1
00:00:07,230 --> 00:00:08,790
Welcome back to recitation.

2
00:00:08,790 --> 00:00:13,050
In this video I'd like us to
consider the following problem.

3
00:00:13,050 --> 00:00:16,770
So the question is for what
simple closed curve C, which

4
00:00:16,770 --> 00:00:19,870
is oriented positively around
the region it encloses,

5
00:00:19,870 --> 00:00:23,590
does the integral over
C of minus the quantity

6
00:00:23,590 --> 00:00:32,550
x squared y plus 3x minus 2y dx
plus 4 y squared x minus 2x dy

7
00:00:32,550 --> 00:00:34,510
achieve its minimum value?

8
00:00:34,510 --> 00:00:37,730
So again, the thing
we want to find here,

9
00:00:37,730 --> 00:00:41,210
the point we want to make,
is that we have this integral

10
00:00:41,210 --> 00:00:43,390
and we're allowed
to vary C. So we're

11
00:00:43,390 --> 00:00:46,860
allowed to change the curve,
the simple closed curve.

12
00:00:46,860 --> 00:00:51,310
And we want to know for what
curve C does this integral

13
00:00:51,310 --> 00:00:52,950
achieve its minimum value?

14
00:00:52,950 --> 00:00:55,400
So why don't you work
on that, think about it,

15
00:00:55,400 --> 00:00:57,410
pause the video, and
when you're feeling

16
00:00:57,410 --> 00:01:00,310
like you're ready to see how I
do it, bring the video back up.

17
00:01:09,570 --> 00:01:10,650
Welcome back.

18
00:01:10,650 --> 00:01:14,440
So again, what we would
like to do for this problem

19
00:01:14,440 --> 00:01:17,760
is we want to take
this quantity, which

20
00:01:17,760 --> 00:01:20,950
is varying in C, and
we want to figure out

21
00:01:20,950 --> 00:01:22,520
a way to minimize it.

22
00:01:22,520 --> 00:01:24,910
To find its minimum value.

23
00:01:24,910 --> 00:01:27,250
And actually what's
interesting to think about,

24
00:01:27,250 --> 00:01:28,950
before we proceed
any further, is

25
00:01:28,950 --> 00:01:32,060
you might think you want to take
the smallest possible simple

26
00:01:32,060 --> 00:01:32,980
closed curve you can.

27
00:01:32,980 --> 00:01:35,680
In other words, you might want
to shrink it down to nothing.

28
00:01:35,680 --> 00:01:40,670
But then this integral would
be 0 and the question is,

29
00:01:40,670 --> 00:01:41,859
can we do better?

30
00:01:41,859 --> 00:01:43,900
Because we could have a
minimum value, of course,

31
00:01:43,900 --> 00:01:44,760
that's negative.

32
00:01:44,760 --> 00:01:46,790
And so we could do
better by having a larger

33
00:01:46,790 --> 00:01:48,360
curve put in the right place.

34
00:01:48,360 --> 00:01:50,750
So I just want to point that
out, if you were thinking

35
00:01:50,750 --> 00:01:52,770
"I'll just make C
not actually a curve,

36
00:01:52,770 --> 00:01:54,430
I just shrink it
down to nothing."

37
00:01:54,430 --> 00:01:55,840
That won't be the
best we can do.

38
00:01:55,840 --> 00:01:57,800
We can actually
find a curve that

39
00:01:57,800 --> 00:02:01,100
has an integral that is not
0, but is in fact negative.

40
00:02:01,100 --> 00:02:03,680
And then we want to make it
the most negative we can.

41
00:02:03,680 --> 00:02:05,380
So that's the idea.

42
00:02:05,380 --> 00:02:08,510
So what we're
going to do here is

43
00:02:08,510 --> 00:02:10,920
we're going to use Green's
theorem to help us.

44
00:02:10,920 --> 00:02:15,200
And so what we want to remember
is that if we have the integral

45
00:02:15,200 --> 00:02:24,490
over C of M*dx plus N*dy we want
to write that as the integral

46
00:02:24,490 --> 00:02:33,055
over R of N sub x
minus M sub y dx dy.

47
00:02:33,055 --> 00:02:35,500
So what I want to
point out, I'm just

48
00:02:35,500 --> 00:02:37,470
going to write down
what these are.

49
00:02:37,470 --> 00:02:41,132
I'm not going to take
the derivatives for you.

50
00:02:41,132 --> 00:02:42,840
I'm just going to show
you what they are.

51
00:02:42,840 --> 00:02:46,650
So in this case, with
the M and N that we have,

52
00:02:46,650 --> 00:02:50,540
we get exactly this
value for the integral.

53
00:02:50,540 --> 00:02:55,560
We get N sub x minus M sub
y is equal to x squared

54
00:02:55,560 --> 00:03:03,154
plus 4 y squared minus 4 dx dy.

55
00:03:03,154 --> 00:03:04,820
You can obviously
compute that yourself.

56
00:03:04,820 --> 00:03:07,370
Just take the derivative
of N with respect to x,

57
00:03:07,370 --> 00:03:10,330
subtract the derivative
of M with respect to y,

58
00:03:10,330 --> 00:03:13,110
you get a little cancellation
and you end up with this.

59
00:03:13,110 --> 00:03:16,210
And so now, instead of
thinking about trying

60
00:03:16,210 --> 00:03:22,320
to minimize this quantity
over here in terms of a curve,

61
00:03:22,320 --> 00:03:25,170
now we can think about trying
to minimize the quantity here

62
00:03:25,170 --> 00:03:27,710
in terms of a region.

63
00:03:27,710 --> 00:03:31,920
And the goal here
is to make the sum

64
00:03:31,920 --> 00:03:33,744
of all of this over
the whole region--

65
00:03:33,744 --> 00:03:36,160
we want to make it as negative
as we can possibly make it.

66
00:03:36,160 --> 00:03:37,870
So essentially
what we want to do

67
00:03:37,870 --> 00:03:41,050
is we want-- on the
boundary of this region,

68
00:03:41,050 --> 00:03:43,690
we would like the
value here to be 0,

69
00:03:43,690 --> 00:03:46,930
and on the inside of the region
we'd like it to be negative.

70
00:03:46,930 --> 00:03:49,210
So let me point that
out again and just

71
00:03:49,210 --> 00:03:50,620
make sure we understand this.

72
00:03:50,620 --> 00:03:53,350
To make this quantity
as small as possible,

73
00:03:53,350 --> 00:03:54,920
what we would like--
let me actually

74
00:03:54,920 --> 00:03:56,960
just draw a little region.

75
00:03:56,960 --> 00:03:59,750
So say this is the region.

76
00:03:59,750 --> 00:04:04,060
To make this integral as small
as possible, what we want

77
00:04:04,060 --> 00:04:07,530
is that x squared plus
4 y squared minus 4 is

78
00:04:07,530 --> 00:04:09,960
negative inside the region.

79
00:04:09,960 --> 00:04:14,770
So if this whole quantity is
less than 0 inside the region,

80
00:04:14,770 --> 00:04:17,030
and we want it to, on the
boundary of the region,

81
00:04:17,030 --> 00:04:18,090
equal 0.

82
00:04:18,090 --> 00:04:20,444
And why do we want it to
equal 0 on the boundary?

83
00:04:20,444 --> 00:04:22,860
Well, that's because then we've
gotten all the negative we

84
00:04:22,860 --> 00:04:25,840
could get and we haven't
added in any positive

85
00:04:25,840 --> 00:04:28,150
and brought the value up.

86
00:04:28,150 --> 00:04:33,640
So that's really why we want
the boundary of the region

87
00:04:33,640 --> 00:04:36,750
to be exactly where
this quantity equals 0.

88
00:04:36,750 --> 00:04:42,980
And so let's think about,
geometrically, what

89
00:04:42,980 --> 00:04:47,480
describes R-- oops,
that should have an s.

90
00:04:47,480 --> 00:04:54,880
What describes R,
where x squared

91
00:04:54,880 --> 00:05:03,220
plus 4 y squared minus 4
is less than 0 inside R,

92
00:05:03,220 --> 00:05:08,110
and that's the same thing as--
in what we're interested in-- x

93
00:05:08,110 --> 00:05:21,960
squared plus 4y squared minus 4
equals 0 on the boundary of R.

94
00:05:21,960 --> 00:05:24,930
And if you look at this,
this is really the expression

95
00:05:24,930 --> 00:05:27,270
that will probably help you.

96
00:05:27,270 --> 00:05:29,470
This expression,
if you rewrite it,

97
00:05:29,470 --> 00:05:34,820
you rewrite it as x squared
plus 4y squared is equal to 4.

98
00:05:34,820 --> 00:05:38,010
And you see that this is
actually the equation that

99
00:05:38,010 --> 00:05:41,000
describes an ellipse.

100
00:05:41,000 --> 00:05:43,830
Maybe you see it more often
if you divide everything by 4,

101
00:05:43,830 --> 00:05:45,780
and so on the right-hand
side you have a 1,

102
00:05:45,780 --> 00:05:48,930
and your coefficients are
fractional, potentially, there.

103
00:05:48,930 --> 00:05:52,170
But this is exactly the
equation for an ellipse.

104
00:05:52,170 --> 00:05:55,790
And so the boundary of R
is an ellipse described

105
00:05:55,790 --> 00:05:58,795
by this equation,
but the boundary of R

106
00:05:58,795 --> 00:06:01,070
is actually just C.

107
00:06:01,070 --> 00:06:04,840
So C we now know is
exactly the curve

108
00:06:04,840 --> 00:06:07,830
that is carved out by this
equation on the plane,

109
00:06:07,830 --> 00:06:08,840
on the xy-plane.

110
00:06:08,840 --> 00:06:11,930
That's an ellipse.

111
00:06:11,930 --> 00:06:15,279
So just to remind you
what we were trying to do.

112
00:06:15,279 --> 00:06:16,820
If you come back
here, we were trying

113
00:06:16,820 --> 00:06:20,340
to figure out a way to
minimize the certain integral

114
00:06:20,340 --> 00:06:23,020
over a path.

115
00:06:23,020 --> 00:06:25,280
And what we did was
instead of trying

116
00:06:25,280 --> 00:06:28,484
to look at a bunch of
paths and figure out

117
00:06:28,484 --> 00:06:29,900
what would minimize
that, we tried

118
00:06:29,900 --> 00:06:31,570
to see if Green's
theorem would help us.

119
00:06:31,570 --> 00:06:33,744
So Green's theorem allowed
us to take something

120
00:06:33,744 --> 00:06:35,660
that was an integral
over a path and change it

121
00:06:35,660 --> 00:06:37,790
to an integral over a region.

122
00:06:37,790 --> 00:06:40,070
And then when we look at
what we ended up with,

123
00:06:40,070 --> 00:06:44,930
we realize that we could
make this the most minimum

124
00:06:44,930 --> 00:06:48,310
if we let this be on the
region where it was negative

125
00:06:48,310 --> 00:06:49,199
everywhere.

126
00:06:49,199 --> 00:06:50,740
So we were looking
for a region where

127
00:06:50,740 --> 00:06:53,890
this quantity was everywhere
negative on the inside and 0

128
00:06:53,890 --> 00:06:55,250
on the boundary.

129
00:06:55,250 --> 00:06:56,950
And that's exactly what we did.

130
00:06:56,950 --> 00:06:59,429
And then we see that
we get to a point

131
00:06:59,429 --> 00:07:00,970
where the boundary
has this equation,

132
00:07:00,970 --> 00:07:03,340
x squared plus 4y
squared equals 4.

133
00:07:03,340 --> 00:07:05,310
We see then the
boundary's an ellipse,

134
00:07:05,310 --> 00:07:07,660
and C is indeed the
boundary of the region.

135
00:07:07,660 --> 00:07:12,420
So we see that C is the ellipse
described by this equation.

136
00:07:12,420 --> 00:07:13,934
So that's where I'll stop.