People asked me if I'm going to go over homework.
Of course I will.
Let me explain.
Out of the four hours you have, three
should be more or less lecture time.
And the fourth hour, which is the instructor's latitude,
where they put it-- it's applications, problems,
homework like problems, all sorts of practice for exams
and so on.
It's not a recitation.
It's some sort of workshop that the instructor conducts himself
personally.
All right.
If you don't have questions, I'm just
going to go ahead and review a little bit of what
we discussed last time.
Something new and exciting was chapter 11, section 11.1.
And we did 11.2.
And what was that about?
That was about functions of several variables.
And we discussed several examples,
but then we focused our attention mainly
to explicit functions, which means z equals f of x, y,
of two variables.
And we call this a graph because it is a graph.
In 3D, it's a surface whose domain is on the floor.
And the altitude is z, and that is the-- this is the-- OK.
How many of you are non-math majors?
Can you raise hands?
Oh, OK.
So you know a little bit about research
from your own classes, science classes
or from science fairs from school.
These are the independent variables, x, y.
And z is the dependent variable.
We don't use this kind of terminology in this class.
But so that you know-- we discussed domain last time.
This was about what?
Domain, range.
After range, what did we do?
We talked about level curves.
What is the level curve?
Level curves are curves x, y in the plane corresponding
to f of x, y equals constant.
These are called level curves in plane,
in the plane called x, y plane.
What else have we discussed?
We went straight into 11.2.
In 11.2, we were very happy to remember
a little bit of Calculus 1, which was practically
a review of limits from Calc 1.
And what did we do?
We did epsilon delta, which was not covered in Calculus 1.
And where is Aaron?
OK.
Thank you, Aaron.
And today, I was thinking, I want to show you actually
an example that is quite easy of how you use epsilon
delta for continuity, to show if the function is continuous,
but for a function of true variables.
And that's not hard.
You may think, oh, my god.
That must be hard.
That's not hard at all.
I'm going to move on to the second part of 11.2, which
is continuity.
11.2, second part.
The first part was what?
It was limits of functions, right, guys?
We discussed properties of limits,
algebraic properties of adding sums and taking a limit
of a sum, taking a limit of a product of functions,
taking the limit of a quotient of function, when it exists,
when it doesn't.
Now the second part of 11.2 is called continuity.
Continuity of what?
Well, I'm too lazy to right down,
but it's continuity of functions of two variables, right?
Now in Calc 1-- you reminded me last time.
I tried to remind you.
You tried to remind me.
Let's remind each other.
This is like a discussion.
What was the meaning of f of x being a continuous function x0,
which is part of the domain?
x0 has to be in the domain.
This is if and only if what?
Well, what kind of function is that?
A one variable function, real value.
It takes values on, let's say, an interval on the real line.
What was the group of properties that
have to be simultaneously satisfied,
satisfied at the same time?
And you told me it has to be at the same time.
And I was very happy because if one of the three conditions
is missing, then goodbye, continuity.
One?
STUDENT: It's defined at that point.
MAGDALENA TODA: Yes, sir. f of x0 is defined.
Actually, I said that here in the domain.
I'll remove it because now I said it better.
Two?
STUDENT: The limit exists.
MAGDALENA TODA: Very good.
The limit, as I approach x0 with any kind of value
closer and closer, exists and is finite.
Let's give it a name.
Let's call it L.
STUDENT: [? The following value ?]
equals the limit.
MAGDALENA TODA: Yes, sir.
That's the last thing.
And I'm glad I didn't have to pull the truth out
of your mouth.
So the limit will-- the limit of f of x when x goes to x0
equals f of x0.
No examples.
You should know Calc 1, and you do.
I'm just going to move on to Calc 3.
And let's see what the definition of continuity
would mean for us in Calc 3.
Can anybody mimic the properties that-- well, f of x, y
is said to be continuous at x0, y0
if and only if the following conditions are-- my arm hurts.
Are simultaneously satisfied.
I don't like professors who use PDF files or slides.
Shh.
OK.
I don't want anything premade.
The class is a construction, is working,
is something like a work in progress.
We are building things together.
This is teamwork.
If I come up with some slides that were
made at home or a PDF file.
First of all, it means I'm lazy.
Second of all, it means that I'm not
willing to take it one step at a time
and show you how the idea's revealed.
One.
Who is telling me?
I'm not going to say it.
It's a work in progress.
STUDENT: [INAUDIBLE]
MAGDALENA TODA: f of--
STUDENT: [INAUDIBLE]
MAGDALENA TODA: Of x0, y0 is defined.
And why not?
Well, just to have a silly [? pun ?].
Two.
Limit as the pair x, y approaches x0, x0-- and guys,
when you close your eyes-- no you close your eyes--
and you imagine x, y going to x0,
y0 by any possible paths in any possible way,
it's not that you have a predetermined path to x0, y0,
because you may be trapped.
You may have-- as you've seen last time, you may have,
coming from this direction, the limit will exist,
will be this one.
Coming from that direction, the limit will exist,
would be another one.
And then you don't have overall limits.
So the limit-- when I call that, that means the overall limit
exists, exists and equals L. It's finite.
That's what I mean.
And three, the value of the function at x0,
y0 must be equal to the limit of the function that value
as you approach it, x0, y0.
And equals L, of course.
So great.
So it's so obvious that we are following
exactly the same type of definition,
the same type of pattern.
I'm going to ask you to help me, to help
me solve a harder problem that involves continuity.
And I'm asking you, if I have the following function--
I'm going to erase the definition of continuity
from Calc 1.
I'm going to ask you, what if I have this funny function?
You've seen it before, and I gave you a little bit
of a warning about it.
Limit as x, y goes to 0, 0 of x squared
plus y squared times sine of 1 over x squared plus y squared.
Does that exist?
And also--
STUDENT: It's actually-- so the limit is actually
approaching a plane rather than a set of [INAUDIBLE].
MAGDALENA TODA: So well, actually, it's
not approaching a plane.
Let's see what's happening when--
STUDENT: Sorry, sorry.
Not a plane, a [? line. ?]
MAGDALENA TODA: Yes.
STUDENT: And is the z-axis-- the entire z-axis is 0, 0?
MAGDALENA TODA: So this is the z-axis.
And that means exactly that x and y-- it will be 0.
Now I am just looking at what happens
in the plane, in the floor plane x, y.
The pairs x, y are wiggly.
They are like little wormy worms.
And they float on the water on the floor.
And these squiggly things approach
x, y from any possible path.
They go like this.
They go like that.
They go in every possible way.
Let's see what happens.
Continuity-- is this continuous?
Well, you say, Magdalena, come on.
You cannot have this continuous at 0, 0,
because it's undefined at 0, 0.
Yes.
But maybe I can extend it by continuity.
So let me introduce-- this is my favorite, f of x, y.
But I'll say, f of x, y is not defined at 0, 0.
But how about g of x, y as being my f of x, y for any x,
y different from 0, 0.
And at the origin, at the very origin, I will say,
I want to have-- when x, y equals 0,
0, I want to have a value.
Which value do you think might extend
this function by continuity?
STUDENT: The limit.
MAGDALENA TODA: The limit if it exists
and if-- well, you know already, I think, what the limit is
because some of you thought about this at home
for extra credit.
So it's not fair, right?
No, I'm just kidding.
So I claim that maybe-- if I put a 0 here,
will this be continuous?
Will g be continuous?
So prove, prove either way, prove, justify your answer
by a proof, a complete proof with epsilon delta.
Proof.
OK.
So now is a worried face.
Like, oh, my god.
This guy is worried because, oh, my god.
Epsilon delta.
Oh, my god.
But the principle-- the intuition
tells us that we should look first at some sort of a graph,
just like Ryan pointed out.
One should close their eyes and imagine a graph of a function
with-- it's hard to visualize in 3D the graph of a function that
is a surface.
This is a surface. z equals the whole shebang.
But when I'm going to look at the one dimensional case
from last time, we remember the sine of 1/x
was a crazy function.
We called it the harmonica, well, 20-something years ago
when I was in high school.
I was in an advanced calculus class.
And our teacher was not funny at all.
He was also not teaching much, gave us a lot of homework,
very challenging.
So in order to make our life a little bit easier,
we always worked in groups, which was allowed.
So we called it a harmonica because it was oscillating
like that to the point that-- you've seen
the harmonica-- the accordion.
When you bring it back to the-- harmonica came to my mind
from the harmonic function.
So the accordion is-- when you actually
squeeze it, all that oscillation things, the cusps are
closer and closer to a line.
So what you have here is this kind of oscillation,
very, very rapid oscillation for sine of 1/x.
When we want to multiply by an x, what's going to happen?
Well, this has not limit at 0 because it takes all the values
infinitesimally close to 0.
It keeps going through all the values between minus 1 and 1,
closer and closer.
So that was no good.
But if we take this guy-- that's going to go to-- well,
I cannot do better.
MATLAB can do better than me.
Mathematica can do better.
You can do that.
In most engineering classes, if you are--
who is an electrical engineering major?
But even if you are not, you are going
to see this type of function a lot.
And you're going to see it again in differential equations.
How can I imagine-- this graph is hard to draw.
Don't ask me to draw that.
But ask me if I can use epsilon delta to prove continuity.
So what would it mean, proving continuity?
I have a feeling--
STUDENT: Well, actually, if this is-- going back to that graph,
doesn't that graph look like--
MAGDALENA TODA: This goes to 0.
The limit exists for x sine of 1/x, and it is 0.
Why?
Ryan?
RYAN: Wouldn't the graph with the x squared plus
y squared times that side-- wouldn't that
just look like a ripple in a circle going out
from the center?
MAGDALENA TODA: Yeah, it will be ripples.
STUDENT: Just like a [INAUDIBLE] from an epicenter
going outwards [INAUDIBLE].
MAGDALENA TODA: And I think-- yes, we managed to-- you
have a concentric image, right?
STUDENT: Yeah.
MAGDALENA TODA: Like those ripples, exactly like--
STUDENT: So that's what that looks like?
MAGDALENA TODA: --when you throw a stone into the water,
this kind of wave.
But it's infinitesimally close.
It's like acting weird.
But then it sort of shrinks here.
And that-- it imposes the limit 0.
How come this goes to 0, you say?
Well, Magdalena, this guy is crazy, right?
Sine of 1/x goes between minus 1 and 1
infinitely many times as I go close, close,
closer and closer, more rapidly, more and more rapidly close
to 0.
This will oscillate more rapidly,
more rapidly, and more rapidly.
This is crazy, right?
How does this guy, x-- how is this guy taming this guy?
STUDENT: Because as 0 [INAUDIBLE].
Something really small times something [INAUDIBLE].
MAGDALENA TODA: Something very small
that shrinks to 0 times something bounded.
Ryan brought the main idea.
If something goes strongly to 0, and that multiplies something
that's bounded, bounded by a finite number,
the whole problem will go to 0.
Actually, you can prove that as a theorem.
And some of you did.
In most honors classes unfortunately,
epsilon delta was not covered.
So let's see how we prove this with epsilon delta.
And, oh, my god.
Many of you read from the book and may be able to help me.
So what am I supposed to show with epsilon delta?
The limit of x squared plus y squared sine of 1 over x
squared plus y squared is 0 as I approach the origin
with my pair, couple, x, y, which can go any one path that
approaches 0.
So you say, oh, well, Magdalena, the Ryan principle-- this
is the Ryan theorem.
It's the same because this guy will be
bounded between minus 1 and 1.
I multiplied with a guy that very determinedly goes
to 0 very strongly.
And he knows where he's going.
x squared plus y squared says, I know what I'm doing.
I'm not going to change my mind.
This is like the guy who changes his major too many times.
And this guy knows what he's doing.
He's going there, and he's a polynomial, goes to 0,
0 very rapidly.
Now it's clear what happens intuitively.
But I'm a mathematician.
And if I don't publish my proof, my article
will be very nicely rejected by all the serious journals
on the market.
This is how it goes in mathematics.
Even before journals existed, mathematicians
had to show a rigorous proof of their work,
of their conjecture.
OK.
So I go, for every epsilon positive, no matter how small,
there must exist a delta positive, which
depends on epsilon-- that depends on epsilon-- such that
as soon as-- how did we write the distance?
I'll write the distance again because I'm lazy.
The distance between the point x, y and the origin
is less than delta.
It follows that the absolute value--
these are all real numbers-- of f of x, y or g of x,
y-- g of x, y is the extension.
f of x, y minus 0, which I claim to be the limit,
will be less than epsilon.
So you go, oh, my god.
What is this woman doing?
It's not hard.
I need your help though.
I need your help to do that.
So it's hard to see how you should-- you take any epsilon.
You pick your favorite epsilon, infinitesimally small,
any small number, but then you go, but then I
have to show this delta exists.
You have to grab that delta and say, you are my delta.
You cannot escape me.
I tell you who you are.
And that's the hardest part in here,
figuring out who that delta must be as a function of epsilon.
Is that hard?
How do you build such a construction?
First of all, understand what proof.
"Choose any positive epsilon."
Then forget about him, because he's your friend,
and he's going to do whatever you want to do with him.
Delta, chasing after delta is going
to be a little bit harder.
"Chasing after delta with that property."
Dot, dot, dot, dot, dot.
What is this distance?
You guys have helped me last time,
you cannot let me down now.
So as soon as this distance, your gradient distance
is less than delta, you must have
that f of x, y [INAUDIBLE].
Could you tell me what that would be?
It was Euclidean, right?
So I had squared root of-- did I?
Square root of x minus 0 squared plus y minus 0 squared.
You say, but that's silly, Magdalena.
So you have to write it down like that?
STUDENT: It's the [INAUDIBLE].
MAGDALENA TODA: Huh?
Yeah.
So square root of this plus square root of that
plus then delta, that means what?
If and only if x squared plus y squared is less than delta
squared.
And what do I want to do, what do I want to build?
So we are thinking how to set up all this thing.
How to choose the delta.
OK, so what do I-- what am I after?
"I am after having" double dot.
F of x, y must be Mr. Ugly.
This one.
So absolute value of x squared plus y squared, sine of 1
over x squared plus y squared minus 0.
Duh.
I'm not going to write it.
We all know what that means.
Less than epsilon.
This is what must follow as a conclusion.
This is what must follow, must happen.
Must happen.
Now I'm getting excited.
Why?
Because I am thinking.
I started thinking.
Once I started thinking, I'm dangerous, man.
So here sine of 1 over x squared plus y squared is your friend.
Why is that your friend?
Sine of 1 over x squared plus y squared, this
is always an absolute value.
The absolute value of that is always less than 1.
OK?
STUDENT: Can't it be 4?
MAGDALENA TODA: So-- so-- so what
shall I take in terms of delta-- this is my question.
What shall I take in terms of delta?
"Delta equals 1 as a function of epsilon
in order to have the conclusion satisfied."
You say, OK.
It's enough to choose delta like that function of epsilon,
and I'm done, because then everything will be fine.
So you chose your own epsilon, positive, small, or God
gave you an epsilon.
You don't care how you got the epsilon.
The epsilon is arbitrary.
You pick positive and small.
Now, it's up to you to find delta.
So what delta would satisfy everything?
What delta would be good enough--
you don't care for all the good--
it's like when you get married.
Do you care for all the people who'd match you?
Hopefully not, because then you would probably
have too large of a pool, and it's hard to choose.
You only need one that satisfies that assumption, that satisfies
all the conditions you have.
So what is the delta that satisfies all the conditions
that I have?
[INTERPOSING VOICES]
MAGDALENA TODA: [INAUDIBLE].
Who?
[INTERPOSING VOICES]
MAGDALENA TODA: For example, delta equals epsilon.
Would that satisfy?
Well, let's see.
If I take delta to be epsilon, then x
squared plus y squared would be less than epsilon squared.
Now the question is is epsilon squared less than epsilon?
Not always.
If epsilon is between 0 and 1, then epsilon squared
is less then epsilon.
But if I choose epsilon to be greater than 1,
then oh, my God.
Then if it's greater than 1, then epsilon squared
is greater than 1-- greater than it.
So what if I choose delta to be what?
STUDENT: 0?
MAGDALENA TODA: No, no, no.
Delta cannot be 0.
So delta-- look, there exists delta strictly bigger than 0,
that depends on epsilon.
Maybe if epsilon is very small, in a way Alexander was right.
But the delta [INAUDIBLE], we don't go with epsilon
greater than 1.
Come on.
Be serious.
Epsilon is always between 0 and 1.
I mean, it's a lot smaller than that.
It's infinitesimal small.
So in the end, yes, in that case epsilon squared
would be less than epsilon, which would be OK for us
and that would be fine.
OK?
So that would be a possibility to say, hey,
since epsilon-- Alexander, if you write that as a proof
I'll be OK.
You say, I took my epsilon to be a very small number,
so anyway it's going to be less than 1.
So epsilon squared is less than epsilon.
So when I take delta to be epsilon,
for sure this guy will be less than epsilon squared, which
is less than epsilon, so I'm satisfied.
I'll give you a 100%.
I'm happy.
Is that the only way?
STUDENT: But what about the sine?
What about [INAUDIBLE].
STUDENT: Yeah.
MAGDALENA TODA: So this doesn't matter.
Let me write it down.
So note that x squared plus y squared sine of 1
over x squared plus y square would always
be less than absolute value of x squared
plus y, which is positive.
Why is that?
Is this true?
Yeah.
Why is that?
STUDENT: Because the sine can only be one of these negatives.
MAGDALENA TODA: So in absolute value,
sine of 1 over x squared plus y squared is always less than 1.
STUDENT: Can't it equal 1?
MAGDALENA TODA: Well, when does it equal 1?
STUDENT: Wouldn't it be x squared plus y squared equals 1
[INAUDIBLE]?
MAGDALENA TODA: Less than or equal to.
For some values it will.
STUDENT: Yeah.
OK.
MAGDALENA TODA: Now, will that be a problem with us?
No.
Let's put it here.
Less than or equal to x squared plus y squared, which
has to be less than epsilon if and only if-- well,
if delta is what?
So, again, Alexander said, well, but if I take delta
to be epsilon, I'm done.
STUDENT: [INAUDIBLE].
MAGDALENA TODA: How about square root?
Can I take delta to be square root of epsilon.
STUDENT: That's what I said.
MAGDALENA TODA: No.
You said epsilon.
STUDENT: I said square root of epsilon.
MAGDALENA TODA: OK.
If delta is square root of epsilon,
then everything will be perfect and it will be a perfect match.
In what case?
STUDENT: If epsilon is in between 0 and 1
and if delta is equal to bigger than epsilon.
MAGDALENA TODA: So that's exactly the same assumption.
Epsilon should be made in less than.
STUDENT: But I thought delta was supposed
to be less than epsilon in every case.
So if epsilon is between 0 and 1, the square root of epsilon
is going to be [INAUDIBLE].
MAGDALENA TODA: So when both of them are small,
delta squared will be-- if I take delta-- so take delta
to be square root of epsilon.
STUDENT: Then anything less than 1 and greater than 0,
epsilon would be great than [INAUDIBLE].
MAGDALENA TODA: "Delta to be square root of epsilon,
then x squared plus y squared less than delta squared equals
epsilon."
Then x squared plus y squared sine of 1
over x squared plus y squared less than
or equal to x squared plus y squared.
I dont' need the absolute value.
I can [INAUDIBLE].
Less than epsilon [INAUDIBLE].
Qed.
STUDENT: Well, but you told us delta
has to be less than epsilon.
Well, if--
MAGDALENA TODA: No, I didn't say that.
I didn't say that delta has to be less than epsilon.
Absolutely--
STUDENT: Yeah.
You said for all the values of epsilon greater than 0,
there's a value of delta that is greater than 0 that [INAUDIBLE]
such that as soon as the distance between is less than
delta-- I don't remember what--
MAGDALENA TODA: OK, so, again--
STUDENT: Such that the distance is less than--
MAGDALENA TODA: So again, for epsilon positive,
there is a delta positive, very small.
Very small means very small, OK?
I'm not threatened by-- what?
For epsilon greater than 0, very small,
there is a delta greater than 0, very small,
which depends on epsilon-- I didn't say it cannot be equal
to epsilon-- that depends on epsilon such that whenever x,
y is within delta distance from origin,
[INAUDIBLE] that f of x, y is within epsilon of from l.
All right?
And now I will actually give you another example where
maybe delta will be epsilon.
And let me challenge you with another problem that's
not hard.
OK?
So let me give you the function g
of x, y equals x sine of 1 over y as x, y.
y is equal [? to delta 0. ?] And let's say 0 for the rest.
Can you show-- can you check if g is continuous at 0, 0?
This is one of the problems in your book.
So how do you check that with epsilon delta?
Again, we recite the poetry.
We have to say that.
"For every epsilon positive, small, very small,
there is a delta positive that depends
on epsilon, such that as soon as--" how is the distance?
Square root of x squared plus y squared is less than delta.
This is the distance between point and origin.
"It follows that absolute value of x sine of 1 over y minus--"
so practically x, y no 0.
x, y different from 0.
OK?
I"m careful here, because if y is 0, then I blow up.
And I don't want to blow up.
So x sine of 1 over y minus who?
Minus 0 is less than epsilon.
So now you're thinking, OK, you want me
to prove there is such a delta?
Yes.
That depends on epsilon?
Yes.
And what would that delta be?
The simplest choice you can have in this case.
So you go, oh, my God.
How do I do that?
You have to always think backwards.
So "we need to satisfy absolute value of x sine of 1
over y less than epsilon."
Is this hard?
What is your advantage here?
Do you have any advantage?
Remark absolute value of x sine of 1 over y
is smaller than who?
Smaller than the product of absolute values.
Say it again?
Yes?
STUDENT: But, like, for example, the only condition
for that equation is that y must not be equal to 0.
What if you used another point for x?
Would the answer for delta be different?
MAGDALENA TODA: Well, x is-- you can
choose-- you were right here.
You can say, OK, can you be more restrictive, Magdelena,
and say, for every point of the type x equals 0
and y not 0, it's still OK?
Yes.
So you could be a professional mathematician.
So practically all I care about is x, y in the disk.
What disk?
What is this disk?
Disk of radius 0 when-- what is the radius?
Delta-- such that your y should not be 0.
So a more rigorous point would be
like take all the couples that are
in this small disk of radius delta,
except for those where y is 0.
So what do you actually remove?
You remove this stinking line.
But everybody else in this disk, every couple in this disk
should be happy, should be analyzed
as part of this thread.
Right?
OK.
x sine of 1 over y less than-- is that true?
Is that less than the absolute value of x?
STUDENT: Yeah.
MAGDALENA TODA: Right.
So it should be-- less than should be made
should be less than epsilon.
When is this happening on that occasion?
If I take delta-- meh?
STUDENT: When delta's epsilon.
MAGDALENA TODA: So if I take-- very good.
So Alex saw that, hey, Magdelena, your proof is over.
And I mean it's over.
Take delta, which is delta of epsilon, to be epsilon.
You're done.
Why?
Let me explain what Alex wants, because he
doesn't want to explain much, but it's not his job.
He's not your teacher.
Right?
So why is this working?
Because in this case, note that if I take delta
to be exactly epsilon, what's going to happen?
x, Mr. x, could be positive or negative.
See, x could be positive or negative.
Let's take this guy and protect him in absolute value.
He's always less than square root of x square plus y
squared.
Why is that, guys?
STUDENT: Because y can't be 0.
MAGDALENA TODA: So this is-- square it in your mind.
You got x squared less than x squared plus y squared.
So this is always true.
Always satisfied.
But we chose this to be less than delta,
and if we choose delta to be epsilon, that's our choice.
So God gave us the epsilon, but delta is our choice,
because you have to prove you can do something
with your life.
Right?
So delta equals epsilon.
If you take delta equals epsilon,
then you're done, because in that case absolute value
of x is less than epsilon, and your conclusion, which is this,
was satisfied.
Now, if a student is really smart--
one time I had a student, I gave him this proof.
That was several years ago in honors,
because we don't do epsilon delta in non-honors.
And we very rarely do it in honors as well.
His proof consisted of this.
Considering the fact that absolute value of sine
is less than 1, if I take delta to be epsilon,
that is sufficient.
I'm done.
And of course I gave him 100%, because this
is the essence of the proof.
He didn't show any details.
And I thought, this is the kind of guy who is great.
He's very smart, but he's not going to make a good teacher.
So he's probably going to be the next researcher,
the next astronaut, the next something else, but not--
And then, years later, he took advanced calculus.
He graduated with a graduate degree
in three years sponsored by the Air Force.
And he works right now for the Air Force.
He came out dressed as a captain.
He came and gave a talk this year at Tech in a conference--
he was rushed.
I mean, if I talk like that, my student
wouldn't be able to follow me.
But he was the same brilliant student that I remember.
So he's working on some very important top secret projects.
Very intelligent guy.
And every now and than going to give talks at conferences.
Like, research talks about what he's doing.
In his class-- he took advanced calculus with me,
which was actually graduate level [INAUDIBLE]--
I explained epsilon delta, and he had it very well understood.
And after I left the classroom he explained it to his peers,
to his classmates.
And he explained it better than me.
And I was there listening, and I remember being jealous,
because although he was very rushed,
he had a very clear understanding of how
you take an epsilon, no matter how small, and then
you take a little ball here, radius delta.
So the image of that little ball will fit in that ball
that you take here.
So even if you shrink on the image,
you can take this ball even smaller
so the image will still fit inside.
And I was going, gosh, this is the essence,
but I wish I could convey it, because no book
will say it just-- or show you how to do it with your hands.
STUDENT: [INAUDIBLE]
MAGDALENA TODA: Right.
So he was rushed, but he had a very clear picture
of what is going on.
OK.
11.3 is a completely new start.
And you are gonna read and be happy about that
because that's partial derivatives.
And you say, Magdalena, finally, this is piece of cake.
You see, I know these things.
I can do them in my-- in my sleep.
So f of x and y is still a graph.
And then you say, how do we introduce
the partial derivative with respect to one variable only.
You think, I draw the graph.
OK.
On this graph, I pick a point x0, y0.
And if I were to take x to be 0, what is-- what is the z
equals f of x0, y?
So I'll try to draw it.
It's not easy.
This is x and y and z, and you want your x0 to be a constant.
STUDENT: [INAUDIBLE]
MAGDALENA TODA: It's a so-called coordinate curve.
Very good.
It's a curve, but I want to be good enough to draw it.
So you guys have to wish me luck,
because I don't-- didn't have enough coffee and I don't feel
like I can draw very well.
x0 is here.
So x is there, so you cut with this board-- are
you guys with me?
You cut with this board at the level x0 over here.
You cut.
When you cut with this board-- you
cut your surface with this board--
you get a curve like that.
And we call that a curve f of x0, y.
Some people who are a little bit in a hurry and smarter than me,
they say x equals x0.
That's called coordinate curve.
So, the thing is, this-- it's a curve in plane.
This is the blue plane.
I don't know how to call it.
Pi.
You know I love to call it pi.
Since I'm in plane with a point in a curve--
a plane curve-- this curve has a slope at x0, y0.
Can I draw that slope?
I'll try.
The slope of the blue line, though.
Let me make it red.
The slope of the red line-- now, if you don't have colors
you can make it a dotted line.
The slope of the dotted line is-- who the heck is that?
The derivative of f with respect to y, because x0 is a constant.
So how do we write that?
Because x0 is sort of in our way, driving us crazy.
Although he was fixed.
We keep him fixed by keeping him in this plane.
x0 is fixed.
We have to write another notation.
We cannot say f prime.
Because f depends on two variables.
f prime were for when we were babies in calculus 1.
We cannot use f prime anymore.
We have two variables.
Life became too complicated.
So we have to say--
STUDENT: Professor?
MAGDALENA TODA: --instead of df dy-- yes, sir.
May you use a subscript?
MAGDALENA TODA: You use-- yeah, you can do that as well.
That's what I do.
Let me do both.
f sub y at-- who was fixed? x0 and y.
But this is my favorite notation.
I'm going to make a face because I love it.
This is what engineers love.
This is what we physicists love.
Mathematicians, though, are crazy people.
They are.
All of them.
And they invented another notation.
Do you remember that Mr. Leibniz,
because he had nothing better to do, when he invented calculus,
he did df dy, or df dx?
What is that?
That was the limit of delta f, delta y, right?
That's what Leibniz did.
He introduced this delta notation,
and then he said if you have delta space over delta time,
then shrink both, and you make a ratio in the limit,
you should read-- you should write it df dy.
And that's the so-called Leibniz notation, right?
That was in calc 1.
But I erased it because that was calc 1.
Now, mathematicians, to imitate the Leibniz notation,
they said, I cannot use df dy.
So what the heck shall I use?
After they thought for about a year,
and I was reading through the history about how
they invented this, they said, let's take
the Greek-- the Greek d.
Which is the del.
That's partial.
The del f, del y, at x0, y.
When I was 20-- no, I was 18 when
I saw this the first time-- I had the hardest time making
this sign.
It's all in the wrist.
It's very-- OK.
Now.
df dy.
If you don't like it, then what do you do?
You can adopt this notation.
And what is the meaning of this by definition?
You say, you haven't even defined it, Magdalena.
It has to be limit of a difference quotient,
just like here.
But we have to be happy and think of that.
What is the delta f versus the delta y?
It has to be like that.
f of Mr. x0 is fixed.
x0, comma, y.
We have an increment in y.
y plus delta y. y plus delta y minus-- that's
the difference quotient.
f of what-- the original point was, well--
STUDENT: x0, y0.
MAGDALENA TODA: x0-- let me put y0
because our original point was x0, y0.
x0, y0 over-- over delta y.
But if I am at x0, y0, I better put x0, y0 fixed point here.
And I would like you to photograph or put this thing--
STUDENT: So is that a delta that's in front of the f?
MAGDALENA TODA: Let me review the whole thing
because it's very important.
Where shall I start, here, or here?
It doesn't matter.
So the limit--
STUDENT: [INAUDIBLE] start at m.
MAGDALENA TODA: At m?
At m.
OK, I'll start at m.
The slopes of this line at x0, y0, right at my point,
will be, my favorite notation is f sub y at x0,
y0, which means partial derivative of f with respect
to y at the point-- fixed point x0, y0.
Or, for most mathematicians, df-- of del-- del f,
del y at x0, y0.
Which is by definition the limit of this difference quotient.
So x0 is held fixed in both cases.
y0 is allowed to deviate a little bit.
So y0 is fixed, but you displace it by a little delta,
or by a little-- how did we denote that in calc 1, h?
Little h?
STUDENT: Yeah.
MAGDALENA TODA: So delta y, sometimes it
was called little h.
And this is the same as little h.
Over that h.
Now you, without my help, because you
have all the knowledge and you're smart,
you should tell me how I define f sub x at x0, y0,
and shut up, Magdalena, let people talk.
This is hard.
[INTERPOSING VOICES]
MAGDALENA TODA: No.
I hope not.
As a limit of a difference quotient,
so it's gonna be an instantaneous rate of change.
That's the limit of a difference quotient.
Limit of what?
Shut up.
I will zip my lips.
STUDENT: Delta x
MAGDALENA TODA: Delta x, excellent.
Delta x going to 0.
So you shrink-- you displace by a small displacement
only in the direction of x.
STUDENT: So f.
MAGDALENA TODA: f.
STUDENT: [INAUDIBLE] this time, x is changing, so--
[INTERPOSING VOICES]
MAGDALENA TODA: X0 plus delta x, y0 is still fixed,
minus f of x0, y0.
Thank God this is always fixed.
I love this guy.
STUDENT: Delta--
MAGDALENA TODA: Delta x, which is
like the h we were talking about.
Now in reality, you never do that.
You would die if for every exercise, derivation exercise,
you would have to compute a limit of a difference quotient.
You will go bananas.
What we do?
We do exactly the same thing.
How can I draw?
Can anybody help me draw?
For y0, I would need to take this other plane through y0.
Where is y0?
Here.
Is my drawing good enough?
I hope so.
So it's something like I have this plane with,
oh, do you see that, guys?
OK.
So what is that, the other curve, coordinate curve, look
like?
Oh my God.
Looks like that.
Through the same point, and then the slope
of the line will be a blue slope and the slope
will be f sub-- well OK.
So here I have in the red one, which was the blue one,
this is f sub y, and for this one, this is f sub x.
Right?
So guys, don't look at the picture.
The picture's confusing.
This is x coming towards me, right?
And y going there and z is going up.
This is the graph.
When I do the derivative with respect
to what is this, y, the derivative with respect to y,
with respect to y, y is my only variable,
so the curve will be like that.
And the slope will be for a curve that depends on y only.
When I do derivative with respect to x,
it's like I'm on top of a hill and I decide to go skiing.
And I'm-- and I point my skis like that,
and the slope is going down, and that's the x direction.
OK?
And what I'm going to describe as a skier
will be a plane curve going down in this direction.
Zzzzsssshh, like that.
And the slope at every point, the slope of the line,
of y trajectory, will be the derivative.
So I have a curve like that, and a curve like this.
And they're called coordinate curves.
Now this is hard.
You'll see how beautiful and easy
it is when you actually compute the partial derivatives
of functions by hand.
Examples?
Let's take f of x, y to be x squared plus y squared.
I'm asking you, who is f sub x at x, y?
Who is f sub x at 1 minus 1, 1, 0, OK.
Who is f sub y at x, y?
And who is f sub y at 3 and 2.
Since I make up my example-- I don't
want to copy the examples from the book,
because you are supposedly going to read the book.
This is-- should be another example, just for you.
So who's gonna help me-- I'm pausing a little bit-- who's
gonna help me here?
What's the answer here?
So how do I think?
I think I got-- when I prime with respect to x, y
is like a held constant.
He's held prisoner.
Poor guy cannot leave his cell.
That's awful.
So you prime with respect to x.
Because x is the only variable.
And he is--
STUDENT: So then it's 2x plus y?
MAGDALENA TODA: 2x plus 0.
Plus 0.
Because y is a constant and when you prime a constant,
you get 0.
STUDENT: So when you take partial derivatives,
you-- when you're taking it with respect
to the first derivative, the first variable [INAUDIBLE]
MAGDALENA TODA: You don't completely
know because it might be multiplied.
But you view it as a constant.
So for you-- very good, Ryan.
So for you, it's like, as if y would be 7.
Imagine that y would be 7.
And then you have x squared plus 7 squared prime is u, right?
STUDENT: So then that means f of 1-- or f x of 1,0
is [INAUDIBLE]
MAGDALENA TODA: Very good.
STUDENT: OK.
And in this case, f sub y, what do you think it is?
STUDENT: 2y.
MAGDALENA TODA: 2y.
And what is f y of 3, 2?
STUDENT: 4.
MAGDALENA TODA: It's 4.
And you say, OK, that makes sense, that was easy.
Let's try something hard.
I'm going to build them on so many examples
that you say, stop, Magdalena, because I became
an expert in partial differentiation
and I-- now everything is so trivial that you have to stop.
So example A, example B. A was f of x, y [INAUDIBLE] x, y plus y
sine x.
And you say, wait, wait, wait, you're
giving me a little bit of trouble.
No, I don't mean to.
It's very easy.
Believe me guys, very, very easy.
We just have to think how we do this.
f sub x at 1 and 2, f sub y at x, y in general,
f sub y at 1 and 2, for God's sake.
OK.
All right.
And now, while you're staring at that,
I take out my beautiful colors that I paid $6 for.
The department told me that they don't buy different colors,
just two or three basic ones.
All right?
So what do we do?
STUDENT: First one will be the y.
MAGDALENA TODA: It's like y would be a constant 7, right,
but you have to keep in mind it's mister called y.
Which for you is a constant.
So you go, I'm priming this with respect to x only--
STUDENT: Then you get y.
MAGDALENA TODA: Very good.
Plus--
STUDENT: y cosine x.
MAGDALENA TODA: y cosine x.
Excellent.
And stop.
Because that's all I have.
You see, it's not hard.
Let me put here a y.
OK.
And then, I plug a different color.
I'm a girl, of course I like different colors.
So 1, 2. x is 1, and y is 2.
2 plus 2 cosine 1.
And you say, oh, wait a minute, what is that cosine of 1?
Never mind.
Don't worry about it.
It's like cosine of 1, [INAUDIBLE]
plug it in the calculator, nobody cares.
Well, in the final, you don't have a calculator,
so you leave it like that.
Who cares?
It's just the perfect-- I would actually hate it
that you gave me-- because all you
could give me would be an approximation, a truncation,
with two decimals.
I prefer you give me the precise answer, which
is an exact answer like that.
f sub y.
Now, Mr. x is held prisoner.
He is a constant.
He cannot move.
Mr. y can move.
He has all the freedom.
So prime with respect to y, what do you have?
STUDENT: x--
[INTERPOSING VOICES]
MAGDALENA TODA: x plus sine x is a constant.
So for God's sake, I'll write it.
So then I get 1, plug in x equals 1. y
doesn't appear in the picture.
I don't care.
1 plus sine 1.
And now comes-- don't erase.
Now comes the-- I mean, you cannot erase it.
I can erase it.
Comes this mean professor who says, wait a minute,
I want more.
Mathematicians always want more.
He goes, I want the second derivative.
f sub x x of x, y.
And you say, what in the world is that?
Even some mathematicians, they denote it
as del 2 f dx 2, which is d of-- d with respect
to x sub d u with respect to x.
What does it mean?
You take the first derivative and you derive it again.
And don't drink and derive because you'll be in trouble.
Right?
So you have d of dx primed again, with-- differentiated
again with respect to x.
Is that hard?
Uh-uh.
What you do?
In the-- don't do it here.
You do it in general, right?
With respect to x as a variable, y is again held as a prisoner,
constant.
So when you prime that y goes away.
You're gonna get 0.
I'll write 0 like a silly because we are just starters.
And what else?
STUDENT: Negative y sine of x.
MAGDALENA TODA: Minus y sine of x.
And I know you've gonna love this process.
You are becoming experts in that.
And in a way I'm a little bit sorry it's so easy,
but I guess not everybody gets it.
There are students who don't get it the first time.
So what do we get here?
Minus--
STUDENT: 0.
MAGDALENA TODA: Please tell me-- sine 1, 0.
Good.
I could do the same thing for f y y.
I could do this thing-- what is f sub x y?
By definition f sub x y--
STUDENT: Is that taking the derivative of the derivative
with respect-- is that taking the second derivative
with respect to y after you take the derivative
of the-- first derivative with respect to x?
MAGDALENA TODA: Right.
So when I write like that, because that's a little bit
confusing, when students ask me, which one is first?
First you do f sub x, and then you do y.
And then f sub y x would be the derivative with respect to y
primed again with respect to x.
Now, let me tell you the good news.
They-- the book doesn't call it any name, because we don't
like to call anybody names.
I'm just kidding.
It's called the Schwartz principle,
or the theorem of Schwartz.
When I told my co-authors, they said, who cares?
Well I care, because I was a student when my professors told
me that this German mathematician made
this discovery, which is so beautiful.
If f is twice differentiable with respect to x and y,
and the partial derivatives-- the second partial
derivatives-- are continuous, then, now in English
it would say it doesn't matter in which order
you differentiate.
The mixed ones are always the same.
Say what?
f sub x y equals f sub y x for every point.
For every-- do you remember what I taught you for every x, y
in the domain.
Or for every x, y where this happens.
So what does this mean?
That means that whether you differentiate
first with respect to x and then with respect to, y, or first
with respect to y and then with respect to x,
it doesn't matter.
The mixed partial derivatives are the same.
Which is wonderful.
I mean, this is one of the best things
that ever happened to us.
Let's see if this is true in our case.
I mean, of course it's true because it's a theorem,
if it weren't true I wouldn't teach it,
but let's verify it on a baby.
Not on a real baby, on a baby example.
Right?
So, f sub x is y plus y equals sine x primed again
with respect to y.
And what do we get out of it?
Cosine of x.
Are you guys with me?
So f sub x was y plus y equals sine x.
Take this guy again, put it here,
squeeze them up a little bit, divide by-- no.
Time with respect to y, x is a constant, what do you think?
Cosine of x, am I right?
STUDENT: 1 plus [INAUDIBLE].
MAGDALENA TODA: That's what it starts with.
Plus [INAUDIBLE].
So cosine of x, [INAUDIBLE] a constant, plus 1.
Another way to have done it is, like, wait a minute,
at this point I go, constant out-- are you with me?--
constant out, prime with respect to y, equals sine x plus 1.
Thank you.
All right.
F sub yx is going to be f sub y.
x plus sine x, but I have to take it from here,
and I prime again with respect to x, and I get the same thing.
I don't know, maybe I'm dyslexic,
I go from the right to the left, what's the matter with me.
Instead of saying 1 plus, I go cosine of x plus 1.
So it's the same thing.
Yes, sir.
STUDENT:I'm looking at the f of xy from the--
MAGDALENA TODA: Which one are you looking at?
Show me.
STUDENT: It's in the purple.
MAGDALENA TODA: It is in the purple.
STUDENT: It's that one right there.
So--
MAGDALENA TODA: This one?
STUDENT: Mmhm.
So, I'm looking at the y plus y cosine x.
You got that from f of x.
MAGDALENA TODA: I got this from f of x,
and I prime it again, with respect to y.
The whole thing.
STUDENT: OK, so you're not writing that as a derivative?
You're just substituting that in for f of x?
MAGDALENA TODA: So, let me write it better,
because I was a little bit rushed, and I don't know,
silly or something.
When I prime this with respect to y--
STUDENT: Then you get the cosine of x plus 1.
MAGDALENA TODA: Yeah.
I could say, I can take out all the constants.
STUDENT: OK.
MAGDALENA TODA: And that constant is this plus 1.
And that's all I'm left with.
Right?
It's the same thing as 1 plus cosine x,
which is a constant times y.
Prime this with respect to y, I get the constant.
It's the same principal as when you have bdy of 7y equals 7.
Right?
OK.
Is this too easy?
I'll give you a nicer function.
I'm imitating the one in WeBWorK [INAUDIBLE]
To make it harder for you.
Nothing I can make at this point is hard for you,
because you're becoming experts in partial differentiation,
and I cannot challenge you on that.
I'm just trying to make it harder for you.
And I'm trying to look up something.
OK, how about that?
This is harder than the ones you have in WeBWorK.
But that was kind of the idea-- that when
you go home, and open those WeBWorK problem sets,
that's a piece of cake.
What we did in class was harder.
When I was a graduate student, one professor said,
the easy examples are the ones that the professor's
supposed to write in class, on the board.
The hard examples are the ones that
are left for the students' homework.
I disagree.
I think it should be the other way around.
So f sub x.
That means bfdx for the pair xy, any xy.
I'm not specifying an x and a y.
I'm not making them a constant.
What am I going to have in this case?
Chain -- if I catch you not knowing the chain rule,
you fail the final.
Not really, but, OK, you get some penalty.
You know it.
Just pay attention to what you do.
I make my own mistakes sometimes.
So 1 over.
What do you do here when you differentiate
with respect to x?
You think, OK, from the outside to the inside, one at a time.
1 over the variable squared plus 1, right?
Whatever that variable, it's like you call variable
of the argument xy, right?
STUDENT: [INAUDIBLE]
MAGDALENA TODA: Square plus 1.
Times-- cover it with your hand-- prime with respect to x.
y, right?
Good!
And you're done.
You see how easy it was.
Just don't forget something because it can cost you points.
Are you guys with me?
So, once we are done with saying, 1 over argument
squared plus 1, I cover this with my hand,
xy prime with respect to 2x is y.
And I'm done.
And here, pause.
What's the easiest way to do that?
You look at it like, she wants me to get
caught in the quotient rule.
She wants to catch me not knowing this rule,
while I can do better.
One way to do it would be numerator prime plus
denominator, minus numerator [INAUDIBLE] What's
the easier way to do it?
STUDENT: x squared plus y squared, all of it
to the negative one.
MAGDALENA TODA: Right.
So you say, hey, you cannot catch me,
I'm the gingerbread man.
Good!
That was a good idea.
Chain rule, and minus 1/2, times--
who tells me what's next?
I'm not going to say a word.
STUDENT: 2x plus y squared.
No, it's 2x.
x squared plus y squared.
MAGDALENA TODA: From the outside to the inside.
From the outside-- to the what?
STUDENT: [INAUDIBLE]
MAGDALENA TODA: Good.
And now I'm done.
I don't see that anymore.
I focus to the core.
2x.
Times 2x.
And that is plenty.
OK, now, let me ask you a question.
What if you would ask a smart kid,
I don't know, somebody who knows that,
can you pose the f sub y of xy without doing the whole thing
all over again?
Can you sort of figure out what it would be?
The beautiful thing about x and y
is that these are symmetric polynomials.
What does it mean, symmetric polynomials?
That means, if you swap x and y, and you swap x and y,
it's the same thing.
Just think of that-- swapping x and y.
Swapping the roles of x and y.
So what do you think you're going to get?
OK, one student said, this is for smart people,
not for people like me.
And I said, well, OK, what's the matter with you?
I'm a hard worker.
I'm the kind of guy who takes the whole thing again, and does
the derivation from scratch.
And thinking back in high school, I think, even
for symmetric polynomials,
I'm sure that being smart and being
able to guess the whole thing-- but I
did the computation many times mechanically,
just in the same way, because I was a hard worker.
So what do you have in that case?
1/xy squared plus 1 times x plus-- the same kind of thing.
Attention, this is the symmetric polynomial, and I go to that.
And then times 2y.
So, see-- that kind of easy, fast thing.
Why is this a good observation when
you have symmetric polynomials?
If you are on the final and you don't have that much time,
or on any kind of exam when you are in a time-crunch.
Now, we want those exams so you are not
going to be in a time-crunch.
If there is something I hate, I hate a final of 2 hours
and a half with 25 serious problems,
and you know nobody can do that.
So, it happens a lot.
I see that-- one of my jobs is also to look at the finals
after people wrote them, and I still
do that every semester-- I see too many people making finals.
The finals are not supposed to be long.
The finals are supposed to be comprehensive, cover
everything, but not extensive.
So maybe you'll have 15 problems that cover practically
the material entirely.
Why?
Because every little problem can have two short questions.
You were done with a section, you
shot half of a chapter only one question.
This is one example just-- not involving [INAUDIBLE]
of an expression like that, no.
That's too time-consuming.
But maybe just tangent of x-squared plus y-squared,
find the partial derivatives.
That's a good exam question, and that's enough
when it comes to testing partials.
By the way, how much-- what is that?
And I'm going to let you go right now.
Use the bathroom.
And when you come back from the bathroom, we'll fill in this.
You know I am horrible in the sense that I want-- I'm greedy.
I need extra time.
I want to use more time.
I will do your problems from now on,
and you can use the bathroom, eat something, wash your hands.
I'll start in about five minutes.
Don't worry.
Alexander?
Are you here?
Come get this.
I apologize.
This is long due back to you.
STUDENT: Oh.
Thank you.
STUDENT: Is there an attendance sheet today?
MAGDALENA TODA: I will-- I'm making up one.
There is already on one side attendance.
Let's use the other side.
Put today's date.
[INAUDIBLE]
[SIDE CONVERSATIONS]
MAGDALENA TODA: They are spoiling me.
They give me new sprays every week.
[INAUDIBLE] take care of this.
[SIDE CONVERSATIONS]
MAGDALENA TODA: So I'm going to ask you something.
And you respond honestly.
Which chapter-- we already browsed through three chapters.
I mean, Chapter 9 was vector spaces,
and it was all review from-- from what?
From Calc 2.
Chapter 10 was curves in [INAUDIBLE] and curves
in space, practically.
And Chapter 11 is functions of several variables.
Now you have a flavor of all of them,
which one was hardest for you?
STUDENT: 9 and 10, both.
MAGDALENA TODA: 9 and 10 both.
STUDENT: [INAUDIBLE].
MAGDALENA TODA: This is so much better than the other.
No, I think you guys actually-- it looks better,
because you've seen a lot more vectors and vector functions.
STUDENT: I didn't understand any of 9 or 10.
STUDENT: [INAUDIBLE].
MAGDALENA TODA: Yes, ma'am.
STUDENT: Could you go over parametrization [INAUDIBLE]?
MAGDALENA TODA: I will go over that again.
And I will go over some other parametrizations today.
And I promised that at the end, in those 20 minutes,
I will do that problem that gave a few of you trouble.
Yes, sir?
STUDENT: Do we take the same final exam
as all the other [INAUDIBLE] classes? [INAUDIBLE]?
MAGDALENA TODA: Well, that's what I was asked yesterday.
So practically, it's at the latitude of the instructor who
teaches honors if they write their own final,
and in general make it harder, or they
take the general final like everybody else.
For your formative purposes, and as a study,
I would like you to take the general final,
because I want to see where you stand compared
to the rest of the population.
So you are my sample, and they are the entire student
population of Calc 3, I want to make
the statistical analysis of your performance compared to them.
STUDENT: So we'll take the regular one?
MAGDALENA TODA: Yeah.
For this one, I just have to make sure
that they also have that extra credit added in.
Because if I have too much extra credit in there,
well they also count that.
So that's what that means.
So we can [INAUDIBLE].
All right.
Let me finish this exercise.
And then [? stop ?] [INAUDIBLE] and go
over some homework problems and some parametrization problems.
And I will see what else.
So tangent of [INAUDIBLE].
Is this hard?
No, it's [INAUDIBLE].
But you have to remind me, because I
pretend that I forgot-- let me pretend
that I forgot what the derivative [INAUDIBLE] notation
of tangent of t was.
STUDENT: Secant squared.
MAGDALENA TODA: You guys love that secant squared thingy.
Why do you like secant squared?
I, as a student, I didn't like expressing it like that.
I liked [INAUDIBLE].
Of course, it's the same thing.
But I always like it like 1 over cosine [INAUDIBLE].
And of course, I have to ask you something,
because I'm curious to see what you remember.
And you say yeah, curiosity killed the cat.
But where did the derivative exist?
Because maybe was that tangent of T--
STUDENT: Wasn't it a quotient rule
of sine and [? cosine x? ?]
MAGDALENA TODA: Good.
I'm proud of you.
That is the answer.
So [? my ?] [? have ?] this blowing up, this blows up--
blows up where cosine T was zero, right?
So where did that blow up?
[INAUDIBLE] blow up of cosine and zero [INAUDIBLE].
The cosine was the shadow on the x-axis.
So here you blow up here, you blow up here, you blow up here,
you blow up here.
So [? what does ?] [INAUDIBLE].
It should not be what?
STUDENT: Pi over 2.
MAGDALENA TODA: Yeah.
And can we express that OK, among 0pi,
let's say you go in between 0 and 2pi only.
I get rid of pi over 2 and 3pi over 2.
But if I express that in general for [INAUDIBLE] T
not restricted to 0 to T, what do I say?
STUDENT: It's k.
STUDENT: So it can [? never be ?] pi over 2
plus pi?
MAGDALENA TODA: 2k plus 1.
2k plus 1.
Odd number over--
STUDENT: Pi over 2.
MAGDALENA TODA: Pi over 2.
Odd number, pi over 2.
And all the odd numbers are 2k plus 1.
Right?
All right.
So you have a not existence and-- OK.
Coming back.
I'm just playing, because we are still in the break.
Now we are ready.
What is dfdx, del f, del x, xy.
And what is del f, del y?
I'm not going to ask you for the second partial derivative.
We've had enough of that.
We also agreed that we have important results in that.
What is the final answer here?
STUDENT: [INAUDIBLE] plus x-squared [INAUDIBLE].
MAGDALENA TODA: 1 over [INAUDIBLE].
I love this one, OK?
Don't tell me what I want to [INAUDIBLE].
I'm just kidding.
[INAUDIBLE] squared times--
STUDENT: 2x.
MAGDALENA TODA: 2x, good.
How about the other one?
The same thing.
Times 2y.
OK.
I want to tell you something that I will repeat.
But you will see it all through the course.
There is a certain notion that Alexander,
who is not talking-- I'm just kidding,
you can talk-- he reminded me of gradient.
We don't talk about gradient until a few sections from now.
But I'd like to anticipate a little bit.
So the gradient of a function, wherever
the partial derivatives exist, with the partial derivative--
that is, f sub x and f sub y exist--
I'm going to have that delta f-- nabla f.
nabla is a [INAUDIBLE].
Nable f at xy represents what?
The vector.
And I know you love vectors.
And that's why I'm going back to the vector notation f sub x
at xy times i, i being the standard vector i
unit along the x axis, f sub y at xy times j.
STUDENT: So it's just like the notation of [INAUDIBLE]?
MAGDALENA TODA: Just the vector notation.
How else could I write it?
Angular bracket, f sub x x at xy, comma, f sub y at xy.
And you know-- people who saw my videos, colleagues
who teach Calc 3 at the same time
said I have a tendency of not going by the book notations
all the time, and just give you the [? round ?] parentheses.
It's OK.
I mean, different books, different notations.
But what I mean is to represent the vector in the standard way
[INAUDIBLE].
All right.
OK.
Can you have this notion for something
like a function of three variables?
Absolutely.
Now I'll give you an easy one.
Suppose that you have x-squared plus y-squared
plus z-squared equals 1.
And that is called-- let's call it names-- f of x, y, z.
Compute the gradient nabla f at any point x, y, z for f.
Find the meaning of that gradient-- of that-- find
the geometric meaning of it.
For this case, not in general, for this case.
So you say, wait, wait, Magdalena.
A-dah-dah, you're confusing me.
This is the gradient.
Hmm.
Depends on how many variables you have.
So you have to show a vector whose coordinates represent
the partial derivatives with respect to all the variables.
If I have n variables, I have f sub x1 comma f sub x2 comma
f sub x3 comma f sub xn, and stop.
Yes, sir.
STUDENT: If the formula was just f of xy,
wouldn't that be implicit?
MAGDALENA TODA: That is implicit.
That's exactly what I meant.
What's the geometric meaning of this animal?
Forget about the left hand side.
I'm going to clean it quickly.
What is that animal?
That is a hippopotamus.
What is that?
STUDENT: It's a sphere.
MAGDALENA TODA: It's a sphere.
But what kind of sphere?
Center 0, 0, 0 with radius 1.
What do we call that?
Unit sphere.
Do you know what notation that mathematicians
use for that object?
You don't know but I'll tell you. s1 is the sphere.
We have s2, I'm sorry, the sphere
of dimension 2, which means the surface.
s1 is the circle.
s1 is a circle.
s2 is a sphere.
So what is this number here for a mathematician?
That's the dimension of that kind of manifold.
So if I have just a circle, we call it s1
because there is only a one independent variable, which
is time, and we parameterize.
Why go clockwise?
Shame on me.
Go counterclockwise.
All right.
That's s1.
For s2, I have two degrees of freedom.
It's a surface.
On earth, what are those two degrees of freedom?
It's a riddle.
No extra credit.
STUDENT: The latitude and longitude?
MAGDALENA TODA: Who said it?
Who said it first?
STUDENT: [INAUDIBLE].
MAGDALENA TODA: How many of you said it at the same time?
Alexander said it.
STUDENT: I know there was one other person.
I wasn't the only one.
STUDENT: I didn't.
STUDENT: [INAUDIBLE], sorry.
[INTERPOSING VOICES]
MAGDALENA TODA: I don't have enough.
STUDENT: I'll take the credit for it.
MAGDALENA TODA: [INAUDIBLE] extra credit.
OK, you choose.
These are good.
They are Valentine's hearts, chocolate [INAUDIBLE].
Wilson.
I heard you saying Wilson.
I have more.
These are cough drops, so I'm [INAUDIBLE].
You set it right next time, Alexander.
STUDENT: [INAUDIBLE].
MAGDALENA TODA: OK.
Anybody else?
Anybody needing cough drops?
OK.
I'll leave them here.
Just let me see.
Do I have more chocolate?
Eh, next time.
I'm going to get some before-- we have-- we
need before Valentine's, right?
So it's Thursday.
I'm going to bring you a lot more.
So in that case, what is the gradient of f?
An x, y, z.
Aha.
I have three variables.
What's the gradient?
I can write it as a bracket, angular notation.
Am I right?
Or I can write it 2xi plus 2ij plus 2zk.
Can anybody tell me why?
What in the world are these, 2x, 2y, 2z?
STUDENT: Those are the partial derivatives.
MAGDALENA TODA: They are exactly the partial derivatives
with respect to x, with respect to y, with respect to z.
Does this have a geometric meaning?
I don't know.
I have to draw.
And maybe when I draw, I get an idea.
Is this a unit vector?
Uh-uh.
It's not.
Nabla s, right.
In a way it is.
It's not a unit vector.
But if I were to [? uniterize ?] it--
and you know very well what it means to [? uniterize it ?].
It means to--
STUDENT: Divide it by--
MAGDALENA TODA: Divide it by its magnitude
and make it a unit vector that would have a meaning.
This is the sphere.
What if I make like this?
n equals nabla f over a magnitude of f.
And what is the meaning of that going to be?
Can you tell me what I'm going to get here?
In your head, compute the magnitude
and divide by the magnitude, and you have exactly 15 seconds
to tell me what it is.
STUDENT: [INAUDIBLE].
MAGDALENA TODA: [? Ryan, ?] [? Ryan, ?]
you are in a Twilight Zone.
But I'm sure once I tell you, once I tell you, [INAUDIBLE].
STUDENT: 1 divided by the square root of 2
for the [? i controller. ?]
STUDENT: [INAUDIBLE].
MAGDALENA TODA: Well, OK.
Say it again, somebody.
STUDENT: x plus y plus z.
MAGDALENA TODA: xi plus yj plus zk, not x plus x, y,
z because that would be a mistake.
It would be a scalar function. [INAUDIBLE] has to be a vector.
If I am to draw this vector, how am I going to draw it?
Well, this is the position vector.
Say it again.
This is the position vector.
When I have a point on this stinking earth, whatever
it is, x, y, z, the position vector is x, y, z.
It's xi plus yj plus zk.
I have this identification between the point
and the vector.
This is our vector.
So I'm going to draw these needles, all these needles,
all these vectors whose tips are exactly on the sphere.
So why?
You say, OK.
I understand that is the position vector,
but why did you put an n here?
And anybody who answers that gets a cough drops.
STUDENT: [INAUDIBLE].
MAGDALENA TODA: Because that is?
STUDENT: The normal to the surface.
MAGDALENA TODA: You get a--
STUDENT: Yeah, cough drop.
MAGDALENA TODA: Two of them.
STUDENT: Aw, yeah.
MAGDALENA TODA: All right.
So that's the normal to the surface, which
would be a continuation of the position vector.
You see, guys?
So imagine you take your position vector.
This is the sphere.
It's like an egg.
And these tips are on the sphere.
If you continue from sitting on the sphere,
another radius vector colinear to that,
that would be the normal to the sphere.
So in topology, we have a name for that.
We call that the hairy ball.
The hairy ball in mathematics, I'm not kidding,
it's a concentrated notations.
You see it in graduate courses, if you're
going to become a graduate student in mathematics,
or you want to do a dual degree or whatever,
you're going to see the hairy ball, all those normal vectors
of length 1.
It's also called the normal field.
So if you ask Dr. Ibragimov, because he
is in this kind of field theory, [INAUDIBLE] normal field
to a surface.
But for the topologists or geometers,
they say, oh, that's the hairy ball.
So if you ask him what the hairy ball is, he will say,
why are you talking nonsense to me?
Right.
Exactly.
So here's where we stopped our intrusion in chapter 11.
It's going to be as fun as it was today
with these partial derivatives.
You're going to love them.
You have a lot of computations like the ones we did today.
Let's go back to something you hated,
which is the parameterizations.
So one of you-- no, three of you--
asked me to redo one problem like the one
with the parameterization of a circle.
But now I have to pay attention to the data
that I come up with.
So write the parameterization of a circle of radius.
Do you want specific data or you want letters?
STUDENT: [INAUDIBLE].
MAGDALENA TODA: OK.
Let's do it [INAUDIBLE] r, and then I'll give an example.
And center x0, y0 in plane where-- what is the point?
Where is the particle moving for time t equals 0?
Where is it located?
All right.
So review.
We had frame that we always picked at the origin.
That was bad because we could pick x0, y0 as a center,
and that has a separate radius.
And now, they want me to write a parameterization of a circle.
How do you achieve it?
You say the circle is x minus x0 squared plus y minus y0 squared
equals r squared.
And one of you asked me by email--
and that was a good question-- you said, come on.
Look, it was [INAUDIBLE].
So you said, I was quite good in math.
I was smart.
Why didn't I know the equations, the parametric equations,
or even this?
I'll tell you why.
This used to be covered in high school.
It's something called college algebra.
We had a chapter, either trigonometry
or college algebra.
We had a chapter called analytic geometry.
This is analytic geometry.
It's the same chapter in which you guys
covered conics, [INAUDIBLE], ellipse, [INAUDIBLE], parabola.
It's no longer covered in most high schools.
I asked around.
The teachers told me that we reduced
the geometric applications a lot,
according to the general standards that are imposed.
That's a pity, because you really need this in college.
All right.
So how do you come up with a parameterization?
You say, I would like to parameterize in such way
that this would be easy to understand
this for Pythagorean theorem.
Oh, OK.
So what is the Pythagorean theorem telling me?
It's telling you that if you are in a unit circle practically,
then this is cosine and theta and this is sine theta,
and the sum of cosine theta squared
plus sine theta squared is 1.
This is 1, so that is the Pythagorean theorem
[INAUDIBLE].
So xy plus x0 should be cosine of theta times an R. Why an R?
Because I want, when I square, I want the R squared up.
And here, this guy inside will be our sine [? thing. ?]
Am I going to be in good shape?
Yes, because when I square this fellow squared
plus this fellow squared will give me exactly R squared.
And here is my [INAUDIBLE] smiley face.
So I want to understand what I'm doing.
x minus x0 must be R cosine theta.
y minus y0 is R sine theta.
Theta in general is an angular velocity, [INAUDIBLE].
But it's also time, right?
It has the meaning of time parameter.
So when we wrote those-- and some of you are bored,
but I think it's not going to harm anybody
that I do this again.
R cosine of t plus x0 y is R sine t plus x0, or plus y0.
Now note, all those examples in web work,
they were not very imaginative.
They didn't mean for you to try other things.
Like if one would put here cosine of 5t or sine of 5t,
that person would move five times faster on the circle.
And instead of being back at 2 pi, in time 2 pi,
they would be there in time 2 pi over 5.
All the examples-- and each of you, it was randomized somehow.
Each of you has a different data set.
Different R, different x0 with 0,
and a different place where the particle is moving.
But no matter what they gave you,
it's a response to the same problem.
And at time t equals 0, you have M. Do
you want me to call it M0?
Yes, from my initial-- M0.
For t equals 0, you're going to have R plus x0.
And for t equals 0, you have y0.
So for example, Ryan had-- Ryan, I don't remember what you had.
You had some where theta R was--
STUDENT: 4 and 8.
MAGDALENA TODA: 7.
You, what did you have?
STUDENT: No, R was 7 and x was 3, y was 1.
MAGDALENA TODA: R was 7 and x0 was--
STUDENT: 3, 1.
MAGDALENA TODA: 3, 1 was x0, y0 so in that case, the point they
gave here was 7 plus 3.
Am I right, Ryan?
You can always check.
I remember.
It was 10 and God knows, and 10 and 1.
So all of the data that you had in that problem
was created so that you have these equations.
And at time 0, you were exactly at the time t equals 0 replaced
the t.
All right.
OK.
STUDENT: What's the M0?
What is--
MAGDALENA TODA: M0 is Magdalena times 0.
I don't know.
I mean, it's the point where you are.
I couldn't come up with a better name.
So I'm going to erase here and I'll
get to another problem, which gave you guys a big headache.
And it's not so hard, but this is the computational problem,
very pretty in itself.
[INAUDIBLE] cosine t i plus e to the 3t sine t j plus e
to the 3tk.
And I think this was more or less in everybody's homework
the same.
There's a position vector given as parameterized form.
So since you love parameterization so much,
I'm going to remind you what that means for x and y and zr.
And what did they want from you?
I forget what number of the problem that was.
They wanted the length of the arc of a curve from t
equals-- I don't know.
STUDENT: 2 to 5.
MAGDALENA TODA: 2 to 5.
Thank you.
[INAUDIBLE] t equals 5.
So this is the beginning and the end of the curve, the beginning
and the end of a curve.
So what is that going to be [INAUDIBLE]?
How does [INAUDIBLE], which we have
to write down 2 to 5 magnitude of r prime at t, dt.
And I don't know.
But I want to review this because-- so what in the world?
Maybe I put this on the midterm or I
make it a little bit easier, but the same what I don't like,
it's time consuming.
But I can give you something a lot easier
that tests the concept, the idea, not
the computational power.
So r prime of t here with a little bit of attention,
of course, most of you computing this correctly.
You are just a little bit scared of what happened after that,
and you should not be scared because now I'll tell you
why you shouldn't be scared.
Chain rule, product rule.
So I have first prime--
STUDENT: 3.
MAGDALENA TODA: 3 into the 3e second and [? time ?]
cosine t plus-- I'm going to do that later.
I know what you're thinking.
STUDENT: e 3t.
MAGDALENA TODA: e to the 3t minus sine.
I'm not worried about this minus now.
I'll take care of that later.
Times i.
Now with your permission-- when you
say, why is she not writing the whole thing in continuation?
Because I don't want to.
No.
Because I want to help you see what's going on.
You do the same kind of stuff for this individual one.
I want to put it right underneath.
If I put it right underneath, it's going to [? agree ?].
Otherwise it's not going to [? agree ?].
E to the 3t times sine t plus e to the 3t cosine t.
You didn't have a problem because you
know how to differentiate.
You started having the problem from this point on.
3 into the 3tk.
The problem came when you were supposed
to identify the coordinates and square them and squeeze them
under the same square root.
And that drove you crazy when you have enough.
Let me put the minus here to make it more obvious what's
going to happen.
When you're going to have problems
like that in differential equations,
you better have the eye for it, [INAUDIBLE].
You should be able to recognize this is like a pattern.
Have you seen the movie A Beautiful Mind?
STUDENT: Yeah.
MAGDALENA TODA: OK, so Nash, when
he was writing with the finger on everything, on the walls
at Princeton, on the window, he was thinking of patterns.
He's actually trying to-- and it's
hard to visualize without drawing,
but this is what most of us recognize all the time when
a mathematician writes down some computations
in a different way.
All we hope for is to get a few steps behind that board
and see a pattern.
And when you do that, you see the pattern.
This is an a minus b and that's an a plus b.
And then you say, OK, if I'm going to square them,
what's going to happen?
When you square an a minus b and you square an a plus b
and you have this giggly guy there-- leave him there.
He's having too much fun.
You actually develop these guys and you put them one
under the other and say wow, what
a beautiful simplification.
When I'm going to add these guys,
this thing in the middle will simply will cancel out,
but the a squared will double and the b squared will double.
And that's the beauty of seeing pattern.
You see how there is something symmetric and magic
in mathematics that make the answer simplified.
And that allows you to compress your equations that originally
seemed to be a mess into something that's
more easily expressed.
So when you're going to compute this r
prime of t magic absolute value of the magnitude, that's
going to be square root of-- instead of writing all the
[INAUDIBLE], I hate writing and rewriting the whole thing
squared plus the whole thing squared plus this squared.
If I love to write so much, I'd be in humanities and not
in mathematics.
So as a mathematician, how am I going to write that?
As a mathematician, I'm going to use some sort of-- like the U
substitution.
So I say, I call this Mr. A, and I call this Mr. B.
And that's A minus B, and that's A plus B.
And that's somebody else.
So when I square the first guy, and I
square the second component, and I square the third component,
and I add them together, I'm going to get what?
Square root of 2A squared plus 2B squared.
Because I know that these are the first two.
This guy squared plus this guy squared
is going to be exactly 2A squared
plus 2B squared, nothing in the middle.
These guys cancel out.
STUDENT: A and B are not the same.
MAGDALENA TODA: Well, yeah, you're right.
Let me call-- you're right, this is the same,
but these are different.
So let me call them A prime plus B prime.
No, that's derivative.
Let me call them C and D-- very good,
thank you-- C squared plus 2CD plus D squared.
But the principle is the same.
So I'm going to have A squared plus C squared.
This goes away.
Why?
Because this times that is the same as this times that.
Say it again.
If we look in the middle, the middle term
will have 3e to the 3t cosine t times e to the 3t sine t.
Middle term here is 3e to the 3t e to the 3t sine and cosine.
So they will cancel out, this and that.
So here I have the sum of the square of A
plus the square of C. And here I'm
going to have the square of B plus the square of D.
OK, now when I square this and that, what do I get?
The beauty of that-- let me write it down then explicitly.
9e to the 3t cosine squared t remains from this guy.
Plus from the square of that, we'll
have 9e to the 3t-- no, just 3, 9 to the 6t, 9 to the 6t sine
squared.
So I take this guy.
I square it.
I take this guy.
I square it.
The middle terms will disappear, thank god.
Then I have this guy, I square it, that guy, I square it,
good.
Plus another parenthesis-- e to the 6t sine squared t plus e
to the 6t cosine squared t.
So even if they don't double because they're not
the same thing, what is the principle
that will make my life easier?
The same pattern of simplification.
What is that same pattern of simplification?
Look at the beauty of this guy and look
at the beauty of this guy.
And then there is something missing,
the happy guy that was quiet because I told him to be quiet.
That's 9e to the 6t.
He was there in the corner.
And you had to square this guy and square this guy
and square this guy and add them on top together.
Now what is the pattern?
The pattern is 9e to the 6t with 9e to the 6t, same guy.
The orange guys-- that's why I love the colors.
Cosine squared cosine squared will be 1.
Another pattern like that, I have e to the 6t, to the 6t,
and the same happy guys sine squared t, sine squared t,
add them together is 1.
So all in all, this mess is not a mess anymore.
So it becomes 9e to the 6t plus e to the 6t plus 9e to the 6t.
Are you guys with me?
All right, now how many e to the 6t's do we have?
9 plus 9 plus 1, 19, square root of 19 e to the 6t.
So when we integrate, we go integral
from 2 to 5 square root of 19.
Kick him out of your life.
He's just making your life harder.
And then you have square root of e to the 6t e to the 3t.
So after you kick the guy out, you
have e to the 3t divided by 3 between t equals 2
and t equals 5.
Actually, I took it right off the WeBWorK problem you had.
So if you type this in your WeBWorK--
you probably already did-- you should get exactly the answer
as being correct.
On the exam, do not expect anything that long.
The idea of simplifying these patterns
by finding the sine cosine, sine squared plus cosine squared is
1, is still going to be there.
But don't expect anything that long.
Also, don't expect-- once you get to this state,
I don't want an answer.
This is the answer.
That's the precise answer.
I don't want any approximation or anything like that.
A few of you did this with a calculator.
Well, you will not have calculators in the final.
You are going to have easy problems.
If you did that with a calculator,
and you truncated your answer later,
and if you were within 0.01 of the correct answer,
you were fine.
But some people approximated too much.
And that's always a problem.
So it's always a good idea to enter something
like that in WeBWorK.
I said I wouldn't do it except in the last 20 minutes.
But I wanted to do something like that.
I want to give you another example, because you love
parametrization so much it just occurred to me that it would
be very, very helpful-- maybe, I don't
know-- to give you another problem similar to this one.
It's not in the book, but it was cooked up
by one of my colleagues for his homework.
So I'd like to show it to you.
e to the t i is a parametrization
of a [INAUDIBLE] space.
Plus e to the minus t j plus square root of 2 tk.
And how do I know?
Well, one of his students came to me
and asked for help with homework.
Well, we don't give help when it comes from another colleague.
So in the end, the student went to the tutoring center.
And the tutoring center helped only in parts.
She came back to me.
So what was the deal here?
Find f prime of t in the most simplified form
and find the absolute value r prime of t
in the most simplified form.
And find the length of the arc of this curve between t
equals 0 and t equals 1.
If this were given by a physicist,
how would that physicist reformulate the problem?
He would say-- he or she-- what is the distance travelled
by the particle between 0 seconds and 1 second?
So how do you write that?
Integral from 0 to 1 of r prime of t [INAUDIBLE].
And you have to do the rest.
So arguably, this is the Chapter 10 review.
It's very useful for the midterm exam.
So although we are just doing this review,
you should not erase it from your memory.
Because I don't like to put surprise problems
on the midterm.
But if you worked a certain type of problem,
you may expect something like that.
Maybe it's different but in the same spirit.
r prime of t, who's going to help me with r prime of t?
This fellow-- e to the t.
And how about that?
Negative e to the negative t.
STUDENT: I thought the arc length was the square root of 1
plus f prime of t squared.
MAGDALENA TODA: For a plane curve.
OK, let me remind you.
If you have a plane curve y equals
f of x, then this thing would become integral from A
to B square root of 1 plus f prime of x dx.
And that, did you do that with your Calc II instructor?
How many of you had Dr. Williams?
That was a wonderful class, wasn't it?
And he taught that.
And of course he was not supposed
to tell you that was the speed of a parametric curve.
If you were to parametrize here, x of t
was t and y of t would be f of t.
He could have told you.
Maybe he told you.
Maybe you don't remember.
OK, let's forget about it.
That was Calc II.
Now, coming back here, I have to list what?
Square root of 2 times t prime is one k.
Who's going to help me compute the speed
and put it in a nice formula?
Well, my god--
STUDENT: [INAUDIBLE]
MAGDALENA TODA: Ahh, you are too smart.
Today you had some what is that called with caffeine
and vitamins and--
STUDENT: You're thinking of Red Bull.
MAGDALENA TODA: I know.
That was very nice.
I try to stay away.
What is that called with the energy booster?
STUDENT: I wouldn't know.
STUDENT: 5-Hour Energy.
MAGDALENA TODA: 5-Hour, OK.
I used to have that.
When I had that, I could anticipate two steps computing.
Just a joke, Alex, don't take it up.
Very good observation.
So Alex saw.
He has a premonition.
He can see two steps in advance.
He said, OK, square that.
You have e to the 2t.
Square this.
The minus doesn't matter.
Plus e to the minus 2t, and square that.
Then he saw patterns.
Because he is the wizard 101 today.
So what is the witchcraft he performed?
Do you see?
Does anybody else see the pattern?
[? Nateesh ?] sees the pattern.
Anybody illuminated?
I didn't see it from the start.
You guys saw it faster than me.
It took me about a minute and a half
when I saw this for the first time.
Is this a perfect square?
Of who?
e to the t plus e to the minus 2 squared
is-- anybody else sees the pattern I don't have candy.
Next time-- Alex, [INAUDIBLE], anybody else?
Do you now see the pattern, e to the 2t plus
e to the minus 2t plus twice the product?
And that's where the student was having the problem.
Where do you see the product?
The product is 1.
The product is 1 doubled.
So you get 2.
So it's indeed exactly the perfect square.
So once-- it was a she.
Once she saw the perfect square, she was so happy.
Because you get square root of the square.
You get e to the t plus e to the minus t.
And that's a trivial thing to integrate that you
have no problem integrating.
It's a positive function, very beautiful.
The professor who gave this was Dr. [INAUDIBLE] from Denmark.
He's one of the best teachers we have.
But he makes up his homework as far as I know.
I think in the sixth edition, this edition,
we actually stole his idea, and we made a problem like that
in the book somewhere.
We doubled the number of problems more or less.
So if you are to compute 0 to 1 of the speed,
what is the speed?
The speed is this beautiful thing.
Because you were able to see the pattern.
If you're not able to see that, do you
realize it's impossible, practically,
for you to integrate by hand?
You have to go to a calculator, Matlab, whatever.
So this is easy.
Why is that easy? e to the t minus e to the minus t at 1
and at 0-- you compare them.
You get at 1 e minus e to the minus 1
minus the fundamental theorem of calc e to the 0 minus
e to the 0.
Well, that's silly.
Why is that silly?
Because I'm going to give it up.
So the answer was e to the minus 1/e.
And she knew what the answer would be.
But she didn't know why.
So she came back to me.
I don't know how the tutoring center helped her figure
out the answer.
But she did not understand the solution.
So I said, I'm not going to take anymore people coming
from Professor [INAUDIBLE].
I was also told it's not OK.
So don't go to another professor with homework coming
for me or the other way around.
Because it's not OK.
But you can go to the tutoring center asking them for hints.
They're open starting 9:00 AM and until around when?
Do you know?
They used to have until 4:00.
But now they're going to work on an extended schedule
until 8:00 PM.
It's going to be something crazy.
Now, the thing is, we want the students to be better,
to do better, to not give up, to be successful,
top one, two, three.
I'm a little bit concerned, but maybe I
shouldn't be, about those hours.
So I don't know if they managed to put a security camera
or not.
But having extended hours may be a problem.
Take advantage of those afternoon hours,
especially if you are busy.
Those late hours will be a big help for you.
Do you know where it is?
Room 106 over there.
Any other questions related to this type of problem
or related to anything else in the material
that maybe I can give you hints on,
at least the hint I'm going to give you?
Sometimes I cannot stop, and I just give the problem away.
I'm not supposed to do that.
Look at your WeBWorK, see what kind of help I can give you.
You still have a little bit of time.
STUDENT: [INAUDIBLE]
MAGDALENA TODA: That's the maximum of what?
It was--
STUDENT: [INAUDIBLE]
MAGDALENA TODA: Was this the problem?
STUDENT: e to the 2x or something like that.
MAGDALENA TODA: Something like that?
I erased it.
STUDENT: You erased that? [INAUDIBLE].
I found an answer.
MAGDALENA TODA: It's very computational I saw.
But before that, I saw that seven of you
guys-- you two also did it.
So I wrote-- you have a brownie waiting for that.
But then I erased it.
STUDENT: You erased the previous one too in the homework one.
MAGDALENA TODA: Because that had a bug in it.
That one, the one in the homework one, had a bug in it.
It only worked for some data.
And for other data it didn't work.
So every time you find a bug, you tell me,
and I will tell the programmer of those problems, who's
really careful.
But one in 1,000 you are bound to find a bug.
And I'm going to give you a chocolate
or something for every bug.
And any other questions?
STUDENT: So are you saying this is too long?
MAGDALENA TODA: Actually, it's very beautiful.
If you have a calculator, it's easier to solve it.
You can do it by hand, write it by hand, also.
But it's a long--
STUDENT: [INAUDIBLE]
MAGDALENA TODA: Right, so let's do it now
for anybody who wants to stay.
You don't have to stay.
So practicing what you do--
[SIDE CONVERSATIONS]