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So, let me remind you,
yesterday we've defined and
8
00:00:28 --> 00:00:34
started to compute line
integrals for work as a vector
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00:00:34 --> 00:00:41
field along a curve.
So, we have a curve in the
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plane, C.
We have a vector field that
11
00:00:48 --> 00:00:55
gives us a vector at every
point.
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00:00:55 --> 00:01:04
And, we want to find the work
done along the curve.
13
00:01:04 --> 00:01:11
So, that's the line integral
along C of F dr,
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00:01:11 --> 00:01:16
or more geometrically,
line integral along C of F.T ds
15
00:01:16 --> 00:01:19
where T is the unit tangent
vector,
16
00:01:19 --> 00:01:23
and ds is the arc length
element.
17
00:01:23 --> 00:01:30
Or, in coordinates,
that they integral of M dx N dy
18
00:01:30 --> 00:01:38
where M and N are the components
of the vector field.
19
00:01:38 --> 00:01:46
OK, so -- Let's do an example
that will just summarize what we
20
00:01:46 --> 00:01:51
did yesterday,
and then we will move on to
21
00:01:51 --> 00:01:57
interesting observations about
these things.
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00:01:57 --> 00:02:03
So, here's an example we are
going to look at now.
23
00:02:03 --> 00:02:11
Let's say I give you the vector
field yi plus xj.
24
00:02:11 --> 00:02:13
So, it's not completely obvious
what it looks like,
25
00:02:13 --> 00:02:16
but here is a computer plot of
that vector field.
26
00:02:16 --> 00:02:21
So, that tells you a bit what
it does.
27
00:02:21 --> 00:02:24
It points in all sorts of
directions.
28
00:02:24 --> 00:02:31
And, let's say we want to find
the work done by this vector
29
00:02:31 --> 00:02:35
field.
If I move along this closed
30
00:02:35 --> 00:02:40
curve, I start at the origin.
But, I moved along the x-axis
31
00:02:40 --> 00:02:43
to one.
That move along the unit circle
32
00:02:43 --> 00:02:46
to the diagonal,
and then I move back to the
33
00:02:46 --> 00:02:54
origin in a straight line.
OK, so C consists of three
34
00:02:54 --> 00:03:06
parts -- -- so that you enclose
a sector of a unit disk -- --
35
00:03:06 --> 00:03:17
corresponding to angles between
zero and 45�.
36
00:03:17 --> 00:03:24
So, to compute this line
integral, all we have to do is
37
00:03:24 --> 00:03:33
we have set up three different
integrals and add that together.
38
00:03:33 --> 00:03:44
OK, so we need to set up the
integral of y dx plus x dy for
39
00:03:44 --> 00:03:54
each of these pieces.
So, let's do the first one on
40
00:03:54 --> 00:03:59
the x-axis.
Well, one way to parameterize
41
00:03:59 --> 00:04:06
that is just use the x variable.
And, say that because we are on
42
00:04:06 --> 00:04:12
the, let's see,
sorry, we are going from the
43
00:04:12 --> 00:04:17
origin to (1,0).
Well, we know we are on the
44
00:04:17 --> 00:04:22
x-axis.
So, y there is actually just
45
00:04:22 --> 00:04:25
zero.
And, the variable will be x
46
00:04:25 --> 00:04:28
from zero to one.
Or, if you prefer,
47
00:04:28 --> 00:04:31
you can parameterize things,
say, x equals t for t from zero
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00:04:31 --> 00:04:36
to one, and y equals zero.
What doesn't change is y is
49
00:04:36 --> 00:04:41
zero, and therefore,
dy is also zero.
50
00:04:41 --> 00:04:48
So, in fact,
we are integrating y dx x dy,
51
00:04:48 --> 00:04:53
but that becomes,
well, zero dx 0,
52
00:04:53 --> 00:04:59
and that's just going to give
you zero.
53
00:04:59 --> 00:05:03
OK, so there's the line
integral.
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00:05:03 --> 00:05:07
Here, it's very easy to compute.
Of course, you can also do it
55
00:05:07 --> 00:05:10
geometrically because
geometrically,
56
00:05:10 --> 00:05:12
you can see in the picture
along the x-axis,
57
00:05:12 --> 00:05:15
the vector field is pointing
vertically.
58
00:05:15 --> 00:05:19
If I'm on the x-axis,
my vector field is actually in
59
00:05:19 --> 00:05:23
the y direction.
So, it's perpendicular to my
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00:05:23 --> 00:05:24
curve.
So, the work done is going to
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00:05:24 --> 00:05:26
be zero.
F dot T will be zero.
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00:05:26 --> 00:05:48
63
00:05:48 --> 00:05:51
OK, so F dot T is zero,
so the integral is zero.
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00:05:51 --> 00:05:57
OK, any questions about this
first part of the calculation?
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00:05:57 --> 00:06:04
No? It's OK?
OK, let's move on to more
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00:06:04 --> 00:06:11
interesting part of it.
Let's do the second part,
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00:06:11 --> 00:06:18
which is a portion of the unit
circle.
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00:06:18 --> 00:06:24
OK, so I should have drawn my
picture.
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00:06:24 --> 00:06:37
And so now we are moving on
this part of the curve that's
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00:06:37 --> 00:06:40
C2.
And, of course we have to
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00:06:40 --> 00:06:43
choose how to express x and y in
terms of a single variable.
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00:06:43 --> 00:06:46
Well, most likely,
when you are moving on a
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00:06:46 --> 00:06:49
circle, you are going to use the
angle along the circle to tell
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00:06:49 --> 00:06:53
you where you are.
OK, so we're going to use the
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00:06:53 --> 00:06:56
angle theta as a parameter.
And we will say,
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00:06:56 --> 00:07:02
we are on the unit circle.
So, x is cosine theta and y is
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00:07:02 --> 00:07:05
sine theta.
What's the range of theta?
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00:07:05 --> 00:07:11
Theta goes from zero to pi over
four, OK?
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00:07:11 --> 00:07:15
So, whenever I see dx,
I will replace it by,
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00:07:15 --> 00:07:19
well, the derivative of cosine
is negative sine.
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00:07:19 --> 00:07:24
So, minus sine theta d theta,
and dy, the derivative of sine
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00:07:24 --> 00:07:29
is cosine.
So, it will become cosine theta
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00:07:29 --> 00:07:34
d theta.
OK, so I'm computing the
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00:07:34 --> 00:07:41
integral of y dx x dy.
That means -- -- I'll be
85
00:07:41 --> 00:07:52
actually computing the integral
of, so, y is sine theta.
86
00:07:52 --> 00:08:01
dx, that's negative sine theta
d theta plus x cosine.
87
00:08:01 --> 00:08:08
dy is cosine theta d theta from
zero to pi/4.
88
00:08:08 --> 00:08:17
OK, so that's integral from
zero to pi / 4 of cosine squared
89
00:08:17 --> 00:08:22
minus sine squared.
And, if you know your trig,
90
00:08:22 --> 00:08:27
then you should recognize this
as cosine of two theta.
91
00:08:27 --> 00:08:33
OK, so that will integrate to
one half of sine two theta from
92
00:08:33 --> 00:08:36
zero to pi over four,
sorry.
93
00:08:36 --> 00:08:44
And, sine pi over two is one.
So, you will get one half.
94
00:08:44 --> 00:08:49
OK, any questions about this
one?
95
00:08:49 --> 00:09:07
No?
OK, then let's do the third one.
96
00:09:07 --> 00:09:15
So, the third guy is when we
come back to the origin along
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00:09:15 --> 00:09:18
the diagonal.
OK, so we go in a straight line
98
00:09:18 --> 00:09:20
from this point.
Where's this point?
99
00:09:20 --> 00:09:25
Well, this point is one over
root two, one over root two.
100
00:09:25 --> 00:09:32
And, we go back to the origin.
OK, so we need to figure out a
101
00:09:32 --> 00:09:38
way to express x and y in terms
of the same parameter.
102
00:09:38 --> 00:09:43
So, one way which is very
natural would be to just say,
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00:09:43 --> 00:09:47
well, let's say we move from
here to here over time.
104
00:09:47 --> 00:09:50
And, at time zero, we are here.
At time one, we are here.
105
00:09:50 --> 00:09:54
We know how to parameterize
this line.
106
00:09:54 --> 00:10:02
So, what we could do is say,
let's parameterize this line.
107
00:10:02 --> 00:10:09
So, we start at one over root
two, and we go down by one over
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00:10:09 --> 00:10:19
root two in time one.
And, same with y.
109
00:10:19 --> 00:10:24
That's actually perfectly fine.
But that's unnecessarily
110
00:10:24 --> 00:10:27
complicated.
OK, why is a complicated?
111
00:10:27 --> 00:10:30
Because we will get all of
these expressions.
112
00:10:30 --> 00:10:34
It would be easier to actually
just look at motion in this
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00:10:34 --> 00:10:38
direction and then say,
well, if we have a certain work
114
00:10:38 --> 00:10:42
if we move from here to here,
then the work done moving from
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00:10:42 --> 00:10:46
here to here is just going to be
the opposite,
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00:10:46 --> 00:10:48
OK?
So, in fact,
117
00:10:48 --> 00:10:55
we can do slightly better by
just saying, well,
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00:10:55 --> 00:10:59
we'll take x = t,
y = t.
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00:10:59 --> 00:11:07
t from zero to one over root
two, and take,
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00:11:07 --> 00:11:15
well, sorry,
that gives us what I will call
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00:11:15 --> 00:11:25
minus C3, which is C3 backwards.
And then we can say the
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00:11:25 --> 00:11:31
integral for work along minus C3
is the opposite of the work
123
00:11:31 --> 00:11:35
along C3.
Or, if you're comfortable with
124
00:11:35 --> 00:11:39
integration where variables go
down,
125
00:11:39 --> 00:11:42
then you could also say that t
just goes from one over square
126
00:11:42 --> 00:11:45
root of two down to zero.
And, when you set up your
127
00:11:45 --> 00:11:49
integral, it will go from one
over root two to zero.
128
00:11:49 --> 00:11:50
And, of course,
that will be the negative of
129
00:11:50 --> 00:11:52
the one from zero to one over
root two.
130
00:11:52 --> 00:11:59
So, it's the same thing.
OK, so if we do it with this
131
00:11:59 --> 00:12:03
parameterization,
we'll get that,
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00:12:03 --> 00:12:08
well of course,
dx is dt, dy is dt.
133
00:12:08 --> 00:12:16
So, the integral along minus C3
of y dx plus x dy is just the
134
00:12:16 --> 00:12:24
integral from zero to one over
root two of t dt plus t dt.
135
00:12:24 --> 00:12:30
Sorry, I'm messing up my
blackboard, OK,
136
00:12:30 --> 00:12:37
which is going to be,
well, the integral of 2t dt,
137
00:12:37 --> 00:12:46
which is t2 between these
bounds, which is one half.
138
00:12:46 --> 00:12:51
That's the integral along minus
C3, along the reversed path.
139
00:12:51 --> 00:13:03
And, if I want to do it along
C3 instead, then I just take the
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00:13:03 --> 00:13:07
negative.
Or, if you prefer,
141
00:13:07 --> 00:13:11
you could have done it directly
with integral from one over root
142
00:13:11 --> 00:13:14
two, two zero,
which gives you immediately the
143
00:13:14 --> 00:13:19
negative one half.
OK, so at the end,
144
00:13:19 --> 00:13:28
we get that the total work --
-- was the sum of the three line
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00:13:28 --> 00:13:32
integrals.
I'm not writing after dr just
146
00:13:32 --> 00:13:36
to save space.
But, zero plus one half minus
147
00:13:36 --> 00:13:39
one half, and that comes out to
zero.
148
00:13:39 --> 00:13:44
So, a lot of calculations for
nothing.
149
00:13:44 --> 00:13:49
OK, so that should give you
overview of various ways to
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00:13:49 --> 00:13:57
compute line integrals.
Any questions about all that?
151
00:13:57 --> 00:14:03
No? OK.
So, next, let me tell you about
152
00:14:03 --> 00:14:07
how to avoid computing like
integrals.
153
00:14:07 --> 00:14:08
Well, one is easy:
don't take this class.
154
00:14:08 --> 00:14:17
But that's not,
so here's another way not to do
155
00:14:17 --> 00:14:20
it, OK?
So, let's look a little bit
156
00:14:20 --> 00:14:24
about one kind of vector field
that actually we've encountered
157
00:14:24 --> 00:14:26
a few weeks ago without saying
it.
158
00:14:26 --> 00:14:30
So, we said when we have a
function of two variables,
159
00:14:30 --> 00:14:32
we have the gradient vector.
Well, at the time,
160
00:14:32 --> 00:14:35
it was just a vector.
But, that vector depended on x
161
00:14:35 --> 00:14:36
and y.
So, in fact,
162
00:14:36 --> 00:14:43
it's a vector field.
OK, so here's an interesting
163
00:14:43 --> 00:14:48
special case.
Say that F, our vector field is
164
00:14:48 --> 00:14:52
actually the gradient of some
function.
165
00:14:52 --> 00:15:01
So, it's a gradient field.
And, so f is a function of two
166
00:15:01 --> 00:15:08
variables, x and y,
and that's called the potential
167
00:15:08 --> 00:15:12
for the vector field.
The reason is,
168
00:15:12 --> 00:15:16
of course, from physics.
In physics, you call potential,
169
00:15:16 --> 00:15:21
electrical potential or
gravitational potential,
170
00:15:21 --> 00:15:25
the potential energy.
This function of position that
171
00:15:25 --> 00:15:29
tells you how much actually
energy stored somehow by the
172
00:15:29 --> 00:15:33
force field, and this gradient
gives you the force.
173
00:15:33 --> 00:15:37
Actually, not quite.
If you are a physicist,
174
00:15:37 --> 00:15:40
that the force will be negative
the gradient.
175
00:15:40 --> 00:15:44
So, that means that physicists'
potentials are the opposite of a
176
00:15:44 --> 00:15:46
mathematician's potential.
Okay?
177
00:15:46 --> 00:15:48
So it's just here to confuse
you.
178
00:15:48 --> 00:15:50
It doesn't really matter all
the time.
179
00:15:50 --> 00:15:55
So to make things simpler we
are using this convention and
180
00:15:55 --> 00:15:59
you just put a minus sign if you
are doing physics.
181
00:15:59 --> 00:16:13
So then I claim that we can
simplify the evaluation of the
182
00:16:13 --> 00:16:21
line integral for work.
Perhaps you've seen in physics,
183
00:16:21 --> 00:16:24
the work done by,
say, the electrical force,
184
00:16:24 --> 00:16:28
is actually given by the change
in the value of a potential from
185
00:16:28 --> 00:16:30
the starting point of the ending
point,
186
00:16:30 --> 00:16:36
or same for gravitational force.
So, these are special cases of
187
00:16:36 --> 00:16:40
what's called the fundamental
theorem of calculus for line
188
00:16:40 --> 00:16:43
integrals.
So, the fundamental theorem of
189
00:16:43 --> 00:16:46
calculus, not for line
integrals, tells you if you
190
00:16:46 --> 00:16:49
integrate a derivative,
then you get back the function.
191
00:16:49 --> 00:16:52
And here, it's the same thing
in multivariable calculus.
192
00:16:52 --> 00:16:54
It tells you,
if you take the line integral
193
00:16:54 --> 00:16:58
of the gradient of a function,
what you get back is the
194
00:16:58 --> 00:16:58
function.
195
00:16:58 --> 00:17:23
196
00:17:23 --> 00:17:30
OK,
so -- -- the fundamental
197
00:17:30 --> 00:17:43
theorem of calculus for line
integrals -- -- says if you
198
00:17:43 --> 00:17:58
integrate a vector field that's
the gradient of a function along
199
00:17:58 --> 00:18:03
a curve,
let's say that you have a curve
200
00:18:03 --> 00:18:06
that goes from some starting
point, P0,
201
00:18:06 --> 00:18:15
to some ending point, P1.
All you will get is the value
202
00:18:15 --> 00:18:21
of F at P1 minus the value of F
at P0.
203
00:18:21 --> 00:18:25
OK, so, that's a pretty nifty
formula that only works if the
204
00:18:25 --> 00:18:28
field that you are integrating
is a gradient.
205
00:18:28 --> 00:18:32
You know it's a gradient,
and you know the function,
206
00:18:32 --> 00:18:35
little f.
I mean, we can't put just any
207
00:18:35 --> 00:18:38
vector field in here.
We have to put the gradient of
208
00:18:38 --> 00:18:41
F.
So, actually on Tuesday we'll
209
00:18:41 --> 00:18:47
see how to decide whether a
vector field is a gradient or
210
00:18:47 --> 00:18:49
not,
and if it is a gradient,
211
00:18:49 --> 00:18:52
how to find the potential
function.
212
00:18:52 --> 00:18:58
So, we'll cover that.
But, for now we need to try to
213
00:18:58 --> 00:19:05
figure out a bit more about
this, what it says,
214
00:19:05 --> 00:19:11
what it means physically,
how to think of it
215
00:19:11 --> 00:19:15
geometrically,
and so on.
216
00:19:15 --> 00:19:18
So, maybe I should say,
if you're trying to write this
217
00:19:18 --> 00:19:21
in coordinates,
because that's also a useful
218
00:19:21 --> 00:19:24
way to think about it,
if I give you the line integral
219
00:19:24 --> 00:19:27
along C,
so, the gradient field,
220
00:19:27 --> 00:19:29
the components are f sub x and
f sub y.
221
00:19:29 --> 00:19:36
So, it means I'm actually
integrating f sub x dx plus f
222
00:19:36 --> 00:19:38
sub y dy.
Or, if you prefer,
223
00:19:38 --> 00:19:42
that's the same thing as
actually integrating df.
224
00:19:42 --> 00:19:46
So, I'm integrating the
differential of a function,
225
00:19:46 --> 00:19:54
f.
Well then, that's the change in
226
00:19:54 --> 00:19:56
F.
And, of course,
227
00:19:56 --> 00:20:02
if you write it in this form,
then probably it's quite
228
00:20:02 --> 00:20:06
obvious to you that this should
be true.
229
00:20:06 --> 00:20:11
I mean, in this form,
actually it's the same
230
00:20:11 --> 00:20:15
statement as in single variable
calculus.
231
00:20:15 --> 00:20:17
OK, and actually that's how we
prove the theorem.
232
00:20:17 --> 00:20:27
So, let's prove this theorem.
How do we prove it?
233
00:20:27 --> 00:20:31
Well, let's say I give you a
curve and I ask you to compute
234
00:20:31 --> 00:20:34
this integral.
How will you do that?
235
00:20:34 --> 00:20:38
Well, the way you compute the
integral actually is by choosing
236
00:20:38 --> 00:20:41
a parameter, and expressing
everything in terms of that
237
00:20:41 --> 00:20:46
parameter.
So, we'll set,
238
00:20:46 --> 00:20:57
well, so we know it's f sub x
dx plus f sub y dy.
239
00:20:57 --> 00:21:03
And, we'll want to parameterize
C in the form x equals x of t.
240
00:21:03 --> 00:21:09
y equals y of t.
So, if we do that,
241
00:21:09 --> 00:21:12
then dx becomes x prime of t
dt.
242
00:21:12 --> 00:21:25
dy becomes y prime of t dt.
So, we know x is x of t.
243
00:21:25 --> 00:21:31
That tells us dx is x prime of
t dt.
244
00:21:31 --> 00:21:38
y is y of t gives us dy is y
prime of t dt.
245
00:21:38 --> 00:21:52
So, now what we are integrating
actually becomes the integral of
246
00:21:52 --> 00:22:05
f sub x times dx dt plus f sub y
times dy dt times dt.
247
00:22:05 --> 00:22:09
OK, but now,
here I recognize a familiar
248
00:22:09 --> 00:22:13
guy.
I've seen this one before in
249
00:22:13 --> 00:22:15
the chain rule.
OK, this guy,
250
00:22:15 --> 00:22:19
by the chain rule,
is the rate of change of f if I
251
00:22:19 --> 00:22:22
take x and y to be functions of
t.
252
00:22:22 --> 00:22:26
And, I plug those into f.
So, in fact,
253
00:22:26 --> 00:22:34
what I'm integrating is df dt
when I think of f as a function
254
00:22:34 --> 00:22:42
of t by just plugging x and y as
functions of t.
255
00:22:42 --> 00:22:51
And so maybe actually I should
now say I have sometimes t goes
256
00:22:51 --> 00:22:59
from some initial time,
let's say, t zero to t one.
257
00:22:59 --> 00:23:03
And now, by the usual
fundamental theorem of calculus,
258
00:23:03 --> 00:23:07
I know that this will be just
the change in the value of f
259
00:23:07 --> 00:23:09
between t zero and t one.
260
00:23:09 --> 00:23:36
261
00:23:36 --> 00:23:44
OK, so integral from t zero to
one of (df /dt) dt,
262
00:23:44 --> 00:23:52
well, that becomes f between t
zero and t one.
263
00:23:52 --> 00:23:55
f of what?
We just have to be a little bit
264
00:23:55 --> 00:23:58
careful here.
Well, it's not quite f of t.
265
00:23:58 --> 00:24:02
It's f seen as a function of t
by putting x of t and y of t
266
00:24:02 --> 00:24:07
into it.
So, let me read that carefully.
267
00:24:07 --> 00:24:15
What I'm integrating to is f of
x of t and y of t.
268
00:24:15 --> 00:24:19
Does that sound fair?
Yeah, and so,
269
00:24:19 --> 00:24:23
when I plug in t1,
I get the point where I am at
270
00:24:23 --> 00:24:26
time t1.
That's the endpoint of my curve.
271
00:24:26 --> 00:24:31
When I plug t0,
I will get the starting point
272
00:24:31 --> 00:24:37
of my curve, p0.
And, that's the end of the
273
00:24:37 --> 00:24:43
proof.
It wasn't that hard, see?
274
00:24:43 --> 00:24:56
OK, so let's see an example.
Well, let's look at that
275
00:24:56 --> 00:25:00
example again.
So, we have this curve.
276
00:25:00 --> 00:25:03
We have this vector field.
Could it be that,
277
00:25:03 --> 00:25:05
by accident,
that vector field was a
278
00:25:05 --> 00:25:08
gradient field?
So, remember,
279
00:25:08 --> 00:25:12
our vector field was y,
x.
280
00:25:12 --> 00:25:16
Can we think of a function
whose derivative with respect to
281
00:25:16 --> 00:25:19
x is y, and derivative with
respect to y is x?
282
00:25:19 --> 00:25:27
Yeah, x times y sounds like a
good candidate where f( x,
283
00:25:27 --> 00:25:30
y) is xy.
OK,
284
00:25:30 --> 00:25:34
so that means that the line
integrals that we computed along
285
00:25:34 --> 00:25:39
these things can be just
evaluated from just finding out
286
00:25:39 --> 00:25:44
the values of f at the endpoint?
So, here's version two of my
287
00:25:44 --> 00:25:50
plot where I've added the
contour plot of a function,
288
00:25:50 --> 00:25:53
x, y on top of the vector
field.
289
00:25:53 --> 00:25:57
Actually, they have a vector
field is still pointing
290
00:25:57 --> 00:26:00
perpendicular to the level
curves that we have seen,
291
00:26:00 --> 00:26:02
just to remind you.
And, so now,
292
00:26:02 --> 00:26:05
when we move,
now when we move,
293
00:26:05 --> 00:26:09
the origin is on the level
curve, f equals zero.
294
00:26:09 --> 00:26:14
And, when we start going along
C1, we stay on f equals zero.
295
00:26:14 --> 00:26:17
So, there's no work.
The potential doesn't change.
296
00:26:17 --> 00:26:21
Then on C2, the potential
increases from zero to one half.
297
00:26:21 --> 00:26:24
The work is one half.
And then, on C3,
298
00:26:24 --> 00:26:27
we go back down from one half
to zero.
299
00:26:27 --> 00:26:40
The work is negative one half.
See, that was much easier than
300
00:26:40 --> 00:26:47
computing.
So, for example,
301
00:26:47 --> 00:26:53
the integral along C2 is
actually just,
302
00:26:53 --> 00:27:00
so, C2 goes from one zero to
one over root two,
303
00:27:00 --> 00:27:09
one over root two.
So, that's one half minus zero,
304
00:27:09 --> 00:27:18
and that's one half,
OK, because C2 was going here.
305
00:27:18 --> 00:27:26
And, at this point, f is zero.
At that point, f is one half.
306
00:27:26 --> 00:27:29
And, similarly for the others,
and of course when you sum,
307
00:27:29 --> 00:27:33
you get zero because the total
change in f when you go from
308
00:27:33 --> 00:27:35
here,
to here, to here, to here,
309
00:27:35 --> 00:27:37
eventually you are back at the
same place.
310
00:27:37 --> 00:27:44
So, f hasn't changed.
OK, so that's a neat trick.
311
00:27:44 --> 00:27:48
And it's important conceptually
because a lot of the forces are
312
00:27:48 --> 00:27:53
gradients of potentials,
namely, gravitational force,
313
00:27:53 --> 00:27:57
electric force.
The problem is not every vector
314
00:27:57 --> 00:28:00
field is a gradient.
A lot of vector fields are not
315
00:28:00 --> 00:28:02
gradients.
For example,
316
00:28:02 --> 00:28:08
magnetic fields certainly are
not gradients.
317
00:28:08 --> 00:28:33
So -- -- a big warning:
everything today only applies
318
00:28:33 --> 00:28:48
if F is a gradient field.
OK, it's not true otherwise.
319
00:28:48 --> 00:29:07
320
00:29:07 --> 00:29:19
OK, still, let's see,
what are the consequences of
321
00:29:19 --> 00:29:30
the fundamental theorem?
So, just to put one more time
322
00:29:30 --> 00:29:39
this disclaimer,
if F is a gradient field -- --
323
00:29:39 --> 00:29:44
then what do we have?
Well, there's various nice
324
00:29:44 --> 00:29:47
features of work done by
gradient fields that are not too
325
00:29:47 --> 00:29:53
far off the vector fields.
So, one of them is this
326
00:29:53 --> 00:30:02
property of path independence.
OK, so the claim is if I have a
327
00:30:02 --> 00:30:05
line integral to compute,
that it doesn't matter which
328
00:30:05 --> 00:30:09
path I take as long as it goes
from point a to point b.
329
00:30:09 --> 00:30:14
It just depends on the point
where I start and the point
330
00:30:14 --> 00:30:18
where I end.
And, that's certainly false in
331
00:30:18 --> 00:30:22
general, but for a gradient
field that works.
332
00:30:22 --> 00:30:25
So if I have a point,
P0,
333
00:30:25 --> 00:30:28
a point, P1,
and I have two different paths
334
00:30:28 --> 00:30:33
that go there,
say, C1 and C2,
335
00:30:33 --> 00:30:39
so they go from the same point
to the same point but in
336
00:30:39 --> 00:30:44
different ways,
then in this situation,
337
00:30:44 --> 00:30:55
the line integral along C1 is
equal to the line integral along
338
00:30:55 --> 00:30:56
C2.
Well, actually,
339
00:30:56 --> 00:31:02
let me insist that this is only
for gradient fields by putting
340
00:31:02 --> 00:31:09
gradient F in here,
just so you don't get tempted
341
00:31:09 --> 00:31:19
to ever use this for a field
that's not a gradient field --
342
00:31:19 --> 00:31:28
-- if C1 and C2 have the same
start and end point.
343
00:31:28 --> 00:31:30
OK, how do you prove that?
Well, it's very easy.
344
00:31:30 --> 00:31:32
We just use the fundamental
theorem.
345
00:31:32 --> 00:31:35
It tells us,
if you compute the line
346
00:31:35 --> 00:31:38
integral along C1,
it's just F at this point minus
347
00:31:38 --> 00:31:41
F at this point.
If you do it for C2,
348
00:31:41 --> 00:31:45
well, the same.
So, they are the same.
349
00:31:45 --> 00:31:48
And for that you don't actually
even need to know what little f
350
00:31:48 --> 00:31:50
is.
You know in advance that it's
351
00:31:50 --> 00:31:53
going to be the same.
So, if I give you a vector
352
00:31:53 --> 00:31:56
field and I tell you it's the
gradient of mysterious function
353
00:31:56 --> 00:31:58
but I don't tell you what the
function is and you don't want
354
00:31:58 --> 00:32:00
to find out,
you can still use path
355
00:32:00 --> 00:32:03
independence,
but only if you know it's a
356
00:32:03 --> 00:32:03
gradient.
357
00:32:03 --> 00:32:25
358
00:32:25 --> 00:32:35
OK, I guess this one is dead.
So, that will stay here forever
359
00:32:35 --> 00:32:40
because nobody is tall enough to
erase it.
360
00:32:40 --> 00:32:49
When you come back next year
and you still see that formula,
361
00:32:49 --> 00:32:53
you'll see.
Yes, but there's no useful
362
00:32:53 --> 00:32:59
information here.
That's a good point.
363
00:32:59 --> 00:33:06
OK, so what's another
consequence?
364
00:33:06 --> 00:33:14
So, if you have a gradient
field, it's what's called
365
00:33:14 --> 00:33:19
conservative.
OK, so what a conservative
366
00:33:19 --> 00:33:21
field?
Well, the word conservative
367
00:33:21 --> 00:33:26
comes from the idea in physics;
if the conservation of energy.
368
00:33:26 --> 00:33:31
It tells you that you cannot
get energy for free out of your
369
00:33:31 --> 00:33:33
force field.
So,
370
00:33:33 --> 00:33:36
what it means is that in
particular,
371
00:33:36 --> 00:33:39
if you take a closed
trajectory,
372
00:33:39 --> 00:33:44
so a trajectory that goes from
some point back to the same
373
00:33:44 --> 00:33:52
point,
so, if C is a closed curve,
374
00:33:52 --> 00:34:03
then the work done along C --
-- is zero.
375
00:34:03 --> 00:34:06
OK, that's the definition of
what it means to be
376
00:34:06 --> 00:34:09
conservative.
If I take any closed curve,
377
00:34:09 --> 00:34:13
the work will always be zero.
On the contrary,
378
00:34:13 --> 00:34:17
not conservative means
somewhere there is a curve along
379
00:34:17 --> 00:34:21
which the work is not zero.
If you find a curve where the
380
00:34:21 --> 00:34:23
work is zero,
that's not enough to say it's
381
00:34:23 --> 00:34:26
conservative.
You have show that no matter
382
00:34:26 --> 00:34:30
what curve I give you,
if it's a closed curve,
383
00:34:30 --> 00:34:34
it will always be zero.
So, what that means concretely
384
00:34:34 --> 00:34:37
is if you have a force field
that conservative,
385
00:34:37 --> 00:34:42
then you cannot build somehow
some perpetual motion out of it.
386
00:34:42 --> 00:34:44
You can't build something that
will just keep going just
387
00:34:44 --> 00:34:47
powered by that force because
that force is actually not
388
00:34:47 --> 00:34:50
providing any energy.
After you've gone one loop
389
00:34:50 --> 00:34:53
around, nothings happened from
the point of view of the energy
390
00:34:53 --> 00:34:57
provided by that force.
There's no work coming from the
391
00:34:57 --> 00:34:59
force,
while if you have a force field
392
00:34:59 --> 00:35:02
that's not conservative than you
can try to actually maybe find a
393
00:35:02 --> 00:35:04
loop where the work would be
positive.
394
00:35:04 --> 00:35:07
And then, you know,
that thing will just keep
395
00:35:07 --> 00:35:08
running.
So actually,
396
00:35:08 --> 00:35:13
if you just look at magnetic
fields and transformers or power
397
00:35:13 --> 00:35:15
adapters,
and things like that,
398
00:35:15 --> 00:35:18
you precisely extract energy
from the magnetic field.
399
00:35:18 --> 00:35:20
Of course, I mean,
you actually have to take some
400
00:35:20 --> 00:35:22
power supply to maintain the
magnetic fields.
401
00:35:22 --> 00:35:25
But, so a magnetic field,
you could actually try to get
402
00:35:25 --> 00:35:31
energy from it almost for free.
A gravitational field or an
403
00:35:31 --> 00:35:35
electric field,
you can't.
404
00:35:35 --> 00:35:41
OK, so and now why does that
hold?
405
00:35:41 --> 00:35:43
Well, if I have a gradient
field,
406
00:35:43 --> 00:35:47
then if I try to compute this
line integral,
407
00:35:47 --> 00:35:50
I know it will be the value of
the function at the end point
408
00:35:50 --> 00:35:52
minus the value at the starting
point.
409
00:35:52 --> 00:36:04
But, they are the same.
So, the value is the same.
410
00:36:04 --> 00:36:09
So, if I have a gradient field,
and I do the line integral,
411
00:36:09 --> 00:36:13
then I will get f at the
endpoint minus f at the starting
412
00:36:13 --> 00:36:23
point.
But, they're the same point,
413
00:36:23 --> 00:36:32
so that's zero.
OK, so just to reinforce my
414
00:36:32 --> 00:36:38
warning that not every field is
a gradient field,
415
00:36:38 --> 00:36:45
let's look again at our
favorite vector field from
416
00:36:45 --> 00:36:49
yesterday.
So, our favorite vector field
417
00:36:49 --> 00:36:54
yesterday was negative y and x.
It's a vector field that just
418
00:36:54 --> 00:36:59
rotates around the origin
counterclockwise.
419
00:36:59 --> 00:37:07
Well, we said,
say you take just the unit
420
00:37:07 --> 00:37:17
circle -- -- for example,
counterclockwise.
421
00:37:17 --> 00:37:23
Well, remember we said
yesterday that the line integral
422
00:37:23 --> 00:37:28
of F dr, maybe I should say F
dot T ds now,
423
00:37:28 --> 00:37:34
because the vector field is
tangent to the circle.
424
00:37:34 --> 00:37:43
So, on the unit circle,
F is tangent to the curve.
425
00:37:43 --> 00:37:49
And so, F dot T is length F
times, well, length T.
426
00:37:49 --> 00:37:53
But, T is a unit vector.
So, it's length F.
427
00:37:53 --> 00:37:57
And, the length of F on the
unit circle was just one.
428
00:37:57 --> 00:38:02
So, that's the integral of 1 ds.
So, it's just the length of the
429
00:38:02 --> 00:38:07
circle that's 2 pi.
And 2 pi is definitely not zero.
430
00:38:07 --> 00:38:13
So, this vector field is not
conservative.
431
00:38:13 --> 00:38:17
And so, now we know actually
it's not the gradient of
432
00:38:17 --> 00:38:22
anything because if it were a
gradient, then it would be
433
00:38:22 --> 00:38:28
conservative and it's not.
So, it's an example of a vector
434
00:38:28 --> 00:38:33
field that is not conservative.
It's not path independent
435
00:38:33 --> 00:38:38
either by the way because,
see, if I go from here to here
436
00:38:38 --> 00:38:43
along the upper half circle or
along the lower half circle,
437
00:38:43 --> 00:38:46
in one case I will get pi.
In the other case I will get
438
00:38:46 --> 00:38:49
negative pi.
I don't get the same answer,
439
00:38:49 --> 00:38:54
and so on, and so on.
It just fails to have all of
440
00:38:54 --> 00:38:59
these properties.
So, maybe I will write that
441
00:38:59 --> 00:39:06
down.
It's not conservative,
442
00:39:06 --> 00:39:21
not path independent.
It's not a gradient.
443
00:39:21 --> 00:39:29
It doesn't have any of these
properties.
444
00:39:29 --> 00:39:38
OK, any questions?
Yes?
445
00:39:38 --> 00:39:40
How do you determine whether
something is a gradient or not?
446
00:39:40 --> 00:39:44
Well, that's what we will see
on Tuesday.
447
00:39:44 --> 00:39:50
Yes?
Is it possible that it's
448
00:39:50 --> 00:39:52
conservative and not path
independent, or vice versa?
449
00:39:52 --> 00:39:54
The answer is no;
these two properties are
450
00:39:54 --> 00:39:57
equivalent, and we are going to
see that right now.
451
00:39:57 --> 00:40:12
At least that's the plan.
OK, yes?
452
00:40:12 --> 00:40:15
Let's see, so you said if it's
not path independent,
453
00:40:15 --> 00:40:19
then we cannot draw level
curves that are perpendicular to
454
00:40:19 --> 00:40:23
it at every point.
I wouldn't necessarily go that
455
00:40:23 --> 00:40:25
far.
You might be able to draw
456
00:40:25 --> 00:40:27
curves that are perpendicular to
it.
457
00:40:27 --> 00:40:32
But they won't be the level
curves of a function for which
458
00:40:32 --> 00:40:34
this is the gradient.
I mean, you might still have,
459
00:40:34 --> 00:40:36
you know,
if you take, say,
460
00:40:36 --> 00:40:40
take his gradient field and
scale it that in strange ways,
461
00:40:40 --> 00:40:42
you know, multiply by two in
some places,
462
00:40:42 --> 00:40:45
by one in other places,
by five and some other places,
463
00:40:45 --> 00:40:48
you will get something that
won't be conservative anymore.
464
00:40:48 --> 00:40:51
And it will still be
perpendicular to the curves.
465
00:40:51 --> 00:40:56
So, it's more subtle than that,
but certainly if it's not
466
00:40:56 --> 00:41:01
conservative then it's not a
gradient, and you cannot do what
467
00:41:01 --> 00:41:04
we said.
And how to decide whether it is
468
00:41:04 --> 00:41:06
or not, they'll be Tuesday's
topic.
469
00:41:06 --> 00:41:17
So, for now,
I just want to figure out again
470
00:41:17 --> 00:41:31
actually, let's now state all
these properties -- Actually,
471
00:41:31 --> 00:41:41
let me first do one minute of
physics.
472
00:41:41 --> 00:41:48
So, let me just tell you again
what's the physics in here.
473
00:41:48 --> 00:42:00
So, it's a force field is the
gradient of a potential -- --
474
00:42:00 --> 00:42:07
so, I'll still keep my plus
signs.
475
00:42:07 --> 00:42:13
So, maybe I should say this is
minus physics.
476
00:42:13 --> 00:42:20
[LAUGHTER]
So, the work of F is the change
477
00:42:20 --> 00:42:31
in value of potential from one
endpoint to the other endpoint.
478
00:42:31 --> 00:42:45
[PAUSE ] And -- -- so,
you know, you might know about
479
00:42:45 --> 00:42:58
gravitational fields,
or electrical -- -- fields
480
00:42:58 --> 00:43:13
versus gravitational -- -- or
electrical potential.
481
00:43:13 --> 00:43:16
And, in case you haven't done
any 8.02 yet,
482
00:43:16 --> 00:43:19
electrical potential is also
commonly known as voltage.
483
00:43:19 --> 00:43:27
It's the one that makes it hurt
when you stick your fingers into
484
00:43:27 --> 00:43:33
the socket.
[LAUGHTER] Don't try it.
485
00:43:33 --> 00:43:45
OK, and so now,
conservativeness means no
486
00:43:45 --> 00:44:01
energy can be extracted for free
-- -- from the field.
487
00:44:01 --> 00:44:05
You can't just have, you know,
a particle moving in that field
488
00:44:05 --> 00:44:08
and going on in definitely,
faster and faster,
489
00:44:08 --> 00:44:11
or if there's actually
friction,
490
00:44:11 --> 00:44:25
then keep moving.
So, total energy is conserved.
491
00:44:25 --> 00:44:29
And, I guess,
that's why we call that
492
00:44:29 --> 00:44:43
conservative.
OK, so let's end with the recap
493
00:44:43 --> 00:44:57
of various equivalent
properties.
494
00:44:57 --> 00:45:04
OK, so the first property that
I will have for a vector field
495
00:45:04 --> 00:45:12
is that it's conservative.
So, to say that a vector field
496
00:45:12 --> 00:45:22
with conservative means that the
line integral is zero along any
497
00:45:22 --> 00:45:26
closed curve.
Maybe to clarify,
498
00:45:26 --> 00:45:32
sorry, along all closed curves,
OK, every closed curve;
499
00:45:32 --> 00:45:36
give me any closed curve,
I get zero.
500
00:45:36 --> 00:45:44
So, now I claim this is the
same thing as a second property,
501
00:45:44 --> 00:45:53
which is that the line integral
of F is path independent.
502
00:45:53 --> 00:45:56
OK, so that means if I have two
paths with the same endpoint,
503
00:45:56 --> 00:45:58
then I will get always the same
answer.
504
00:45:58 --> 00:46:03
Why is that equivalent?
Well, let's say that I am path
505
00:46:03 --> 00:46:06
independent.
If I am path independent,
506
00:46:06 --> 00:46:10
then if I take a closed curve,
well, it has the same endpoints
507
00:46:10 --> 00:46:13
as just the curve that doesn't
move at all.
508
00:46:13 --> 00:46:16
So, path independence tells me
instead of going all around,
509
00:46:16 --> 00:46:21
I could just stay where I am.
And then, the work would just
510
00:46:21 --> 00:46:24
be zero.
So, if I path independent,
511
00:46:24 --> 00:46:28
tonight conservative.
Conversely, let's say that I'm
512
00:46:28 --> 00:46:32
just conservative and I want to
check path independence.
513
00:46:32 --> 00:46:37
Well, so I have two points,
and then I had to paths between
514
00:46:37 --> 00:46:39
that.
I want to show that the work is
515
00:46:39 --> 00:46:43
the same.
Well, how I do that?
516
00:46:43 --> 00:46:49
C1 and C2, well,
I observe that if I do C1 minus
517
00:46:49 --> 00:46:54
C2, I get a closed path.
If I go first from here to
518
00:46:54 --> 00:46:57
here, and then back along that
one, I get a closed path.
519
00:46:57 --> 00:47:01
So, if I am conservative,
I should get zero.
520
00:47:01 --> 00:47:05
But, if I get zero on C1 minus
C2, it means that the work on C1
521
00:47:05 --> 00:47:10
and the work on C2 are the same.
See, so it's the same.
522
00:47:10 --> 00:47:19
It's just a different way to
think about the situation.
523
00:47:19 --> 00:47:24
More things that are
equivalent, I have two more
524
00:47:24 --> 00:47:29
things to say.
The third one,
525
00:47:29 --> 00:47:38
it's equivalent to F being a
gradient field.
526
00:47:38 --> 00:47:46
OK, so this is equivalent to
the third property.
527
00:47:46 --> 00:47:59
F is a gradient field.
Why?
528
00:47:59 --> 00:48:02
Well, if we know that it's a
gradient field,
529
00:48:02 --> 00:48:05
that we've seen that we get
these properties out of the
530
00:48:05 --> 00:48:08
fundamental theorem.
The question is,
531
00:48:08 --> 00:48:12
if I have a conservative,
or path independent vector
532
00:48:12 --> 00:48:15
field, why is it the gradient of
something?
533
00:48:15 --> 00:48:25
OK, so this way is a
fundamental theorem.
534
00:48:25 --> 00:48:31
That way, well,
so that actually,
535
00:48:31 --> 00:48:43
let me just say that will be
how we find the potential.
536
00:48:43 --> 00:48:46
So, how do we find potential?
Well, let's say that I know the
537
00:48:46 --> 00:48:49
value of my potential here.
Actually, I get to choose what
538
00:48:49 --> 00:48:51
it is.
Remember, in physics,
539
00:48:51 --> 00:48:53
the potential is defined up to
adding or subtracting a
540
00:48:53 --> 00:48:56
constant.
What matters is only the change
541
00:48:56 --> 00:48:58
in potential.
So, let's say I know my
542
00:48:58 --> 00:49:01
potential here and I want to
know my potential here.
543
00:49:01 --> 00:49:04
What do I do?
Well, I take my favorite
544
00:49:04 --> 00:49:07
particle and I move it from here
to here.
545
00:49:07 --> 00:49:10
And, I look at the work done.
And that tells me how much
546
00:49:10 --> 00:49:14
potential has changed.
So, that tells me what the
547
00:49:14 --> 00:49:18
potential should be here.
And, this does not depend on my
548
00:49:18 --> 00:49:21
choice of path because I've
assumed that I'm path
549
00:49:21 --> 00:49:25
independence.
So, that's who we will do on
550
00:49:25 --> 00:49:29
Tuesday.
And, let me just state the
551
00:49:29 --> 00:49:36
fourth property that's the same.
So, all that stuff is the same
552
00:49:36 --> 00:49:42
as also four.
If I look at M dx N dy is
553
00:49:42 --> 00:49:48
what's called an exact
differential.
554
00:49:48 --> 00:49:52
So, what that means,
an exact differential,
555
00:49:52 --> 00:49:56
means that it can be put in the
form df for some function,
556
00:49:56 --> 00:49:58
f,
and just reformulating this
557
00:49:58 --> 00:50:02
thing, right,
because I'm saying I can just
558
00:50:02 --> 00:50:05
put it in the form f sub x dx
plus f sub y dy,
559
00:50:05 --> 00:50:08
which means my vector field was
a gradient field.
560
00:50:08 --> 00:50:13
So, these things are really the
same.
561
00:50:13 --> 00:50:16
OK, so after the weekend,
on Tuesday we will actually
562
00:50:16 --> 00:50:19
figure out how to decide whether
these things hold or not,
563
00:50:19 --> 00:50:22
and how to find the potential.
564
00:50:22 --> 00:50:27