PROFESSOR: We will pick up from where we left.
I hope the attendance will get a little bit better today.
It's not even Friday, it's Thursday night.
So last time we talked a little bit about chapter 13,
we started 13-1.
I wanted to remind you that we revisited the notion of work.
Now, if you notice what the book does,
it doesn't give you any specifics
about the force field.
May the force be with you.
They don't say what kind of animal this f is.
We sort of informally said I'm going to have
some sort of path integral.
And I didn't say what conditions I was assuming about f.
And I just said that r is the position vector.
It's important for us to imagine that is plus c1, what
does that mean, c1?
It means that this function, let's write it R of t, equals.
Let's say we are implying not in space, so we have x of t,
y of t, the parametrization of this position vector.
Of course we wrote that last time as well, we
said x is x of t, y is y of t.
But why I took c1 and not continuous?
Could anybody tell me?
If I'm going to go ahead and differentiate it,
of course I'd like it to be differentiable.
And its derivatives should be continuous.
But that's actually not enough for my purposes.
So if I want R of t to be c1, that's good,
I'm going to smile.
But when we did that in chapter-- was it chapter 10?
It was chapter 10, Erin, am I right?
We assumed this was a regular curve.
A regular curve is not just as differentiable
with the derivative's continuous with respect to time.
x prime of t, y prime of t, both must exist
and must be continuous.
We wanted something else about the velocity.
Do you remember the drunken bug?
The drunken but was fine and he was flying.
As long as he was flying, everything was fine.
When did the drunken bug have a problem?
When the velocity field became 0,
at the instant where the bug lost his velocity, right?
So we said regular means c1 and R prime of t at any value of t
should be different from 0.
We do not allow the particle to stop on it's way.
We don't allow it, whether it is a photon, a drunken bug,
an airplane, or whatever it is.
We don't want it to stop in it's trajectory.
Is that good for other reasons as well?
Very good for the reason that we want
to think later in arc length.
[INAUDIBLE] came up with this idea last time.
I didn't want to tell you the truth, but he was right.
One can define certain path integrals with respect
to s, with respect to arc length parameter.
But as you remember very well, he
had this correspondence between an arbitrary parameter type t
and s, and this is s of t.
And also going back and forth, that
means from s you head back to t.
So here's s of t and this is t of s, right?
So we have this correspondence and everything worked fine
in terms of being able to invert that.
And having some sort of equal morphisms
as long as the velocity was non-0.
OK, do you remember who s of t was?
S of t was defined-- It was a long time ago.
So I'm reminding you s of t was integral
from 0 to t-- or from t0 to t.
Your favorite initial moment in time.
Of the speed, uh-huh, and what the heck was the speed?
The speed was the norm or the length of the R prime of t.
This is called speed, that we assume different from 0,
for a good purpose.
We can go back and forth between t and s, t to s,
s to t, with differentiable functions.
Good, so now we can apply the inverse mapping through them.
We can do all sorts of stuff with that.
On this one we did not quite define it rigorously.
What did they say is?
We said f would be a good enough function,
but know that I do not need f to be c1.
This is too strong, too strong.
So in Calc 1 when you had to integrate a function of one
variable you just assumed that- in Calc 1
I remember- you assume that continuous.
It doesn't even have to be continuous
but let's assume that f would be continuous.
OK, so you have, in one sense, that the composition with R,
if you have f of x of t, y of t, z of t.
In terms of time will be a functions of one variable,
and this will be continuous.
All right?
OK, now what if it's not continuous?
Can't I have a piecewise, continuous function?
Like in Calc 1, do you guys remember we had some of this?
And from here like that and from here like this
and from here like that.
And we had these continuities, and this was piecewise
continuous.
Yeah, for god sake, I can integrate that.
Why do we assume integral of a continuous function?
Just to make our lives easier and also
because we are in freshman and sophomore level Calculus.
If we were in advanced Calculus we would say,
I want this function to be integrable.
This is a lot weaker than continuous,
maybe the set of discontinuities is also very large.
Who told you that you have finitely many jumps
these continuities?
Maybe you have a much larger set.
And this is what you learn in advanced Calculus.
But you are not at the level of a senior
yet so we'll just assume, for the time being,
that f is continuous.
All right, and we say, what is this animal?
We called it w and be baptised it.
We said, just give it some sort of name
and we say that is work.
And by definition, by definition,
this is going to be integral from-- Now,
the thing is, we define this as a simple integral with respect
to time as a definition.
That doesn't mean that I introduced
the notion of path integral the way I should,
I was cheating on that.
So the way we introduced it was like,
let f be a function of the spatial coordinates
in terms of time.
x, y, z are space coordinates, t is time.
So I have f of R of t here.
Dot Product, who the heck is the R?
This is nothing but a vector art drawing.
These are both vectors, sometimes
I should put them in bold like they do in the book.
To make it clear I can put a bar on top of them,
they are free vectors.
So, f of R times R prime of t dt.
And your favorite moments of time are-- let's say on my arc
that I'm describing from time, t, equals a, to time equals b.
Therefore, I'm going to take time for a to b.
And this is how we define the work of a force.
The work of a force that's acting
on a particle that is moving between time, a, and time, b,
on this arc of a curve which is called c.
Do you like this c?
Okay, and the force is different.
So we have a force field.
So I cheated, I knew a lot in the sense
that I didn't tell you how you actually
introduce the path integral.
Now this is more or less where I stop and [INAUDIBLE].
But couldn't we actually introduce this integral
and even define it with respect to some arc length grammar?
Maybe if everything goes fine in terms of theory?
And the answer is yes.
And I'm going to show you how one can do that.
I'm going to go ahead and clean here a little bit.
I'm going to leave this on by comparison for awhile.
And then I will assume something that we have not
defined whatsoever, which is an animal called path integral.
So the path integral of a vector field along a trajectory, c.
I don't know how to draw.
I will draw some skewed curve, how about that?
Some pretty skewed curve, c, it's not self intersecting,
not necessarily.
You guys have to imagine this is like the trajectory
of an airplane in the sky, right?
OK, and I have it on d equals a, to d equals b.
But I said forget about the time, t,
maybe I can do everything in arc length forever.
So if that particle, or airplane,
or whatever it is has a continuous motion,
that's also differentiable.
And the velocity never becomes zero.
Then I can parametrize an arc length
and I can say, forget about it, I have integral over c.
See, this is c, it's not f, okay?
But f of x of s, y of s, z of s, okay?
And this is going to be a ds.
And you'll say, yes Magdelina-- this is little s, I'm sorry.
Yes, Magdelina, but what the heck is this animal,
you've never introduced it.
I have not introduced it because I have to discuss about it.
When we introduced Riemann sums, then we took the limit.
We always have to think how to partition our domains.
So this curve can be partitioned in as many as n, this is s k.
S k, this is s1, and this is s n, the last of the Mohicans.
I have n sub intervals, pieces of the art.
And how am I going to introduce this?
As the limit, if it exists.
Because I can be in trouble, maybe this limit
is not going to exist.
The sum of what?
For every [? seg ?] partition I will take a little arbitrary
point inside the subarc.
Subarc?
STUDENT: Yeah.
PROFESSOR: Subarc, it's a little arc.
Contains a-- let's take it here.
What am I going to define in terms of wind?
s k, y k, and z k, some people put a star on it
to make it obvious.
But I'm going to go ahead and say
x star k, y star k, z star k, is my arbitrary point in the k
subarc.
Times, what shall I multiply by?
A delta sk, and then I take k from one to n
and I press to the limit with respect n.
But actually I could also say in some other ways
that the partitions length goes to 0, delta s goes to zero.
And you say but, now wait a minute,
you have s1, s2 s3, s4, s k, little tiny subarc,
what the heck is delta s?
Delta s is the largest subarc.
So the length of the largest subarc, length of the largest
subarc in the partition.
So the more points I take, the more I refine this.
I take the points closer and closer and closer
in this partition.
What happens to the length of this partition?
It shrinks to-- it goes to 0.
Assuming that this would be the largest
one, well if the largest one goes to 0,
everybody else goes to 0.
So this is a Riemann sum, can we know for sure
that this limit exists?
No, we hope to god that this limit exists.
And if the limit exists then I will introduce this notion
of integral around the back.
And you said, OK I believe you, but look,
what is the connection between the work- the way
you introduced it as a simple Calculus 1 integral here-
and this animal that looks like an alien coming from the sky.
We don't know how to look at it.
Actually guys it's not so bad, you do the same thing
as you did before.
In a sense that, s is connected to any time parameter.
So Mr. ds says, I'm your old friend,
trust me, I know who I am. ds was the speed times dt.
Who can tell me if we are in R three, and we are drunken bugs,
ds will become what?
A long square root times dt, and what's inside here?
I want to see if you guys are awake.
[INTERPOSING VOICES]
PROFESSOR: Very good, x prime of t squared, I'm so lazy
but I'll write it down. y prime of t squared plus z prime of t
squared.
And this is going to be the speed.
So I can always do that, and in this case
this is going to become always some-- let's say from time, t0,
to time t1.
Some in the integrals of-- some of the limit
points for the time.
I'm going to have f of R of s of t,
in the end everything will depend on t.
And this is my face being happy.
It's not part of the integral.
Saying what?
Saying that, guys, if I plug in everything
back in terms of t- I'm more familiar to that type
of integral- then I have what?
Square root of-- that's the arc length element
x prime then t squared, plus y prime then t
squared, plus z prime then t squared, dt.
So in the end it is-- I think the video doesn't see me
but it heard me, presumably.
This is our old friend from Calc 1,
which is the simple integral with respect to t from a to b.
OK, all right, and we believe that the work
can be expressed like that.
I introduced it last time, I even
proved it on some particular cases
last time when Alex wasn't here because, I know why.
Were you sick?
ALEX: I'll talk to you about it later.
I'm a bad person.
PROFESSOR: All right, then I'm dragging an object like,
the f was parallel to the direction of displacement.
And then I said the work would be
the magnitude of f times the magnitude of the displacement.
And then we proved that is just a particular case of this,
we proved that last time, it was a piece of cake.
Actually, we proved the other one.
It proved that if force is going to be oblique and at an angle
theta with the displacement direction, then
the work will be the magnitude of the force times
cosine of theta, times the magnitude of displacement,
all right?
And that was all an application of this beautiful warp formula.
Let's see something more interesting from an application
viewpoint.
Assume that you are looking at the washer,
you are just doing laundry.
And you are looking at this centrifugal force.
We have two forces, one is centripetal towards the center
of the motion, circular motion, one is centrifugal.
I will take a centrifugal force f,
and I will say I want to measure at the work
that this force is producing in the circular motion
of my dryer.
My poor dryer died so I had to buy another one
and it cost me a lot of money.
And I was thinking, such a simple thing,
you pay hundreds of dollars on it
but, anyway, we take some things for granted.
I will take the washer because the washer is a simpler
case in the sense that the motion-- I can assume it's
a circular motion of constant velocity.
And let's say this is the washer.
And centrifugal force is acting here.
Let's call that-- what should it be?
Well, it's continuing the position vector
so let's call that lambda x I, plus lambda y j.
In the sense that it's collinear to the vector that
starts at origin, and here is got to be x of t, y.
X and y are the special components
at any point on my circular motion.
If it's a circular motion I have x
squared plus y squared equals r squared,
where the radius is the radius of my washer.
You have to compute the work produced
by the centrifugal force in one full rotation.
It doesn't matter, I can have infinitely many rotations.
I can have a hundred rotations, I couldn't care less.
But assume that the motion has constant speed.
So if I wanted, I could parametrize in our things
but it doesn't bring a difference.
Because guys, when this speed is already constant,
like for the circular motion you are familiar with.
Or the helicoidal case you are familiar with,
you also saw the case when the speed was constant.
Practically, you were just rescaling the time
to get to your speed, to your time parameter s, arc length.
So whether you work with t, or you work with s,
it's the same thing if the speed is a constant.
So I'm not going to use my imagination to go and do it
with respect to s.
I could, but I couldn't give a damn
because I'm going to have a beautiful t that you
are going to help me recover.
From here, what is the parametrization
that comes to mind?
Can you guys help me?
I know you can after all the review of chapter 10
and-- this is what?
R what?
You should whisper cosine t.
Say it out loud, be proud of what you know.
This is R sine t.
And let's take t between 0 and 2 pi.
One, the revolution only, and then I say, good.
The speed is what?
Speed, square root of x prime the t squared
plus y prime the t squared.
Which is the same as writing R prime of the t in magnitude.
Thank God we know that.
How much is this?
R, very good, this is R, very good.
So life is not so hard, it's-- hopefully I'll be able to do
the w.
What is the w?
It's the path integral all over the circle
I have here, that I traveled counterclockwise from any point
to any point.
Let's say this would be the origin of my motion,
then I go back.
And I have this force, F, that I have
to redistribute in terms of R. So this notation
is giving me a little bit of a headache,
but in reality it's going to be very simple.
This is the dot product, R prime dt, which was the R.
Which some other people asked me, how can you write that?
Well, read the review, R of x equals
x of t, i plus y of t, j.
Also, read the next side plus y j.
It short, the dR differential out of t,
sorry, I'll put R. dR is dx i plus dy j.
And if somebody wants to be expressing
this in terms of speeds, we'll say this is x prime dt,
this is y prime dt.
So we can rewrite this x prime then t i, plus y prime then t
j, dt.
All right?
OK, which is the same thing as R prime of t, dt.
[INAUDIBLE]
This looks awfully theoretical.
I say, I don't like it, I want to put my favorite guys
in the picture.
So I have to think, when I do the dot product
I have the dot product between the vector that
has components f1 and f2.
How am I going to do that?
Well, if I multiply with this guy, dot product, the boss guy,
with this boss guy.
Are you guys with me?
What am I going to do?
First component times first component,
plus second component times second component of a vector.
So I have to be smart and understand how I do that.
Lambda is a constant.
Lambda, you're my friend, you stay there.
x is x of t, x of t, but I multiplied
with the first component here, so I
multiplied by x prime of t.
Plus lambda times y of t, times y prime of t.
And who gets out of the picture is dt at the end.
I have integrate with respect to that dt.
This would be incorrect, why?
Because t has to move between some specific limits
when I specify what a path integral is.
I cannot leave a c-- very good, from 0 to 2 pi, excellent.
Is this hard?
No, It's going to be a piece of cake.
Why is that a piece of cake?
Because I can keep writing.
You actually are faster than me.
STUDENT: Your chain rule is already done for you.
PROFESSOR: Right, and then lambda
gets out just because-- well you remember
you kick the lambda out, right?
And then I've put R cosine t times minus R sine t.
I'm done with who?
I'm done with this fellow and that fellow.
And plus y, R sine t, what is R prime?
R cosine, thank you guys, dt.
And now I'm going to ask you, what is this animal?
Stare at that, what is the integrand?
Is a friend of yours, he's so cute.
He's staring at you and saying you are done.
Why are you done?
What happens to the integrand?
It's zero, it's a blessing, it's zero.
How come it's zero?
Because these two terms simplify, they cancel out.
They cancel out, thank god they cancel out, I got a zero.
So we discover something that a physicist
or a mechanical engineer would have told you already.
And do you think he would have actually plugged
in the path integral?
No, they wouldn't think like this.
He has a simpler explanation for that because he's experienced
with linear experiments.
And says, if I drag this like that I know what to work with.
If I drag in, like at an angle, I
know that I have the magnitude of this,
cosine theta, the angle.
So, he knows for linear cases what we have.
For a circular case he can smell the result
without doing the path integral.
So how do you think the guy, if he's a mechanical engineer,
would think in a second?
Say well, think of your trajectory, right?
It's a circle.
The problem is that centrifugal force being perpendicular
to the circle all the time.
And you say, how can a line be perpendicular to a circle?
It is, it's the normal to the circle.
So when you say this is normal to the circle
you mean it's normal to the tangent of the circle.
So if you measure the angle made by the normal at every point
to the trajectory of a circle, it's always lines.
So he goes, gosh, I got cosine of 90, that's zero.
So if you have some sort of work produced
by something perpendicular to your trajectory,
that must be zero.
So he or she has very good intuition.
Of course, how do we prove it?
We are mathematicians, we prove the path integral,
we got zero for the work, all right?
But he could sense that kind of stuff from the beginning.
Now, there is another example where maybe you don't have
90 degrees for your trajectory.
Well, I'm going to just take-- what if I change the force
and I make a difference problem?
Make it into a different problem.
I will do that later, I won't go and erase it.
Last time we did one that was, compute
the work along a parabola from something to something.
Let's do that again.
For some sort of a nice force field
I'll take your vector valued function to be nice to you.
I'll change it, y i plus x j.
And then we are in plane and we move
along this parabola between 0, 0 and-- what is this guys?
1, 1-- well let make it into a one, it's cute.
And I'd like you to measure the work along the parabola
and also along the arc of a-- along the segment of a line
between the two points.
So I want you to compute w1 along the parabola,
and w2 along this thingy, the segment.
Should it be hard?
No, this was old session for many finals.
I remember, I think it was 2003, 2006, 2008,
and very recently, I think a year and 1/2 ago.
A problem like that was given.
Compute the path integrals correspond
to work for both parametrization and compare them.
Is it hard?
I have no idea, let me think.
For the first one we have parametrization
that we need to distinguish from the other one.
The first parametrization for a parabola,
we discussed it last time, was of course
the simplest possible one you can think of.
And we did this last time but I'm
repeating this because I didn't want Alex to miss that.
And I'm going to say integral from some time to some time.
Now, if I'm between 0 and 1, time of course
will be between 0 and 1 because x is time.
All right, good, that means what else?
This Is f1 and this is f2.
So I'm going to have f1 of t, x prime of t, plus f2 of t,
y prime of t, all the [? sausage ?] times dt.
Is this going to be hard?
Hopefully not, I'm going to have to identify everybody.
Identify this guys prime of t with respect to t
is 1, piece of cake, right?
This fellow is-- you told me last time you got 2t
and you got it right.
This guy, I have to be a little bit careful
because y is the fourth guy.
This is t squared and this is t.
So my integral will be a joke.
0 to 1, 2t squared plus t squared equals 3t squared.
Is it hard to integrate?
No, for God's sake, this is integral-- this
is t cubed between 0 and 1, right, right?
So I should get 1, and if I get, I
think I did it right, if I get the other parametrization you
have to help me write it again.
The parametrization of this straight line
between 0, 0 and 1, 1.
Now on the actual exam, I'm never
going to forgive you if you don't know how to parametrize.
Now you know it but two months ago you didn't, many of you
didn't.
So if somebody gives you 2 points, OK, in plane I ask you,
how do you write that symmetric equation of the line between?
You were a little bit hesitant, now
you shouldn't be hesitant because it's a serious thing.
So how did we write that?
We memorized it. x minus x1, over x2 minus x1
equals y minus y1, over y2 minus y1.
This can also be written as-- you know, guys,
that this over that is the actual slope.
This over that, so it can be written
as a [INAUDIBLE] formula.
It can be written in many ways.
And if we put a, t, we transform it into a parametric equation.
So you should be able, on the final,
to do that for any segment of a line with your eyes closed.
Like, you see the numbers, you plug them in,
you get the parametric equations.
We are nice on the exams because we usually give you
a line that's easy to write.
Like in this case you would have x equals t and y equal,
let's see if you are asleep yet, t.
Why is that?
Because the line that joins 0, 0 and 1, 1 is y equals x.
So y equals x is called also, first bicycle.
It's the old friend of yours from trigonometry,
from Pre-Calc, from algebra, I don't know where, college
algebra.
Alrighty, is this hard to do?
No, it's easier than before. w2 is integral from 0 to 1,
this is t and this is t, this is t and this is t, good.
So we have t times 1 plus t times 1,
it's like a funny, nice, game that's too simple, 2t, 2t.
So the fundamental theorem of Calc
says t squared between 0 and 1, the answer is 1.
Am I surprised?
Look at me, do I look surprised at all
that I got the same answer?
No, I told you a secret last time.
I didn't prove it.
I said that R times happy times.
When depending on the force that is with you,
you have the same work no matter what path
you are taking between a and b.
Between the origin and finish line.
So I'm claiming that if I give you this zig-zag line
and I asked you what-- look, it could be any crazy path
but it has to be a nice differentiable path.
Along this differentiable path, no matter
how you compute the work, that's your business,
I claim I still get 1.
Can you even think why, some of you remember maybe,
the force was key?
STUDENT: It's a conservative force.
PROFESSOR: It had to be good conservative.
Now this is conservative but why is that conservative?
What the heck is a conservative force?
So let's write it down on the-- we
say that the vector valued function, f,
valued in R2 or R3, is conservative if there exists
a smooth function.
Little f, it actually has to be just c1,
called scalar potential.
Called scalar potential, such that big F as a vector
field will be not little f.
That means it will be the gradient of the scalar
potential.
Definition, that was the definition,
and then criterion for a f in R2 to be conservative.
I claim that f equals f1 i, plus f two eyes, no, f2 j.
I'm just making silly puns, I don't
know if you guys follow me.
If and only if f sub 1 prime, with respect to y,
is f sub 2 prime, with respect to x.
Can I prove this?
Prove, prove Magdelina, don't just stare at it, prove.
Why would that be necessary and sufficient?
Well, for big F to be conservative
it means that it has to be the gradient
of some little function, little f, some scalar potential.
Alrighty, so let me write it down, proof.
f conservative if and only if there
exists f, such that gradient of f is F. If and only
if-- what does it mean about f1 and f2?
f1 and f2 are the f sub of x and the f sub
y of some scalar potential.
So if f is the gradient, that means that the first component
has to be little f sub x And the second component should
have to be little f sub y.
But that is if and only if f sub 1 prime, with respect to y,
is the same as f sub 2 prime, with respect to x.
What is that?
The red thing here is called a compatibility condition
of this system.
This is a system of two OD's.
You are going to study ODEs in 3350.
And you are going to remember this
and say, Oh, I know that because she taught me that in Calc 3.
Not all instructors will teach you this in Calc 3.
Some of them fool you and skip this material
that's very important to understand in 3350.
So guys, what's going to happened when you prime this
with respect to y?
You get f sub x prime, with respect to y.
When you prime this with respect to x
you get f sub y prime, with respect to x.
Why are they the same thing?
I'm going to remind you that they
are the same thing for a smooth function.
Who said that?
A crazy German mathematician whose name was Schwartz.
Which means black, that's what I'm painting it in black.
Because is the Schwartz guy, the first criterion
saying that no matter in what order
you differentiate the smooth function you
get the same answer for the mixed derivative.
So you see we prove if and only if that you
have to have this criterion, otherwise
it's not going to be conservative.
So I'm asking you, for your old friend, f
equals- example one or example two,
I don't know- y i plus x j.
Is the conservative?
You can prove it in two ways.
Prove in two differently ways that it is conservative.
a, find the criteria.
What does this criteria say?
Take your first component, prime it with respect to y.
So y prime with respect to y.
Take your second component, x, prime it with respect to x.
Is this true?
Yes, and this is me, happy that it's true.
So this is 1 equals 1, so it's true.
So it must be conservative, so it must be conservative.
Could I have done it another way?
By definition, by definition, to prove that a force field
is conservative by definition, that it a matter
of the smart people.
There are people who- unlike me when
I was 18- are able to see the scalar potential in just
about any problem I give them.
I'm not going to make this experiment with a bunch of you
and I'm going to reward you for the correct answers.
But, could anybody see the existence
of the scalar potential?
So these there, this exists.
That's there exists a little f scalar potential such
that nabla f equals F.
And some of you may see it and say, I see it.
So, can you see a little function
f scalar function so that f sub of x i is y
and f sub- Magdelena-- x of y j is this x j?
STUDENT: [INAUDIBLE]?
PROFESSOR: No, you need to drink some coffee first.
You can get this, x times what?
Why is that?
I'll teach you how to get it.
Nevertheless, there are some people
who can do it with their naked eye
because they have a little computer in their head.
But how did I do it?
It's just a matter of experience, I said,
if I take f to be x y, I sort of guessed it.
f sub x would be y and f sub y will be x so this should be it,
and this is going to do.
And f is a nice function, polynomial in two variables,
it's a smooth function.
I'm very happy I'm over the domain,
open this or whatever, open domain in plane.
I'm very happy, I have no problem with it.
So I can know that this is conservative in two ways.
Either I get to the source of the problem
and I find the little scalar potential whose
gradient is my force field.
Or I can verify the criterion and I say,
the derivative of this with respect to y
is the derivative of this with respect
to x is-- one is the same.
The same thing, you're going to see it again in math 3350.
All right, that I taught many times,
I'm not going to teach that in the fall.
But I know of some very good people
who teach that in the fall.
In any case, they would reteach it to you
because good teachers don't assume that you know much.
But when you will see it you'll remember me.
Hopefully fondly, not cursing me or anything, right?
OK, how do we actually get to compute f by hand
if we're not experienced enough to guess it like I was
experienced enough to guess?
So let me show you how you solve a system of two differential
equations like that.
So how I got-- how you are supposed
to get the scalar potential.
f sub x equals F1, f sub y equals F2.
So by integration, 1 and 2.
And you say, what you mean 1 and 2?
I'll show you in a second.
So for my case, example 2, I'll take
my f sub x must be y, right?
Good.
My f sub y must be x, right?
Right.
Who is f?
Solve this property.
Oh, I have to start integrating from the first guy.
What kind of information am I going to squeeze?
I'm going to say I have to go backwards,
I have to get-- f is going to be what?
Integral of y with respect to x, say it again, Magdelina.
Integral of y with respect to x, but attention,
this may come because, for me, the variable is x here,
and y is like, you cannot stay in this picture.
So I have a constant c that depends on y.
Say what?
Yes, because if you go backwards and prime this with respect
to x, what do you get? f sub x will
be y because this is the anti-derivative.
Plus this prime with respect to x, zero.
So this c of y may, a little bit, ruin your plans.
I've had students who forgot about it
and then they got in trouble because they couldn't get
the scalar potential correctly.
All right?
OK, so from this one you say, OK I have some--
what is the integral of y dx?
xy, plus some guy c constant that depends on y.
From this fellow I go, but I have
to verify the second condition, if I don't I'm dead meat.
There are two coupled equations, these
are coupled equations that have to be
verified at the same time.
So f sub y will be prime with respect to y. x plus prime
with respect to y. c prime of y, God gave me x here.
So I'm really lucky in that sense that c prime of y
will be 0 because I have an x here and an x here.
So c of y will simply be any constant k. c of y
is just a constant k, it's not going to depend on y,
it's a constant k.
So my answer was not correct.
The best answer would have been f of xy must be xy plus k.
But any function like xy will work, I just need one to work.
I just need a scalar potential, not all of them.
This will work, x2, xy plus 7 will work, xy plus 3 will work,
xy minus 1,033,045 will work.
But I only need one so I'll take xy.
Now that I trained your mind a little bit,
maybe you don't need to actually solve
the system because your brain wasn't ready before.
But you'd be amazed, we are very trainable people.
And in the process of doing something completely new,
we are learning.
And your brain next, will say, I think
I know how to function a little bit backwards.
And try to integrate and see and guess a potential
because it's not so hard.
So let me give you example 3.
Somebody give you over a domain in plane x i plus y j,
and says, over D, simply connected domain in plane,
open, doesn't matter.
Is this conservative?
Find a scalar potential.
This is again, we do section 13-2,
so today we did 13-1 and 13-2 jointly.
Find the scalar potential.
Do you see it now?
STUDENT: [INAUDIBLE]?
PROFESSOR: Excellent, we teach now, got it.
He says, I know where this comes from.
I've got it, x squared plus y squared over 2.
How did he do it?
He's a genius.
No he's not, he's just learning from the first time
when he failed.
And now he knows what he has to do and his brain says,
oh, I got it.
Now, [INAUDIBLE] could have applied this method
and solved the coupled system and do it slowly
and it would have taken him another 10 minutes.
And he's in the final, he doesn't have time to spare.
If he can guess the potential and then verify that,
it's going to be easy for him.
Why is that?
This is going to be 2x over 2 x i, and this is 2y over 2 y j.
So yeah, he was right.
All right, let me give you another one.
Let's see who gets this one.
F is a vector valued function, maybe a force field,
that is this.
Of course there are many ways-- maybe somebody's
going to ask you to prove this is
conservative by the criterion, but they shouldn't tell you
how to do it.
So show this is conservative.
If somebody doesn't want the scalar potential because they
don't need it, let's say.
Well, prime f1 with respect to y, I'll prime this with respect
to x.
f1 prime with respect to y equals 2x is the same
as f2 prime with respect with.
Yeah, it is conservative, I know it from the criterion.
But [INAUDIBLE] knows that later I will ask him
for the scalar potential.
And I wonder if he can find it for me without computing it
by solving the system.
Just from his mathematical intuition
that is running in the background of your--
STUDENT: x squared, multiply y [INAUDIBLE].
PROFESSOR: x squared y, excellent.
Zach came up with it and anybody else?
Alex?
So all three of you, OK?
Squared y, very good.
Was it hard?
Yeah, it's hard for most people.
It was hard for me when I first saw
that in the first 30 minutes of becoming familiar
with the scalar potential, I was 18 or 19.
But then I got it in about half an hour
and I was able to do them mentally.
Most of the examples I got were really nice.
Were on purpose made nice for us for the exam to work fast.
And now let's see why would the work really not
depend on the trajectory you are taking
if your force is conservative.
If the force is conservative there
is something magic that's going to happen.
And we really don't know what that is,
but we should be able to prove.
So theorem, actually this is funny.
It's called the fundamental theorem of path integrals
but it's the fundamental theorem of calculus 3.
I'm going to write it like this, the fundamental theorem
of calc 3, path integrals.
It's also called- 13.3, section- Independence of path.
So remember you have a work, w, over a path, c.
F dot dR where there R is the regular parametrized curve
overseen.
This is called a supposition vector.
Regular meaning c1, and never vanishing speed,
the velocity never vanishes.
Velocity times 0 such that f is continuous,
or a nice enough integral.
If F is conservative of scalar potential, little f,
then the work, w, equals little f at the endpoint
minus little f at the origin.
Where, by origin and endpoint are those for the path,
are those for the arc, are those for the curve, c.
So the work, the w, will be independent of time.
So w will be independent of f.
And you saw an example when I took
a conservative function that was really nice,
y times i plus x times j.
That was the force field.
Because that was conservative, we got w being 1
no matter what path we took.
We took a parabola, we took a straight line,
and we could have taken a zig-zag
and we still get w equals 1.
So no matter what path you are taking.
Can we prove this?
Well, regular classes don't prove anything, almost nothing.
But we are honor students so lets see what we can do.
We have to understand what's going on.
Why do we have this fundamental theorem of calculus 3?
The work, w, can be expressed-- assume f is conservative
which means it's going to come from a potential little f.
Where f is [INAUDIBLE] scalar function over my domain, omega.
Now, the curve, c, is part of this omega
so I don't have any problems on the curve.
w will be rewritten beautifully.
So I'm giving you a sketch of a proof.
But you would be able to do this, maybe even better than me
because I have taught you what you need to do.
So this is going to be f1 i, plus f2 j.
And I'm going to write it. f1 times-- what is this guys?
dR, I taught you, you taught me, x prime of t, right?
Plus f2 times y prime of t, all dt, and time from t0 to t1.
I start my motion along the curve at t equals t0
and I finished my motion at t equals t1.
Do I know where f1 and f2 are?
This is the point, that's the whole point,
I know who they are, thank God.
And now I have to again apply some magical think,
I'll ask you in a minute what that is.
So, what is f1?
df dx, or f sub base.
If you don't like f sub base, if you don't like my notation,
you put f sub x, right?
And this df dy.
Why?
Because it's conservative and that
was the gradient of little f.
Of course I'm using the fact that the first component would
be the partial of little f with respect to x.
The second component would be the partial of little
f with respect to y.
Have you seen this formula before?
What in the world is this formula?
STUDENT: It's the chain rule?
PROFESSOR: It's the chain rule.
I don't have a dollar but I will give you a dollar, OK?
Imagine a virtual dollar.
This is the chain rule.
So by the chain rule we can write
this to be the derivative with respect
to t of little f of x of t, and y of t.
Alrighty, so I know what I'm doing.
I know that by chain rule I had little f evaluated at x of t,
y of t, and time t, prime with respect to t.
Now when we take the fundamental theorem of calculus, FTC.
That reminds me, I was teaching calc 1 a few years ago
and I said, that's the Federal Trade Commission.
Federal Trade Commission, fundamental theorem
of calculus.
So coming back to what I have, I prove
that w is the Federal Trade Commission, no.
w is the application of something
that we knew from calc 1, which is beautiful.
f of xt, y of t dt, this is nothing but what?
Little f evaluated at-- I'm going to have to write it down,
this whole sausage.
f of x of t1, y of t1, minus f of x of t0, y of t0.
For somebody as lazy as I am, that they effort.
How can I write It better?
f at the endpoint minus f at the origin.
And of course, we are trying to be quite rigorous in the book.
We would never say that in the book.
We actually denote the first point
with p, the origin, and the endpoint with q.
So we say, f of q minus f of p.
And we proved q e d, we proved the fundamental theorem
of path integrals, the independence of that.
So that means the work is independent of path
when the force is conservative.
Now attention, if f is not conservative you are dead meat.
You cannot say what I just said.
So I'll give you two separate examples
and let's see how we solve each of them.
A final exam type of problem-- every final exam
contains an application like that.
Even the force field, f, or the vector value, f.
Is it conservative?
Prove what is proved and after that-- so the path
integral in any way you can.
If it's conservative you're really lucky because you're
in business.
You don't have to do any work.
You just find the little scalar potential evaluated
at the endpoints and subtract, and that's your answer.
So I'm going to give you an example of a final exam problem
that happened in the past year.
So, a final type exam problem.
f of xy equals 2xy i plus x squared j,
over r over r squared.
STUDENT: Didn't we just do [INAUDIBLE] that?
PROFESSOR: Well, I just did that but I changed the problem.
I wanted to keep the same force field.
STUDENT: Alright.
PROFESSOR: OK.
Compute the work, w, performed by f along the arc
of the circle in the picture.
And they draw a picture.
And they do a picture for you, and you stare at this picture
and-- So you say, oh my God, if I
were to parametrize it would be a little bit of-- I
could, but it would be a little bit of work.
I would have x equals 2 cosine t, y equals sine t.
I would have to plug in and do that whole work, definition
with parametrization.
Do you have to parametrize?
Not in this case, why?
Because the f is conservative.
If they ask you- some professors give hints, most of them
are nice and give hints- show f is conservative.
So that's a big hint in the sense
that you see it immediately, how you do it.
You have 2xy prime with respect to y,
is 2x, which is x squared prime with respect to x.
So it is conservative.
But he or she told you more.
He said, I'm selling you something here,
you have to get your own scalar potential.
And you did, and you got x squared y.
Now, most of the scalar potentials
that we are giving you on the exam
can be seen with naked eyes.
You wouldn't have to do all the integration of that
coupled system with respect to x, with respect to y,
integrate backwards, and things like that.
What do I need to do in that case guys?
Say in words.
Since the force is conservative, just two lines.
I'm applying the fundamental theorem of calculus.
I'm applying the fundamental theorem of path integrals.
I know the work is independent of that.
So w, in this case is already there,
is going to be f at the point.
I didn't say how I'm going to travel it, in what direction.
f of q minus f of p.
And you'll say, well what does it mean?
How do we do that?
That means x squared y, evaluated at q,
who the heck is q?
Attention, negative 2, 0, Matt, you got it?
OK, right?
So, are you guys with me?
Right?
And p is 2, 0.
So at negative 2, 0, minus x squared y at 2, 0.
So, if you set 0, big, as you knew that you got 0,
that is the answer.
Now if somebody would give you a wiggly look that this guy's
wrong, then he took this past.
He's going to do exactly the same work
if he's under influence the same conservative force.
If the force acting on it is the same.
No matter what path you are taking-- yes, sir?
STUDENT: Is it still working for your self-intersecting.
PROFESSOR: Yeah, because you're not stopping,
it works for any parametrization.
So if you're able to parametrize that as a differentiable
function so that the derivative would never vanish,
it's going to work, right?
All right, it can work also for this piecewise contour
or any other path point.
As long as it starts and it ends at the same point
and as long as your conservative force is the same.
So that force is very [INAUDIBLE], yes?
STUDENT: Do [INAUDIBLE] graphs but the endpoints the same.
And if they are conservative then [INTERPOSING VOICES].
PROFESSOR: If everything is conservative
the work along this path is the same.
There were people who played games
like that to catch if the student knows what
they are talking-- yes, sir?
STUDENT: So, in a problem, if they wanted
to find the work couldn't we simplify it
by just saying, we need to find-- because,
since we're in R2, couldn't we just say it's a straight line?
Because that's the-- Like, instead of a curve
we could just set the straight line [INTERPOSING VOICES].
PROFESSOR: If it were moving from here
to here in a straight line you would still
get the same answer.
And you could have-- if you did that, actually,
if you were to compute this kind of work on a straight line
it has-- let me show you something.
You see, when you compute dx y dx, plus x squared dy.
If y is 0 like Matthew said, I'm walking on-- I'm not drunk.
I'm walking straight.
y will be 0 here, and 0 here, and the integral will be 0.
So he would have noticed that from the beginning.
But unless you know the force is conservative,
there is no guarantee that on another path
you don't have a different answer, right?
So, let me give you another example because now Matthew
brought this up.
A catchy example that a professor
gave just to make the students life miserable,
and I'll show it to you in a second.
He said, for the picture-- very similar to this one,
just to make people confused.
Somebody gives you this arc of a circle
and you travel from a to b.
And this the thing.
And he says, compute w for the force given by y i plus j.
And the students said OK, I guess
I'm going to get 0 because I'm going to get something
like y dx plus x dy.
And if y is 0, I get 0, and that way you wouldn't be 0
and I'm done.
No, it's not how you think because this is not
conservative.
So you cannot say I can change my path and it's still going
to be the same.
No, why is this not conservative?
Quickly, this prime with respect to y
is 1, this prime with respect to x is 0.
So 1 different from 0 so, oh my God, no.
In that case, why do we do?
We have no other choice but say, x equals 2 cosine t,
y equals 2 sine t, and t between 0 and pi.
And then I get integral of f1, y, what the heck is y?
2 sine t, times x prime of t.
Yeah, minus 2 sine t, this is x prime.
Plus 1, are you guys with me?
Times y prime, which is 2 cosine t, dt.
And t between 0 and pi.
And you get something ugly, you get 0 to pi.
What is the nice thing?
When you integrate this with respect to t, you get sine t.
And thank God, sine, whether you are at 0 or if pi is still 0.
So this part will disappear.
So all you have left is minus 4 sine squared dt.
But you are not done so compute this at home
because we are out of time.
So don't jump to conclusions unless you know
that the force is conservative.
If your force is not conservative
then things are going to look very ugly
and your only chance is to go back to the parametrization,
to the basics.
So we are practically done with 13.3
but I want to watch more examples next time.
And I'll send you the homework.
Over the weekend you would be able to start doing homework.
Now, when shall I grab the homework?
What if I closed it right before the final, is that?
STUDENT: Yeah.
PROFESSOR: Yeah?