1
00:00:01,000 --> 00:00:03,000
The following content is
provided under a Creative

2
00:00:03,000 --> 00:00:05,000
Commons license.
Your support will help MIT

3
00:00:05,000 --> 00:00:08,000
OpenCourseWare continue to offer
high quality educational

4
00:00:08,000 --> 00:00:13,000
resources for free.
To make a donation or to view

5
00:00:13,000 --> 00:00:18,000
additional materials from
hundreds of MIT courses,

6
00:00:18,000 --> 00:00:23,000
visit MIT OpenCourseWare at
ocw.mit.edu.

7
00:00:24,000 --> 00:00:28,000
The other thing is the last
lecture before the break ended a

8
00:00:28,000 --> 00:00:31,000
bit abruptly because I ran out
of time,

9
00:00:31,000 --> 00:00:34,000
so just to summarize what the
main point was,

10
00:00:34,000 --> 00:00:37,000
I mean probably you figured
this out if you looked at the

11
00:00:37,000 --> 00:00:39,000
notes.
It is not that important,

12
00:00:39,000 --> 00:00:43,000
but anyway.
I just wanted to remind you,

13
00:00:43,000 --> 00:00:48,000
just to clarify what happened
at the end, we got the diffusion

14
00:00:48,000 --> 00:00:52,000
equation from two bits of
information.

15
00:00:52,000 --> 00:00:55,000
I mean the unknown,
in this partial differential

16
00:00:55,000 --> 00:00:59,000
equation, is a function that we
call u that corresponds to the

17
00:00:59,000 --> 00:01:04,000
concentration of some substance.
And we used a vector field that

18
00:01:04,000 --> 00:01:12,000
represents the flow of whatever
the substance is whose diffusion

19
00:01:12,000 --> 00:01:17,000
we are studying.
And so we got two bits of

20
00:01:17,000 --> 00:01:25,000
information, one that came from
physics that said the flow goes

21
00:01:25,000 --> 00:01:31,000
from high concentration to
smaller concentration.

22
00:01:31,000 --> 00:01:37,000
And that told us that the flow
is proportional to a negative

23
00:01:37,000 --> 00:01:42,000
gradient of a concentration.
And the second piece of

24
00:01:42,000 --> 00:01:46,000
information that we got was from
the divergence theorem,

25
00:01:46,000 --> 00:01:50,000
and that was the one I spent
time trying to explain.

26
00:01:50,000 --> 00:01:57,000
And that one told us that the
divergence of F is actually

27
00:01:57,000 --> 00:02:02,000
negative partial u over partial
t.

28
00:02:02,000 --> 00:02:07,000
When you combine these two
relations together that is how

29
00:02:07,000 --> 00:02:11,000
you get the diffusion equation.
Sorry, I should say this is not

30
00:02:11,000 --> 00:02:12,000
the statement of a divergence
theorem.

31
00:02:12,000 --> 00:02:17,000
This is something that would
derive from it with quite a few

32
00:02:17,000 --> 00:02:21,000
steps involved.
And so what we got out of that

33
00:02:21,000 --> 00:02:27,000
is a diffusion equation because
we end up getting that partial u

34
00:02:27,000 --> 00:02:33,000
over partial t is minus div F,
which is therefore positive k

35
00:02:33,000 --> 00:02:39,000
times divergence of grad u,
which is what we denoted by del

36
00:02:39,000 --> 00:02:42,000
square u,
the Laplacian.

37
00:02:42,000 --> 00:02:59,000
So, that is how we got the
diffusion equation.

38
00:02:59,000 --> 00:03:04,000
Anyway, I will let you have a
look at the notes that were

39
00:03:04,000 --> 00:03:08,000
handed out in case you really
want to see more.

40
00:03:08,000 --> 00:03:16,000
I just wanted to give the
missing part of the last

41
00:03:16,000 --> 00:03:19,000
lecture.
Let me just switch gears

42
00:03:19,000 --> 00:03:24,000
completely and switch to today's
topic, which is line integrals

43
00:03:24,000 --> 00:03:27,000
and work in 3D.
That is going to look a lot

44
00:03:27,000 --> 00:03:29,000
like what we did in the plane,
except, of course,

45
00:03:29,000 --> 00:03:32,000
there is a z coordinate.
You will see it doesn't change

46
00:03:32,000 --> 00:03:35,000
things much when it comes to
computing a line integral.

47
00:03:35,000 --> 00:03:39,000
It changes things quite a bit,
however, when it comes to

48
00:03:39,000 --> 00:03:42,000
testing whether a field is a
gradient field.

49
00:03:42,000 --> 00:03:45,000
That is why we have to be more
careful.

50
00:03:45,000 --> 00:03:57,000
Let's start right away with
line integrals in space.

51
00:03:57,000 --> 00:04:08,000
Let's say that we have a vector
field F with components P,

52
00:04:08,000 --> 00:04:12,000
Q and R.
We should think of it maybe as

53
00:04:12,000 --> 00:04:18,000
representing a force.
And let's say that we have a

54
00:04:18,000 --> 00:04:24,000
curve C in space.
Then the work done by the field

55
00:04:24,000 --> 00:04:29,000
will be the line integral along
C of F dot dr.

56
00:04:29,000 --> 00:04:34,000
That is a familiar formula.
And what we do with that

57
00:04:34,000 --> 00:04:37,000
formula is also familiar,
except now, of course,

58
00:04:37,000 --> 00:04:44,000
we have a z coordinate.
We are going to think of vector

59
00:04:44,000 --> 00:04:50,000
dr as a space vector with
components dx,

60
00:04:50,000 --> 00:04:56,000
dy and dz.
When we do the dot product of F

61
00:04:56,000 --> 00:05:04,000
with dr that will tell us that
we have to integrate Pdx Qdy

62
00:05:04,000 --> 00:05:07,000
Rdz.
But it is still a line integral

63
00:05:07,000 --> 00:05:12,000
so it is still going to turn
into a single integral when you

64
00:05:12,000 --> 00:05:16,000
plug in the correct values.
So the method will be exactly

65
00:05:16,000 --> 00:05:19,000
the same as in the plane,
namely we will find some way to

66
00:05:19,000 --> 00:05:22,000
parameterize our curve,
x plus x, y z in terms of a

67
00:05:22,000 --> 00:05:30,000
single variable,
and then we will integrate with

68
00:05:30,000 --> 00:05:44,000
respect to that variable.
The way that we evaluate is by

69
00:05:44,000 --> 00:05:56,000
parameterizing C and express x,
y, z, dx, dy,

70
00:05:56,000 --> 00:06:07,000
dz in terms of the parameter.
Let's do an example just to

71
00:06:07,000 --> 00:06:09,000
convince you that you actually
know how to do this,

72
00:06:09,000 --> 00:06:12,000
or at least you should know how
to do this.

73
00:06:12,000 --> 00:06:21,000
Let's say that I give you the
vector field with components yz,

74
00:06:21,000 --> 00:06:27,000
xz and xy.
And let's say that we have a

75
00:06:27,000 --> 00:06:33,000
curve given by x equals t^3,
y equals t^2,

76
00:06:33,000 --> 00:06:40,000
z equals t for t going from
zero to one.

77
00:06:40,000 --> 00:06:45,000
The way we will set up the line
integral for the work done will

78
00:06:45,000 --> 00:06:49,000
be -- Well, sorry.
Before we actually set up the

79
00:06:49,000 --> 00:06:51,000
line integral,
we need to know how we will

80
00:06:51,000 --> 00:06:54,000
express everything in terms of t
and dt.

81
00:06:54,000 --> 00:06:56,000
x, y and z, in terms of t,
are given here.

82
00:06:56,000 --> 00:07:01,000
We just need to do also dx,
dy and dz.

83
00:07:01,000 --> 00:07:05,000
By differentiating you get dx
is 3t^2 dt.

84
00:07:05,000 --> 00:07:14,000
That is the derivative of t^3,
dy will be 2t dt and dz will

85
00:07:14,000 --> 00:07:23,000
just be dt.
And we will evaluate the line

86
00:07:23,000 --> 00:07:36,000
integral for work.
That will be the integral of yz

87
00:07:36,000 --> 00:07:48,000
dx xz dy xy dz,
which will become -- yz is t^3

88
00:07:48,000 --> 00:08:04,000
times dx is 3t^2 dt plus xz is
t^4 times dy is 2t dt plus xy is

89
00:08:04,000 --> 00:08:08,000
t^5 dt.
That just becomes the integral

90
00:08:08,000 --> 00:08:11,000
from, well, I guess t goes from
zero to one, actually.

91
00:08:11,000 --> 00:08:14,000
And we are integrating three
plus two plus one.

92
00:08:14,000 --> 00:08:22,000
That is 6t^5 dt which,
I am sure you know,

93
00:08:22,000 --> 00:08:30,000
integrates to t^6,
so we will just get one.

94
00:08:30,000 --> 00:08:33,000
It is the same method as usual.
And if you are being given a

95
00:08:33,000 --> 00:08:36,000
geometric description of a curve
then, of course,

96
00:08:36,000 --> 00:08:39,000
you will have to decide for
yourself what the best parameter

97
00:08:39,000 --> 00:08:43,000
will be.
It might be some time parameter

98
00:08:43,000 --> 00:08:45,000
t like here.
It might be one of the

99
00:08:45,000 --> 00:08:49,000
coordinates.
Here we could have used z as

100
00:08:49,000 --> 00:08:53,000
our parameter because,
in fact, this curve is x equals

101
00:08:53,000 --> 00:08:57,000
x3 and y equals z2.
And we could also have used

102
00:08:57,000 --> 00:08:59,000
maybe some angle.
Well, not here,

103
00:08:59,000 --> 00:09:04,000
but if we had been moving on a
circle or something like that.

104
00:09:04,000 --> 00:09:11,000
Any questions so far?
No.

105
00:09:11,000 --> 00:09:26,000
OK.
Well,

106
00:09:26,000 --> 00:09:31,000
because we can do a bit more
practice,

107
00:09:31,000 --> 00:09:40,000
let's do another one where we
do the same vector field F but

108
00:09:40,000 --> 00:09:49,000
our curve C will be going from
the origin to the point (1,0,

109
00:09:49,000 --> 00:09:54,000
0) along the x-axis.
Let's call that C1.

110
00:09:54,000 --> 00:10:02,000
Then to (1,1, 0).
Let's call that C2 by moving

111
00:10:02,000 --> 00:10:08,000
parallel to the y-axis.
And then up to (1,1,

112
00:10:08,000 --> 00:10:14,000
1) parallel to the z-axis,
let's call that C3.

113
00:10:14,000 --> 00:10:19,000
I am sure that some of you at
least are suspecting what I am

114
00:10:19,000 --> 00:10:23,000
getting at here,
but let's not spoil it for

115
00:10:23,000 --> 00:10:27,000
those who don't see it yet.
OK.

116
00:10:27,000 --> 00:10:33,000
If we want to compute the line
integral along this guy then we

117
00:10:33,000 --> 00:10:38,000
have to break it into a sum of
three terms.

118
00:10:38,000 --> 00:10:42,000
Well, maybe I should call that
C', not C, because that is not

119
00:10:42,000 --> 00:10:48,000
the same C anymore.
I want to do the sum of the

120
00:10:48,000 --> 00:10:55,000
line integrals along C1,
C2 and C3.

121
00:10:55,000 --> 00:11:13,000
And, well, if I look at C1 and
C2 they take place inside the x,

122
00:11:13,000 --> 00:11:19,000
y plane.
In fact, you know that z will

123
00:11:19,000 --> 00:11:25,000
be zero and dz will also be zero
on both of these.

124
00:11:25,000 --> 00:11:30,000
And if you just look at the
formula for line integral in the

125
00:11:30,000 --> 00:11:34,000
role of yz dx plus xz dy plus xy
dz,

126
00:11:34,000 --> 00:11:37,000
well, it looks like if you plug
z equals zero and dz equals zero

127
00:11:37,000 --> 00:11:47,000
you will just get zero.
These are actually very fast.

128
00:11:47,000 --> 00:12:00,000
Let me write it.
This is going to be zero,

129
00:12:00,000 --> 00:12:04,000
this is going to be zero,
this is going to zero,

130
00:12:04,000 --> 00:12:07,000
and we will get zero.
Now, if we do C3,

131
00:12:07,000 --> 00:12:11,000
well, there we might have to do
some calculation,

132
00:12:11,000 --> 00:12:18,000
but it won't be all that bad.
C3, well, x and y are both

133
00:12:18,000 --> 00:12:22,000
equal to one.
And, of course,

134
00:12:22,000 --> 00:12:26,000
because they are constant that
means dx is zero and dy is zero.

135
00:12:26,000 --> 00:12:35,000
On the other hand,
z varies from zero to one.

136
00:12:35,000 --> 00:12:42,000
If I look at the line integral
on C3 -- The first two terms,

137
00:12:42,000 --> 00:12:48,000
yz, dx and xz dy go away
because the dx and dy are zero,

138
00:12:48,000 --> 00:12:55,000
so I am just left with xy dz.
But because x and y are one it

139
00:12:55,000 --> 00:13:01,000
is just the integral of dz from
zero to one, and that will just

140
00:13:01,000 --> 00:13:07,000
end up being one.
If add these numbers together,

141
00:13:07,000 --> 00:13:13,000
zero plus zero plus one,
I get one again.

142
00:13:13,000 --> 00:13:15,000
And, of course,
it is not a coincidence because

143
00:13:15,000 --> 00:13:18,000
this vector field is a gradient
field.

144
00:13:18,000 --> 00:13:20,000
I am sure some of you have
already figured out what it is

145
00:13:20,000 --> 00:13:24,000
the gradient of.
Otherwise, we will figure it

146
00:13:24,000 --> 00:13:28,000
out together.
And so that is why we get the

147
00:13:28,000 --> 00:13:33,000
same answer for these two paths
going both from the origin to

148
00:13:33,000 --> 00:13:36,000
(1,1, 1).
Maybe I should point out,

149
00:13:36,000 --> 00:13:40,000
to make it clear,
that if you plug t equals zero

150
00:13:40,000 --> 00:13:42,000
in up there you will get (0,0,
0).

151
00:13:42,000 --> 00:13:45,000
If you plug t equals one,
you will get (1,1,

152
00:13:45,000 --> 00:13:56,000
1).
In fact -- F that we have here

153
00:13:56,000 --> 00:14:11,000
happens to be conservative.
And, if you plug the two curves

154
00:14:11,000 --> 00:14:20,000
together -- Well,
I am not really sure if I know

155
00:14:20,000 --> 00:14:28,000
how to plot this correctly.
It is not exactly how it looks.

156
00:14:28,000 --> 00:14:33,000
Whatever.
The first curve C goes from the

157
00:14:33,000 --> 00:14:39,000
origin to this point,
and so does C ',

158
00:14:39,000 --> 00:14:45,000
just in a slightly more
roundabout way.

159
00:14:45,000 --> 00:14:51,000
They both go from the origin to
(1,1, 1).

160
00:14:51,000 --> 00:15:00,000
It is not a surprise that you
will get the same answer for

161
00:15:00,000 --> 00:15:05,000
both line integrals.
And how do we see that?

162
00:15:05,000 --> 00:15:11,000
Well, actually here it is not
very hard to find a function

163
00:15:11,000 --> 00:15:15,000
whose gradient is this vector
field.

164
00:15:15,000 --> 00:15:21,000
Namely, the gradient of x,
y, z looks like it should be

165
00:15:21,000 --> 00:15:26,000
exactly what we want.
If you take partial of this

166
00:15:26,000 --> 00:15:29,000
with respect to x,
you will get yz,

167
00:15:29,000 --> 00:15:33,000
then with respect to y,
xz, and with respect to z,

168
00:15:33,000 --> 00:15:35,000
xy.
And so, in fact,

169
00:15:35,000 --> 00:15:38,000
what was the easier way to
compute these line integrals was

170
00:15:38,000 --> 00:15:41,000
to use the fundamental theorem
of calculus.

171
00:15:41,000 --> 00:15:45,000
Once we have this remark,
we don't need to compute these

172
00:15:45,000 --> 00:15:56,000
line integrals anymore.
We can just use the fundamental

173
00:15:56,000 --> 00:16:08,000
theorem.
If we know this fundamental

174
00:16:08,000 --> 00:16:20,000
theorem -- -- for line
integrals,

175
00:16:20,000 --> 00:16:27,000
that tells us that the line
integral of a gradient field is

176
00:16:27,000 --> 00:16:33,000
equal to the value of the
potential at the final point

177
00:16:33,000 --> 00:16:40,000
minus the value of the potential
at the starting point.

178
00:16:40,000 --> 00:16:43,000
And that, of course,
only applies if you have a

179
00:16:43,000 --> 00:16:46,000
potential.
So, in particular,

180
00:16:46,000 --> 00:16:52,000
only if you have a conservative
field, a gradient field.

181
00:16:52,000 --> 00:16:58,000
Here, in our example,
we have to look at,

182
00:16:58,000 --> 00:17:06,000
let's call little f of x,
y, z that potential xyz,

183
00:17:06,000 --> 00:17:12,000
then we take f(1,1,
1) - f(0,0, 0).

184
00:17:12,000 --> 00:17:17,000
And that indeed is one minus
zero which is one.

185
00:17:17,000 --> 00:17:24,000
Everything is consistent.
All this stuff so far works

186
00:17:24,000 --> 00:17:32,000
exactly as in the plane.
Any questions?

187
00:17:32,000 --> 00:17:35,000
No.
OK.

188
00:17:35,000 --> 00:17:40,000
Let's try to see where things
do get a little bit different.

189
00:17:40,000 --> 00:17:44,000
And the first such place is
when we try to test whether a

190
00:17:44,000 --> 00:17:47,000
vector field is a gradient
field.

191
00:17:47,000 --> 00:17:53,000
Remember when we had a vector
field in the plane,

192
00:17:53,000 --> 00:17:56,000
to know whether it was a
gradient of a function of two

193
00:17:56,000 --> 00:17:59,000
variables we just had to check
one condition,

194
00:17:59,000 --> 00:18:04,000
N sub x equals M sub y.
Now we actually have three

195
00:18:04,000 --> 00:18:06,000
different conditions to check,
and that means,

196
00:18:06,000 --> 00:18:07,000
of course, more work.

197
00:18:23,000 --> 00:18:34,000
OK.
So what is our test for

198
00:18:34,000 --> 00:18:44,000
gradient fields?
We want to know whether a given

199
00:18:44,000 --> 00:18:50,000
vector field with components P,
Q and R can be written as f sub

200
00:18:50,000 --> 00:18:55,000
x, f sub y and f sub z for a
same function F.

201
00:18:55,000 --> 00:18:59,000
And for that to possibly
happen, well,

202
00:18:59,000 --> 00:19:03,000
we need certainly some
relations between P,

203
00:19:03,000 --> 00:19:05,000
Q and R.
And, as before,

204
00:19:05,000 --> 00:19:11,000
this comes from the fact that
the mixed second derivatives are

205
00:19:11,000 --> 00:19:16,000
the same, no matter in which
order you take them.

206
00:19:16,000 --> 00:19:21,000
If that is the case then I can
compute f sub xy,

207
00:19:21,000 --> 00:19:28,000
which is the same as f sub yx
in two different ways.

208
00:19:28,000 --> 00:19:35,000
F sub xy should be P sub y.
F sub yx, well,

209
00:19:35,000 --> 00:19:39,000
since f sub y is Q,
that should be Q sub x.

210
00:19:39,000 --> 00:19:43,000
That is a part of a criterion
that we already had when we had

211
00:19:43,000 --> 00:19:46,000
only two variables.
But now, of course,

212
00:19:46,000 --> 00:19:50,000
we need to do the same thing
when we look at x and z or y and

213
00:19:50,000 --> 00:19:53,000
z.
That gives us two more

214
00:19:53,000 --> 00:19:58,000
conditions.
P sub z is f sub xz,

215
00:19:58,000 --> 00:20:10,000
which is the same as f sub zx,
so it should be the same as R

216
00:20:10,000 --> 00:20:14,000
sub x.
Finally, Q sub z,

217
00:20:14,000 --> 00:20:22,000
which is f sub yz,
equals f sub zy equals R sub y.

218
00:20:22,000 --> 00:20:33,000
We have three conditions,
so our criterion -- Vector

219
00:20:33,000 --> 00:20:41,000
field F equals <P,
Q, R>.

220
00:20:41,000 --> 00:20:46,000
And here, to be completely
truthful, I have to say defined

221
00:20:46,000 --> 00:20:50,000
in a simply connected region.
Otherwise, we might have the

222
00:20:50,000 --> 00:20:53,000
same kind of strange things
happening as before.

223
00:20:53,000 --> 00:21:03,000
Let's not worry too much about
it.

224
00:21:03,000 --> 00:21:07,000
For accuracy we need our vector
field to be defined in a simply

225
00:21:07,000 --> 00:21:10,000
connected region.
And example is just if it is

226
00:21:10,000 --> 00:21:14,000
defined everywhere.
If you don't have any evil

227
00:21:14,000 --> 00:21:21,000
eliminators then you can just go
ahead and there is no problem.

228
00:21:21,000 --> 00:21:36,000
It is a gradient field.
We need three conditions.

229
00:21:36,000 --> 00:21:43,000
Let's do it in order.
P sub y equals Q sub x.

230
00:21:43,000 --> 00:21:52,000
And we have P sub z equals R
sub x and Q sub z equals R sub

231
00:21:52,000 --> 00:21:55,000
y.
How do you remember these three

232
00:21:55,000 --> 00:21:56,000
conditions?
Well, it is pretty easy.

233
00:21:56,000 --> 00:22:01,000
You pick any two components,
say the x and the z component,

234
00:22:01,000 --> 00:22:04,000
and you take the partial of the
x component with respect to z,

235
00:22:04,000 --> 00:22:07,000
the partial of the z component
with respect to x and you must

236
00:22:07,000 --> 00:22:12,000
make them equal.
And the same with every pair of

237
00:22:12,000 --> 00:22:15,000
variables.
In fact, if you had a function

238
00:22:15,000 --> 00:22:17,000
of many more variables the
criterion would still look

239
00:22:17,000 --> 00:22:21,000
exactly like that.
For every pair of components

240
00:22:21,000 --> 00:22:24,000
the mixed partials must be the
same.

241
00:22:24,000 --> 00:22:27,000
But we are not going to go
beyond three variables so you

242
00:22:27,000 --> 00:22:32,000
don't need to know that.
This you need to know so let me

243
00:22:32,000 --> 00:22:33,000
box it.

244
00:22:51,000 --> 00:23:02,000
That is pretty straightforward.
Let's do an example just to see

245
00:23:02,000 --> 00:23:11,000
how it goes.
By the way, we can also think

246
00:23:11,000 --> 00:23:19,000
of it in terms of differentials.
Before I do the example,

247
00:23:19,000 --> 00:23:22,000
let me just say in a different
language.

248
00:23:22,000 --> 00:23:29,000
If we have a differential given
to us of a form Pdx Qdy Rdz is

249
00:23:29,000 --> 00:23:33,000
going to be an exact
differential,

250
00:23:33,000 --> 00:23:40,000
which means it is equal to df
for some function F exactly and

251
00:23:40,000 --> 00:23:43,000
of the same conditions.
That is the same thing.

252
00:23:43,000 --> 00:23:51,000
Just in the language of
differentials.

253
00:23:51,000 --> 00:23:57,000
The example that I promised.
Of course, I could do again the

254
00:23:57,000 --> 00:24:00,000
same one over there and check
that it satisfies the condition,

255
00:24:00,000 --> 00:24:02,000
but then it wouldn't be much
fun.

256
00:24:02,000 --> 00:24:10,000
So let's do a better one.
Actually, let's do it in a way

257
00:24:10,000 --> 00:24:18,000
that looks like an exam problem.
Let's say for which a and b is

258
00:24:18,000 --> 00:24:29,000
a xy dx plus -- Oh,
it is not going to fit here.

259
00:24:29,000 --> 00:24:39,000
But it will fit here.
a xy dx ( x^2 z^3) dy (byz^2 -

260
00:24:39,000 --> 00:24:51,000
4z^3) dz, an exact differential.
Or, if you don't like exact

261
00:24:51,000 --> 00:24:54,000
differentials,
for which a and b is the

262
00:24:54,000 --> 00:24:58,000
corresponding vector field with
i, j and k instead,

263
00:24:58,000 --> 00:25:02,000
a gradient field.
Let's just apply the criterion.

264
00:25:02,000 --> 00:25:06,000
And, of course,
you can guess that what will

265
00:25:06,000 --> 00:25:11,000
follow is figuring out how to
find the potential when there is

266
00:25:11,000 --> 00:25:12,000
one.

267
00:25:33,000 --> 00:25:40,000
Let's do it one by one.
We want to compare P sub y with

268
00:25:40,000 --> 00:25:45,000
Q sub x,
we want to compare P sub z with

269
00:25:45,000 --> 00:25:53,000
R sub x and we want to compare Q
sub z with R sub y where we call

270
00:25:53,000 --> 00:25:57,000
P, Q and R these guys.
Let's see.

271
00:25:57,000 --> 00:26:04,000
What is P sub y?
That seems to be ax.

272
00:26:04,000 --> 00:26:09,000
What is Q sub x?
2x.

273
00:26:09,000 --> 00:26:12,000
Q is this one.
Actually, let me write them

274
00:26:12,000 --> 00:26:16,000
down.
Because otherwise I am going to

275
00:26:16,000 --> 00:26:21,000
get confused myself.
This guy here,

276
00:26:21,000 --> 00:26:32,000
that is P, this guy here,
that is Q and that guy here,

277
00:26:32,000 --> 00:26:38,000
that is R.
This one tells us that a should

278
00:26:38,000 --> 00:26:43,000
be equal to two of the first
product that you hold.

279
00:26:43,000 --> 00:26:49,000
OK. Let's look at P sub z.
That is just zero.

280
00:26:49,000 --> 00:26:51,000
R sub x?
Well, R doesn't have any x

281
00:26:51,000 --> 00:26:56,000
either so that is zero.
This one is not a problem.

282
00:26:56,000 --> 00:27:02,000
Q sub z?
Well, that seems to be 3z2.

283
00:27:02,000 --> 00:27:14,000
R sub y seems to be bz2,
so b should be equal to three.

284
00:27:14,000 --> 00:27:19,000
We need to have a equals two
and, this is an and,

285
00:27:19,000 --> 00:27:23,000
not or, b equals three for this
to be exact.

286
00:27:23,000 --> 00:27:27,000
For those values of a and b,
we can look for a potential

287
00:27:27,000 --> 00:27:30,000
using the method that we are
going to see right now.

288
00:27:30,000 --> 00:27:33,000
For any other values of a and b
we cannot.

289
00:27:33,000 --> 00:27:38,000
If we have to compute a line
integral, we have to do it by

290
00:27:38,000 --> 00:27:42,000
finding a parameter and setting
up everything.

291
00:27:42,000 --> 00:27:57,000
Any questions at this point?
Yes?

292
00:27:57,000 --> 00:28:00,000
I see.
Well, if I got the same answer,

293
00:28:00,000 --> 00:28:05,000
oh, did say bz^2 or 3bz^2?
Well, 3bz^2, for example,

294
00:28:05,000 --> 00:28:09,000
I would need b to be zero
because the only time that 3bz2

295
00:28:09,000 --> 00:28:13,000
equals bz2 as not just at one
point but everywhere,

296
00:28:13,000 --> 00:28:15,000
I need them to be the same
function of x,

297
00:28:15,000 --> 00:28:18,000
y, z.
Well, if a coefficient of z2 is

298
00:28:18,000 --> 00:28:22,000
the same that would be give b
equals 3b, that would give me b

299
00:28:22,000 --> 00:28:25,000
equals zero.
If you got bz2 on both sides

300
00:28:25,000 --> 00:28:29,000
then it would mean for any value
of b it works,

301
00:28:29,000 --> 00:28:35,000
and you wouldn't have to worry
about what the value of b is.

302
00:28:35,000 --> 00:28:41,000
Any other questions?
No.

303
00:28:41,000 --> 00:28:50,000
OK.
Now, how do we find the

304
00:28:50,000 --> 00:28:58,000
potential?
Well, there are two methods as

305
00:28:58,000 --> 00:28:59,000
before.
One of them,

306
00:28:59,000 --> 00:29:02,000
I don't remember if it was the
first one or the second one last

307
00:29:02,000 --> 00:29:04,000
time, but it really doesn't
matter.

308
00:29:04,000 --> 00:29:10,000
One of them was just to say
that the value of F at the

309
00:29:10,000 --> 00:29:15,000
point,
let me call that x1, y1, z1,

310
00:29:15,000 --> 00:29:22,000
is equal to the line integral
of my field along a well-chosen

311
00:29:22,000 --> 00:29:25,000
curve plus,
of course, a constant,

312
00:29:25,000 --> 00:29:30,000
which is going to be the
integration constant.

313
00:29:30,000 --> 00:29:39,000
And the kind of curve that I
will take to do this calculation

314
00:29:39,000 --> 00:29:48,000
will just be my favorite curve
going from the origin to the

315
00:29:48,000 --> 00:29:51,000
point x1, y1,
z1.

316
00:29:51,000 --> 00:29:57,000
And so, typically the most
common choice would be to go

317
00:29:57,000 --> 00:30:04,000
just first along the x-axis,
then parallel to the y-axis and

318
00:30:04,000 --> 00:30:11,000
then parallel to the z-axis all
the way to my point x1,

319
00:30:11,000 --> 00:30:14,000
y1, z1.
I would just calculate three

320
00:30:14,000 --> 00:30:18,000
easy line integrals.
Add them together and that

321
00:30:18,000 --> 00:30:21,000
would give me the value of my
function.

322
00:30:21,000 --> 00:30:27,000
That method works exactly the
same way as it did in two

323
00:30:27,000 --> 00:30:30,000
variables.
Now, I seem to recall that you

324
00:30:30,000 --> 00:30:32,000
guys mostly preferred the other
method.

325
00:30:32,000 --> 00:30:34,000
I am going to tell you about
the other method as well,

326
00:30:34,000 --> 00:30:37,000
but I just want to point out
this one actually doesn't become

327
00:30:37,000 --> 00:30:40,000
more complicated.
The other one has actually more

328
00:30:40,000 --> 00:30:42,000
steps.
I mean, of course,

329
00:30:42,000 --> 00:30:45,000
here there are also a bit more
steps because you have three

330
00:30:45,000 --> 00:30:47,000
parts to your path instead of
two.

331
00:30:47,000 --> 00:30:54,000
You have three line integrals
to compute instead of two,

332
00:30:54,000 --> 00:30:59,000
but conceptually it remains
exactly the same idea.

333
00:30:59,000 --> 00:31:07,000
I should say it works the same
way as in 2D.

334
00:31:07,000 --> 00:31:17,000
Not much changes.
Let's look at the other method

335
00:31:17,000 --> 00:31:24,000
using anti-derivatives.
Remember we want to find a

336
00:31:24,000 --> 00:31:29,000
function little f whose partials
are exactly the things we have

337
00:31:29,000 --> 00:31:31,000
been given.
We want to solve,

338
00:31:31,000 --> 00:31:36,000
well, let me plug in the values
of a and b that will work.

339
00:31:36,000 --> 00:31:47,000
We said a should be two,
so f sub x should be 2xy,

340
00:31:47,000 --> 00:31:59,000
f sub y should be x2 plus z3,
and f sub z should be 3yz^2

341
00:31:59,000 --> 00:32:04,000
minus 4z^3.
We are going to look at them

342
00:32:04,000 --> 00:32:07,000
one at a time and get partial
information on the function.

343
00:32:07,000 --> 00:32:11,000
And then we will compare with
the others to get more

344
00:32:11,000 --> 00:32:15,000
information until we are
completely done.

345
00:32:15,000 --> 00:32:25,000
The first thing we will do,
we know that f sub x is 2xy.

346
00:32:25,000 --> 00:32:27,000
That should tell us something
about f.

347
00:32:27,000 --> 00:32:31,000
Well, let's just integrate that
with respect to x.

348
00:32:31,000 --> 00:32:36,000
Let me write integral dx next
to that.

349
00:32:36,000 --> 00:32:41,000
That tells us that f should be,
well, if we integrate that with

350
00:32:41,000 --> 00:32:44,000
respect to x,
2x integrates to x^2,

351
00:32:44,000 --> 00:32:47,000
so we should get x2y.
Plus, of course,

352
00:32:47,000 --> 00:32:50,000
an integration constant.
Now, what do we mean by

353
00:32:50,000 --> 00:32:53,000
integration constant.
It means that for given values

354
00:32:53,000 --> 00:32:58,000
of y and z we will get a term
that does not depend on x.

355
00:32:58,000 --> 00:33:02,000
It still depends on y and z.
In fact, what we get is a

356
00:33:02,000 --> 00:33:07,000
function of y and z.
See, if you took the derivative

357
00:33:07,000 --> 00:33:12,000
of this with respect to x you
will get 2xy and this guy will

358
00:33:12,000 --> 00:33:16,000
go away because there is no x in
it.

359
00:33:16,000 --> 00:33:19,000
That is the first step.
Now we need to get some

360
00:33:19,000 --> 00:33:21,000
information on g.
How do we do that?

361
00:33:21,000 --> 00:33:28,000
Well, we look at the other
partials.

362
00:33:28,000 --> 00:33:35,000
F sub y, we want that to be x^2
z^3.

363
00:33:35,000 --> 00:33:41,000
But we have another way to find
it, which is starting from this

364
00:33:41,000 --> 00:33:50,000
and differentiating.
Let me try to use color for

365
00:33:50,000 --> 00:33:54,000
this.
Now, if I take the partial of

366
00:33:54,000 --> 00:33:58,000
this with respect to y,
I am going to get a different

367
00:33:58,000 --> 00:34:06,000
formula for f sub y.
That will be x^2 plus g sub y.

368
00:34:06,000 --> 00:34:15,000
Well, if I compare these two
expressions that tells me that g

369
00:34:15,000 --> 00:34:23,000
sub y should be z3.
Now, if I have this I can

370
00:34:23,000 --> 00:34:32,000
integrate with respect to y.
That will tell me that g is

371
00:34:32,000 --> 00:34:39,000
actually yz^3 plus an
integration constant.

372
00:34:39,000 --> 00:34:42,000
That constant,
again, does not depend on y,

373
00:34:42,000 --> 00:34:46,000
but it can still depend on z
because we still have not said

374
00:34:46,000 --> 00:34:49,000
anything about partial with
respect to z.

375
00:34:49,000 --> 00:34:54,000
In fact, that constant I will
write as a function h of z.

376
00:34:54,000 --> 00:35:00,000
If I have this function of z
and I take its partial with

377
00:35:00,000 --> 00:35:04,000
respect to y,
I will still get z^3 no matter

378
00:35:04,000 --> 00:35:06,000
what h was.
Now, how do I find h?

379
00:35:06,000 --> 00:35:09,000
Well, obviously,
I have to look at f sub z.

380
00:35:41,000 --> 00:35:47,000
F sub z.
We know from the given vector

381
00:35:47,000 --> 00:35:53,000
field that we want it to be
3yz^2 minus 4z^3.

382
00:35:53,000 --> 00:35:56,000
In case you are wondering where
that came from,

383
00:35:56,000 --> 00:36:01,000
that was R.
But that is also obtained by

384
00:36:01,000 --> 00:36:09,000
differentiating with respect to
z what we had so far.

385
00:36:09,000 --> 00:36:15,000
Sorry.
What did we have so far?

386
00:36:15,000 --> 00:36:20,000
Well, we had f equals x^2y plus
g.

387
00:36:20,000 --> 00:36:30,000
And we said g is actually yz^3
plus h of z.

388
00:36:30,000 --> 00:36:37,000
That is what we have so far.
If we take the derivative of

389
00:36:37,000 --> 00:36:44,000
that with respect to z,
we will get zero plus 3yz^2

390
00:36:44,000 --> 00:36:51,000
plus h prime of z,
or dh dz as you want.

391
00:36:51,000 --> 00:36:59,000
Now, if we compare these two,
we will get the derivative of

392
00:36:59,000 --> 00:37:03,000
h.
It will tell us that h prime is

393
00:37:03,000 --> 00:37:08,000
negative for z3.
That means that h is negative

394
00:37:08,000 --> 00:37:13,000
z^4 plus a constant.
And this it is at last an

395
00:37:13,000 --> 00:37:17,000
actual constant.
Because it does not depend on z

396
00:37:17,000 --> 00:37:21,000
and there is nothing else for it
to depend on.

397
00:37:21,000 --> 00:37:33,000
Now we plug this into what we
had before, and that will give

398
00:37:33,000 --> 00:37:44,000
us our function f.
We get that f=x^2y yz^3 - z^4

399
00:37:44,000 --> 00:37:49,000
plus constant.
If you just wanted to find one

400
00:37:49,000 --> 00:37:51,000
potential, you can just forget
the constant.

401
00:37:51,000 --> 00:37:55,000
This guy was a potential.
If you want all the potentials

402
00:37:55,000 --> 00:37:58,000
they differ by this constant.
OK.

403
00:37:58,000 --> 00:38:01,000
Just to recap the method what
did we do?

404
00:38:01,000 --> 00:38:04,000
We started with -- And,
of course, you can do it in

405
00:38:04,000 --> 00:38:07,000
whichever order you prefer,
but you have to still follow

406
00:38:07,000 --> 00:38:11,000
the systematic method.
You start with f sub x and you

407
00:38:11,000 --> 00:38:13,000
integrate that with respect to
x.

408
00:38:13,000 --> 00:38:19,000
That gives you f up to a
function of y and z only.

409
00:38:19,000 --> 00:38:25,000
Now you compare f sub y as
given to you by the vector field

410
00:38:25,000 --> 00:38:32,000
with the formula you get from
this expression for f.

411
00:38:32,000 --> 00:38:36,000
And, of course,
this one will involve g sub y.

412
00:38:36,000 --> 00:38:41,000
Out of this,
you will get the value of g sub

413
00:38:41,000 --> 00:38:44,000
y.
When you have g sub y that

414
00:38:44,000 --> 00:38:48,000
gives you g up to a function of
z only.

415
00:38:48,000 --> 00:38:52,000
And so now you have f up to a
function of z only.

416
00:38:52,000 --> 00:38:57,000
And what you will do is look at
the derivative with respect to

417
00:38:57,000 --> 00:38:59,000
z,
the one you want coming from

418
00:38:59,000 --> 00:39:02,000
the vector field and the one you
have coming from this formula

419
00:39:02,000 --> 00:39:05,000
for f,
match them and that will tell

420
00:39:05,000 --> 00:39:09,000
you h prime.
You will get h and then you

421
00:39:09,000 --> 00:39:20,000
will get f.
Any questions?

422
00:39:20,000 --> 00:39:25,000
Who still prefers this method?
OK, still most of you.

423
00:39:25,000 --> 00:39:29,000
Who is thinking that maybe the
other method was not so bad

424
00:39:29,000 --> 00:39:33,000
after all?
OK. That is still a minority.

425
00:39:33,000 --> 00:39:36,000
You can choose whichever one
you prefer.

426
00:39:36,000 --> 00:39:42,000
I would encourage you to get
some practice by trying both on

427
00:39:42,000 --> 00:39:47,000
least a couple of examples just
to make sure that you know how

428
00:39:47,000 --> 00:39:53,000
to do them both and then stick
to whichever one you prefer.

429
00:39:53,000 --> 00:39:59,000
Any questions on that?
No. I guess I already asked.

430
00:39:59,000 --> 00:40:07,000
Still no questions?
OK.

431
00:40:07,000 --> 00:40:13,000
The next logical thing is going
to be curl.

432
00:40:13,000 --> 00:40:17,000
And the theorem that is going
to replace Green's theorem for

433
00:40:17,000 --> 00:40:21,000
work in this setting is going to
be called Stokes' theorem.

434
00:40:21,000 --> 00:40:34,000
Let me start by telling you
about curl in 3D.

435
00:40:34,000 --> 00:40:39,000
Here is the statement.
The curl is just going to

436
00:40:39,000 --> 00:40:43,000
measure how much your vector
field fails to be conservative.

437
00:40:43,000 --> 00:40:47,000
And, if you want to think about
it in terms of motions,

438
00:40:47,000 --> 00:40:51,000
that also will measure the
rotation part of the motion.

439
00:40:51,000 --> 00:40:55,000
Well, let me first give a
definition.

440
00:40:55,000 --> 00:41:00,000
Let's say that my vector field
has components P,

441
00:41:00,000 --> 00:41:06,000
Q and R.
Then we define the curl of F to

442
00:41:06,000 --> 00:41:16,000
be R sub y minus Q sub z times i
plus P sub z minus R sub x times

443
00:41:16,000 --> 00:41:21,000
j plus Q sub x minus P sub y
times k.

444
00:41:21,000 --> 00:41:25,000
And of course nobody can
remember this formula,

445
00:41:25,000 --> 00:41:28,000
so what is the structure of
this formula?

446
00:41:28,000 --> 00:41:35,000
Well, you see,
each of these guys is one of

447
00:41:35,000 --> 00:41:43,000
the things that have to be zero
for our field to be

448
00:41:43,000 --> 00:41:52,000
conservative.
If F is defined in a simply

449
00:41:52,000 --> 00:42:05,000
connected region then we have
that F is conservative and is

450
00:42:05,000 --> 00:42:15,000
equivalent to if and only if
curl F is zero.

451
00:42:15,000 --> 00:42:19,000
Now, an important difference
between curl here and curl in

452
00:42:19,000 --> 00:42:23,000
the plane is that now the curl
of a vector field is again a

453
00:42:23,000 --> 00:42:27,000
vector field.
These expressions are functions

454
00:42:27,000 --> 00:42:31,000
of x, y, z and together you form
a vector out of them.

455
00:42:31,000 --> 00:42:36,000
The curl of a vector field in
space is actually a vector

456
00:42:36,000 --> 00:42:43,000
field, not a scalar function.
I have delayed the inevitable.

457
00:42:43,000 --> 00:42:47,000
I have to really tell you how
to remember this evil formula.

458
00:42:47,000 --> 00:42:55,000
The secret is that,
in fact, you can think of this

459
00:42:55,000 --> 00:43:01,000
as del cross f.
Maybe you have seen that in

460
00:43:01,000 --> 00:43:04,000
physics.
This is really where this del

461
00:43:04,000 --> 00:43:08,000
notation becomes extremely
useful, because that is

462
00:43:08,000 --> 00:43:13,000
basically the only way to
remember the formula for curl.

463
00:43:30,000 --> 00:43:34,000
Remember we introduced the dell
operator.

464
00:43:34,000 --> 00:43:42,000
That was this symbolic vector
operator in which the components

465
00:43:42,000 --> 00:43:47,000
are the partial derivative
operators.

466
00:43:47,000 --> 00:43:59,000
We have seen that if you apply
this to a scalar function then

467
00:43:59,000 --> 00:44:08,000
that will give you the gradient.
And we have seen that if you do

468
00:44:08,000 --> 00:44:13,000
the dot product between dell and
a vector field,

469
00:44:13,000 --> 00:44:19,000
maybe I should give it
components P,

470
00:44:19,000 --> 00:44:24,000
Q and R,
you will get partial P over

471
00:44:24,000 --> 00:44:30,000
partial x plus partial Q over
partial y plus partial R over

472
00:44:30,000 --> 00:44:36,000
partial z,
which is the divergence.

473
00:44:36,000 --> 00:44:47,000
And so now what is new is that
if I try to do dell cross F,

474
00:44:47,000 --> 00:44:53,000
well, what is dell cross F?
I have to set up a

475
00:44:53,000 --> 00:44:58,000
cross-product between this
strange thing that is not really

476
00:44:58,000 --> 00:45:02,000
a vector.
I mean, I cannot really think

477
00:45:02,000 --> 00:45:05,000
of partial over partial x as a
number.

478
00:45:05,000 --> 00:45:10,000
And my vector field <P,
Q, R>.

479
00:45:10,000 --> 00:45:14,000
See, that is really a
completely perverted use of a

480
00:45:14,000 --> 00:45:18,000
determinant notation.
Initially, determinants were

481
00:45:18,000 --> 00:45:21,000
just supposed to be you had a
three by three table of numbers

482
00:45:21,000 --> 00:45:23,000
and you computed a number out of
them.

483
00:45:23,000 --> 00:45:28,000
These guys are functions so
they count as numbers,

484
00:45:28,000 --> 00:45:33,000
but these are vectors and these
are partial derivatives.

485
00:45:33,000 --> 00:45:37,000
It doesn't really make much
sense, except this notation.

486
00:45:37,000 --> 00:45:43,000
If you try to enter this into a
calculator or computer,

487
00:45:43,000 --> 00:45:48,000
it will just yell back at you
saying are you crazy.

488
00:45:48,000 --> 00:45:50,000
[LAUGHTER]
We just use that as a notation

489
00:45:50,000 --> 00:45:53,000
to remember what is in there.
Let's try and see how that

490
00:45:53,000 --> 00:45:55,000
works.
The component of i in this

491
00:45:55,000 --> 00:45:58,000
cross-product,
remember that is this smaller

492
00:45:58,000 --> 00:46:01,000
determinant,
that smaller determinant is

493
00:46:01,000 --> 00:46:07,000
partial over partial y of R
minus partial over partial z of

494
00:46:07,000 --> 00:46:10,000
Q,
the coefficient of i.

495
00:46:10,000 --> 00:46:14,000
And that seems to be what I had
over there.

496
00:46:14,000 --> 00:46:18,000
If not then I made a mistake.
Minus the next determinant

497
00:46:18,000 --> 00:46:20,000
times z.
Remember there is always a

498
00:46:20,000 --> 00:46:23,000
minus sign in front of a j
component when you do a

499
00:46:23,000 --> 00:46:27,000
cross-product.
The other one is partial over

500
00:46:27,000 --> 00:46:33,000
partial x R minus partial over
partial z of P plus the

501
00:46:33,000 --> 00:46:39,000
component of z which is going to
be partial over partial x Q

502
00:46:39,000 --> 00:46:46,000
minus partial over partial y P.
And that is indeed going to be

503
00:46:46,000 --> 00:46:48,000
the curl of F.
In practice,

504
00:46:48,000 --> 00:46:51,000
if you have to compute the curl
of a vector field,

505
00:46:51,000 --> 00:46:53,000
you know, don't try to remember
this formula.

506
00:46:53,000 --> 00:46:59,000
Just set up this cross-product
with whatever formulas you have

507
00:46:59,000 --> 00:47:04,000
for the components of a field
and then compute it.

508
00:47:04,000 --> 00:47:15,000
Don't bother to try to remember
the general formula,

509
00:47:15,000 --> 00:47:25,000
just remember this.
What is the geometric

510
00:47:25,000 --> 00:47:35,000
interpretation of curl,
just to finish?

511
00:47:35,000 --> 00:47:48,000
In a way, I will say just curl
measures the rotation component

512
00:47:48,000 --> 00:47:55,000
in a velocity field.
An exercise that you can do,

513
00:47:55,000 --> 00:47:59,000
which is actually pretty easy
to check, is say that we have a

514
00:47:59,000 --> 00:48:04,000
fluid that is just rotating
about the x-axis uniformly.

515
00:48:04,000 --> 00:48:10,000
Your fluid is just rotating
like that about the z-axis.

516
00:48:10,000 --> 00:48:15,000
If I take a rotation about the
z-axis.

517
00:48:15,000 --> 00:48:28,000
That is given by a velocity
field with components at angular

518
00:48:28,000 --> 00:48:35,000
velocity omega.
That will be negative omega

519
00:48:35,000 --> 00:48:41,000
times y, then omega x and zero.
And the curl of that you can

520
00:48:41,000 --> 00:48:45,000
compute, and you will find two
omega times k.

521
00:48:45,000 --> 00:48:48,000
Concretely,
this curl gives you the angular

522
00:48:48,000 --> 00:48:51,000
velocity of the rotation,
well, with a factor two but

523
00:48:51,000 --> 00:48:53,000
that doesn't matter,
and the axis of rotation,

524
00:48:53,000 --> 00:48:56,000
the direction of the axis of
rotation.

525
00:48:56,000 --> 00:48:59,000
It tells you it is rotating
about a vertical axis.

526
00:48:59,000 --> 00:49:01,000
And, in general,
if you have a complicated

527
00:49:01,000 --> 00:49:03,000
motion some of it might be,
you know, there is a

528
00:49:03,000 --> 00:49:05,000
translation.
And then within that

529
00:49:05,000 --> 00:49:08,000
translation there is maybe
expansion and rotation and

530
00:49:08,000 --> 00:49:11,000
sharing and everything.
And the curl will compute how

531
00:49:11,000 --> 00:49:14,000
much rotation is taking place.
It will tell you,

532
00:49:14,000 --> 00:49:16,000
say that you have a very small
solid,

533
00:49:16,000 --> 00:49:19,000
I don't know like a ping pong
ball in your flow,

534
00:49:19,000 --> 00:49:21,000
and it is just going with the
flow,

535
00:49:21,000 --> 00:49:25,000
it tells you how it is going to
start rotating.

536
00:49:25,000 --> 00:49:32,000
That is what curl measures.
On Thursday we will see Stokes'

537
00:49:32,000 --> 00:49:37,000
theorem, which will be the last
ingredient before the next exam.

538
00:49:37,000 --> 00:49:40,000
And then on Friday we will
review stuff.