1
00:00:00,000 --> 00:00:08,480
CHRISTINE BREINER: Welcome
back to recitation.

2
00:00:08,480 --> 00:00:11,600
In this video, I want us to work
on the following problem, which

3
00:00:11,600 --> 00:00:15,720
is to show that this vector
field is not conservative.

4
00:00:15,720 --> 00:00:20,140
And the vector field
is minus y*i plus x*j,

5
00:00:20,140 --> 00:00:22,220
all divided by x
squared plus y squared.

6
00:00:22,220 --> 00:00:24,740
So you can think about this
in two separate components,

7
00:00:24,740 --> 00:00:27,590
if you need to, as minus y
divided by x squared plus y

8
00:00:27,590 --> 00:00:32,760
squared i plus x over x
squared plus y squared j.

9
00:00:32,760 --> 00:00:35,029
So that's really
exactly the same thing.

10
00:00:35,029 --> 00:00:37,320
So your object is to show
that this vector field is not

11
00:00:37,320 --> 00:00:38,320
conservative.

12
00:00:38,320 --> 00:00:41,529
And why don't you work on that
for awhile, pause the video,

13
00:00:41,529 --> 00:00:43,570
and then when you're ready
to see my solution you

14
00:00:43,570 --> 00:00:44,736
can bring the video back up.

15
00:00:53,850 --> 00:00:54,820
So welcome back.

16
00:00:54,820 --> 00:00:58,082
Well, maybe some of you thought
I had a typo in this problem

17
00:00:58,082 --> 00:00:59,540
initially, and I
wanted you to show

18
00:00:59,540 --> 00:01:03,004
it, in fact, was conservative,
but it actually is not

19
00:01:03,004 --> 00:01:04,170
a conservative vector field.

20
00:01:04,170 --> 00:01:08,180
And let me explain how we can
show it is not conservative

21
00:01:08,180 --> 00:01:11,540
and why probably what
you did initially to try

22
00:01:11,540 --> 00:01:13,985
and show it was not didn't work.

23
00:01:13,985 --> 00:01:15,860
So maybe that wording
was a little confusing,

24
00:01:15,860 --> 00:01:17,270
but let me take you through it.

25
00:01:17,270 --> 00:01:19,730
So the first thing
I would imagine

26
00:01:19,730 --> 00:01:23,710
you tried is you looked
at M sub y and N sub x.

27
00:01:23,710 --> 00:01:27,690
So M in this case is negative y
over x squared plus y squared.

28
00:01:27,690 --> 00:01:30,050
And N in this case,
capital N in this case,

29
00:01:30,050 --> 00:01:32,760
is x divided by x
squared plus y squared.

30
00:01:32,760 --> 00:01:35,120
So if you worked that
out, you probably did

31
00:01:35,120 --> 00:01:37,480
or maybe you didn't,
and I'll just show you.

32
00:01:37,480 --> 00:01:40,690
M sub y-- let me just
double check-- is y

33
00:01:40,690 --> 00:01:44,910
squared minus x squared over
x squared plus y squared,

34
00:01:44,910 --> 00:01:46,410
I think with an
extra squared on it.

35
00:01:46,410 --> 00:01:48,130
Yeah.

36
00:01:48,130 --> 00:01:51,210
And that's also
equal to N sub x.

37
00:01:51,210 --> 00:01:51,710
Right?

38
00:01:51,710 --> 00:01:54,680
So what you know so far,
what you might have thought

39
00:01:54,680 --> 00:01:59,880
immediately, was well, N sub x
minus M sub y is the curl of F

40
00:01:59,880 --> 00:02:02,390
and that's equal to 0, and
therefore this vector field

41
00:02:02,390 --> 00:02:03,730
is conservative.

42
00:02:03,730 --> 00:02:06,070
But the problem is the
theorem you were thinking

43
00:02:06,070 --> 00:02:08,480
about referencing doesn't hold.

44
00:02:08,480 --> 00:02:10,000
And the reason is
because there are

45
00:02:10,000 --> 00:02:11,330
two hypotheses in that theorem.

46
00:02:11,330 --> 00:02:14,980
And one is that if I
define this vector field,

47
00:02:14,980 --> 00:02:18,350
if I call it capital F, the
vector field-- or the theorem

48
00:02:18,350 --> 00:02:30,110
is that capital F defined
everywhere, and curl

49
00:02:30,110 --> 00:02:37,275
of F equal to 0,
implies F conservative.

50
00:02:40,231 --> 00:02:40,730
OK.

51
00:02:40,730 --> 00:02:44,350
So that's the theorem you
might have been trying to use.

52
00:02:44,350 --> 00:02:46,930
You see from this the
curl of F equals 0,

53
00:02:46,930 --> 00:02:50,070
but the problem is the first
part of this statement,

54
00:02:50,070 --> 00:02:52,370
that F being defined
everywhere, is not true.

55
00:02:52,370 --> 00:02:55,990
In fact, there's one place in
R^2 where this vector field is

56
00:02:55,990 --> 00:02:58,829
not defined, and that is
when x is 0 and y is 0.

57
00:02:58,829 --> 00:03:00,870
Because at that point,
obviously, the denominator

58
00:03:00,870 --> 00:03:03,120
is zero and we run into trouble.

59
00:03:03,120 --> 00:03:07,050
So you cannot use this theorem
to say F is conservative

60
00:03:07,050 --> 00:03:09,760
because it's not
defined everywhere.

61
00:03:09,760 --> 00:03:11,460
Or I should be careful
how I say that.

62
00:03:11,460 --> 00:03:13,890
There is somewhere
that it is not defined.

63
00:03:13,890 --> 00:03:16,647
So even though the
curl of F equals 0,

64
00:03:16,647 --> 00:03:18,480
the first part of the
statement is not true.

65
00:03:18,480 --> 00:03:22,220
So you cannot get anything
out of this theorem.

66
00:03:22,220 --> 00:03:25,510
So knowing the curl of F equals
0 doesn't tell you whether it's

67
00:03:25,510 --> 00:03:26,610
conservative or not.

68
00:03:26,610 --> 00:03:27,750
OK?

69
00:03:27,750 --> 00:03:29,270
So now what I'm
going to do is I'm

70
00:03:29,270 --> 00:03:30,911
going to show-- I
told you we want

71
00:03:30,911 --> 00:03:32,160
to show it's not conservative.

72
00:03:32,160 --> 00:03:34,250
I'm going to show you
how we can show that.

73
00:03:34,250 --> 00:03:37,900
And what we're going to do is
we're going to find a loop,

74
00:03:37,900 --> 00:03:41,277
a closed loop, so a
closed curve in R^2,

75
00:03:41,277 --> 00:03:43,610
that when I integrate this
vector field over that closed

76
00:03:43,610 --> 00:03:45,910
curve, I don't get zero.

77
00:03:45,910 --> 00:03:48,550
And then we would know that
the vector field is not

78
00:03:48,550 --> 00:03:49,830
conservative.

79
00:03:49,830 --> 00:03:51,540
So that's what
we're going to do.

80
00:03:51,540 --> 00:03:56,274
So let me write
it out explicitly

81
00:03:56,274 --> 00:03:58,690
and then we'll figure out the
curve we want and then we'll

82
00:03:58,690 --> 00:04:00,860
parameterize the
curve appropriately.

83
00:04:00,860 --> 00:04:04,610
So I'm going to show,
for some closed curve--

84
00:04:04,610 --> 00:04:06,160
I'm going to pick
my curve and I'm

85
00:04:06,160 --> 00:04:07,826
going to integrate
over the closed curve

86
00:04:07,826 --> 00:04:09,330
and I'm going to integrate this.

87
00:04:09,330 --> 00:04:16,790
Minus y over x squared
plus y squared dx plus x

88
00:04:16,790 --> 00:04:20,610
over x squared
plus y squared dy.

89
00:04:20,610 --> 00:04:23,390
And I'm going to show that
if I pick a certain curve,

90
00:04:23,390 --> 00:04:26,710
I'm going to get something
that's not equal to zero, OK?

91
00:04:26,710 --> 00:04:31,070
And the curve I'm going to
pick is the unit circle.

92
00:04:31,070 --> 00:04:33,120
So we're going to let
C be the unit circle.

93
00:04:36,310 --> 00:04:41,280
Let C equal-- I'll just
write the unit circle.

94
00:04:41,280 --> 00:04:44,180
But how can I parameterize
the unit circle easily?

95
00:04:44,180 --> 00:04:46,980
I can parameterize
the unit circle easily

96
00:04:46,980 --> 00:04:51,380
by x equal to cosine theta
and y equal to sine theta.

97
00:04:51,380 --> 00:04:52,850
So let me do that.

98
00:04:52,850 --> 00:04:54,510
And why am I picking
the unit circle?

99
00:04:54,510 --> 00:04:56,660
We'll see why that
is in a second.

100
00:04:56,660 --> 00:05:00,750
So we're going to let x equal
cosine theta and y equal sine

101
00:05:00,750 --> 00:05:03,046
theta.

102
00:05:03,046 --> 00:05:04,420
And so now we know
what goes here

103
00:05:04,420 --> 00:05:05,700
and we know what goes here.

104
00:05:05,700 --> 00:05:08,100
By the way, what is x
squared plus y squared?

105
00:05:08,100 --> 00:05:10,260
It's cosine squared
theta plus sine squared

106
00:05:10,260 --> 00:05:12,160
theta, which is equal to 1.

107
00:05:12,160 --> 00:05:13,850
This is exactly the
square of the radius

108
00:05:13,850 --> 00:05:16,160
and since we're on the
unit circle, that's 1.

109
00:05:16,160 --> 00:05:17,710
That's why I picked
the unit circle

110
00:05:17,710 --> 00:05:21,790
because I wanted the
denominator to be very simple.

111
00:05:21,790 --> 00:05:23,590
So I've got the x's and the y's.

112
00:05:23,590 --> 00:05:26,310
Now what's dx?

113
00:05:26,310 --> 00:05:30,920
dx is going to be equal to
minus sine theta d theta.

114
00:05:30,920 --> 00:05:32,010
And what's dy?

115
00:05:32,010 --> 00:05:33,970
I'll just write it
right underneath here.

116
00:05:33,970 --> 00:05:36,980
dy is going to equal
cosine theta d theta.

117
00:05:40,140 --> 00:05:43,620
So let me just point out
again what we were doing here.

118
00:05:43,620 --> 00:05:45,600
We want to parameterize
the unit circle.

119
00:05:45,600 --> 00:05:48,650
I chose to parameterize it in
theta, which I haven't told you

120
00:05:48,650 --> 00:05:51,030
what my bounds are yet, but
I've done everything else.

121
00:05:51,030 --> 00:05:53,230
I needed to know
what x and y were

122
00:05:53,230 --> 00:05:55,510
and also what dx and dy were.

123
00:05:55,510 --> 00:05:58,210
And so now I can
start substituting in.

124
00:05:58,210 --> 00:06:02,380
So let's figure out what I get
when I start substituting in.

125
00:06:02,380 --> 00:06:03,581
I'm integrating now.

126
00:06:03,581 --> 00:06:05,330
Again, I said I didn't
mention the bounds.

127
00:06:05,330 --> 00:06:08,240
What are the bounds on theta
to get the whole unit circle?

128
00:06:08,240 --> 00:06:11,200
I'm just going to
integrate from 0 to 2*pi.

129
00:06:11,200 --> 00:06:13,170
So I integrate from 0 to 2*pi.

130
00:06:13,170 --> 00:06:15,650
That takes me all the
way around the circle.

131
00:06:15,650 --> 00:06:19,310
Minus y is negative sine theta.

132
00:06:19,310 --> 00:06:21,660
This part is 1 as I
mentioned earlier.

133
00:06:21,660 --> 00:06:25,230
And then dx is minus
sine theta d theta.

134
00:06:25,230 --> 00:06:28,990
So I have a minus sine theta
times a minus sine theta.

135
00:06:28,990 --> 00:06:33,560
That's going to give me a
sine squared theta d theta.

136
00:06:33,560 --> 00:06:36,720
And then this term,
the second term,

137
00:06:36,720 --> 00:06:38,600
when I integrate
over the curve, I'm

138
00:06:38,600 --> 00:06:41,090
going to just rewrite
another one here separately

139
00:06:41,090 --> 00:06:42,404
momentarily.

140
00:06:42,404 --> 00:06:45,970
x is cosine theta.

141
00:06:45,970 --> 00:06:47,285
x squared plus y squared is 1.

142
00:06:47,285 --> 00:06:50,010
And dy is cosine theta d theta.

143
00:06:50,010 --> 00:06:52,215
So I get cosine
squared theta d theta.

144
00:06:52,215 --> 00:06:52,970
All right?

145
00:06:57,100 --> 00:06:58,420
Now here we are.

146
00:06:58,420 --> 00:07:01,010
If I tried to integrate
both of these separately,

147
00:07:01,010 --> 00:07:02,860
it would take potentially
a very long time

148
00:07:02,860 --> 00:07:04,400
and be kind of annoying.

149
00:07:04,400 --> 00:07:07,720
But if you notice, because
I can add these integrals,

150
00:07:07,720 --> 00:07:10,660
I can add over-- they
have the same bounds,

151
00:07:10,660 --> 00:07:12,660
so I can put them back together.

152
00:07:12,660 --> 00:07:17,490
Sine squared theta plus cosine
squared theta always equals 1.

153
00:07:17,490 --> 00:07:19,350
That's a trig identity
that's good to know.

154
00:07:19,350 --> 00:07:24,050
So this in fact is equal to the
integral from 0 to 2*pi of d

155
00:07:24,050 --> 00:07:25,960
theta.

156
00:07:25,960 --> 00:07:27,934
So let me come
back one more time

157
00:07:27,934 --> 00:07:29,100
and make sure we understand.

158
00:07:29,100 --> 00:07:32,320
We're integrating from 0 to 2*pi
sine squared theta d theta plus

159
00:07:32,320 --> 00:07:33,730
cosine squared theta d theta.

160
00:07:33,730 --> 00:07:35,940
That's just sine squared
theta plus cosine squared

161
00:07:35,940 --> 00:07:37,920
theta d theta.

162
00:07:37,920 --> 00:07:39,040
So that's 1.

163
00:07:39,040 --> 00:07:41,090
So 1 times d theta.

164
00:07:41,090 --> 00:07:41,920
But what is this?

165
00:07:41,920 --> 00:07:45,420
Well, this integral from 0
to 2 pi of d theta is theta

166
00:07:45,420 --> 00:07:46,980
evaluated at 2*pi and 0.

167
00:07:46,980 --> 00:07:51,770
I actually get 2*pi, which is
in particular not equal to 0,

168
00:07:51,770 --> 00:07:52,660
right?

169
00:07:52,660 --> 00:07:55,950
So that actually shows you
that this vector field is not

170
00:07:55,950 --> 00:07:57,060
conservative.

171
00:07:57,060 --> 00:07:58,940
Now why does this make sense?

172
00:07:58,940 --> 00:08:00,680
This makes sense
because if I really

173
00:08:00,680 --> 00:08:02,960
think about what I'm
doing-- actually,

174
00:08:02,960 --> 00:08:05,532
this is the place where maybe
it'll ultimately make the most

175
00:08:05,532 --> 00:08:06,990
sense-- what you're
doing is you're

176
00:08:06,990 --> 00:08:10,120
looking at how theta
changes as you go all

177
00:08:10,120 --> 00:08:12,400
the way around the origin once.

178
00:08:12,400 --> 00:08:14,240
And theta changes by 2*pi.

179
00:08:14,240 --> 00:08:16,620
If you go all the
way around one time,

180
00:08:16,620 --> 00:08:19,750
the theta value starts at
0 and then goes up to 2*pi.

181
00:08:19,750 --> 00:08:23,120
And so that's exactly where
this 2*pi is coming from.

182
00:08:23,120 --> 00:08:25,002
So that actually
is ultimately how

183
00:08:25,002 --> 00:08:27,460
you were going to be able to
show that this vector field is

184
00:08:27,460 --> 00:08:28,880
not conservative.

185
00:08:28,880 --> 00:08:32,430
So let me go back and just
remind you where we came from.

186
00:08:32,430 --> 00:08:35,510
We started off
with a vector field

187
00:08:35,510 --> 00:08:38,284
and we wanted to know if
it was not conservative.

188
00:08:38,284 --> 00:08:40,450
We wanted to show-- sorry--
it was not conservative.

189
00:08:40,450 --> 00:08:42,241
So the first thing you
might want to check,

190
00:08:42,241 --> 00:08:45,880
which maybe you did, was
if the curl was zero.

191
00:08:45,880 --> 00:08:48,920
And in fact the curl is zero.

192
00:08:48,920 --> 00:08:50,470
And so maybe you
thought, well, she

193
00:08:50,470 --> 00:08:51,685
might have done something
wrong, or she might

194
00:08:51,685 --> 00:08:52,893
have written something wrong.

195
00:08:52,893 --> 00:08:56,510
But we can't actually appeal to
the theorem you want to appeal

196
00:08:56,510 --> 00:08:59,210
to to make any conclusion
about the vector field,

197
00:08:59,210 --> 00:09:02,390
because the vector field in
our example is not defined

198
00:09:02,390 --> 00:09:04,570
for every value of (x,y).

199
00:09:04,570 --> 00:09:08,090
So for every value in the
xy-plane, we cannot define F.

200
00:09:08,090 --> 00:09:10,890
There's one value for
which we cannot define F.

201
00:09:10,890 --> 00:09:13,050
And so we cannot say
that if the curl's 0,

202
00:09:13,050 --> 00:09:15,540
then the vector field
is conservative.

203
00:09:15,540 --> 00:09:18,130
We can't draw any conclusions
from this theorem.

204
00:09:18,130 --> 00:09:19,890
So then we had to
actually find a way

205
00:09:19,890 --> 00:09:22,820
to show it was not conservative
without looking at the curl.

206
00:09:22,820 --> 00:09:25,120
And that amounts to showing
there is a closed curve

207
00:09:25,120 --> 00:09:27,274
that when I integrate
over that closed curve--

208
00:09:27,274 --> 00:09:29,690
when I look at what the vector
field does over that closed

209
00:09:29,690 --> 00:09:32,300
curve-- I get
something non-zero.

210
00:09:32,300 --> 00:09:34,610
And we picked an easy example.

211
00:09:34,610 --> 00:09:36,270
This is actually
what the integral

212
00:09:36,270 --> 00:09:39,240
will look like over the
closed curve in x and y.

213
00:09:39,240 --> 00:09:41,080
We let our closed curve
be the unit circle,

214
00:09:41,080 --> 00:09:43,070
then we parameterized in theta.

215
00:09:43,070 --> 00:09:46,020
And we see that actually what
this vector field is doing

216
00:09:46,020 --> 00:09:48,630
is it's looking at
what is d theta?

217
00:09:48,630 --> 00:09:50,440
It's finding out
what d theta is.

218
00:09:50,440 --> 00:09:53,890
And so we find out the integral
from 0 to 2*pi of d theta is

219
00:09:53,890 --> 00:09:55,385
obviously not zero.

220
00:09:55,385 --> 00:09:57,260
And that gives us that
the vector field's not

221
00:09:57,260 --> 00:09:58,420
conservative.

222
00:09:58,420 --> 00:09:59,900
So that's where I'll stop.