 0, you don't get 0.
For god's sake, it's 1, right?
So z0 is 1.
Now you're getting the sense that you
have z minus 1 equals f sub x, computed as 0,
0 will be 0, lucky you.
f sub y computed at 0, 0, you were expected to say that.
So the answer for this problem was z equals 1.
Still, if you messed up, I gave some partial credit,
because I didn't want to punish you too much, too harshly.
On number four, find the direction u-- now
you're using number three, so I should not
erase number three completely.
On number four, you use number three, so
the same type of function.
But it says find the direction in which
this function increases most rapidly, at the point 1, 1.
OK.
So what do you do? you compute the gradient at the point 1,1,
and you say, this is a piece of cake.
It's going to be ei plus ej, ee.
Wonderful.
So what you do is a u is the gradient f over the length
of the gradient of f at p.
Which is ee divided by the length of it.
But you say, I don't have to compute the length of it.
I know what is pulling your two e's is what?
And no matter what you have here you get the same unique result.
Remember we talked about that uniqueness?
This is what I tried to emphasize,
that you can have 77, ee, 99, 55, 100 and 100.
If you divide by the norm, you still get the same answer.
Not 11, but 1 over root 2, 1 over root 2.
So no matter what you had there--
it could have had a million, or something instead of e,
you still have the same u.
Yes, put it back.
Give yourself points, modify that.
OK, so let me tell you.
Normally I should penalize, because I say write the answer
in the space provided.
And thank god you had enough space, right?
Look, this person wrote-- I shouldn't show you who
he is, he's not in here anyway.
He has space and he provided last year
with no square root of 2, because only two
rows are enough to write that.
It's OK, I understand you forgot to copy.
My son did the same thing.
He got a scantron at the UIL.
Come to visit my son, I wanted to kill him, but it's OK.
He got all the answers right, and then
the teacher-- that reminds me of a movie with Mr. Bean.
So the teacher comes to him and says,
wow your scantron is blank.
So what was I supposed to do?
Adjust for all the answers you got in the box,
put them in the scantron.
Oh, really?
So he goes quickly.
And then he got only 75% of them transferred.
The rest of them were not transferred.
I don't know what they did.
I have no idea.
But the professor would have given full credit,
even for the answers that he had in the box.
From what I understood, the rule for scantrons,
exams like you I only say, if you don't have them
on the scantron, they don't count.
This is very harsh, because we don't do that.
For example, the final-- if you--
that's why I'm trying to read everything.
Suppose you box you answer and it's 1 over square root of 2.
If that's the right answer.
Then if have the multiple choice,
and they forgot to circle 1 over square root of 2.
I still give you 100% on that problem.
Some professor do not.
So this is at the latitude at whoever makes the rules,
or whoever writes the exam.
OK.
So again for the final, even for the multiple choice problems,
I still need the solutions.
I'm going to ask you to use a bluebook.
Some professors do not ask you to use a bluebook.
They say, as long as you can write on the sheet, circle
the answer, I'm fine.
I'm not fine.
I want to keep what's in the bluebook.
So buy-- how much is it?
Like a dollar?
Buy the books ahead of time, make sure you have them.
Now, number five was a piece of cake once you did number four.
You have a question?
STUDENT: What size bluebook do you need for the final?
PROFESSOR: The big one.
Bigger than that, right?
The direction u for five.
With the problem four was i plus j over root 2, right?
This is what you remember that you did in problem four.
If you didn't do problem four, you cannot do problem five.
Problem five says, this is parallel to one line.
This is parallel to-- what is i plus j?
Of course, you don't have to draw that.
I'm not expecting you to draw that.
y equals x is the first bisection.
All you had to do was circle C, and that
was-- once you circled C, you get full credit.
If you don't do that, you don't get credit for anything.
Now six.
What is the maximum rate of increase of the function
z the same of your friend, your fellow z equals
e to the xy at p0, coordinates 1, 1?
Then the value of the maximum rate of change is?
A noun.
What's the simplest way to do it?
There are two ways to do it.
One is the long way, one is the short way.
What's the short way, guys?
Just compute the length of the gradient.
The length of the gradient at the point P.
So you have whatever that was, ee in length.
So the answer was e root 2.
Am I right?
What was the long way?
I saw somebody do it.
This is a lot more work, but of course,
would be to compute the directional derivative
at the point p for this function.
In the direction of u, where u is the gradient
divided by the length.
at the point p.
And you get, of course, the same answer.
Why?
Because we proved that actually the maximum rate of change
represented directional derivative exactly
in the direction given by the gradient.
This is something we proved.
One of the few things we proved in this class.
Alright.
So the answer was e root 2.
Let's move on to number seven.
Number seven-- and remind me of your five points.
Can you email me, so I have an Excel sheet,
and I'll put it in.
Consider the function f of x, y e to the negative x squared,
y squared.
What can you tell me about this type of function?
It's the headache function.
If I would ask you to do an anti-derivative of each
of the negative squares, you would say Magdalene,
didn't you say that this is impossible?
While the anti-derivative exists, it cannot be expressed.
It cannot be expressed as an elementary function.
And that's a big headache.
This problem is beautiful, why is it beautiful?
Because in the end, it becomes magic.
So it's a positive function.
It's like a bell on top of the church something.
And then, you have to compute double integral
over the unit disk of centers of 0 and radius 1.
Of e to the negative x squared minus y squared dx/dy.
Well then you say, well I've done this kind of thing
before, but not with Cartesian coordinates.
We did it with the Jacobian r, that changes everything
into polar coordinates.
So this guy becomes e to the minus r squared.
Each of the numbers are squared dr, d theta.
D on the unit [INAUDIBLE] disk means
the radius goes from 0 to 1.
This is a blessing for us, because it's easy data.
Then we have 0 to 2 pi.
You could have put it in any order.
For u, it's easier to close your eyes when it comes to theta.
Say, theta is independent.
He is like a partition that has to do nothing
with what's inside here.
So let's pull him out of this picture.
And he wants to live by himself.
An integral from 0 to 2 pi of d theta was of course 2 pi.
He's happy to go out, having fun.
This guy inside has to be thoroughly computed.
In the sense that you perform the substitution.
I was actually amused that half of you did u equals r squared,
and half of you did u equals minus r squared.
It really doesn't matter which one.
But the problem is that some of you
made a mess when you put the limit points back in place,
and you made mistakes.
Somebody even got negative answers,
I was about to fall off the chair.
Of course, I was in a good mood because it was a holiday,
I graded them.
Fortunately, I graded them over the break.
So after I came back from Georgia.
I have minus r dr. rdr was minus a half du.
This fellow is just into the u, and he's
a blessing because the [INAUDIBLE]
So into the u, however, take it between 1 and what?
Not 0 and 1.
But when you have 0 here, you have 0 here.
When you have 1, you have minus 1.
So pay attention to that, otherwise, you
get something that makes no sense.
Times minus a half.
That, you will have to be careful about.
Why?
Because there will be a minus from here and here, in the end,
the answer will be positive.
And that's reminding me of that city plumber joke
when he doesn't pay attention to the limits of integration.
And you can get a minus volume, or a minus area.
So e to the minus 1 minus 1.
But that leaves a negative number,
but when you multiply it by a minus, you have 1 minus 1/e.
1 minus 1/e.
Good, thank god.
This is a nice guy, less than 1.
And this is key to your answer, because 2 goes away
and pi stays in place and this is less than pi.
So the answer to this question was an answer less than pi.
And if you didn't get it, I'm very sorry,
if you didn't get less than pi, you didn't get any points.
But, there are enough chances for you to get another point.
I was brokenhearted for 10 people or more out of 25
did not remember what I taught in class about the area
of a collateral triangle.
And it broke my heart, and I was about to cry,
but I said, c'mon, they'll do it better in the final.
Honestly, I was so brokenhearted.
So this is 1, 0, 0.
This was 0, 1, 0.
This was 0, 0, 1.
On number eight.
Thank you.
Beautiful.
It's an equilateral triangle, and the l side
of that equilateral triangle is the square root of 2.
I even taught you how to cheat.
That's why I was mad.
I taught you how to cheat, and you didn't take advantage.
So the area was l squared, square root of 2, 4.
Which we did this together in fifth or sixth grade
by multiplying that height and the width and divided by 2.
And then we came up with this formula with the Pythagorean
theorem in the classroom.
If eligible to, you can very quickly get an answer.
So that's going to be 2 root 2 over 4, just root 3 over 2.
And when I saw that people got something else except root 3
over 2, that broke my heart.
Really.
You have plenty of time to catch up
with that being on your final.
Did I expect you to really do the surface integral?
Some people again, need to write integral over the shaded domain
d a square root of f sub x squared
plus f sub y squared plus 1.
That was the right track, because this is root 3.
And then the area, you get the area of the 1 times 1 over 2,
right?
1/3 is the area.
Root 3 gets out of this, so you have-- when you integrate,
you have the area of the shaded base that I have.
And you get the same answer.
No matter how you do it, with calculators
or without calculators, you still could have passed.
Am I if you didn't get the answer?
No.
Absolutely.
But it hurts me as if, I don't know, a relative of mine
messed up some task.
That's why it's better that you don't know your students,
because when you know your students,
you know that they could have done better,
because you know them.
So we can say, OK, it really hurts
when you know that they messed up, not because they are not
smart or educated, but because they just either didn't
pay attention or they were stressed out.
However, my substitute, the guy who came here,
was my Ph.D. student.
He got a doctoral degree mathematics with me last year.
And he told me you were not stressed out at all.
And I said, thank god.
I'm glad that they were calm.
And he said, I didn't look at the exam,
but it seemed like they did very well and they were comfortable.
And I was so happy.
I was in Athens, Georgia.
And reading this email I said, yay!
Everybody's going to get an A!
So I come home and I start grading it.
I was sad to see that my prediction was not correct.
But anyway, [INAUDIBLE] with an average of B.
For an honors class, it's OK.
I just expected a lot better.
And I know it's going to be a lot better in the final.
Number nine.
This was done by almost everybody,
except for a few people who messed up on the limits.
I don't know why.
When they compute-- when they drew, they drew x squared,
and they drew square root of xn.
Of course, you were supposed-- the answer was 0 to 1,
integral of.
Now, if you do first x, you have x from y
squared to square root of y.
You guys with me?
Because this is smaller than that.
OK?
So you have 1 and dx dy equals to integral from 0 to 1,
integral x squared to square root of x1 dy dx.
Now, what a few people did-- and I just forgave them.
They just-- one put this like that.
And here, he put root 2.
Root y and y squared.
Don't do that.
It's like chasing that a positive number equals
a negative number, which is all complete nonsense.
So the correct answer was we put y squared down,
and square root of y because this guy is
bigger than this guy for something between 0 and 1.
Because I told you.
Square root of 0.04 is bigger than the square of that.
OK.
Now am I happy with that?
I'm quite happy.
In general, people understood the vertical strip method
compared to the horizontal strip method.
And why am I happy?
Because I was asked by three people from other classes
to help them, over there, on the corridor.
And I asked them, who is your teacher?
This and that.
But we did not understand reversing the order
of integration in class.
And I said, how come?
Well, they didn't explain it very well.
So I started explaining it to them.
And then I realized that it's a conflict of interest.
I'm not allowed to do that.
And then I go, oh my god, I cannot do the homework for you.
I'm not allowed.
But I was already talking.
So I said, guys, can you do it?
I don't know.
I said, do you draw?
Why would we draw?
They didn't teach us how to draw.
I said, but how do you know about vertical strips
and horizontal strips?
No.
And how do you do this?
We don't know.
We felt like we have to figure it out.
Without drawing, without understanding how the vertical
strips are drawn between two functions,
and how you switch the horizontal strips,
you cannot do this problem, period.
So if you don't have-- maybe some people have
enough imagination-- but that's very rare--
That they can close their eyes and they
can see a picture with their eyes closed
and they can solve that.
But that's not the way to learn.
The way to learn is a very visual learning thing.
So that's why we draw all the time.
STUDENT: Professor, you can cheat these with Cal 2.
PROFESSOR: Yes.
You can do that with Cal 2.
What's the problem?
You have integral from 0 to 1.
Square root of y minus y squared.
Well, they learn to do the other one.
The one with square root x minus x squared, 0,1 and so on.
But they were told explicitly to write-- the professor even
left these empty and put spaces, fill in the spaces.
And they say, how the heck do we fill in those spaces?
Plus the whiteboard problems have the empty spaces.
And they couldn't believe that at all.
And one of them went to the tutoring center and was lucky.
Because he got-- this is like when
you go to a medical doctor, sometimes you
are lucky and get a good doctor who takes care of you,
figures out what your problem is.
And sometimes, they give you the wrong medicine.
So one of them got the right tutor who knew how to explain
and sort of knew something.
But the other one got a tutor who never took Calculus 3
and said, I don't know what the heck these multiple snakes are.
So I'm not going to be able to help you.
So he was very disappointed.
OK.
Compute the area of the domain D from the previous problem.
This was something that nobody's telling you,
hey, you have to do it with the double snakes.
You can do it with just with a simple snake
and you're still fine.
So in Calc 1-- this Calc 1, whatever it is.
In Calc 1, you learn that you have to integrate this
and you'll get 2/3 x to the 3/2 minus 1/3 x
cubed at x equals 1 minus whatever you have with 0.
But at 0, you have 0, so you say, forget about it.
And you have 2/3 minus 1/3 equals 1/3, then you're done.
OK?
Did I expect you to show me work?
No.
For everybody who wrote 1.3-- and there
were many people who did this mentally,
and they came up with 1/3.
They got 10 pionts on the problem.
Finally, number 11.
Without computing the volume inside the sphere,
x squared plus y squared plus z squared equals 2.
Set up a triple integral corresponding to it
in the space provided below.
Some people, a few people, messed up.
They forgot the Jacobian.
So they put the 1 instead of r squared [? side-side. ?]
When you work in three components,
they do fine setting up the limits.
[INAUDIBLE] 1 here.
Don't look at it in the final.
You can ruin your life this way.
So we have r squared sine phi.
Phi was the latitude from the North Pole.
it doesn't matter in which order you do it.
But I would do to er b phi b theta.
You tell me what the end points are, and we are done.
STUDENT: From 0 to 5.
PROFESSOR: 0 to--
STUDENT: No, on the first one.
PROFESSOR: 0 to--
STUDENT: Dr?
It's the square root of 2.
PROFESSOR: Mm-hmm.
STUDENT: And b theta--
PROFESSOR: 0.
2pi.
And theta, all around.
STUDENT: 2pi.
PROFESSOR: Longitude 360 meridian degrees.
OK.
0 to 2pi.
So good.
So we are done.
Did I expect you to write it down?
No.
I had three people who were nice and wrote down 4.
I mean, they actually did the work.
Maybe they had nothing better to do.
I have no idea why.
4pi i cubed over 3, right?
And then they proved the formula in general using the Jacobian.
Using the formula, they got the correct formula for r
equals square root of 2.
And I was very happy.
But did I ask you to do that?
No.
Did I give you extra credit.
No.
So all the extra credit was just one problem to
asked to do exactly what you were told to do.
I don't know about how you feel about this exam,
but it wasn't a hard exam.
It was not an easy exam.
It was an exam that was supposed to test
what you learned until now all through the course.
And that was the whole idea.
I think you've learned very much,
and I think you did fine, the majority of you.
And that should ease the pressure on you
when it comes to preparing for the final.
I was thinking last night, I'm going to send you, probably
by email or in-person in class, two
or three samples of the final from old finals
that inspire us when we write the final.
A few of us will provide problems and comments
and suggestions when we write out the departmental final.
But the final will be departmental for all sections.
I don't expect more than 15 problems on the final.
I have yet to think and decide if I want to [? lift ?]
probably the same policy.
I mean, the final is the same for everybody.
But the policy about how to give partial credit
or not give partial credit. [INAUDIBLE].
And I already decided that I'm going to read everything,
so in case that you mess up at the end with your miracle
answer, you still get partial credit
for your integrals [INAUDIBLE] shown.
Also, one of those 15 problems. might be for extra credit.
I have to think a little bit better
how-- what is the maximum weight I want to put.
What I would say, since I never [INAUDIBLE] open a homework,
and I never curve exams, I would think
I could make 110% as the possible maximum.
In this case, you have some cushion
to make a mistake or two and still get a perfect score.
OK.
I'm going to move on to a new chapter.
I have actually moved on already,
but nobody believed me.
Last time, I started Chapter 13.
Chapter 13 is a mixture of mathematics and physics.
You will be surprised how many things
are coming from solid mechanics, fluid mechanics.
Yes, Regan.
STUDENT: [INAUDIBLE]
PROFESSOR: For a job?
You want me to come with you?
[LAUGHTER]
STUDENT: Because I tried to talk to you [INAUDIBLE].
PROFESSOR: Yes, yes.
Yes.
Yeah.
And you have to sign up.
Start a [? sheet, ?] attend [? the sheet, ?] and sign
your name and good luck with the interview.
You should have told me before!
I could have said a prayer for you.
This things are very stressful!
I remember my own interviews.
There were several.
I didn't know anything about it, and my hands were all sweaty.
And you know you should never shake hands with somebody
when your hands are sweaty.
You have to do like this first.
Be confident and don't be nervous.
Don't sweat or anything.
Because they can see that.
All right.
You just be yourself.
Do you have earrings?
Because after my several job interviews--
those are good earrings-- I was told that I should never
wear dangling earrings at the interviews, which I did not,
because I didn't have any.
But I love dangling earrings.
And I was asking some academics why that was [? our ?] problem.
And they say they are distracting.
Because mathematicians are like cats.
[LAUGHTER]
PROFESSOR: --pendulum, and then they
get hypnotized by the dangling.
So I don't know.
I think most of the interviewers have some problems
and they find some things distracting or annoying.
Otherwise, I think you are fine.
You're dressed fine for an interview.
OK.
So now serious job.
We have to remember some of the things we don't remember.
Which are the gradient for a function of let's say
three variables.
Let's grow up a little bit.
And that was what the vector field
F sub xi plus F sub [? I j ?] plus F sub z k.
Right?
At an arbitrary point xyz in your domain.
So where xyz is in some domain, you are in a potato.
And the meaning of the gradient, the geometric meaning of this,
doesn't look like a theta [INAUDIBLE].
It's some sort of solid that it corresponds
to a closed surface.
And this closed surface that closes up on its own
is having a hard time [INAUDIBLE].
It has a normal.
And this normal is given by the gradient of this function,
we can increase [? it ?] like that.
You remember that.
And that was a long time ago.
But you should still master that.
Last time, I gave you the z equals f of xy,
z equals little f of xy, as a graph
of the function of two variables over a domain in plane.
We computed the gradient of that.
But that's what we did all through the [? meter ?].
So that's no fun.
We know that too well.
On this problem, I gave you some new piece
of information last time.
So I said, if you have a vector field that
looks F 1i plus F 2j plus F 3k, where
Fi is C1, that means that the differentiable
and the derivatives are continuous,
what was the divergence of it?
Well, that was before the Easter break.
And I know we had a long break.
I cannot recover from this break so easily, because it was long.
And I also traveled last week.
But before I traveled, I remember that I gave you this.
And you memorized it.
Most of you memorised it.
How was it?
The first component differentiated with respect
to the first variable plus the second component
differentiated with respect to the second variable.
Plus the third component differentiated with respect
to the third variable.
So I'm asking you, as an exercise, like I
did last time, the same thing.
Exercise one for this section.
Compute divergence of the gradient
of F, where F is a C1 function of xyz.
That means F is [? like this ?] differentiable
and with continuous derivatives.
What does it mean?
It means that you have to compute divergence of F sub xi
plus F sub yj plus F sub zk.
And you're thinking, I can do that!
By definition, I take the first component-- who was that?
Hmm?
STUDENT: Brian.
PROFESSOR: Oh, right.
I thought that somebody wanted to come in
and then he heard me and changed his mind.
[LAUGHTER]
PROFESSOR: F sub x parentheses [INAUDIBLE]
x plus F sub-- like when you go on a blind date
and you see, change your mind.
OK.
F sub y y plus F sub z z.
Do you remember that I gave away 95 cents
for this type of question?
So what was this operator?
We can write it better.
We can write it using the second partial derivatives
with respect to z, y, and z.
And we gave a name to this one.
We called this names--
STUDENT: Laplacian.
PROFESSOR: Laplacian.
Laplace operator.
Laplace.
Laplacian.
That's how you spell it.
Laplac-ian.
OK?
Of F. And then what do you have?
You have to introduce a new notation in.
When you see this triangle that looks
like an equilateral triangle, this
means Laplacian of something.
So if you have a function of two variables-- so
let's say z equals F of xy.
What is the Laplacian of this little f?
Little f x x plus little f y y.
So we could be second partial with respect
to x plus the second partial with respect to y.
What if I have something else?
Like let me give you a more general function.
Let's say I have a differentiable function
of N variables with continuous derivatives.
And it looks like crazy.
It looks like that.
x1, x2, x n minus what?
Well, the Laplace operator in this case will be F sub x1 x1
plus [? A of ?] sub x2 x2.
Which means the partial of F, the second derivative
with respect to x2.
And plus the last derivative with respect-- two [INAUDIBLE]
with respect to the same variable.
The last variable is xm minus 1.
This could be one million and 1.
I don't know.
You can have this as many variables as you want.
Now, actually in engineering, there
are functions that have many parameters.
You have three special opponents.
Then you have time.
Then you have temperature, then you have pressure,
then you have god knows what.
The surface tension of the membrane.
Many things.
You really have a million parameters.
Actually, it's impossible.
It's even hard to work with 10 parameters.
Imagine always working with equations
that have lots of variables and having do deal with that.
In fluid flows, hydrodynamical problems,
most the time in 3D turbulent flows,
for example, then you have xyz spatial coordinates
and time T. So even with four variables,
once you get those operators, you could have something like F
sub x x x t plus g sub x x t plus and so on.
All sorts of ugly components.
Sometimes you'll have equations of fluid flows
in dynamic software.
Fluid flows with turbulence are really
an area of mathematics in itself,
of really complicated equations with most of the operators.
I was looking at them in Georgia,
where I went to this conference.
Most of those equations were order 4.
Of course, most of them you cannot even think about solving
by hand, or with any known methods.
You can solve them numerically with computational software.
That is the only [INAUDIBLE] that modern mathematics
has in some areas right now.
The right software, in order to find solutions
to a fluid flow with turbulence.
That is the solution to this type of equation.
Like [INAUDIBLE], for example.
Now we are going to see-- well, you are going to see.
I'm too old and I saw that 20 years ago.
When you're going 3350 [INAUDIBLE]
differential equations.
And then, if you do PD 3350 one in engineering,
You're going to see lots of equations
that are hard to solve.
But in many of them, you're going to see partials,
like that.
And you're going to say, oh, thank god
that I like partials in Calc Three
so they became my friends.
And you'll never have headaches-- [? you know what ?]
would be easy, if you understood that notion of differential
well, the notion of partial derivatives very well.
So I'm going to erase this one.
OK.
And then I'll say, I don't how many of you--
I'll try to make this formula more visible.
Some of you maybe, who are engineering majors
know about curl.
Have you heard about curl?
Curl of a vector value function.
No.
You haven't.
Suppose that you have a vector value function.
That is F of coordinates x, y, z, the coordinates.
The C1 of over seven domain omega.
Omega is the domain that your special coordinates live in.
Xyz living some potato.
That's it.
Whose solid body enclosed by a closed surface.
In that potato, F is a differentiable function
with respect to xyz, and the derivatives are continuous.
Now, in most cases, if you work with Laplacian,
this is not enough C1.
If you work with Laplacian, what do you want?
What do you need?
You have F sub x x plus F sub y1.
So you need C2.
You work with at least C2.
Many examples have C infinity.
That means you're having really beautiful functions that
are elementary.
Some of them even polynomial approximations.
And then you really can differentiate
them ad infinitum and all the derivatives [INAUDIBLE],
and then you can call yourself lucky.
How do you introduce the notion of curl of it?
And it sounds funny, and this is why they made this fun.
And my hair used to be curly, but I shaved my head
over the holiday, and now it's between.
So curl of F is something that looks horrible
when you try to memorize it.
So you say, OK, if I'm going to get this on the final,
you better wear this T-shirt.
No, there is something better than that.
One time I was the wearing-- OK.
My students got no permission from the [INAUDIBLE]
to come in with a cheat sheet.
But I was wearing a T-shirt that had Green's theorem.
I don't know how many of you have
heard about Green's theorem.
We are going to learn it in two weeks.
And I was wearing that T-shirt.
And it was by accident, OK?
I didn't do it on purpose to help my students cheat.
So one student at some point goes like, well, I
don't remember Green's theorem.
And then he looked my T-shirt.
Oh, all right.
Never mind.
So I had Green's theorem on my shirt, [INAUDIBLE].
But it's hard to wear like 10 T-shirts, one for the-- I
have one for the formula of the curvature of a curve in space.
Remember that one, how it is so nasty?
OK, I have this one.
I have Green's theorem.
I have [INAUDIBLE], all the important formulas actually.
I have 10 T-shirts.
And then I was thinking, how will I
be if I were like taking ten T-shirts on top of the other
and taking them one off at a time during the final.
There is no cheat sheet.
There are no formula sheets, no nothing.
But I would look like Joey from "Friends."
Remember Joey, when he was dressed in many layers.
So rather than that, I say ask me.
Say oh, you know, I'm freaking out.
I'm taking this final, and I forgot curl.
Rather than not attempting the complex problem at all,
ask me before the exam, and I will remind everybody
how to set up the curl formula.
So you simply have to think in terms
of operators-- ddx, ddy, ddz.
What are these?
These are derivative operators.
So if you take this and multiply it by a function,
that means df, ds-- [INAUDIBLE].
All right, so in this case, if F is-- I'll
go by my T-shirt-- PI plus QJ plus RK, where PQ and R are all
scalar functions of xyz.
STUDENT: Then we will not forget it.
PROFESSOR: Then we are no longer forget it,
and you'll no longer need my T-shirt.
All right, so how do you do that?
You go expand along your first row,
I times whoever the minor will be, which is this guy.
How do you do the [? cowboy ?] problem?
These guys multiply each other.
So you go dr, dy.
Plus or minus?
Minus dq, dz.
Close times I. So the I is the corresponding element
to the minor that I just completed.
This minor is the determinant, which is exactly this guy.
And this is exactly what my T-shirt says.
Right, precisely.
OK.
The second term, if we put the minus-- no,
they changed the signs.
That's the thing.
I would put minus, because I am expanding along the first row.
And the second that I'm in minus something minor times
J. Which minor?
Let me make in the lime.
Lime is a nice color.
And then I'll take this, this, this, and that-- dr,
dx shooting [? cowboys ?] there-- minus dq, dz.
And of course they wrote dq, dz minus dr, dx.
So I would leave it like that.
It doesn't matter.
You can put the minus in if you want.
Plus the k dot.
k goes at the end.
All right, now k goes at the end.
And then k multiplies this determinant-- dq, dx minus dp,
dy.
dq, dx minus dp, dy.
Is it hard?
No.
It is not going to be hard to memorize.
So then how did we do that?
We set up the first row to be I, J, K, the second row
to be ddx, ddy, and ddz.
And then all in order the components of your vector value
function in the exact order they are with respect
to the standard basis i j k.
All right, now there are other names
and other symbols for curl of F. They use
curl because it's in English.
Well actually, in Great Britain I
saw that they used [INAUDIBLE], or else they use both.
In my language, in Romanian, we call it [? rotore. ?]
And I saw that in French it's very similar.
They use the same.
Now in the mechanical engineering notation
it's funny.
They use another symbol and a cross [? broad dot ?] symbol F.
And by that they mean curl F. So if you
talk to a professor who's in mechanical engineering,
or fluid mechanics, or something,
when they talk about curl, they will use this notation.
When they use this other notation,
what do you think this is again?
Divergence, yes.
I told you last time that is divergence of F.
So make the distinction between-- again,
when are you leaving?
Huh?
OK, so you have been in [INAUDIBLE].
And then we have this distinction
we use here, like for dot product
and you use here as a cross product.
Now you have to understand the conceptual difference is
huge between these guys.
This is a scalar function.
This is a vector function-- vector, scalar-- vector,
scalar, vector scalar.
Because I've had to do it on so [INAUDIBLE].
It makes [INAUDIBLE].
And I heard of colleagues complaining
while grading the final that the students did not
understand that this is a vector, and this is a scalar.
OK, a few simple exercises-- I'm going to go ahead and do
some of them.
We tried to make the data on the final exam
very accessible and very easy to apply in problems.
And one of the problems that-- we'll start with example 2--
would be this one.
And you may think, why?
Sometimes we put it in disguise.
And we said assume you have a sphere-- that's
the unit sphere-- of origin O. And say compute.
What is the equation of the unit sphere, guys?
X squared plus y squared plus z squared equals one, right?
From [INAUDIBLE], F equals normal-- external
normal-- to the unit sphere pointing out, [? through ?]
than N is the same at a different point as the position
vector.
Then compute.
[? Now follow. ?] Gradient of F, divergence of F,
and curl of F. Now that should be a piece of cake.
Now one is not [INAUDIBLE] so much of a piece of cake
if you don't understand what the problem wants from you.
It is to actually graph the expression of this one.
So you're going to say what is the normal
to a function like that?
First of all, we just talked today about it.
If you have a function, even if it's implicitly as F of x,
y, z equals c, in that case N is your friend from the past.
If it's a unit normal, unit normal to a surface
happens all the time in engineering.
Whether you do solid mechanics or fluid mechanics,
you always have to complete these things.
This is going to be hard.
The gradient of F divided by the length
of-- but here I have a problem.
I have to put G here, because G will be my position vector.
This is the point x,y,z.
Or you prefer big R. But I think I prefer big G,
because big R looks like a scalar radius,
and I don't like that.
So the position vector will be the circle middle
that starts at the origin and whose N is on the surface,
right?
And this is the equation, xy equals yj plus [? ek1. ?]
Because my point x,y,z has a corresponding vector xi plus yj
plus zk-- big deal.
Now I'm trying to convince you that, for the unit
normal for the sphere, I have the same kind of thing.
So how do we compute this normally?
I take the function F that implicitly defines the surface.
All right, so in my case F is something else.
What is it? x squared plus y squared plus z squared.
Let's compute it.
N is going to be [INAUDIBLE].
It's very nice.
2x comma 2y comma 2z divided by the square root of the sums.
Do I like this?
Uh, no, but I'll have to do it whether I like it or not.
I want to simplify up and down via 2.
Can I do that?
Of course I can.
I'm going to get x,y,z divided by square root of x squared
plus y squared plus z squared.
And this was 1.
STUDENT: Wouldn't there still be a 2
there, because it's 2 squared [INAUDIBLE]?
PROFESSOR: No, I pulled it out.
That's exactly what I said.
There was a 4 inside.
I pulled out with the forceps.
I put it up here, square root of 4.
And I have a 2 here, and that cancels out.
So I got something much simpler than you guys
expected at first.
I got xi plus yj plus zk as being the normal.
Did you expect this?
And you were supposed to expect that this is y,
because this is the position vector that has one length.
The length of a root vector is 1,
and the point is on the sphere.
The normal will be exactly the continuation.
Take your root vector, and continue
in the same direction-- this is the beauty
of the normal to a surface, that it continues the radius.
It continues the radius of the sphere in the same direction.
So you copy and paste your vector here.
Position vector G will be the same as the normal N.
All you do is you shift, but it's the same vector
at the different point.
Instead of starting at O, it starts at P. So
[? that ?] is the same vector.
So you take the radius vector from inside the sphere--
the position vector-- and you shift it out,
and that's the normal to the sphere.
So the equation is still xi plus yj plus zk.
Yes, sir.
STUDENT: Does it remain the same for any other functions,
like [INAUDIBLE]?
PROFESSOR: For the unit sphere, yes it is.
But for a general sphere, no.
For example, what if my sphere will be
of center origin and radius R?
And its position vector v is x,y,z-- like that.
[INAUDIBLE] I don't know.
G, right?
That's the position normal.
STUDENT: [INAUDIBLE] just divide them by the R.
PROFESSOR: You just divide by the R.
So instead of radius being big R,
your unit vector will be this one.
And you take this one and shift it here,
and that's all you have.
For the sphere, it's beautiful.
For any surface in general, no.
Let me show you.
You have a bunch of [INAUDIBLE], and your position vectors
look like crazies like that.
And the normals could be-- they don't have
to continue their position.
They could be-- it depends how the tangent planes look like.
And the tangent plane at the point
has to be perpendicular to the normal.
So the normal field is the N of [INAUDIBLE] vectors.
But the little thingies that look like rectangles
or whatever they are-- those are the tangent planes
of those points.
So in general there is no obvious relationship
between the position and the normal for the surface.
You are really lucky for this [? field. ?]
And for many reasons, like how beautiful the sphere is,
these functions will be easy to compute.
Can you tell me what they are without computing?
Because that should be a piece of cake.
What is the gradient field?
STUDENT: [INAUDIBLE] to that one?
That's the x, y, and z.
PROFESSOR: For the sphere.
STUDENT: 2x, 2y--
PROFESSOR: Actually, let's do it for both divergence G and curl
G. And you say wait, they will be-- so gradient-- no,
I meant here.
You don't have gradient.
When F is a scalar function, then you have gradient.
Then for that gradient you're going to have divergence.
And for that-- I changed notations, that's shy
I have to fix it.
Because F used to be that, and it's not a vector anymore.
So big F is not a vector anymore.
It's a scalar function, and now I have to change the problem.
What is the gradient there?
What's divergence of the gradient?
[INAUDIBLE] gradient of F. And for the G that I gave you,
I want the divergence in the curve?
So I made the problem fluffier that it was before.
More things to confuse for practice.
What's the gradient?
We did it before.
2x--
STUDENT: 2xi, 2--
PROFESSOR: 2y, 2z-- we are at a [? 93 ?] point p on the sphere.
It could be anywhere-- anywhere in space.
What's the divergence of this individual?
So remember guys, what I told you?
First component differentiated with a straight 2x
plus second component differentiated with respect
to y plus third component differentiated
with respect to z.
2 plus 2 plus 2 equals 6-- piece of cake.
And curl of the gradient of F-- is that hard?
[? STUDENT: Yeah. ?]
PROFESSOR: No, but we have to know the definition.
And without looking at the T-shirt, how do we do that?
The determinant-- I, J, K. Operators-- ddx, ddy, and ddz.
STUDENT: [INAUDIBLE] 2x, 2y, 2z, correct?
PROFESSOR: And we copy and paste the three components.
[INAUDIBLE] in the trash.
I'll take the blue.
So we put 2x, 2y, 2z.
Do you think it's going to be easy or hard?
Do you see the answer?
Some of are very sharp, and you may see the answer.
For example, when the cowboys shoot at each other
like this, dz, dy is here.
dy, dz is here.
So this, as a minor, is 0-- 0I, an eye for an eye.
And what else?
dz, dx-- dx dz, same thing, minus 0j.
Is this meant to say minus 0j?
Yes it is.
But I did it because I want you to have the good habit
of saying plus minus plus.
And that's finally the same kind of thing
that'll give you 0k if you think that when
you do partial derivative of y with respect to [? f ?],
you get 0.
You have 0.
So some student of mine asked, so this is the 0 vector,
how in the world do I write a 0 vector on short?
Let me show you how.
You're going to laugh at me.
Some people write 0 bar, which means the 0 vector.
Some other people don't like it, it's silly.
Some people write O with double like that, meaning that,
hey, this is a vector element, the vector with its components
of 0, 0, 0-- to distinguish that vector from the number 0,
which is not in bold-- So the notations for the vector are 0.
So I'm going to write here 0, 0, 0.
How about Mr. G?
Mr. G will act similarly.
When you do the divergence it's going to be-- 1
plus 1 plus 1 equals 3.
You should remember this thing.
We are going to do the divergence 3,
and they will ask you to do a triple integral of a divergence
of a vector field.
And when you do that, you are going
to get a triple integer of something
like 3, which is a custom, which will make your life very easy.
So you will very easily compute those triple integrals
of constants.
Curl of G, G being of [? a. ?] OK?
I should make the distinction between a scalar function
and a vector function by putting a G bar on the vector function.
How about this?
Is it hard?
No, because it's the same fellow.
Instead of that, I have just x, y, z.
The answer will be the same.
So I still want to get 0, 0, 0-- the vector 0.
So the point was that we will give you enough.
You may expect them to be very hard,
but they are not going to be very hard.
Let's do one more like the ones we have in the book.
What do you think this one will be?
I'm making you a new vector value function.
That's maybe two little exercises
we can do just working exercise three, four,
I don't know what they are.
Let me give you R vector of x, y, z
equals yzI plus xzj plus xyk.
Compute the curl.
Let me write it like engineers do just for fun-- [INAUDIBLE]
cross.
R is the same as curl R, which is I,
J, K-- oh my god-- ddx, ddy, ddz.
Why is z-- xz-- xy.
Are you saying oh, that's not so easy anymore.
You-- you will see that it becomes easy,
OK? i times what is the minor?
This times-- x, right?
Minus x plus minus j times 1 minus what?
Minor will be the red thingie.
And the red thingie is beautiful,
because it's gonna be y minus y plus k times--
who do you think it's gonna be? z z minus.
So it's still 0.
Do we expect something like that on the final?
An easy computation.
Somebody says, find me the curve of this function.
And the functions usually we give
you are nice and significant.
Something where the result will be pretty.
OK.
Let me see what else I wanted.
I'm gonna-- I have space here.
So compute the curl and Laplace operator of f of xyz
equals x squared yzi plus x y squared zj plus xy z squared k.
Of divergence.
Sorry, guys.
This is not a-- it's not a scalar function.
I want the divergence and the curl.
The curl will be a vector.
The divergence will be a scalar function.
Later on I'll give you a nice function where you can
compute the Laplace operator.
That's gonna have to be a scalar function.
And although the Laplace operator
can be generalized to vector functions,
and I'll tell you later how-- what that is.
It's very easy.
It's practically the Laplace operators in every direction.
OK.
So let's see the curl.
i j k, d dx, d dy, d dz.
Today I'm gonna cook up the homework.
And with all the practice that we are doing now,
you should have absolutely no problem doing the homework
for the first two sections.
At least for the section-- today's section, 13.1.
x squared yz, x y squared z, xy z squared.
You see there is some sort of symmetry.
I'm playing a game here.
So I have i.
I want you to tell me [INAUDIBLE], see now,
I don't work much in groups.
I don't make you work in groups, but I
want you to answer my question.
So what is going to be this minor that I-- the first thing
is gonna be?
STUDENT: x z squared.
PROFESSOR: Very good.
STUDENT: Minus x y squared.
PROFESSOR: Minus j.
Potential [? plus or ?] minus.
OK, what is the next guy?
STUDENT: y z squared.
PROFESSOR: y z squared, thank you.
STUDENT: x squared.
PROFESSOR: x squared, right?
Plus k times--
STUDENT: y squared z.
PROFESSOR: So, you see what I'm doing?
I'm doing this from respect to x. y squared z.
You said it, right.
Minus this guy, x squared z.
Can I write it more-- I don't really like the way I wrote it.
But I'll write like that.
How about x times z squared minus y squared i minus j.
Or maybe better plus j.
I'll change this up.
Plus j at the end.
Because it's the vector. y times x
squared minus z squared j plus-- who gets out?
z-- z times y squared minus x squared k.
OK.
Good There is some symmetry in there.
The break in the symmetry is in the middle.
Because, as you see, x is separate,
and then z is followed by y, and then
x squared-- x is followed by z and y is followed by x.
So I have some-- some symmetry of some sort.
What else did I want?
Divergence operator.
And that will be the last example of the kind.
[INAUDIBLE]
How do you write the divergence?
Is this hard?
Very easy?
I'm going go ask you to simplify because I don't like it,
like, as a sum.
2xyz--
STUDENT: 2xyz plus 2xyz plus 2xyz.
PROFESSOR: And now you see why I don't like it as a sum.
Because it's 6xyz and it's very pretty like that.
I'd like you-- on the exam, I'd like
you to take the function and box the answer,
and that's all I want you to do.
All right.
I'm gonna go ahead and erase.
I'm going to move on to 13.2, but I'd
like to review some physics a little bit with you
and see what you remember from physics.
It's a little bit messy.
I'll use this instead because I like the board to be clean.
If I were to ask you to remember work in physics,
I would say-- I'm changing a little bit the order in 13.2.
I'd like you to go back in time and see
what work was in physics class.
STUDENT: [INAUDIBLE] force dx.
PROFESSOR: What if you didn't know any calculus?
Let's go a long time.
The section is 13.2 preliminaries.
STUDENT: Force multiplied distance.
PROFESSOR: Very good.
Preliminary work.
[INAUDIBLE] The notion of work from physics-- hey, come on.
Physics, or engineering, mechanics, whatever you study,
work.
Imagine that you're taking a-- this
is your body that you're playing with-- not your own body,
but the body you are acting on in physics.
And you are dragging this object from a place A
to a place B, another position.
A B is the distance.
And the force is parallel to the trajectory.
This is very important.
This is a simpler case.
In general, it's not so simple.
So this force is acting, and it's a constant force.
And you pull the object from one place to another.
That's case one.
In case two, life is harder.
You actually pull the poor object
with the force in this direction.
Actually, most of us do that, right?
If I were to have a gliding object on the surface,
I would actually act on that object in the direction
of my arm by pulling it.
So when I displace this body from point a to point b,
I still travel the distance d, but-- so d
is a displacement vector that can be written like--
or it can be drawn like that.
I have to be smart in both cases,
figure out what I want from life,
because it's not so clear.
When they taught me I think the first time
I was in-- oh my God-- eighth grade,
and they-- that was a long time ago.
In this case, w is gonna be a scalar
and I'm gonna have the magnitude of the force F. F is a vector,
but to indicate it in Newtons or whatever I measure it in,
it's gonna be in magnitude.
Times the little d, but instead of little d
I should be a little smarter and say,
Magdalena, this is the magnitude of the vector A B, which
is a displacement vector.
So this is called work.
And if the force is 10 Newtons, and the distance
is 10 meters, because we want to go international.
We want to be global, right, at Texas Tech, so good.
So we have 100 Newton-meters.
Now you can measure-- well, you can have another example.
I'm thinking gravity and then you can say it in pounds,
and that measures force.
And you have other units that are not international.
I'm not gonna mess up.
When you have the work in this case, though,
it's more complicated.
And I'm not gonna be mad at you. [INAUDIBLE] is
trying to tell me what it is.
I'm not going to be mad at the people who don't know
what the work is in this case.
Although, I was looking at-- I am the person
who has run a committee to oversee
the finals for different math classes, all the math
classes we offer here.
And every semester I see the [? streak ?] pre-calculus,
calc 1, calc 2, calc 3.
In trig and pre-calculus, they'll
always have a work there.
And I was wondering how many of you took pre-calculus,
and how many of you remember that you
studied this in pre-calculus.
It's a little bit awkward.
I'm thinking, how do they do it, but I gave you the formula.
And they say the force in itself as a vector dot product
the displacement vector.
So they are both forces in dot product.
And I was surprised to see that they gave you [INAUDIBLE]
If I were to express it, how would I express it?
I'll say the magnitude of F, of course,
in Newtons, whatever it is, times the magnitude
of the displacement vector--
STUDENT: Multiply those cosines.
PROFESSOR: Cosine of the angle between.
And I'm too lazy, I don't know, theta.
Let's call it angle theta.
Because I don't want to include that in locations, OK?
It really doesn't matter in which direction
I'm going, because cosine theta, thank God, is an even function.
It's equal to cosine of minus theta.
So whether I go this way or that way, it's the same cosine.
All right.
So the cosine of the angle between the two.
It's very easy when you don't need calculus.
But when you use calculus, because your trajectory is
no longer a line, life is becoming more complicated.
So we have to come up with a different formula,
with a different notion of work.
I'm gonna erase-- are you guys done with that?
Is it visible?
You're done.
OK.
So again, life is not so easy in reality anymore.
I have a particle in physics-- a photon enters a four-star hotel
and says-- talks to the bellboy, and the bellboy,
can I help you with your luggage.
No, I'm traveling light.
[LAUGHTER]
So the particle, the photon-- whatever.
A particle is moving-- is moving on a trajectory.
Suppose that trajectory is planar,
just to make your life easier at first.
It's in the plane x y.
And this is the little particle that's moving.
And this is R. And that is x i plus yj.
Good.
And this is the point x, y.
And that's the position, the current position
of my particle, right now.
Not in the past, not in the future.
My particle is moving, and this is now.
Suppose time doesn't even exist.
We think of the movies that we saw lately,
in The Theory of Everything.
So then, they say OK, we only care about now. x, y is now
and that is the current position vector.
Well, what would be the work between now--
whatever now-- and the next, let's say, this is gonna be x1,
y1.
And this is x0, y0.
That's the general formula, will be x i plus yj.
So I actually cannot forget about time.
Not as much as I want.
So x and y-- x and y are both changing in time.
We're gonna have x equals x sub t, y equals y sub t.
Do you guys remember what we call that kind of equation
for a curve from here to here?
Para--
STUDENT: Parametrization.
PROFESSOR: Parametrization, or parametric equations.
Parametric equations.
Good.
So I have may the force be with you.
I have a force.
I have a force, and I have some sort of displacement.
But I cannot express that displacement linearly anymore.
I'm moving along a [INAUDIBLE] of a curve.
So I have to think differently.
And the work will be defined, whether you like it or not,
as F vector field, dot product with dR over c.
And you say, what in the world is that?
What would c be?
How do I integrate along a path?
And I will tell you in a second what we mean by that.
Meaning that-- this is by definition if you want.
This is like an application of calculus 1.
It can be proved, but we don't-- we do a rigorous job
in the book about introducing and proving that along
a curvilinear path.
This is gonna be-- where am I here, at time t0?
And this is at time t1.
That means x0 is x of t0.
y0 is y of t0.
And at the finish point, I'm at x1, which is x of t1,
and y1 equals y of t1.
So between t0 and t1, I have traveled
and I have F where measure at x of t y of t,
where t is between-- moving between t0 and t1.
I'm done with this is the F part.
What is the dR?
Now you guys know about differential.
Thank God you know about differential,
because this is gonna help you very much.
OK.
So instead of dR, I'm going to write dot, and let's
see how I write-- what I write in terms of dR. You may say,
well, what does she mean?
dR was dxi plus dyj.
And you say, why is that?
I don't understand.
Because R itself is x i plus yj.
And x is a function of t, and y is a function of t.
That means that when you apply the differential,
you are gonna apply the differentials to dx and dy,
and these are gonna be infinitesimal displacement.
Infinitesimal displacement.
What is an infinitesimal displacement in terms of time?
Well, we have our parametric equations.
So Mr. dx as a differential is just x prime dt.
It's like in the [INAUDIBLE] substitution.
dy is just y prime dt.
So let me write this down again.
This is x prime of t i plus y prime of t j times dt.
So Mr. dt is like a common factor.
If he wants to go out for a walk,
he says, I'm gonna go out for a walk.
I go out for a walk.
So dR is actually x prime of t times i
plus y prime of t times j dt.
And this will represent the derivative of R
with respect to pi.
So that will be what?
The differential.
Differential of R with respect to pi.
This is the same as writing dx i plus dy j.
And it's the same as writing dR. Why is this happening?
Because it's [INAUDIBLE] Because x and y
themselves are functions of one variable only, which is time.
This is why it happens.
Oh, so we will simply have to do--
to learn new things, right?
We are gonna have to learn new things,
like integral from a time-- fixed time 0 to t1, which
is 10 seconds, of a dot product between a certain vector that
depends on time and another vector that depends on time,
and dt.
So we are gonna have to learn how
to compute the work through this type of curvilinear integral.
And this is-- this is called either path integral-- path
integral along the curve c, or curvilinear integral along c.
Yes.
STUDENT: Let's say if I move the force this is
[INAUDIBLE] function, correct?
So if I can find [? arc length ?] that
is between the x--
PROFESSOR: Yeah--
STUDENT: [INAUDIBLE] points.
PROFESSOR: Yeah, we will do the one with that length next.
The [? reason ?] so--
STUDENT: Is it harder?
PROFESSOR: No.
No, you can pass through a plane.
And you can-- we'll do that next time.
You will have an integral with respect to S.
So the integration will be with respect to dS,
they are correct.
And then you will have a function that depends on S
[INAUDIBLE] So I'll-- for today, I'll only teach you that.
Next time I'll teach you that with respect
to arc length, which is also very-- it's not hard at all.
STUDENT: OK.
PROFESSOR: So I will work with you on [INAUDIBLE]
Now assume that we have-- I will spray all this thing.
Assume that I have a problem.
I have a parabola-- arc of a parabola, all right?
Between-- let's say the parabola is
y equals x squared between two points.
And I'll ask you to compute some work,
and I'll tell you in a second what [INAUDIBLE] to do.
So exercise-- assume the parabola y
equals x squared between points A of coordinates 0, 0
and point B of coordinates 1, 1.
a, Parametrized this parabola in the simplest way you can.
And b, compute the work along this arc of a parabola,
arc AB of this parabola.
For [INAUDIBLE] the function big F of t,
you see that-- I'm going to say, no, big F of the point x, y,
because you haven't parametrized that yet, big F of x,
y being xi plus yg.
So you say, OK, wait a minute.
W will be integral over the arc of a parabola.
Do you want to draw that first?
Yes, I need to draw that first.
So I have this parabola from A to B. A is of coordinates 0, 0.
B is of coordinates 1, 1.
And this is y equals x squared.
So what kind of parametrization is
the simplest one, the regular one that people take?
Take x to be t?
And of course, take y in that case. y will be t squared.
And for 1 you have 1.
For 0 you have 0.
When you have that work by definition, what was that?
It was written as integral or on the graph C. Let's call
this path C a curvilinear path.
Look, script C-- so beautiful.
Let me [INAUDIBLE] red and draw the C of what is work?
F force-- may the force be with us-- dot dR. All righty, that's
a little bit of a headache.
This F is going to be-- can I write an alternative formula
that I have not written yet but I will write in a second?
dR will be dxi plus dyj.
So I can also write that F dot dR
as the dot product will seem to be-- what was the dot product
guys, do you remember?
First component times first component, F1dx
plus second scalar component times second scalar component,
F2dy.
I'll write it down.
Along the path C I'll have F1dx plus F2dy.
But god knows what it's going to be in terms of time.
So I have to change variable thinking.
Okey-dokey, Mr. dx by substitution
was x prime to T. Mr. dy by substitution was y prime dt.
So I'd rather write this in a simpler way.
This is a new object, path integral.
But we know this object from Calc I
as being a simple integral from time t0--
I'll write it down-- time t1, F1x prime of t plus F2y
prime of t.
This is the integral dt.
This would be a piece of cake for us
to apply in this problem.
Equals-- now you tell me what I'm supposed to write.
Because if you don't, I'm going to not write anything.
t0 for me is what time?
When did we leave this?
0.
And when did we arrive?
At 1 o'clock.
We arrived when t is 1, or every one second or whatever
depending on [INAUDIBLE].
OK, from 0 to 1, now who is F1?
F1 is this.
But it drives me crazy.
Because I need this to be expressed in t.
So I think of x and y as functions of t.
So if 1 is not x, not [INAUDIBLE]
right here, but t, which is the same thing in parametrization--
this is t, t times.
Who is x prime?
1, thank god.
That is easy, times 1, plus F2.
Who is F2?
t squared.
I'll have to write it down.
Times who is y prime?
2t.
y prime is t2.
So I write it down-- 2t, dt.
So that's how I compute this integral back.
Is it hard?
No, because it's just a simple integral from Calculus I.
So I have to integrate what function?
A polynomial, 2t cubed plus t with respect
to t between 0, time 0 and time 1.
Good, let's do it.
Because that's a piece of cake-- 2 times t to the 4 over 4
plus t squared over 2.
I take the whole thing between, I apply the fundamental theorem
of calculus, and I have between t equals 1 up and t equals 0
down.
What's the final answer?
It's a single final answer.
And again, on the exam, on the final,
do not expect a headache computation.
Do expect something simple like that
where you don't need a calculator.
You just have either integers only or simple fractions
to add, and you should get the answer.
What is the answer, guys?
1-- 1/2 plus 1/2 equals 1.
So 1 is the value of the work in what?
Measured in newtons times meters,
whatever your units are.
When you drag the object from this point
to this point, on which the acting force is the only
acting force-- it could be the result
that there are several forces.
That is that force that you have here.
Is it useful?
It's very useful for engineers.
It's very useful for physicists.
It's very useful for anybody who works
in applied mathematics, this notion of work given like that.
I'm going to go ahead and erase.
And I'll ask you one thing here that is not
in the book I think as far as I remember.
Can you guys prove that this sophisticated formula becomes
your formula of the one you claimed,
the first formula you gave me?
Is it hard?
Do you think it's hard to prove this?
OK, what if we have the simplest possible case.
Let's think of--
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah, I'm thinking maybe I
should do, well, A to B, right?
AB, what kind of expression do I have [INAUDIBLE]?
If I take this to be-- I could have any line, right?
I could have any line.
But if I have any line, I can pick my frame
according to my preference.
Nobody's going to tell me, well, you
have to take the frame like that,
and then your line will be of the form ax plus by equals.
No, I'll just take the frame to be this one, where
AB will be x axis, and A will be of coordinates 0, 0
and B will be of coordinates B and 0.
And this is just my line.
So x will be moving between 0 and B. And y is 0, right?
It should be, at least.
And then F, let's say, should be this function, this.
I'll assume the angle is constant,
just like I had it with theta.
And then it's acting all the way on your object.
You have the same angle here always.
So F is F1i plus F2j.
dR will be dxi plus dyj.
But then you say, wait a minute, but didn't you say, Magdalena,
that you are along this line?
Didn't you say that y is 0?
So which y?
So there is no y.
So this is 0, right?
OK, Mr. x, I want to parametrize my trajectory.
How do I parametrize it the simplest way?
I'll take x to be t.
And time will be exactly between 0 and d.
And y will be 0.
And thank you god, because that's easy.
And so all you need to give me is
W is integral of F dR C in that case.
So what am I going to have in that case?
I'll have this formula.
I'll skip a step, and I'll have that formula.
And that means I have integral from t0 equals 0 to t1
equals B.
F1-- now you have to tell me what F1 will be. x prime
[? noted ?] is 1.
The second guy is 0, thank you very much, and [INAUDIBLE].
F1 will be what?
Well, life is nice.
F1 will be the projection of the vector F on my x-axis.
So F1 is the length of this blue vector, I'll say.
So F1 is a scalar.
Let's say F1 is a scalar component.
That means it's F length cosine theta.
Because it's hypotenuse times cosine theta.
So it's easy.
So you have length of F, how much it is,
how big this vector is, times cosine theta, times what
when you integrate it, guys?
When you integrate 1 with respect to t, what do you get?
t between d and 0.
So you have t between d and 0.
We got the formula.
So we got that F length times [INAUDIBLE]
times cosine theta times d, this is the displacement.
This is the cosine.
This is the magnitude of the force that I'm-- look,
this is the force.
My force is along my arm.
I'm just dragging this poor object.
The force I'm acting with, suppose
it's always the same parallel to that that I can feel.
So that's what I have, F cosine theta, and it was easy.
So as a particular case of this nasty integral,
I have my old work from school that I had to believe.
I tell you guys, I did not believe a word.
Because my teacher in eighth grade
came up with this out of nothing,
and we were supposed to be good students preparing
for a high school like this kind of scientific-- back home,
there are different kinds of high school.
There is scientific high school with emphasis
in math and physics.
There is one for chemistry/biology.
There is one for language, linguistics, [INAUDIBLE],
and so on.
And I was for the math and physics one.
And I had to solve this formula without understanding it.
And it took me many other years to understand
that it's just a little piece of a big picture,
and that there's something bigger than what
we were taught in eighth grade.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah, yeah, it's true.
Now I want to ask you a question.
So do you think that I would get any kind of conservation laws
in physics that apply to calculus?
I mean, how hard is it really to compute the work?
And I'm making an announcement now.
Since I have not given you a break,
I have to let you go in a few minutes.
But I'm making a big announcement
without proving it.
So we will, in about one week at the maximum,
in maximum one week, study the independence of path of work
if that work is performed by a conservative force.
And you're going to say, wait a minute,
what the heck is a conservative force and what does she mean?
Well, I just showed you that the work is a path integral.
We don't know what that is.
I'll introduce more.
I just introduced the definition of a path integral with respect
to parametrization, general parametrization with respect
to t.
So that becomes an integral with respect to dt,
like the one in Calc I. This is how
you have to view it at first.
But guys, if this force is not just any force,
it's something magic, if F comes from a scalar potential that
is F represents the gradient of a scalar function F--
this is called scalar potential-- then
F is called-- now let's see how much money I
have for just the last two or three minutes that I have left.
I don't have money or I have money?
Come on, big money.
No, I have $5.
I was looking for $1.
Here, I'll give you $5 if you give me $4 back
if you guess-- I don't know.
So maybe in your engineering courses-- maybe
I give you some candy instead.
So if there is a scalar function little f of coordinates x,
y, whatever you have in the problem,
so that big F will be the nabla.
F nabla means the gradient.
We say that F comes from a scalar potential.
But it has also a name, which is called-- god.
It starts with a C, ends with an E. In that case,
if this is going to be equal to nabla F, in that case,
there is a magic theorem that I'm anticipating.
I'm not proving.
I'm doing exercises right now.
We'll see it in two sections, that the work does not depend
on the path you are taking.
So you can go from A to B like that, or you can go like this.
You can go like this.
You can go like that.
You can go on a parabola, on a line, on anything.
The result is always the same.
And it's like the fundamental theorem
of Calc III in plane for the work.
So you have little f endpoint.
STUDENT: Is that [INAUDIBLE].
PROFESSOR: Little f of [INAUDIBLE].
So all that matters is computing this scalar potential
here and here, making the difference,
and that will be your work.
It's a magic thing.
In mechanical engineering maybe you
met it, in physics-- in mechanical engineering,
because that's where you guys drag all sorts of objects
around.
STUDENT: Conservative.
PROFESSOR: Ah, thank god.
Rachel, you're a math major I think.
You're an engineering major.
STUDENT: [INAUDIBLE]
PROFESSOR: Wow, OK, and who else said conservative?
And were there other people who said conservative?
I'm sorry I don't have.
Well, next time I'll bring a bunch of dollars,
and I'll start giving prizes as dollar bills
like I used to give in differential equations.
Everybody was so happy in my class.
Because for everything that they got quickly and right,
they got $1.
So conservative-- very good.
Remember that for the next few lessons.
We will show that when this f is magical, that is conservative,
you guys don't have to compute the integral at all.
There's no parametrization, no nothing.
It really doesn't depend on what path you take.
All you would need is to figure who this little f will be,
this scalar potential.
Our future work can do that.
And then you compute the values of that scalar potential here
and here, make the difference.
And for sure you'll have such a problem in the final.
So I'm just anticipating it, because I
want this to be absorbed in time into your system.
When we will do the final exam review,
you should be baptized in this kind of problem
so that everybody will get 100% on that for the final.
OK, now I'll let you go.
Sorry I didn't give you a break.
But now I give you more time.
And enjoy the day.
I'll see you Thursday.
I'm moving to my office.
If you have questions, you can come to my office.
Maybe you were getting close.
How did-- did you know, or it just came to you?
[BACKGROUND CHATTER]
STUDENT: Do you know what section it would be?
Because I don't even think he's listed or anything.
PROFESSOR: Send me an email if you don't figure it out.
But for sure [INAUDIBLE].
STUDENT: OK, because I was going to do the honors,
but it was with [INAUDIBLE].
I don't know if he's good, or she's good.
PROFESSOR: She's good.
But he's fantastic in the sense that he will help you
whenever you stumble.
He's an extremely good teacher.
He explains really well.
He has a talent.
STUDENT: I'll look for him.
Thank you.
PROFESSOR: And if you don't get him,
she is good as well-- not exceptional like he is.
He's an exceptional teacher.
STUDENT: I'll go to the office.
PROFESSOR: Yes, yes, [INAUDIBLE].