0:00:00.000,0:00:00.499


0:00:00.499,0:00:04.934
PROFESSOR: We will pick[br]up from where we left.

0:00:04.934,0:00:08.406
I hope the attendance will[br]get a little bit better today.

0:00:08.406,0:00:13.862
It's not even Friday,[br]it's Thursday night.

0:00:13.862,0:00:17.334
So last time we talked a[br]little bit about chapter 13,

0:00:17.334,0:00:21.302
we started 13-1.

0:00:21.302,0:00:27.750
I wanted to remind you that we[br]revisited the notion of work.

0:00:27.750,0:00:40.700


0:00:40.700,0:00:44.226
Now, if you notice[br]what the book does,

0:00:44.226,0:00:45.980
it doesn't give[br]you any specifics

0:00:45.980,0:00:48.860
about the force field.

0:00:48.860,0:00:50.550
May the force be with you.

0:00:50.550,0:00:56.031
They don't say what kind[br]of animal this f is.

0:00:56.031,0:01:02.570
We sort of informally[br]said I'm going to have

0:01:02.570,0:01:06.290
some sort of path integral.

0:01:06.290,0:01:10.880
And I didn't say what conditions[br]I was assuming about f.

0:01:10.880,0:01:15.030
And I just said that r[br]is the position vector.

0:01:15.030,0:01:18.170


0:01:18.170,0:01:22.825
It's important for us to[br]imagine that is plus c1, what

0:01:22.825,0:01:24.290
does that mean, c1?

0:01:24.290,0:01:30.200
It means that this function,[br]let's write it R of t, equals.

0:01:30.200,0:01:35.360
Let's say we are implying not[br]in space, so we have x of t,

0:01:35.360,0:01:40.450
y of t, the parametrization[br]of this position vector.

0:01:40.450,0:01:42.900
Of course we wrote that[br]last time as well, we

0:01:42.900,0:01:46.280
said x is x of t, y is y of t.

0:01:46.280,0:01:51.640
But why I took c1[br]and not continuous?

0:01:51.640,0:01:53.240
Could anybody tell me?

0:01:53.240,0:01:56.290
If I'm going to go ahead[br]and differentiate it,

0:01:56.290,0:01:59.290
of course I'd like it[br]to be differentiable.

0:01:59.290,0:02:03.520
And its derivatives[br]should be continuous.

0:02:03.520,0:02:07.660
But that's actually not[br]enough for my purposes.

0:02:07.660,0:02:11.170
So if I want R of t[br]to be c1, that's good,

0:02:11.170,0:02:13.150
I'm going to smile.

0:02:13.150,0:02:18.082
But when we did that in[br]chapter-- was it chapter 10?

0:02:18.082,0:02:20.560
It was chapter 10,[br]Erin, am I right?

0:02:20.560,0:02:23.615
We assumed this was[br]a regular curve.

0:02:23.615,0:02:27.650
A regular curve is not[br]just as differentiable

0:02:27.650,0:02:31.390
with the derivative's[br]continuous with respect to time.

0:02:31.390,0:02:34.380
x prime of t, y prime[br]of t, both must exist

0:02:34.380,0:02:36.030
and must be continuous.

0:02:36.030,0:02:39.850
We wanted something[br]else about the velocity.

0:02:39.850,0:02:42.200
Do you remember the drunken bug?

0:02:42.200,0:02:46.520
The drunken but was[br]fine and he was flying.

0:02:46.520,0:02:49.820
As long as he was flying,[br]everything was fine.

0:02:49.820,0:02:54.080
When did the drunken[br]bug have a problem?

0:02:54.080,0:02:56.895
When the velocity[br]field became 0,

0:02:56.895,0:03:02.860
at the instant where the bug[br]lost his velocity, right?

0:03:02.860,0:03:10.740
So we said regular means c1 and[br]R prime of t at any value of t

0:03:10.740,0:03:13.520
should be different from 0.

0:03:13.520,0:03:16.620
We do not allow the particle[br]to stop on it's way.

0:03:16.620,0:03:20.680
We don't allow it, whether it[br]is a photon, a drunken bug,

0:03:20.680,0:03:23.690
an airplane, or whatever it is.

0:03:23.690,0:03:27.570
We don't want it to[br]stop in it's trajectory.

0:03:27.570,0:03:31.990
Is that good for[br]other reasons as well?

0:03:31.990,0:03:35.260
Very good for the[br]reason that we want

0:03:35.260,0:03:39.150
to think later in arc length.

0:03:39.150,0:03:43.320
[INAUDIBLE] came up with[br]this idea last time.

0:03:43.320,0:03:46.480
I didn't want to tell you[br]the truth, but he was right.

0:03:46.480,0:03:53.010
One can define certain[br]path integrals with respect

0:03:53.010,0:03:56.540
to s, with respect to[br]arc length parameter.

0:03:56.540,0:03:58.400
But as you remember[br]very well, he

0:03:58.400,0:04:03.260
had this correspondence between[br]an arbitrary parameter type t

0:04:03.260,0:04:07.641
and s, and this is s of t.

0:04:07.641,0:04:10.200
And also going back[br]and forth, that

0:04:10.200,0:04:15.370
means from s you head back to t.

0:04:15.370,0:04:19.820
So here's s of t and[br]this is t of s, right?

0:04:19.820,0:04:26.500
So we have this correspondence[br]and everything worked fine

0:04:26.500,0:04:31.450
in terms of being[br]able to invert that.

0:04:31.450,0:04:34.320
And having some sort[br]of equal morphisms

0:04:34.320,0:04:40.620
as long as the[br]velocity was non-0.

0:04:40.620,0:04:43.720
OK, do you remember[br]who s of t was?

0:04:43.720,0:04:47.900
S of t was defined--[br]It was a long time ago.

0:04:47.900,0:04:50.710
So I'm reminding you[br]s of t was integral

0:04:50.710,0:04:55.350
from 0 to t-- or from t0 to t.

0:04:55.350,0:04:59.200
Your favorite initial[br]moment in time.

0:04:59.200,0:05:03.570
Of the speed, uh-huh, and[br]what the heck was the speed?

0:05:03.570,0:05:10.660
The speed was the norm or the[br]length of the R prime of t.

0:05:10.660,0:05:16.646
This is called speed, that[br]we assume different from 0,

0:05:16.646,0:05:19.160
for a good purpose.

0:05:19.160,0:05:24.600
We can go back and forth[br]between t and s, t to s,

0:05:24.600,0:05:30.440
s to t, with[br]differentiable functions.

0:05:30.440,0:05:35.300
Good, so now we can apply the[br]inverse mapping through them.

0:05:35.300,0:05:38.170
We can do all sorts[br]of stuff with that.

0:05:38.170,0:05:42.430
On this one we did not[br]quite define it rigorously.

0:05:42.430,0:05:43.690
What did they say is?

0:05:43.690,0:05:46.490
We said f would be a[br]good enough function,

0:05:46.490,0:05:50.330
but know that I do[br]not need f to be c1.

0:05:50.330,0:05:54.424
This is too strong, too strong.

0:05:54.424,0:05:58.870
So in Calc 1 when you had to[br]integrate a function of one

0:05:58.870,0:06:02.972
variable you just[br]assumed that- in Calc 1

0:06:02.972,0:06:04.700
I remember- you assume[br]that continuous.

0:06:04.700,0:06:07.580
It doesn't even have[br]to be continuous

0:06:07.580,0:06:09.862
but let's assume that[br]f would be continuous.

0:06:09.862,0:06:15.650


0:06:15.650,0:06:21.105
OK, so you have, in one sense,[br]that the composition with R,

0:06:21.105,0:06:27.006
if you have f of x[br]of t, y of t, z of t.

0:06:27.006,0:06:33.600
In terms of time will be a[br]functions of one variable,

0:06:33.600,0:06:35.580
and this will be continuous.

0:06:35.580,0:06:39.060


0:06:39.060,0:06:41.050
All right?

0:06:41.050,0:06:44.610
OK, now what if[br]it's not continuous?

0:06:44.610,0:06:47.850
Can't I have a piecewise,[br]continuous function?

0:06:47.850,0:06:51.425
Like in Calc 1, do you guys[br]remember we had some of this?

0:06:51.425,0:06:54.170
And from here like that[br]and from here like this

0:06:54.170,0:06:55.170
and from here like that.

0:06:55.170,0:06:58.000
And we had these continuities,[br]and this was piecewise

0:06:58.000,0:06:59.860
continuous.

0:06:59.860,0:07:02.660
Yeah, for god sake,[br]I can integrate that.

0:07:02.660,0:07:06.530
Why do we assume integral[br]of a continuous function?

0:07:06.530,0:07:08.680
Just to make our[br]lives easier and also

0:07:08.680,0:07:12.500
because we are in freshman[br]and sophomore level Calculus.

0:07:12.500,0:07:16.690
If we were in advanced[br]Calculus we would say,

0:07:16.690,0:07:21.650
I want this function[br]to be integrable.

0:07:21.650,0:07:24.540
This is a lot weaker[br]than continuous,

0:07:24.540,0:07:28.360
maybe the set of discontinuities[br]is also very large.

0:07:28.360,0:07:31.510
Who told you that you[br]have finitely many jumps

0:07:31.510,0:07:32.590
these continuities?

0:07:32.590,0:07:34.530
Maybe you have a[br]much larger set.

0:07:34.530,0:07:39.070
And this is what you learn[br]in advanced Calculus.

0:07:39.070,0:07:41.710
But you are not at[br]the level of a senior

0:07:41.710,0:07:44.880
yet so we'll just assume,[br]for the time being,

0:07:44.880,0:07:47.070
that f is continuous.

0:07:47.070,0:07:49.480
All right, and we say,[br]what is this animal?

0:07:49.480,0:07:52.330
We called it w and[br]be baptised it.

0:07:52.330,0:07:56.880
We said, just give[br]it some sort of name

0:07:56.880,0:07:58.550
and we say that is work.

0:07:58.550,0:08:02.600
And by definition,[br]by definition,

0:08:02.600,0:08:09.375
this is going to be[br]integral from-- Now,

0:08:09.375,0:08:14.890
the thing is, we define this as[br]a simple integral with respect

0:08:14.890,0:08:17.900
to time as a definition.

0:08:17.900,0:08:19.830
That doesn't mean[br]that I introduced

0:08:19.830,0:08:24.660
the notion of path[br]integral the way I should,

0:08:24.660,0:08:26.340
I was cheating on that.

0:08:26.340,0:08:28.550
So the way we[br]introduced it was like,

0:08:28.550,0:08:33.100
let f be a function of[br]the spatial coordinates

0:08:33.100,0:08:34.650
in terms of time.

0:08:34.650,0:08:37.789
x, y, z are space[br]coordinates, t is time.

0:08:37.789,0:08:42.068
So I have f of R of t here.

0:08:42.068,0:08:45.443
Dot Product, who[br]the heck is the R?

0:08:45.443,0:08:48.280
This is nothing but[br]a vector art drawing.

0:08:48.280,0:08:50.510
These are both[br]vectors, sometimes

0:08:50.510,0:08:53.440
I should put them in bold[br]like they do in the book.

0:08:53.440,0:08:56.460
To make it clear I can[br]put a bar on top of them,

0:08:56.460,0:08:57.890
they are free vectors.

0:08:57.890,0:09:02.990
So, f of R times[br]R prime of t dt.

0:09:02.990,0:09:07.180
And your favorite moments of[br]time are-- let's say on my arc

0:09:07.180,0:09:12.270
that I'm describing from time,[br]t, equals a, to time equals b.

0:09:12.270,0:09:15.220
Therefore, I'm going to[br]take time for a to b.

0:09:15.220,0:09:19.535
And this is how we define[br]the work of a force.

0:09:19.535,0:09:22.786


0:09:22.786,0:09:25.610
The work of a[br]force that's acting

0:09:25.610,0:09:32.770
on a particle that is moving[br]between time, a, and time, b,

0:09:32.770,0:09:36.890
on this arc of a curve[br]which is called c.

0:09:36.890,0:09:38.273
Do you like this c?

0:09:38.273,0:09:41.500
Okay, and the[br]force is different.

0:09:41.500,0:09:44.055
So we have a force field.

0:09:44.055,0:09:46.360
So I cheated, I knew[br]a lot in the sense

0:09:46.360,0:09:48.510
that I didn't tell[br]you how you actually

0:09:48.510,0:09:55.280
introduce the path integral.

0:09:55.280,0:10:01.220
Now this is more or less[br]where I stop and [INAUDIBLE].

0:10:01.220,0:10:04.650
But couldn't we actually[br]introduce this integral

0:10:04.650,0:10:11.320
and even define it with respect[br]to some arc length grammar?

0:10:11.320,0:10:14.580
Maybe if everything goes[br]fine in terms of theory?

0:10:14.580,0:10:16.250
And the answer is yes.

0:10:16.250,0:10:20.100
And I'm going to show[br]you how one can do that.

0:10:20.100,0:10:25.670
I'm going to go ahead and[br]clean here a little bit.

0:10:25.670,0:10:29.170
I'm going to leave this on[br]by comparison for awhile.

0:10:29.170,0:10:33.510
And then I will assume[br]something that we have not

0:10:33.510,0:10:40.044
defined whatsoever, which is[br]an animal called path integral.

0:10:40.044,0:10:54.260
So the path integral of a vector[br]field along a trajectory, c.

0:10:54.260,0:10:55.790
I don't know how to draw.

0:10:55.790,0:10:58.630
I will draw some skewed[br]curve, how about that?

0:10:58.630,0:11:04.130
Some pretty skewed curve, c,[br]it's not self intersecting,

0:11:04.130,0:11:05.360
not necessarily.

0:11:05.360,0:11:08.770
You guys have to imagine[br]this is like the trajectory

0:11:08.770,0:11:13.220
of an airplane in[br]the sky, right?

0:11:13.220,0:11:17.740
OK, and I have it on d[br]equals a, to d equals b.

0:11:17.740,0:11:21.150
But I said forget[br]about the time, t,

0:11:21.150,0:11:24.520
maybe I can do everything[br]in arc length forever.

0:11:24.520,0:11:26.320
So if that particle,[br]or airplane,

0:11:26.320,0:11:31.120
or whatever it is has[br]a continuous motion,

0:11:31.120,0:11:32.750
that's also differentiable.

0:11:32.750,0:11:35.710
And the velocity[br]never becomes zero.

0:11:35.710,0:11:38.650
Then I can parametrize[br]an arc length

0:11:38.650,0:11:44.160
and I can say, forget about[br]it, I have integral over c.

0:11:44.160,0:11:46.970
See, this is c,[br]it's not f, okay?

0:11:46.970,0:11:54.752
But f of x of s, y[br]of s, z of s, okay?

0:11:54.752,0:11:57.570
And this is going to be a ds.

0:11:57.570,0:12:00.690
And you'll say, yes Magdelina--[br]this is little s, I'm sorry.

0:12:00.690,0:12:03.310
Yes, Magdelina, but what[br]the heck is this animal,

0:12:03.310,0:12:05.040
you've never introduced it.

0:12:05.040,0:12:09.000
I have not introduced it because[br]I have to discuss about it.

0:12:09.000,0:12:14.740
When we introduced Riemann[br]sums, then we took the limit.

0:12:14.740,0:12:19.890
We always have to think how[br]to partition our domains.

0:12:19.890,0:12:30.620
So this curve can be partitioned[br]in as many as n, this is s k.

0:12:30.620,0:12:36.220
S k, this is s1, and this is[br]s n, the last of the Mohicans.

0:12:36.220,0:12:42.170
I have n sub intervals,[br]pieces of the art.

0:12:42.170,0:12:44.410
And how am I going[br]to introduce this?

0:12:44.410,0:12:47.220
As the limit, if it exists.

0:12:47.220,0:12:49.480
Because I can be in[br]trouble, maybe this limit

0:12:49.480,0:12:51.580
is not going to exist.

0:12:51.580,0:12:54.220
The sum of what?

0:12:54.220,0:12:58.760
For every [? seg ?] partition[br]I will take a little arbitrary

0:12:58.760,0:13:00.190
point inside the subarc.

0:13:00.190,0:13:03.150


0:13:03.150,0:13:03.783
Subarc?

0:13:03.783,0:13:04.550
STUDENT: Yeah.

0:13:04.550,0:13:06.320
PROFESSOR: Subarc,[br]it's a little arc.

0:13:06.320,0:13:09.450
Contains a-- let's take it here.

0:13:09.450,0:13:13.000
What am I going to[br]define in terms of wind?

0:13:13.000,0:13:20.726
s k, y k, and z k, some[br]people put a star on it

0:13:20.726,0:13:23.890
to make it obvious.

0:13:23.890,0:13:27.040
But I'm going to[br]go ahead and say

0:13:27.040,0:13:35.600
x star k, y star k, z star k,[br]is my arbitrary point in the k

0:13:35.600,0:13:38.140
subarc.

0:13:38.140,0:13:42.095
Times, what shall I multiply by?

0:13:42.095,0:13:48.110
A delta sk, and then[br]I take k from one to n

0:13:48.110,0:13:52.140
and I press to the[br]limit with respect n.

0:13:52.140,0:13:57.150
But actually I could also[br]say in some other ways

0:13:57.150,0:14:05.170
that the partitions length goes[br]to 0, delta s goes to zero.

0:14:05.170,0:14:07.050
And you say but,[br]now wait a minute,

0:14:07.050,0:14:11.600
you have s1, s2 s3, s4,[br]s k, little tiny subarc,

0:14:11.600,0:14:13.416
what the heck is delta s?

0:14:13.416,0:14:21.820
Delta s is the largest subarc.

0:14:21.820,0:14:25.980
So the length of the largest[br]subarc, length of the largest

0:14:25.980,0:14:29.180
subarc in the partition.

0:14:29.180,0:14:34.274
So the more points I take,[br]the more I refine this.

0:14:34.274,0:14:36.190
I take the points closer[br]and closer and closer

0:14:36.190,0:14:37.600
in this partition.

0:14:37.600,0:14:41.010
What happens to the[br]length of this partition?

0:14:41.010,0:14:43.850
It shrinks to-- it goes to 0.

0:14:43.850,0:14:45.960
Assuming that this[br]would be the largest

0:14:45.960,0:14:48.690
one, well if the[br]largest one goes to 0,

0:14:48.690,0:14:51.830
everybody else goes to 0.

0:14:51.830,0:14:55.750
So this is a Riemann[br]sum, can we know for sure

0:14:55.750,0:14:57.970
that this limit exists?

0:14:57.970,0:15:02.690
No, we hope to god[br]that this limit exists.

0:15:02.690,0:15:08.150
And if the limit exists then[br]I will introduce this notion

0:15:08.150,0:15:09.773
of integral around the back.

0:15:09.773,0:15:15.450


0:15:15.450,0:15:18.100
And you said, OK I[br]believe you, but look,

0:15:18.100,0:15:22.120
what is the connection[br]between the work- the way

0:15:22.120,0:15:25.850
you introduced it as a simple[br]Calculus 1 integral here-

0:15:25.850,0:15:30.500
and this animal that looks like[br]an alien coming from the sky.

0:15:30.500,0:15:33.280
We don't know how to look at it.

0:15:33.280,0:15:37.830
Actually guys it's not so[br]bad, you do the same thing

0:15:37.830,0:15:40.240
as you did before.

0:15:40.240,0:15:44.180
In a sense that, s is connected[br]to any time parameter.

0:15:44.180,0:15:47.750
So Mr. ds says, I'm[br]your old friend,

0:15:47.750,0:15:55.584
trust me, I know who I am.[br]ds was the speed times dt.

0:15:55.584,0:16:00.640
Who can tell me if we are in R[br]three, and we are drunken bugs,

0:16:00.640,0:16:02.880
ds will become what?

0:16:02.880,0:16:07.550
A long square root times[br]dt, and what's inside here?

0:16:07.550,0:16:10.220
I want to see if[br]you guys are awake.

0:16:10.220,0:16:11.654
[INTERPOSING VOICES]

0:16:11.654,0:16:17.180
PROFESSOR: Very good, x prime[br]of t squared, I'm so lazy

0:16:17.180,0:16:20.875
but I'll write it down. y prime[br]of t squared plus z prime of t

0:16:20.875,0:16:21.685
squared.

0:16:21.685,0:16:24.210
And this is going[br]to be the speed.

0:16:24.210,0:16:28.980
So I can always do[br]that, and in this case

0:16:28.980,0:16:33.200
this is going to become always[br]some-- let's say from time, t0,

0:16:33.200,0:16:36.240
to time t1.

0:16:36.240,0:16:39.245
Some in the integrals[br]of-- some of the limit

0:16:39.245,0:16:40.580
points for the time.

0:16:40.580,0:16:45.460
I'm going to have[br]f of R of s of t,

0:16:45.460,0:16:47.670
in the end everything[br]will depend on t.

0:16:47.670,0:16:49.900
And this is my face being happy.

0:16:49.900,0:16:51.970
It's not part of the integral.

0:16:51.970,0:16:52.710
Saying what?

0:16:52.710,0:16:55.290
Saying that, guys, if[br]I plug in everything

0:16:55.290,0:16:58.565
back in terms of t- I'm[br]more familiar to that type

0:16:58.565,0:17:00.770
of integral- then I have what?

0:17:00.770,0:17:04.540
Square root of-- that's[br]the arc length element

0:17:04.540,0:17:07.450
x prime then t squared,[br]plus y prime then t

0:17:07.450,0:17:12.510
squared, plus z prime[br]then t squared, dt.

0:17:12.510,0:17:17.790
So in the end it is-- I think[br]the video doesn't see me

0:17:17.790,0:17:21.569
but it heard me, presumably.

0:17:21.569,0:17:24.220
This is our old[br]friend from Calc 1,

0:17:24.220,0:17:29.990
which is the simple integral[br]with respect to t from a to b.

0:17:29.990,0:17:34.370
OK, all right, and we[br]believe that the work

0:17:34.370,0:17:36.870
can be expressed like that.

0:17:36.870,0:17:39.610
I introduced it[br]last time, I even

0:17:39.610,0:17:41.860
proved it on some[br]particular cases

0:17:41.860,0:17:45.470
last time when Alex wasn't[br]here because, I know why.

0:17:45.470,0:17:46.927
Were you sick?

0:17:46.927,0:17:48.510
ALEX: I'll talk to[br]you about it later.

0:17:48.510,0:17:49.850
I'm a bad person.

0:17:49.850,0:17:56.915
PROFESSOR: All right, then[br]I'm dragging an object like,

0:17:56.915,0:18:02.610
the f was parallel to the[br]direction of displacement.

0:18:02.610,0:18:04.800
And then I said[br]the work would be

0:18:04.800,0:18:09.150
the magnitude of f times the[br]magnitude of the displacement.

0:18:09.150,0:18:13.450
And then we proved that is[br]just a particular case of this,

0:18:13.450,0:18:15.950
we proved that last time,[br]it was a piece of cake.

0:18:15.950,0:18:17.860
Actually, we proved[br]the other one.

0:18:17.860,0:18:23.280
It proved that if force is going[br]to be oblique and at an angle

0:18:23.280,0:18:27.180
theta with the displacement[br]direction, then

0:18:27.180,0:18:31.030
the work will be the[br]magnitude of the force times

0:18:31.030,0:18:35.796
cosine of theta, times the[br]magnitude of displacement,

0:18:35.796,0:18:36.730
all right?

0:18:36.730,0:18:42.190
And that was all an application[br]of this beautiful warp formula.

0:18:42.190,0:18:46.215
Let's see something more[br]interesting from an application

0:18:46.215,0:18:48.000
viewpoint.

0:18:48.000,0:18:53.030
Assume that you are[br]looking at the washer,

0:18:53.030,0:18:56.760
you are just doing laundry.

0:18:56.760,0:19:01.460
And you are looking at[br]this centrifugal force.

0:19:01.460,0:19:08.170
We have two forces, one is[br]centripetal towards the center

0:19:08.170,0:19:11.590
of the motion, circular[br]motion, one is centrifugal.

0:19:11.590,0:19:15.220
I will take a[br]centrifugal force f,

0:19:15.220,0:19:19.440
and I will say I want[br]to measure at the work

0:19:19.440,0:19:25.310
that this force is producing[br]in the circular motion

0:19:25.310,0:19:27.700
of my dryer.

0:19:27.700,0:19:31.710
My poor dryer died so I[br]had to buy another one

0:19:31.710,0:19:33.135
and it cost me a lot of money.

0:19:33.135,0:19:36.180
And I was thinking,[br]such a simple thing,

0:19:36.180,0:19:38.630
you pay hundreds[br]of dollars on it

0:19:38.630,0:19:41.570
but, anyway, we take[br]some things for granted.

0:19:41.570,0:19:45.990


0:19:45.990,0:19:54.610
I will take the washer because[br]the washer is a simpler

0:19:54.610,0:19:59.780
case in the sense that the[br]motion-- I can assume it's

0:19:59.780,0:20:04.180
a circular motion of[br]constant velocity.

0:20:04.180,0:20:07.490
And let's say this[br]is the washer.

0:20:07.490,0:20:10.840


0:20:10.840,0:20:16.940
And centrifugal[br]force is acting here.

0:20:16.940,0:20:21.060
Let's call that--[br]what should it be?

0:20:21.060,0:20:28.710
Well, it's continuing[br]the position vector

0:20:28.710,0:20:36.440
so let's call that lambda[br]x I, plus lambda y j.

0:20:36.440,0:20:42.360
In the sense that it's[br]collinear to the vector that

0:20:42.360,0:20:46.580
starts at origin, and here[br]is got to be x of t, y.

0:20:46.580,0:20:50.040
X and y are the[br]special components

0:20:50.040,0:20:52.560
at any point on my[br]circular motion.

0:20:52.560,0:20:55.460
If it's a circular[br]motion I have x

0:20:55.460,0:20:57.550
squared plus y squared[br]equals r squared,

0:20:57.550,0:21:01.660
where the radius is the[br]radius of my washer.

0:21:01.660,0:21:06.520
You have to compute[br]the work produced

0:21:06.520,0:21:10.960
by the centrifugal force[br]in one full rotation.

0:21:10.960,0:21:14.306
It doesn't matter, I can have[br]infinitely many rotations.

0:21:14.306,0:21:19.030
I can have a hundred rotations,[br]I couldn't care less.

0:21:19.030,0:21:23.310
But assume that the[br]motion has constant speed.

0:21:23.310,0:21:26.650
So if I wanted, I could[br]parametrize in our things

0:21:26.650,0:21:30.270
but it doesn't[br]bring a difference.

0:21:30.270,0:21:33.710
Because guys, when this[br]speed is already constant,

0:21:33.710,0:21:36.330
like for the circular motion[br]you are familiar with.

0:21:36.330,0:21:39.760
Or the helicoidal case[br]you are familiar with,

0:21:39.760,0:21:44.120
you also saw the case when[br]the speed was constant.

0:21:44.120,0:21:47.810
Practically, you were[br]just rescaling the time

0:21:47.810,0:21:52.750
to get to your speed, to your[br]time parameter s, arc length.

0:21:52.750,0:21:54.840
So whether you work with[br]t, or you work with s,

0:21:54.840,0:21:58.230
it's the same thing if[br]the speed is a constant.

0:21:58.230,0:22:02.030
So I'm not going to use my[br]imagination to go and do it

0:22:02.030,0:22:02.900
with respect to s.

0:22:02.900,0:22:05.500
I could, but I[br]couldn't give a damn

0:22:05.500,0:22:09.565
because I'm going to have[br]a beautiful t that you

0:22:09.565,0:22:13.000
are going to help me recover.

0:22:13.000,0:22:15.640
From here, what is[br]the parametrization

0:22:15.640,0:22:18.696
that comes to mind?

0:22:18.696,0:22:19.570
Can you guys help me?

0:22:19.570,0:22:23.640
I know you can after all[br]the review of chapter 10

0:22:23.640,0:22:25.860
and-- this is what?

0:22:25.860,0:22:29.470


0:22:29.470,0:22:32.970
R what?

0:22:32.970,0:22:34.970
You should whisper cosine t.

0:22:34.970,0:22:38.680
Say it out loud, be[br]proud of what you know.

0:22:38.680,0:22:40.640
This is R sine t.

0:22:40.640,0:22:43.080
And let's take t[br]between 0 and 2 pi.

0:22:43.080,0:22:50.730
One, the revolution only,[br]and then I say, good.

0:22:50.730,0:22:53.996
The speed is what?

0:22:53.996,0:23:00.930
Speed, square root of[br]x prime the t squared

0:23:00.930,0:23:03.430
plus y prime the t squared.

0:23:03.430,0:23:08.260
Which is the same as writing[br]R prime of the t in magnitude.

0:23:08.260,0:23:09.480
Thank God we know that.

0:23:09.480,0:23:11.550
How much is this?

0:23:11.550,0:23:14.266
R, very good, this[br]is R, very good.

0:23:14.266,0:23:18.790
So life is not so hard, it's--[br]hopefully I'll be able to do

0:23:18.790,0:23:21.660
the w.

0:23:21.660,0:23:22.410
What is the w?

0:23:22.410,0:23:25.910
It's the path integral[br]all over the circle

0:23:25.910,0:23:30.380
I have here, that I traveled[br]counterclockwise from any point

0:23:30.380,0:23:31.270
to any point.

0:23:31.270,0:23:34.559
Let's say this would be[br]the origin of my motion,

0:23:34.559,0:23:37.340
then I go back.

0:23:37.340,0:23:43.320
And I have this[br]force, F, that I have

0:23:43.320,0:23:47.540
to redistribute in terms[br]of R. So this notation

0:23:47.540,0:23:49.222
is giving me a little[br]bit of a headache,

0:23:49.222,0:23:51.410
but in reality it's[br]going to be very simple.

0:23:51.410,0:23:59.740
This is the dot product, R[br]prime dt, which was the R.

0:23:59.740,0:24:05.390
Which some other people asked[br]me, how can you write that?

0:24:05.390,0:24:12.090
Well, read the[br]review, R of x equals

0:24:12.090,0:24:16.810
x of t, i plus y of t, j.

0:24:16.810,0:24:19.990
Also, read the[br]next side plus y j.

0:24:19.990,0:24:26.590
It short, the dR[br]differential out of t,

0:24:26.590,0:24:32.920
sorry, I'll put R.[br]dR is dx i plus dy j.

0:24:32.920,0:24:35.950
And if somebody wants[br]to be expressing

0:24:35.950,0:24:40.390
this in terms of speeds,[br]we'll say this is x prime dt,

0:24:40.390,0:24:43.010
this is y prime dt.

0:24:43.010,0:24:48.130
So we can rewrite this x prime[br]then t i, plus y prime then t

0:24:48.130,0:24:49.836
j, dt.

0:24:49.836,0:24:54.620


0:24:54.620,0:24:55.800
All right?

0:24:55.800,0:25:01.350
OK, which is the same[br]thing as R prime of t, dt.

0:25:01.350,0:25:04.203


0:25:04.203,0:25:04.703
[INAUDIBLE]

0:25:04.703,0:25:07.580


0:25:07.580,0:25:11.760
This looks awfully theoretical.

0:25:11.760,0:25:16.550
I say, I don't like it, I[br]want to put my favorite guys

0:25:16.550,0:25:18.290
in the picture.

0:25:18.290,0:25:22.140
So I have to think, when[br]I do the dot product

0:25:22.140,0:25:26.380
I have the dot product[br]between the vector that

0:25:26.380,0:25:29.890
has components f1 and f2.

0:25:29.890,0:25:31.870
How am I going to do that?

0:25:31.870,0:25:38.700
Well, if I multiply with this[br]guy, dot product, the boss guy,

0:25:38.700,0:25:39.800
with this boss guy.

0:25:39.800,0:25:43.180
Are you guys with me?

0:25:43.180,0:25:44.810
What am I going to do?

0:25:44.810,0:25:48.560
First component times[br]first component,

0:25:48.560,0:25:53.740
plus second component times[br]second component of a vector.

0:25:53.740,0:25:58.830
So I have to be smart and[br]understand how I do that.

0:25:58.830,0:26:01.060
Lambda is a constant.

0:26:01.060,0:26:04.110
Lambda, you're my[br]friend, you stay there.

0:26:04.110,0:26:09.570
x is x of t, x of[br]t, but I multiplied

0:26:09.570,0:26:11.490
with the first[br]component here, so I

0:26:11.490,0:26:14.860
multiplied by x prime of t.

0:26:14.860,0:26:19.600
Plus lambda times y of[br]t, times y prime of t.

0:26:19.600,0:26:23.750
And who gets out of the[br]picture is dt at the end.

0:26:23.750,0:26:27.530
I have integrate with[br]respect to that dt.

0:26:27.530,0:26:29.570
This would be incorrect, why?

0:26:29.570,0:26:34.600
Because t has to move[br]between some specific limits

0:26:34.600,0:26:38.180
when I specify what[br]a path integral is.

0:26:38.180,0:26:41.870
I cannot leave a c-- very good,[br]from 0 to 2 pi, excellent.

0:26:41.870,0:26:45.300


0:26:45.300,0:26:46.240
Is this hard?

0:26:46.240,0:26:48.670
No, It's going to[br]be a piece of cake.

0:26:48.670,0:26:50.336
Why is that a piece of cake?

0:26:50.336,0:26:54.190
Because I can keep writing.

0:26:54.190,0:26:58.519
You actually are faster than me.

0:26:58.519,0:27:00.560
STUDENT: Your chain rule[br]is already done for you.

0:27:00.560,0:27:03.930
PROFESSOR: Right,[br]and then lambda

0:27:03.930,0:27:06.920
gets out just because--[br]well you remember

0:27:06.920,0:27:09.600
you kick the lambda out, right?

0:27:09.600,0:27:20.140
And then I've put R cosine[br]t times minus R sine t.

0:27:20.140,0:27:21.310
I'm done with who?

0:27:21.310,0:27:23.625
I'm done with this[br]fellow and that fellow.

0:27:23.625,0:27:27.340


0:27:27.340,0:27:37.550
And plus y, R sine[br]t, what is R prime?

0:27:37.550,0:27:48.580
R cosine, thank you guys, dt.

0:27:48.580,0:27:50.910
And now I'm going to ask[br]you, what is this animal?

0:27:50.910,0:27:53.660


0:27:53.660,0:27:56.020
Stare at that, what[br]is the integrand?

0:27:56.020,0:27:59.130
Is a friend of[br]yours, he's so cute.

0:27:59.130,0:28:01.540
He's staring at you and[br]saying you are done.

0:28:01.540,0:28:04.895
Why are you done?

0:28:04.895,0:28:08.346
What happens to the integrand?

0:28:08.346,0:28:12.365
It's zero, it's a[br]blessing, it's zero.

0:28:12.365,0:28:14.000
How come it's zero?

0:28:14.000,0:28:21.390
Because these two terms[br]simplify, they cancel out.

0:28:21.390,0:28:26.290
They cancel out, thank god[br]they cancel out, I got a zero.

0:28:26.290,0:28:28.530
So we discover something[br]that a physicist

0:28:28.530,0:28:32.152
or a mechanical engineer[br]would have told you already.

0:28:32.152,0:28:34.110
And do you think he would[br]have actually plugged

0:28:34.110,0:28:36.230
in the path integral?

0:28:36.230,0:28:39.700
No, they wouldn't[br]think like this.

0:28:39.700,0:28:45.540
He has a simpler explanation for[br]that because he's experienced

0:28:45.540,0:28:47.260
with linear experiments.

0:28:47.260,0:28:51.740
And says, if I drag this like[br]that I know what to work with.

0:28:51.740,0:28:54.770
If I drag in, like[br]at an angle, I

0:28:54.770,0:28:57.950
know that I have the[br]magnitude of this,

0:28:57.950,0:28:59.390
cosine theta, the angle.

0:28:59.390,0:29:03.740
So, he knows for linear[br]cases what we have.

0:29:03.740,0:29:10.060
For a circular case he[br]can smell the result

0:29:10.060,0:29:12.880
without doing the path integral.

0:29:12.880,0:29:18.080
So how do you think the guy,[br]if he's a mechanical engineer,

0:29:18.080,0:29:20.040
would think in a second?

0:29:20.040,0:29:24.315
Say well, think of[br]your trajectory, right?

0:29:24.315,0:29:27.360
It's a circle.

0:29:27.360,0:29:31.620
The problem is that centrifugal[br]force being perpendicular

0:29:31.620,0:29:33.560
to the circle all the time.

0:29:33.560,0:29:37.140
And you say, how can a line[br]be perpendicular to a circle?

0:29:37.140,0:29:39.570
It is, it's the[br]normal to the circle.

0:29:39.570,0:29:43.054
So when you say this[br]is normal to the circle

0:29:43.054,0:29:45.470
you mean it's normal to[br]the tangent of the circle.

0:29:45.470,0:29:50.840
So if you measure the angle made[br]by the normal at every point

0:29:50.840,0:29:53.800
to the trajectory of a[br]circle, it's always lines.

0:29:53.800,0:30:01.520
So he goes, gosh, I got[br]cosine of 90, that's zero.

0:30:01.520,0:30:04.270
So if you have some[br]sort of work produced

0:30:04.270,0:30:06.470
by something perpendicular[br]to your trajectory,

0:30:06.470,0:30:07.800
that must be zero.

0:30:07.800,0:30:12.140
So he or she has[br]very good intuition.

0:30:12.140,0:30:13.875
Of course, how do we prove it?

0:30:13.875,0:30:16.535
We are mathematicians, we[br]prove the path integral,

0:30:16.535,0:30:19.460
we got zero for the[br]work, all right?

0:30:19.460,0:30:25.200
But he could sense that kind[br]of stuff from the beginning.

0:30:25.200,0:30:34.740
Now, there is another example[br]where maybe you don't have

0:30:34.740,0:30:38.360
90 degrees for your trajectory.

0:30:38.360,0:30:47.970
Well, I'm going to just take--[br]what if I change the force

0:30:47.970,0:30:50.570
and I make a difference problem?

0:30:50.570,0:30:52.844
Make it into a[br]different problem.

0:30:52.844,0:31:01.293


0:31:01.293,0:31:04.772
I will do that later, I[br]won't go and erase it.

0:31:04.772,0:31:15.710


0:31:15.710,0:31:19.620
Last time we did one[br]that was, compute

0:31:19.620,0:31:25.330
the work along a parabola[br]from something to something.

0:31:25.330,0:31:26.830
Let's do that again.

0:31:26.830,0:31:30.668


0:31:30.668,0:31:35.585
For some sort of[br]a nice force field

0:31:35.585,0:31:39.740
I'll take your vector valued[br]function to be nice to you.

0:31:39.740,0:31:42.380
I'll change it, y i plus x j.

0:31:42.380,0:31:45.330


0:31:45.330,0:31:48.550
And then we are in[br]plane and we move

0:31:48.550,0:31:55.020
along this parabola between[br]0, 0 and-- what is this guys?

0:31:55.020,0:31:59.348
1, 1-- well let make it[br]into a one, it's cute.

0:31:59.348,0:32:04.690
And I'd like you to measure[br]the work along the parabola

0:32:04.690,0:32:11.410
and also along the arc of a--[br]along the segment of a line

0:32:11.410,0:32:13.440
between the two points.

0:32:13.440,0:32:16.728
So I want you to compute[br]w1 along the parabola,

0:32:16.728,0:32:22.990
and w2 along this[br]thingy, the segment.

0:32:22.990,0:32:24.070
Should it be hard?

0:32:24.070,0:32:27.230
No, this was old[br]session for many finals.

0:32:27.230,0:32:33.180
I remember, I think it[br]was 2003, 2006, 2008,

0:32:33.180,0:32:36.320
and very recently, I[br]think a year and 1/2 ago.

0:32:36.320,0:32:38.020
A problem like that was given.

0:32:38.020,0:32:40.750
Compute the path[br]integrals correspond

0:32:40.750,0:32:45.000
to work for both parametrization[br]and compare them.

0:32:45.000,0:32:45.650
Is it hard?

0:32:45.650,0:32:48.861
I have no idea, let me think.

0:32:48.861,0:32:53.260
For the first one we[br]have parametrization

0:32:53.260,0:32:55.600
that we need to distinguish[br]from the other one.

0:32:55.600,0:32:58.706
The first parametrization[br]for a parabola,

0:32:58.706,0:33:01.430
we discussed it last[br]time, was of course

0:33:01.430,0:33:04.080
the simplest possible[br]one you can think of.

0:33:04.080,0:33:05.830
And we did this[br]last time but I'm

0:33:05.830,0:33:10.360
repeating this because I[br]didn't want Alex to miss that.

0:33:10.360,0:33:18.250
And I'm going to say integral[br]from some time to some time.

0:33:18.250,0:33:21.090
Now, if I'm between 0[br]and 1, time of course

0:33:21.090,0:33:25.780
will be between 0 and[br]1 because x is time.

0:33:25.780,0:33:28.554
All right, good,[br]that means what else?

0:33:28.554,0:33:32.140
This Is f1 and this is f2.

0:33:32.140,0:33:39.720
So I'm going to have f1 of t,[br]x prime of t, plus f2 of t,

0:33:39.720,0:33:46.310
y prime of t, all the[br][? sausage ?] times dt.

0:33:46.310,0:33:47.930
Is this going to be hard?

0:33:47.930,0:33:52.830
Hopefully not, I'm going to[br]have to identify everybody.

0:33:52.830,0:33:55.970
Identify this guys prime[br]of t with respect to t

0:33:55.970,0:33:58.110
is 1, piece of cake, right?

0:33:58.110,0:34:02.850
This fellow is-- you told[br]me last time you got 2t

0:34:02.850,0:34:05.360
and you got it right.

0:34:05.360,0:34:08.219
This guy, I have to be[br]a little bit careful

0:34:08.219,0:34:10.679
because y is the fourth guy.

0:34:10.679,0:34:15.920
This is t squared and this is t.

0:34:15.920,0:34:19.889


0:34:19.889,0:34:22.960
So my integral will be a joke.

0:34:22.960,0:34:29.630
0 to 1, 2t squared plus t[br]squared equals 3t squared.

0:34:29.630,0:34:30.980
Is it hard to integrate?

0:34:30.980,0:34:33.110
No, for God's sake,[br]this is integral-- this

0:34:33.110,0:34:39.270
is t cubed between 0[br]and 1, right, right?

0:34:39.270,0:34:44.080
So I should get[br]1, and if I get, I

0:34:44.080,0:34:48.600
think I did it right, if I get[br]the other parametrization you

0:34:48.600,0:34:52.540
have to help me write it again.

0:34:52.540,0:34:57.190
The parametrization[br]of this straight line

0:34:57.190,0:34:59.380
between 0, 0 and 1, 1.

0:34:59.380,0:35:03.300
Now on the actual[br]exam, I'm never

0:35:03.300,0:35:08.500
going to forgive you if you[br]don't know how to parametrize.

0:35:08.500,0:35:13.140
Now you know it but two months[br]ago you didn't, many of you

0:35:13.140,0:35:14.100
didn't.

0:35:14.100,0:35:21.510
So if somebody gives you 2[br]points, OK, in plane I ask you,

0:35:21.510,0:35:27.160
how do you write that symmetric[br]equation of the line between?

0:35:27.160,0:35:29.030
You were a little[br]bit hesitant, now

0:35:29.030,0:35:34.030
you shouldn't be hesitant[br]because it's a serious thing.

0:35:34.030,0:35:36.290
So how did we write that?

0:35:36.290,0:35:40.466
We memorized it. x minus[br]x1, over x2 minus x1

0:35:40.466,0:35:44.440
equals y minus y1,[br]over y2 minus y1.

0:35:44.440,0:35:47.300
This can also be written[br]as-- you know, guys,

0:35:47.300,0:35:49.840
that this over that[br]is the actual slope.

0:35:49.840,0:35:52.275
This over that, so[br]it can be written

0:35:52.275,0:35:53.560
as a [INAUDIBLE] formula.

0:35:53.560,0:35:55.250
It can be written in many ways.

0:35:55.250,0:35:59.920
And if we put a, t, we transform[br]it into a parametric equation.

0:35:59.920,0:36:02.140
So you should be[br]able, on the final,

0:36:02.140,0:36:05.690
to do that for any segment of[br]a line with your eyes closed.

0:36:05.690,0:36:07.550
Like, you see the[br]numbers, you plug them in,

0:36:07.550,0:36:10.640
you get the[br]parametric equations.

0:36:10.640,0:36:14.820
We are nice on the exams[br]because we usually give you

0:36:14.820,0:36:18.750
a line that's easy to write.

0:36:18.750,0:36:25.830
Like in this case you would[br]have x equals t and y equal,

0:36:25.830,0:36:29.940
let's see if you[br]are asleep yet, t.

0:36:29.940,0:36:31.500
Why is that?

0:36:31.500,0:36:37.290
Because the line that joins[br]0, 0 and 1, 1 is y equals x.

0:36:37.290,0:36:40.095
So y equals x is called[br]also, first bicycle.

0:36:40.095,0:36:43.100
It's the old friend of[br]yours from trigonometry,

0:36:43.100,0:36:46.300
from Pre-Calc, from algebra,[br]I don't know where, college

0:36:46.300,0:36:47.820
algebra.

0:36:47.820,0:36:49.160
Alrighty, is this hard to do?

0:36:49.160,0:36:55.850
No, it's easier than before.[br]w2 is integral from 0 to 1,

0:36:55.850,0:37:00.380
this is t and this is t, this[br]is t and this is t, good.

0:37:00.380,0:37:07.990
So we have t times[br]1 plus t times 1,

0:37:07.990,0:37:15.362
it's like a funny, nice, game[br]that's too simple, 2t, 2t.

0:37:15.362,0:37:18.340


0:37:18.340,0:37:23.100
So the fundamental[br]theorem of Calc

0:37:23.100,0:37:27.042
says t squared between 0[br]and 1, the answer is 1.

0:37:27.042,0:37:28.500
Am I surprised?

0:37:28.500,0:37:31.360
Look at me, do I[br]look surprised at all

0:37:31.360,0:37:34.240
that I got the same answer?

0:37:34.240,0:37:37.224
No, I told you a[br]secret last time.

0:37:37.224,0:37:39.390
I didn't prove it.

0:37:39.390,0:37:43.100
I said that R times happy times.

0:37:43.100,0:37:46.740
When depending on the[br]force that is with you,

0:37:46.740,0:37:51.300
you have the same work[br]no matter what path

0:37:51.300,0:37:53.830
you are taking between a and b.

0:37:53.830,0:37:57.720
Between the origin[br]and finish line.

0:37:57.720,0:38:01.030
So I'm claiming that if I[br]give you this zig-zag line

0:38:01.030,0:38:08.010
and I asked you what-- look,[br]it could be any crazy path

0:38:08.010,0:38:11.090
but it has to be a nice[br]differentiable path.

0:38:11.090,0:38:14.630
Along this differentiable[br]path, no matter

0:38:14.630,0:38:17.420
how you compute the work,[br]that's your business,

0:38:17.420,0:38:19.040
I claim I still get 1.

0:38:19.040,0:38:21.840


0:38:21.840,0:38:26.440
Can you even think why,[br]some of you remember maybe,

0:38:26.440,0:38:28.986
the force was key?

0:38:28.986,0:38:30.444
STUDENT: It's a[br]conservative force.

0:38:30.444,0:38:32.780
PROFESSOR: It had to[br]be good conservative.

0:38:32.780,0:38:37.320
Now this is conservative but[br]why is that conservative?

0:38:37.320,0:38:39.900
What the heck is a[br]conservative force?

0:38:39.900,0:38:46.710
So let's write it[br]down on the-- we

0:38:46.710,0:39:01.650
say that the vector[br]valued function, f,

0:39:01.650,0:39:19.084
valued in R2 or R3, is[br]conservative if there exists

0:39:19.084,0:39:25.460
a smooth function.

0:39:25.460,0:39:28.990
Little f, it actually[br]has to be just c1,

0:39:28.990,0:39:31.660
called scalar potential.

0:39:31.660,0:39:45.380
Called scalar potential,[br]such that big F as a vector

0:39:45.380,0:39:48.100
field will be not little f.

0:39:48.100,0:39:52.830
That means it will be the[br]gradient of the scalar

0:39:52.830,0:39:54.838
potential.

0:39:54.838,0:39:57.370
Definition, that[br]was the definition,

0:39:57.370,0:40:08.440
and then criterion for a f[br]in R2 to be conservative.

0:40:08.440,0:40:13.390


0:40:13.390,0:40:21.370
I claim that f equals f1 i,[br]plus f two eyes, no, f2 j.

0:40:21.370,0:40:23.370
I'm just making[br]silly puns, I don't

0:40:23.370,0:40:27.880
know if you guys follow me.

0:40:27.880,0:40:33.470
If and only if f sub 1[br]prime, with respect to y,

0:40:33.470,0:40:37.890
is f sub 2 prime,[br]with respect to x.

0:40:37.890,0:40:40.166
Can I prove this?

0:40:40.166,0:40:45.560
Prove, prove Magdelina, don't[br]just stare at it, prove.

0:40:45.560,0:40:48.030
Why would that be[br]necessary and sufficient?

0:40:48.030,0:40:52.020


0:40:52.020,0:40:58.890
Well, for big F[br]to be conservative

0:40:58.890,0:41:02.180
it means that it has[br]to be the gradient

0:41:02.180,0:41:06.970
of some little function, little[br]f, some scalar potential.

0:41:06.970,0:41:13.204
Alrighty, so let me[br]write it down, proof.

0:41:13.204,0:41:20.584
f conservative if[br]and only if there

0:41:20.584,0:41:27.964
exists f, such that gradient[br]of f is F. If and only

0:41:27.964,0:41:32.990
if-- what does it[br]mean about f1 and f2?

0:41:32.990,0:41:36.540
f1 and f2 are the f[br]sub of x and the f sub

0:41:36.540,0:41:39.460
y of some scalar potential.

0:41:39.460,0:41:44.300
So if f is the gradient, that[br]means that the first component

0:41:44.300,0:41:48.600
has to be little f sub x And[br]the second component should

0:41:48.600,0:41:51.906
have to be little f sub y.

0:41:51.906,0:41:58.240
But that is if and only if f[br]sub 1 prime, with respect to y,

0:41:58.240,0:42:01.710
is the same as f sub 2[br]prime, with respect to x.

0:42:01.710,0:42:04.144
What is that?

0:42:04.144,0:42:14.052
The red thing here is called[br]a compatibility condition

0:42:14.052,0:42:16.440
of this system.

0:42:16.440,0:42:22.660
This is a system of two OD's.

0:42:22.660,0:42:27.426
You are going to[br]study ODEs in 3350.

0:42:27.426,0:42:29.730
And you are going[br]to remember this

0:42:29.730,0:42:33.020
and say, Oh, I know that because[br]she taught me that in Calc 3.

0:42:33.020,0:42:36.310
Not all instructors will[br]teach you this in Calc 3.

0:42:36.310,0:42:38.880
Some of them fool you[br]and skip this material

0:42:38.880,0:42:44.410
that's very important[br]to understand in 3350.

0:42:44.410,0:42:49.490
So guys, what's going to[br]happened when you prime this

0:42:49.490,0:42:51.580
with respect to y?

0:42:51.580,0:42:55.440
You get f sub x prime,[br]with respect to y.

0:42:55.440,0:42:57.110
When you prime this[br]with respect to x

0:42:57.110,0:42:59.860
you get f sub y prime,[br]with respect to x.

0:42:59.860,0:43:01.680
Why are they the same thing?

0:43:01.680,0:43:04.716
I'm going to remind[br]you that they

0:43:04.716,0:43:07.220
are the same thing[br]for a smooth function.

0:43:07.220,0:43:09.250
Who said that?

0:43:09.250,0:43:19.060
A crazy German mathematician[br]whose name was Schwartz.

0:43:19.060,0:43:22.134
Which means black, that's[br]what I'm painting it in black.

0:43:22.134,0:43:29.090
Because is the Schwartz[br]guy, the first criterion

0:43:29.090,0:43:31.750
saying that no[br]matter in what order

0:43:31.750,0:43:34.320
you differentiate the[br]smooth function you

0:43:34.320,0:43:37.800
get the same answer for[br]the mixed derivative.

0:43:37.800,0:43:41.620
So you see we prove if[br]and only if that you

0:43:41.620,0:43:43.900
have to have this[br]criterion, otherwise

0:43:43.900,0:43:46.210
it's not going to[br]be conservative.

0:43:46.210,0:43:49.630
So I'm asking you,[br]for your old friend, f

0:43:49.630,0:43:52.840
equals- example[br]one or example two,

0:43:52.840,0:43:56.200
I don't know- y i plus x j.

0:43:56.200,0:43:58.510
Is the conservative?

0:43:58.510,0:44:00.440
You can prove it in two ways.

0:44:00.440,0:44:06.685
Prove in two differently[br]ways that it is conservative.

0:44:06.685,0:44:13.210


0:44:13.210,0:44:16.716
a, find the criteria.

0:44:16.716,0:44:20.450
What does this criteria say?

0:44:20.450,0:44:25.430
Take your first component,[br]prime it with respect to y.

0:44:25.430,0:44:27.780
So y prime with respect to y.

0:44:27.780,0:44:32.870
Take your second component,[br]x, prime it with respect to x.

0:44:32.870,0:44:34.750
Is this true?

0:44:34.750,0:44:37.460
Yes, and this is me,[br]happy that it's true.

0:44:37.460,0:44:40.740
So this is 1 equals[br]1, so it's true.

0:44:40.740,0:44:44.780
So it must be conservative,[br]so it must be conservative.

0:44:44.780,0:44:46.780
Could I have done[br]it another way?

0:44:46.780,0:44:49.610


0:44:49.610,0:44:57.530
By definition, by definition,[br]to prove that a force field

0:44:57.530,0:45:01.640
is conservative by[br]definition, that it a matter

0:45:01.640,0:45:04.075
of the smart people.

0:45:04.075,0:45:07.100
There are people[br]who- unlike me when

0:45:07.100,0:45:11.670
I was 18- are able to see[br]the scalar potential in just

0:45:11.670,0:45:13.852
about any problem I give them.

0:45:13.852,0:45:18.070
I'm not going to make this[br]experiment with a bunch of you

0:45:18.070,0:45:22.320
and I'm going to reward you[br]for the correct answers.

0:45:22.320,0:45:25.430
But, could anybody[br]see the existence

0:45:25.430,0:45:27.710
of the scalar potential?

0:45:27.710,0:45:30.750
So these there, this exists.

0:45:30.750,0:45:37.590
That's there exists a little[br]f scalar potential such

0:45:37.590,0:45:43.081
that nabla f equals F.

0:45:43.081,0:45:47.680
And some of you may see[br]it and say, I see it.

0:45:47.680,0:45:50.070
So, can you see[br]a little function

0:45:50.070,0:45:54.660
f scalar function so[br]that f sub of x i is y

0:45:54.660,0:46:00.810
and f sub- Magdelena--[br]x of y j is this x j?

0:46:00.810,0:46:02.220
STUDENT: [INAUDIBLE]?

0:46:02.220,0:46:07.330
PROFESSOR: No, you need to[br]drink some coffee first.

0:46:07.330,0:46:12.990
You can get this, x times what?

0:46:12.990,0:46:14.165
Why is that?

0:46:14.165,0:46:17.270
I'll teach you how to get it.

0:46:17.270,0:46:18.880
Nevertheless, there[br]are some people

0:46:18.880,0:46:21.140
who can do it with[br]their naked eye

0:46:21.140,0:46:26.066
because they have a little[br]computer in their head.

0:46:26.066,0:46:28.710
But how did I do it?

0:46:28.710,0:46:30.970
It's just a matter of[br]experience, I said,

0:46:30.970,0:46:36.690
if I take f to be x y,[br]I sort of guessed it.

0:46:36.690,0:46:42.000
f sub x would be y and f sub y[br]will be x so this should be it,

0:46:42.000,0:46:44.440
and this is going to do.

0:46:44.440,0:46:49.040
And f is a nice function,[br]polynomial in two variables,

0:46:49.040,0:46:50.420
it's a smooth function.

0:46:50.420,0:46:55.200
I'm very happy I'm[br]over the domain,

0:46:55.200,0:46:57.770
open this or whatever,[br]open domain in plane.

0:46:57.770,0:47:00.160
I'm very happy, I have[br]no problem with it.

0:47:00.160,0:47:04.756
So I can know that this is[br]conservative in two ways.

0:47:04.756,0:47:07.750
Either I get to the[br]source of the problem

0:47:07.750,0:47:10.380
and I find the little[br]scalar potential whose

0:47:10.380,0:47:13.480
gradient is my force field.

0:47:13.480,0:47:16.630
Or I can verify the[br]criterion and I say,

0:47:16.630,0:47:19.100
the derivative of[br]this with respect to y

0:47:19.100,0:47:20.972
is the derivative[br]of this with respect

0:47:20.972,0:47:23.140
to x is-- one is the same.

0:47:23.140,0:47:29.145
The same thing, you're going[br]to see it again in math 3350.

0:47:29.145,0:47:30.710
All right, that I[br]taught many times,

0:47:30.710,0:47:33.020
I'm not going to teach[br]that in the fall.

0:47:33.020,0:47:34.820
But I know of some[br]very good people

0:47:34.820,0:47:36.065
who teach that in the fall.

0:47:36.065,0:47:38.670
In any case, they[br]would reteach it to you

0:47:38.670,0:47:42.060
because good teachers don't[br]assume that you know much.

0:47:42.060,0:47:45.970
But when you will see[br]it you'll remember me.

0:47:45.970,0:47:49.970
Hopefully fondly, not cursing[br]me or anything, right?

0:47:49.970,0:47:54.890
OK, how do we actually[br]get to compute f by hand

0:47:54.890,0:47:59.150
if we're not experienced[br]enough to guess it like I was

0:47:59.150,0:48:01.740
experienced enough to guess?

0:48:01.740,0:48:08.240
So let me show you how you solve[br]a system of two differential

0:48:08.240,0:48:11.290
equations like that.

0:48:11.290,0:48:15.625
So how I got-- how[br]you are supposed

0:48:15.625,0:48:23.010
to get the scalar potential.

0:48:23.010,0:48:28.790
f sub x equals F1,[br]f sub y equals F2.

0:48:28.790,0:48:35.358
So by integration, 1 and 2.

0:48:35.358,0:48:38.740


0:48:38.740,0:48:40.910
And you say, what[br]you mean 1 and 2?

0:48:40.910,0:48:42.670
I'll show you in a second.

0:48:42.670,0:48:47.820
So for my case,[br]example 2, I'll take

0:48:47.820,0:48:50.570
my f sub x must be y, right?

0:48:50.570,0:48:51.070
Good.

0:48:51.070,0:48:54.025
My f sub y must be x, right?

0:48:54.025,0:48:55.270
Right.

0:48:55.270,0:48:58.160
Who is f?

0:48:58.160,0:49:00.690
Solve this property.

0:49:00.690,0:49:05.580
Oh, I have to start[br]integrating from the first guy.

0:49:05.580,0:49:08.390
What kind of information[br]am I going to squeeze?

0:49:08.390,0:49:11.572
I'm going to say I[br]have to go backwards,

0:49:11.572,0:49:17.581
I have to get-- f[br]is going to be what?

0:49:17.581,0:49:23.156
Integral of y with respect to[br]x, say it again, Magdelina.

0:49:23.156,0:49:27.440
Integral of y with respect[br]to x, but attention,

0:49:27.440,0:49:31.620
this may come because, for[br]me, the variable is x here,

0:49:31.620,0:49:35.480
and y is like, you cannot[br]stay in this picture.

0:49:35.480,0:49:40.590
So I have a constant[br]c that depends on y.

0:49:40.590,0:49:41.510
Say what?

0:49:41.510,0:49:46.290
Yes, because if you go backwards[br]and prime this with respect

0:49:46.290,0:49:49.658
to x, what do you[br]get? f sub x will

0:49:49.658,0:49:51.840
be y because this is[br]the anti-derivative.

0:49:51.840,0:49:55.460
Plus this prime with[br]respect to x, zero.

0:49:55.460,0:49:59.520
So this c of y may, a[br]little bit, ruin your plans.

0:49:59.520,0:50:02.630
I've had students[br]who forgot about it

0:50:02.630,0:50:04.950
and then they got in trouble[br]because they couldn't get

0:50:04.950,0:50:08.130
the scalar potential correctly.

0:50:08.130,0:50:08.630
All right?

0:50:08.630,0:50:14.340
OK, so from this one you[br]say, OK I have some--

0:50:14.340,0:50:16.980
what is the integral of y dx?

0:50:16.980,0:50:22.246
xy, plus some guy c[br]constant that depends on y.

0:50:22.246,0:50:25.610
From this fellow[br]I go, but I have

0:50:25.610,0:50:29.520
to verify the second condition,[br]if I don't I'm dead meat.

0:50:29.520,0:50:32.320
There are two coupled[br]equations, these

0:50:32.320,0:50:33.980
are coupled equations[br]that have to be

0:50:33.980,0:50:36.320
verified at the same time.

0:50:36.320,0:50:45.120
So f sub y will be prime with[br]respect to y. x plus prime

0:50:45.120,0:50:52.750
with respect to y. c prime[br]of y, God gave me x here.

0:50:52.750,0:50:56.760
So I'm really lucky in that[br]sense that c prime of y

0:50:56.760,0:51:01.300
will be 0 because I have[br]an x here and an x here.

0:51:01.300,0:51:05.540
So c of y will simply be[br]any constant k. c of y

0:51:05.540,0:51:09.487
is just a constant k, it's[br]not going to depend on y,

0:51:09.487,0:51:10.980
it's a constant k.

0:51:10.980,0:51:14.510
So my answer was not correct.

0:51:14.510,0:51:21.320
The best answer would have[br]been f of xy must be xy plus k.

0:51:21.320,0:51:26.650
But any function like xy will[br]work, I just need one to work.

0:51:26.650,0:51:29.330
I just need a scalar[br]potential, not all of them.

0:51:29.330,0:51:33.846
This will work, x2, xy plus 7[br]will work, xy plus 3 will work,

0:51:33.846,0:51:38.860
xy minus 1,033,045 will work.

0:51:38.860,0:51:43.830
But I only need one[br]so I'll take xy.

0:51:43.830,0:51:46.220
Now that I trained[br]your mind a little bit,

0:51:46.220,0:51:48.370
maybe you don't need[br]to actually solve

0:51:48.370,0:51:53.720
the system because your[br]brain wasn't ready before.

0:51:53.720,0:51:57.300
But you'd be amazed, we[br]are very trainable people.

0:51:57.300,0:52:03.550
And in the process of doing[br]something completely new,

0:52:03.550,0:52:05.400
we are learning.

0:52:05.400,0:52:10.305
And your brain next,[br]will say, I think

0:52:10.305,0:52:15.330
I know how to function[br]a little bit backwards.

0:52:15.330,0:52:20.460
And try to integrate and[br]see and guess a potential

0:52:20.460,0:52:24.080
because it's not so hard.

0:52:24.080,0:52:25.580
So let me give you example 3.

0:52:25.580,0:52:29.650


0:52:29.650,0:52:36.240
Somebody give you over a[br]domain in plane x i plus y j,

0:52:36.240,0:52:40.605
and says, over D, simply[br]connected domain in plane,

0:52:40.605,0:52:44.000
open, doesn't matter.

0:52:44.000,0:52:45.455
Is this conservative?

0:52:45.455,0:52:48.370


0:52:48.370,0:52:51.880
Find a scalar potential.

0:52:51.880,0:53:00.710


0:53:00.710,0:53:03.960
This is again, we[br]do section 13-2,

0:53:03.960,0:53:07.650
so today we did 13-1[br]and 13-2 jointly.

0:53:07.650,0:53:10.630


0:53:10.630,0:53:11.930
Find the scalar potential.

0:53:11.930,0:53:14.994
Do you see it now?

0:53:14.994,0:53:16.392
STUDENT: [INAUDIBLE]?

0:53:16.392,0:53:19.560
PROFESSOR: Excellent,[br]we teach now, got it.

0:53:19.560,0:53:28.450
He says, I know where[br]this comes from.

0:53:28.450,0:53:31.650
I've got it, x squared[br]plus y squared over 2.

0:53:31.650,0:53:32.850
How did he do it?

0:53:32.850,0:53:33.900
He's a genius.

0:53:33.900,0:53:39.560
No he's not, he's just[br]learning from the first time

0:53:39.560,0:53:40.760
when he failed.

0:53:40.760,0:53:44.180
And now he knows what he has[br]to do and his brain says,

0:53:44.180,0:53:47.230
oh, I got it.

0:53:47.230,0:53:50.110
Now, [INAUDIBLE] could[br]have applied this method

0:53:50.110,0:53:54.640
and solved the coupled[br]system and do it slowly

0:53:54.640,0:53:56.820
and it would have taken[br]him another 10 minutes.

0:53:56.820,0:54:00.100
And he's in the final, he[br]doesn't have time to spare.

0:54:00.100,0:54:06.030
If he can guess the potential[br]and then verify that,

0:54:06.030,0:54:07.720
it's going to be easy for him.

0:54:07.720,0:54:08.315
Why is that?

0:54:08.315,0:54:15.490
This is going to be 2x over 2[br]x i, and this is 2y over 2 y j.

0:54:15.490,0:54:17.838
So yeah, he was right.

0:54:17.838,0:54:21.806


0:54:21.806,0:54:26.634
All right, let me[br]give you another one.

0:54:26.634,0:54:29.104
Let's see who gets this one.

0:54:29.104,0:54:34.080
F is a vector valued[br]function, maybe a force field,

0:54:34.080,0:54:35.341
that is this.

0:54:35.341,0:54:38.047


0:54:38.047,0:54:41.640
Of course there are many[br]ways-- maybe somebody's

0:54:41.640,0:54:44.180
going to ask you[br]to prove this is

0:54:44.180,0:54:49.320
conservative by the criterion,[br]but they shouldn't tell you

0:54:49.320,0:54:51.550
how to do it.

0:54:51.550,0:54:53.050
So show this is conservative.

0:54:53.050,0:54:56.550
If somebody doesn't want the[br]scalar potential because they

0:54:56.550,0:54:57.890
don't need it, let's say.

0:54:57.890,0:55:01.945
Well, prime f1 with respect to[br]y, I'll prime this with respect

0:55:01.945,0:55:03.400
to x.

0:55:03.400,0:55:07.560
f1 prime with respect to[br]y equals 2x is the same

0:55:07.560,0:55:09.720
as f2 prime with respect with.

0:55:09.720,0:55:13.600
Yeah, it is conservative, I[br]know it from the criterion.

0:55:13.600,0:55:16.570
But [INAUDIBLE] knows[br]that later I will ask him

0:55:16.570,0:55:19.150
for the scalar potential.

0:55:19.150,0:55:25.610
And I wonder if he can find[br]it for me without computing it

0:55:25.610,0:55:27.030
by solving the system.

0:55:27.030,0:55:30.265
Just from his[br]mathematical intuition

0:55:30.265,0:55:33.036
that is running in the[br]background of your--

0:55:33.036,0:55:35.880
STUDENT: x squared,[br]multiply y [INAUDIBLE].

0:55:35.880,0:55:38.405
PROFESSOR: x squared[br]y, excellent.

0:55:38.405,0:55:42.770


0:55:42.770,0:55:47.140
Zach came up with[br]it and anybody else?

0:55:47.140,0:55:49.090
Alex?

0:55:49.090,0:55:51.610
So all three of you, OK?

0:55:51.610,0:55:54.060
Squared y, very good.

0:55:54.060,0:55:56.030
Was it hard?

0:55:56.030,0:55:59.480
Yeah, it's hard for most people.

0:55:59.480,0:56:02.720
It was hard for me[br]when I first saw

0:56:02.720,0:56:05.530
that in the first 30[br]minutes of becoming familiar

0:56:05.530,0:56:09.283
with the scalar[br]potential, I was 18 or 19.

0:56:09.283,0:56:11.928
But then I got it in[br]about half an hour

0:56:11.928,0:56:15.536
and I was able to[br]do them mentally.

0:56:15.536,0:56:20.430
Most of the examples I[br]got were really nice.

0:56:20.430,0:56:28.730
Were on purpose made nice for[br]us for the exam to work fast.

0:56:28.730,0:56:35.790
And now let's see why[br]would the work really not

0:56:35.790,0:56:38.620
depend on the trajectory[br]you are taking

0:56:38.620,0:56:41.070
if your force is conservative.

0:56:41.070,0:56:42.955
If the force is[br]conservative there

0:56:42.955,0:56:46.440
is something magic[br]that's going to happen.

0:56:46.440,0:56:50.600
And we really don't[br]know what that is,

0:56:50.600,0:56:53.160
but we should be able to prove.

0:56:53.160,0:56:57.600


0:56:57.600,0:57:19.890
So theorem, actually[br]this is funny.

0:57:19.890,0:57:25.511
It's called the fundamental[br]theorem of path integrals

0:57:25.511,0:57:28.457
but it's the fundamental[br]theorem of calculus 3.

0:57:28.457,0:57:36.313
I'm going to write it like[br]this, the fundamental theorem

0:57:36.313,0:57:45.170
of calc 3, path integrals.

0:57:45.170,0:57:54.708
It's also called- 13.3,[br]section- Independence of path.

0:57:54.708,0:58:01.980


0:58:01.980,0:58:11.733
So remember you have a[br]work, w, over a path, c.

0:58:11.733,0:58:22.610
F dot dR where there R is the[br]regular parametrized curve

0:58:22.610,0:58:23.110
overseen.

0:58:23.110,0:58:33.850


0:58:33.850,0:58:36.850
This is called a[br]supposition vector.

0:58:36.850,0:58:40.850


0:58:40.850,0:58:45.730
Regular meaning c1, and[br]never vanishing speed,

0:58:45.730,0:58:49.400
the velocity never vanishes.

0:58:49.400,0:58:55.025
Velocity times 0 such[br]that f is continuous,

0:58:55.025,0:59:01.580
or a nice enough integral.

0:59:01.580,0:59:07.950


0:59:07.950,0:59:25.920
If F is conservative of[br]scalar potential, little f,

0:59:25.920,0:59:39.430
then the work, w, equals[br]little f at the endpoint

0:59:39.430,0:59:41.560
minus little f at the origin.

0:59:41.560,0:59:44.730


0:59:44.730,0:59:56.682
Where, by origin and endpoint[br]are those for the path,

0:59:56.682,1:00:00.618
are those for the arc, are[br]those for the curve, c.

1:00:00.618,1:00:05.060


1:00:05.060,1:00:09.576
So the work, the w, will[br]be independent of time.

1:00:09.576,1:00:18.720
So w will be independent of f.

1:00:18.720,1:00:21.120
And you saw an[br]example when I took

1:00:21.120,1:00:24.260
a conservative function[br]that was really nice,

1:00:24.260,1:00:27.620
y times i plus x times j.

1:00:27.620,1:00:29.440
That was the force field.

1:00:29.440,1:00:33.470
Because that was[br]conservative, we got w being 1

1:00:33.470,1:00:34.780
no matter what path we took.

1:00:34.780,1:00:37.970
We took a parabola, we[br]took a straight line,

1:00:37.970,1:00:39.770
and we could have[br]taken a zig-zag

1:00:39.770,1:00:41.860
and we still get w equals 1.

1:00:41.860,1:00:44.550
So no matter what[br]path you are taking.

1:00:44.550,1:00:46.572
Can we prove this?

1:00:46.572,1:00:53.000
Well, regular classes don't[br]prove anything, almost nothing.

1:00:53.000,1:00:56.900
But we are honor students[br]so lets see what we can do.

1:00:56.900,1:00:59.210
We have to understand[br]what's going on.

1:00:59.210,1:01:06.520
Why do we have this fundamental[br]theorem of calculus 3?

1:01:06.520,1:01:15.610
The work, w, can be expressed--[br]assume f is conservative

1:01:15.610,1:01:19.700
which means it's going to come[br]from a potential little f.

1:01:19.700,1:01:29.077
Where f is [INAUDIBLE] scalar[br]function over my domain, omega.

1:01:29.077,1:01:33.280


1:01:33.280,1:01:36.670
Now, the curve, c,[br]is part of this omega

1:01:36.670,1:01:40.490
so I don't have any[br]problems on the curve.

1:01:40.490,1:01:44.680
w will be rewritten beautifully.

1:01:44.680,1:01:46.625
So I'm giving you a[br]sketch of a proof.

1:01:46.625,1:01:50.335
But you would be able to do[br]this, maybe even better than me

1:01:50.335,1:01:56.310
because I have taught[br]you what you need to do.

1:01:56.310,1:02:03.580
So this is going to[br]be f1 i, plus f2 j.

1:02:03.580,1:02:09.750
And I'm going to write it.[br]f1 times-- what is this guys?

1:02:09.750,1:02:14.420
dR, I taught you, you taught[br]me, x prime of t, right?

1:02:14.420,1:02:23.010
Plus f2 times y prime of t,[br]all dt, and time from t0 to t1.

1:02:23.010,1:02:27.480
I start my motion along[br]the curve at t equals t0

1:02:27.480,1:02:30.728
and I finished my[br]motion at t equals t1.

1:02:30.728,1:02:33.620


1:02:33.620,1:02:36.770
Do I know where f1 and f2 are?

1:02:36.770,1:02:38.860
This is the point,[br]that's the whole point,

1:02:38.860,1:02:41.780
I know who they are, thank God.

1:02:41.780,1:02:47.310
And now I have to again[br]apply some magical think,

1:02:47.310,1:02:49.650
I'll ask you in a[br]minute what that is.

1:02:49.650,1:02:53.510
So, what is f1?

1:02:53.510,1:02:56.100
df dx, or f sub base.

1:02:56.100,1:02:59.260
If you don't like f sub base,[br]if you don't like my notation,

1:02:59.260,1:03:00.600
you put f sub x, right?

1:03:00.600,1:03:02.551
And this df dy.

1:03:02.551,1:03:03.050
Why?

1:03:03.050,1:03:05.340
Because it's[br]conservative and that

1:03:05.340,1:03:08.080
was the gradient of little f.

1:03:08.080,1:03:11.060
Of course I'm using the fact[br]that the first component would

1:03:11.060,1:03:13.610
be the partial of little[br]f with respect to x.

1:03:13.610,1:03:16.032
The second component would[br]be the partial of little

1:03:16.032,1:03:16.865
f with respect to y.

1:03:16.865,1:03:19.270
Have you seen this[br]formula before?

1:03:19.270,1:03:23.599
What in the world[br]is this formula?

1:03:23.599,1:03:25.502
STUDENT: It's the chain rule?

1:03:25.502,1:03:26.828
PROFESSOR: It's the chain rule.

1:03:26.828,1:03:30.630
I don't have a dollar but I[br]will give you a dollar, OK?

1:03:30.630,1:03:33.960
Imagine a virtual dollar.

1:03:33.960,1:03:35.190
This is the chain rule.

1:03:35.190,1:03:38.110
So by the chain[br]rule we can write

1:03:38.110,1:03:41.350
this to be the[br]derivative with respect

1:03:41.350,1:03:48.840
to t of little f of[br]x of t, and y of t.

1:03:48.840,1:03:52.630
Alrighty, so I know[br]what I'm doing.

1:03:52.630,1:03:57.860
I know that by chain rule I had[br]little f evaluated at x of t,

1:03:57.860,1:04:01.640
y of t, and time t,[br]prime with respect to t.

1:04:01.640,1:04:07.750
Now when we take the fundamental[br]theorem of calculus, FTC.

1:04:07.750,1:04:13.540
That reminds me, I was[br]teaching calc 1 a few years ago

1:04:13.540,1:04:17.390
and I said, that's the[br]Federal Trade Commission.

1:04:17.390,1:04:20.755
Federal Trade Commission,[br]fundamental theorem

1:04:20.755,1:04:22.450
of calculus.

1:04:22.450,1:04:27.988
So coming back to[br]what I have, I prove

1:04:27.988,1:04:36.890
that w is the Federal[br]Trade Commission, no.

1:04:36.890,1:04:43.450
w is the application[br]of something

1:04:43.450,1:04:49.719
that we knew from calc[br]1, which is beautiful.

1:04:49.719,1:05:00.460
f of xt, y of t dt, this[br]is nothing but what?

1:05:00.460,1:05:05.530
Little f evaluated at-- I'm[br]going to have to write it down,

1:05:05.530,1:05:07.450
this whole sausage.

1:05:07.450,1:05:13.700
f of x of t1, y of t1,[br]minus f of x of t0, y of t0.

1:05:13.700,1:05:17.240
For somebody as lazy as[br]I am, that they effort.

1:05:17.240,1:05:19.490
How can I write It better?

1:05:19.490,1:05:26.060
f at the endpoint[br]minus f at the origin.

1:05:26.060,1:05:29.450
And of course, we are trying to[br]be quite rigorous in the book.

1:05:29.450,1:05:31.480
We would never say[br]that in the book.

1:05:31.480,1:05:34.650
We actually denote[br]the first point

1:05:34.650,1:05:37.690
with p, the origin, and[br]the endpoint with q.

1:05:37.690,1:05:41.808
So we say, f of q minus f of p.

1:05:41.808,1:05:49.920
And we proved q e d, we[br]proved the fundamental theorem

1:05:49.920,1:05:53.000
of path integrals, the[br]independence of that.

1:05:53.000,1:06:00.200
So that means the work[br]is independent of path

1:06:00.200,1:06:03.000
when the force is conservative.

1:06:03.000,1:06:07.355
Now attention, if f is not[br]conservative you are dead meat.

1:06:07.355,1:06:12.500
You cannot say what I just said.

1:06:12.500,1:06:16.390
So I'll give you two[br]separate examples

1:06:16.390,1:06:19.692
and let's see how we[br]solve each of them.

1:06:19.692,1:06:27.400


1:06:27.400,1:06:30.400
A final exam type of[br]problem-- every final exam

1:06:30.400,1:06:36.100
contains an[br]application like that.

1:06:36.100,1:06:41.940
Even the force field, f,[br]or the vector value, f.

1:06:41.940,1:06:44.056
Is it conservative?

1:06:44.056,1:06:51.600
Prove what is proved and[br]after that-- so the path

1:06:51.600,1:06:54.642
integral in any way you can.

1:06:54.642,1:06:58.020
If it's conservative you're[br]really lucky because you're

1:06:58.020,1:06:58.520
in business.

1:06:58.520,1:06:59.900
You don't have to do any work.

1:06:59.900,1:07:03.585
You just find the little[br]scalar potential evaluated

1:07:03.585,1:07:08.500
at the endpoints and subtract,[br]and that's your answer.

1:07:08.500,1:07:13.070
So I'm going to give you an[br]example of a final exam problem

1:07:13.070,1:07:15.545
that happened in the past year.

1:07:15.545,1:07:18.515
So, a final type exam problem.

1:07:18.515,1:07:25.445


1:07:25.445,1:07:37.242
f of xy equals 2xy[br]i plus x squared j,

1:07:37.242,1:07:43.500
over r over r squared.

1:07:43.500,1:07:45.750
STUDENT: Didn't we just[br]do [INAUDIBLE] that?

1:07:45.750,1:07:49.150
PROFESSOR: Well, I just did[br]that but I changed the problem.

1:07:49.150,1:07:51.242
I wanted to keep the[br]same force field.

1:07:51.242,1:07:51.950
STUDENT: Alright.

1:07:51.950,1:07:52.810
PROFESSOR: OK.

1:07:52.810,1:08:18.319
Compute the work, w,[br]performed by f along the arc

1:08:18.319,1:08:23.725
of the circle in the picture.

1:08:23.725,1:08:26.710


1:08:26.710,1:08:27.930
And they draw a picture.

1:08:27.930,1:08:31.649
And they do a picture for you,[br]and you stare at this picture

1:08:31.649,1:08:49.290
and-- So you say,[br]oh my God, if I

1:08:49.290,1:08:53.046
were to parametrize it[br]would be a little bit of-- I

1:08:53.046,1:08:56.259
could, but it would be[br]a little bit of work.

1:08:56.259,1:09:00.340
I would have x equals 2[br]cosine t, y equals sine t.

1:09:00.340,1:09:05.029
I would have to plug in and[br]do that whole work, definition

1:09:05.029,1:09:06.270
with parametrization.

1:09:06.270,1:09:08.210
Do you have to parametrize?

1:09:08.210,1:09:10.149
Not in this case, why?

1:09:10.149,1:09:12.729
Because the f is conservative.

1:09:12.729,1:09:19.470
If they ask you- some professors[br]give hints, most of them

1:09:19.470,1:09:22.558
are nice and give hints-[br]show f is conservative.

1:09:22.558,1:09:28.859


1:09:28.859,1:09:30.729
So that's a big[br]hint in the sense

1:09:30.729,1:09:33.095
that you see it[br]immediately, how you do it.

1:09:33.095,1:09:36.149
You have 2xy prime[br]with respect to y,

1:09:36.149,1:09:39.710
is 2x, which is x squared[br]prime with respect to x.

1:09:39.710,1:09:41.700
So it is conservative.

1:09:41.700,1:09:44.350
But he or she told you more.

1:09:44.350,1:09:46.859
He said, I'm selling[br]you something here,

1:09:46.859,1:09:50.080
you have to get your[br]own scalar potential.

1:09:50.080,1:09:56.920
And you did, and[br]you got x squared y.

1:09:56.920,1:09:58.970
Now, most of the[br]scalar potentials

1:09:58.970,1:10:01.620
that we are giving[br]you on the exam

1:10:01.620,1:10:04.196
can be seen with naked eyes.

1:10:04.196,1:10:06.770
You wouldn't have to do[br]all the integration of that

1:10:06.770,1:10:09.540
coupled system with respect[br]to x, with respect to y,

1:10:09.540,1:10:14.160
integrate backwards,[br]and things like that.

1:10:14.160,1:10:16.700
What do I need to do[br]in that case guys?

1:10:16.700,1:10:19.018
Say in words.

1:10:19.018,1:10:22.480
Since the force is[br]conservative, just two lines.

1:10:22.480,1:10:25.800
I'm applying the fundamental[br]theorem of calculus.

1:10:25.800,1:10:29.650
I'm applying the fundamental[br]theorem of path integrals.

1:10:29.650,1:10:33.050
I know the work is[br]independent of that.

1:10:33.050,1:10:38.750
So w, in this case[br]is already there,

1:10:38.750,1:10:43.230
is going to be f at the point.

1:10:43.230,1:10:48.006
I didn't say how I'm going to[br]travel it, in what direction.

1:10:48.006,1:10:52.028
f of q minus f of p.

1:10:52.028,1:10:57.360


1:10:57.360,1:11:00.630
And you'll say, well[br]what does it mean?

1:11:00.630,1:11:02.490
How do we do that?

1:11:02.490,1:11:05.890
That means x squared[br]y, evaluated at q,

1:11:05.890,1:11:09.940
who the heck is q?

1:11:09.940,1:11:13.710
Attention, negative 2,[br]0, Matt, you got it?

1:11:13.710,1:11:14.920
OK, right?

1:11:14.920,1:11:17.770
So, are you guys with me?

1:11:17.770,1:11:19.550
Right?

1:11:19.550,1:11:22.520
And p is 2, 0.

1:11:22.520,1:11:31.690
So at negative 2, 0,[br]minus x squared y at 2, 0.

1:11:31.690,1:11:35.170
So, if you set 0, big, as[br]you knew that you got 0,

1:11:35.170,1:11:38.235
that is the answer.

1:11:38.235,1:11:43.830
Now if somebody would give you[br]a wiggly look that this guy's

1:11:43.830,1:11:49.640
wrong, then he took this past.

1:11:49.640,1:11:52.740
He's going to do[br]exactly the same work

1:11:52.740,1:11:57.800
if he's under influence the[br]same conservative force.

1:11:57.800,1:12:01.254
If the force acting[br]on it is the same.

1:12:01.254,1:12:04.265
No matter what path you[br]are taking-- yes, sir?

1:12:04.265,1:12:07.320
STUDENT: Is it still working[br]for your self-intersecting.

1:12:07.320,1:12:11.420
PROFESSOR: Yeah, because[br]you're not stopping,

1:12:11.420,1:12:13.120
it works for any[br]parametrization.

1:12:13.120,1:12:16.390
So if you're able to parametrize[br]that as a differentiable

1:12:16.390,1:12:20.640
function so that the[br]derivative would never vanish,

1:12:20.640,1:12:23.090
it's going to work, right?

1:12:23.090,1:12:27.412
All right, it can work also[br]for this piecewise contour

1:12:27.412,1:12:29.300
or any other path point.

1:12:29.300,1:12:32.720
As long as it starts and[br]it ends at the same point

1:12:32.720,1:12:38.020
and as long as your[br]conservative force is the same.

1:12:38.020,1:12:39.890
So that force is very[br][INAUDIBLE], yes?

1:12:39.890,1:12:42.536
STUDENT: Do [INAUDIBLE] graphs[br]but the endpoints the same.

1:12:42.536,1:12:44.840
And if they are conservative[br]then [INTERPOSING VOICES].

1:12:44.840,1:12:46.506
PROFESSOR: If everything[br]is conservative

1:12:46.506,1:12:49.550
the work along this[br]path is the same.

1:12:49.550,1:12:52.080
There were people[br]who played games

1:12:52.080,1:12:54.254
like that to catch if[br]the student knows what

1:12:54.254,1:12:55.420
they are talking-- yes, sir?

1:12:55.420,1:12:59.067
STUDENT: So, in a[br]problem, if they wanted

1:12:59.067,1:13:00.733
to find the work[br]couldn't we simplify it

1:13:00.733,1:13:02.539
by just saying, we[br]need to find-- because,

1:13:02.539,1:13:05.080
since we're in R2, couldn't we[br]just say it's a straight line?

1:13:05.080,1:13:07.132
Because that's the--[br]Like, instead of a curve

1:13:07.132,1:13:08.660
we could just set the straight[br]line [INTERPOSING VOICES].

1:13:08.660,1:13:10.260
PROFESSOR: If it[br]were moving from here

1:13:10.260,1:13:12.270
to here in a straight[br]line you would still

1:13:12.270,1:13:13.940
get the same answer.

1:13:13.940,1:13:17.810
And you could have-- if[br]you did that, actually,

1:13:17.810,1:13:21.170
if you were to compute this[br]kind of work on a straight line

1:13:21.170,1:13:26.120
it has-- let me[br]show you something.

1:13:26.120,1:13:32.470
You see, when you compute[br]dx y dx, plus x squared dy.

1:13:32.470,1:13:38.460
If y is 0 like Matthew said,[br]I'm walking on-- I'm not drunk.

1:13:38.460,1:13:41.350
I'm walking straight.

1:13:41.350,1:13:46.810
y will be 0 here, and 0 here,[br]and the integral will be 0.

1:13:46.810,1:13:50.580
So he would have noticed[br]that from the beginning.

1:13:50.580,1:13:54.240
But unless you know the[br]force is conservative,

1:13:54.240,1:13:57.890
there is no guarantee[br]that on another path

1:13:57.890,1:14:01.490
you don't have a[br]different answer, right?

1:14:01.490,1:14:05.230
So, let me give you another[br]example because now Matthew

1:14:05.230,1:14:06.610
brought this up.

1:14:06.610,1:14:09.230


1:14:09.230,1:14:13.620
A catchy example[br]that a professor

1:14:13.620,1:14:18.470
gave just to make the[br]students life miserable,

1:14:18.470,1:14:22.740
and I'll show it[br]to you in a second.

1:14:22.740,1:14:29.060


1:14:29.060,1:14:36.860
He said, for the picture--[br]very similar to this one,

1:14:36.860,1:14:39.924
just to make people confused.

1:14:39.924,1:14:43.270
Somebody gives you[br]this arc of a circle

1:14:43.270,1:14:45.660
and you travel from a to b.

1:14:45.660,1:14:49.500
And this the thing.

1:14:49.500,1:15:01.850
And he says, compute w for[br]the force given by y i plus j.

1:15:01.850,1:15:07.760
And the students[br]said OK, I guess

1:15:07.760,1:15:12.440
I'm going to get 0 because[br]I'm going to get something

1:15:12.440,1:15:16.610
like y dx plus x dy.

1:15:16.610,1:15:21.699
And if y is 0, I get 0, and[br]that way you wouldn't be 0

1:15:21.699,1:15:22.240
and I'm done.

1:15:22.240,1:15:27.489
No, it's not how you[br]think because this is not

1:15:27.489,1:15:30.210
conservative.

1:15:30.210,1:15:33.800
So you cannot say I can change[br]my path and it's still going

1:15:33.800,1:15:35.820
to be the same.

1:15:35.820,1:15:38.260
No, why is this[br]not conservative?

1:15:38.260,1:15:39.980
Quickly, this prime[br]with respect to y

1:15:39.980,1:15:42.580
is 1, this prime with[br]respect to x is 0.

1:15:42.580,1:15:46.055
So 1 different from[br]0 so, oh my God, no.

1:15:46.055,1:15:48.160
In that case, why do we do?

1:15:48.160,1:15:52.480
We have no other choice but[br]say, x equals 2 cosine t,

1:15:52.480,1:15:57.770
y equals 2 sine t, and[br]t between 0 and pi.

1:15:57.770,1:16:04.110
And then I get integral of[br]f1, y, what the heck is y?

1:16:04.110,1:16:08.948
2 sine t, times x prime of t.

1:16:08.948,1:16:15.360


1:16:15.360,1:16:20.850
Yeah, minus 2 sine[br]t, this is x prime.

1:16:20.850,1:16:24.650
Plus 1, are you guys with me?

1:16:24.650,1:16:30.140
Times y prime, which[br]is 2 cosine t, dt.

1:16:30.140,1:16:33.370
And t between 0 and pi.

1:16:33.370,1:16:36.190


1:16:36.190,1:16:42.242
And you get something[br]ugly, you get 0 to pi.

1:16:42.242,1:16:43.660
What is the nice thing?

1:16:43.660,1:16:48.845
When you integrate this with[br]respect to t, you get sine t.

1:16:48.845,1:16:53.480
And thank God, sine, whether you[br]are at 0 or if pi is still 0.

1:16:53.480,1:16:55.600
So this part will disappear.

1:16:55.600,1:17:01.800
So all you have left is[br]minus 4 sine squared dt.

1:17:01.800,1:17:04.760
But you are not done[br]so compute this at home

1:17:04.760,1:17:07.438
because we are out of time.

1:17:07.438,1:17:11.075
So don't jump to[br]conclusions unless you know

1:17:11.075,1:17:12.894
that the force is conservative.

1:17:12.894,1:17:14.878
If your force is[br]not conservative

1:17:14.878,1:17:17.688
then things are going[br]to look very ugly

1:17:17.688,1:17:20.830
and your only chance is to go[br]back to the parametrization,

1:17:20.830,1:17:24.302
to the basics.

1:17:24.302,1:17:27.310
So we are practically[br]done with 13.3

1:17:27.310,1:17:31.320
but I want to watch[br]more examples next time.

1:17:31.320,1:17:33.120
And I'll send you the homework.

1:17:33.120,1:17:38.220
Over the weekend you would be[br]able to start doing homework.

1:17:38.220,1:17:40.620
Now, when shall I[br]grab the homework?

1:17:40.620,1:17:43.320
What if I closed it right[br]before the final, is that?

1:17:43.320,1:17:44.220
STUDENT: Yeah.

1:17:44.220,1:17:46.070
PROFESSOR: Yeah?

1:17:46.070,2:39:17.402