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So, let me remind you,
yesterday we've defined and

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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

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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.

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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

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00:01:16 --> 00:01:19
where T is the unit tangent
vector,

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00:01:19 --> 00:01:23
and ds is the arc length
element.

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00:01:23 --> 00:01:30
Or, in coordinates,
that they integral of M dx N dy

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00:01:30 --> 00:01:38
where M and N are the components
of the vector field.

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00:01:38 --> 00:01:46
OK, so -- Let's do an example
that will just summarize what we

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00:01:46 --> 00:01:51
did yesterday,
and then we will move on to

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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.

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00:02:03 --> 00:02:11
Let's say I give you the vector
field yi plus xj.

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00:02:11 --> 00:02:13
So, it's not completely obvious
what it looks like,

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00:02:13 --> 00:02:16
but here is a computer plot of
that vector field.

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00:02:16 --> 00:02:21
So, that tells you a bit what
it does.

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00:02:21 --> 00:02:24
It points in all sorts of
directions.

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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

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00:02:40 --> 00:02:43
to one.
That move along the unit circle

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00:02:43 --> 00:02:46
to the diagonal,
and then I move back to the

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00:02:46 --> 00:02:54
origin in a straight line.
OK, so C consists of three

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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

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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

48
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,

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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
 
 

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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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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

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00:07:41 --> 00:07:52
actually computing the integral
of, so, y is sine theta.

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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,

103
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.

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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?

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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

115
00:10:42 --> 00:10:46
here to here is just going to be
the opposite,

116
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,

118
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,

120
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

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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

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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,

132
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

145
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

150
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