1
00:00:00,000 --> 00:00:07,230

2
00:00:07,230 --> 00:00:08,790
Welcome back to recitation.

3
00:00:08,790 --> 00:00:10,720
In this video I'd like
us to consider

4
00:00:10,720 --> 00:00:13,050
the following problem.

5
00:00:13,050 --> 00:00:16,770
So the question is for what
simple closed curve C, which

6
00:00:16,770 --> 00:00:20,510
is oriented positively around
the region it encloses, does

7
00:00:20,510 --> 00:00:25,120
the integral over C of minus the
quantity x squared y plus

8
00:00:25,120 --> 00:00:33,050
3x minus 2y dx plus 4y squared
x minus 2x dy achieve its

9
00:00:33,050 --> 00:00:34,510
minimum value?

10
00:00:34,510 --> 00:00:38,212
So again, the thing we want to
find here, the point we want

11
00:00:38,212 --> 00:00:41,950
to make, is that we have this
integral and we're allowed to

12
00:00:41,950 --> 00:00:45,760
vary C. So we're allowed to
change the curve, the simple

13
00:00:45,760 --> 00:00:46,860
closed curve.

14
00:00:46,860 --> 00:00:51,310
And we want to know for what
curve C does this integral

15
00:00:51,310 --> 00:00:52,950
achieve its minimum value?

16
00:00:52,950 --> 00:00:55,800
So why don't you work on that,
think about it, pause the

17
00:00:55,800 --> 00:00:58,880
video, and when you're feeling
like you're ready to see how I

18
00:00:58,880 --> 00:01:00,310
do it, bring the
video back up.

19
00:01:00,310 --> 00:01:09,570

20
00:01:09,570 --> 00:01:10,650
Welcome back.

21
00:01:10,650 --> 00:01:14,710
So again, what we would like to
do for this problem is we

22
00:01:14,710 --> 00:01:19,580
want to take this quantity which
is varying in C and we

23
00:01:19,580 --> 00:01:22,520
want to figure out a
way to minimize it.

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

25
00:01:24,910 --> 00:01:27,250
And actually what's interesting
to think about,

26
00:01:27,250 --> 00:01:30,170
before we proceed any further,
is you might think you want to

27
00:01:30,170 --> 00:01:32,980
take the smallest possible
simple closed curve you can.

28
00:01:32,980 --> 00:01:35,680
In other words, you might want
to shrink it down to nothing.

29
00:01:35,680 --> 00:01:40,790
But then this integral would be
0 and the question is, can

30
00:01:40,790 --> 00:01:41,940
we do better?

31
00:01:41,940 --> 00:01:43,900
Because we could have a minimum
value, of course,

32
00:01:43,900 --> 00:01:44,760
that's negative.

33
00:01:44,760 --> 00:01:47,440
And so we could do better by
having a larger curve put in

34
00:01:47,440 --> 00:01:48,360
the right place.

35
00:01:48,360 --> 00:01:50,980
So I just want to point that
out, if you were thinking why

36
00:01:50,980 --> 00:01:53,590
just make C not actually a
curve, I just shrink it down

37
00:01:53,590 --> 00:01:54,430
to nothing.

38
00:01:54,430 --> 00:01:55,840
That won't be the
best we can do.

39
00:01:55,840 --> 00:01:59,270
We can actually find a curve
that has an integral that is

40
00:01:59,270 --> 00:02:01,100
not 0, but is in
fact negative.

41
00:02:01,100 --> 00:02:03,680
And then we want to make it
the most negative we can.

42
00:02:03,680 --> 00:02:05,380
So that's the idea.

43
00:02:05,380 --> 00:02:09,500
So what we're going to do here
is we're going to use Green's

44
00:02:09,500 --> 00:02:10,920
theorem to help us.

45
00:02:10,920 --> 00:02:14,170
And so what we want to remember
is that if we have

46
00:02:14,170 --> 00:02:22,330
the integral over C of M dx
plus N dy we want to write

47
00:02:22,330 --> 00:02:33,055
that as the integral over R of
N sub x minus M sub y dx dy.

48
00:02:33,055 --> 00:02:36,560
So what I want to point out, I'm
just going to write down

49
00:02:36,560 --> 00:02:37,470
what these are.

50
00:02:37,470 --> 00:02:41,320
I'm not going to take the
derivatives for you.

51
00:02:41,320 --> 00:02:42,840
I'm just going to show
you what they are.

52
00:02:42,840 --> 00:02:46,950
So in this case, with the M
and N that we have, we get

53
00:02:46,950 --> 00:02:50,540
exactly this value
for the integral.

54
00:02:50,540 --> 00:02:56,740
We get N sub x minus M sub y is
equal to x squared plus 4y

55
00:02:56,740 --> 00:03:03,160
squared minus 4 dx dy.

56
00:03:03,160 --> 00:03:04,820
You can obviously compute
that yourself.

57
00:03:04,820 --> 00:03:07,840
Just take the derivative of N
with respect to x, subtract

58
00:03:07,840 --> 00:03:10,930
the derivative of M with respect
to y, you get a little

59
00:03:10,930 --> 00:03:13,110
cancellation and you
end up with this.

60
00:03:13,110 --> 00:03:18,430
And so now, instead of thinking
about trying to

61
00:03:18,430 --> 00:03:22,630
minimize this quantity over here
in terms of a curve, now

62
00:03:22,630 --> 00:03:25,280
we can think about trying to
minimize the quantity here in

63
00:03:25,280 --> 00:03:27,710
terms of a region.

64
00:03:27,710 --> 00:03:33,080
And the goal here is to make
the sum of all of this over

65
00:03:33,080 --> 00:03:33,810
the whole region--

66
00:03:33,810 --> 00:03:36,160
we want to make it as negative
as we can possibly make it.

67
00:03:36,160 --> 00:03:39,575
So essentially what we want to
do is we want on the boundary

68
00:03:39,575 --> 00:03:43,940
of this region, we would like
the value here to be 0 and on

69
00:03:43,940 --> 00:03:46,930
the inside of the region we'd
like it to be negative.

70
00:03:46,930 --> 00:03:49,630
So let me point that out again
and just make sure we

71
00:03:49,630 --> 00:03:50,620
understand this.

72
00:03:50,620 --> 00:03:53,920
To make this quantity as small
as possible, what we would

73
00:03:53,920 --> 00:03:56,960
like-- let me actually just
draw a little region.

74
00:03:56,960 --> 00:03:59,750
So say this is the region.

75
00:03:59,750 --> 00:04:04,290
To make this integral as small
as possible, what we want is

76
00:04:04,290 --> 00:04:07,530
that x squared plus 4y
squared minus 4 is

77
00:04:07,530 --> 00:04:09,960
negative inside the region.

78
00:04:09,960 --> 00:04:14,770
So if this whole quantity is
less than 0 inside the region,

79
00:04:14,770 --> 00:04:16,560
and we want it to, on
the boundary of the

80
00:04:16,560 --> 00:04:18,090
region, equal 0.

81
00:04:18,090 --> 00:04:20,490
And why do we want it to equal
0 on the boundary?

82
00:04:20,490 --> 00:04:22,720
Well, that's because then we've
gotten all the negative

83
00:04:22,720 --> 00:04:26,740
we could get and we haven't
added in any positive and

84
00:04:26,740 --> 00:04:28,150
brought the value up.

85
00:04:28,150 --> 00:04:33,730
So that's really why we want the
boundary of the region to

86
00:04:33,730 --> 00:04:36,750
be exactly where this
quantity equals 0.

87
00:04:36,750 --> 00:04:42,700
And so let's think about
geometrically

88
00:04:42,700 --> 00:04:44,900
what describes R--

89
00:04:44,900 --> 00:04:47,480
oops, that should have an s.

90
00:04:47,480 --> 00:04:57,320
What describes R where x squared
plus 4y squared minus

91
00:04:57,320 --> 00:05:06,720
4 is less than 0 inside R, and
that's the same thing as in

92
00:05:06,720 --> 00:05:09,870
what we're interested in x
squared plus 4y squared minus

93
00:05:09,870 --> 00:05:21,960
4 equals 0 on the
boundary of R.

94
00:05:21,960 --> 00:05:25,040
And if you look at this, this is
really the expression that

95
00:05:25,040 --> 00:05:27,270
will probably help you.

96
00:05:27,270 --> 00:05:30,250
This expression, if you rewrite
it, you rewrite it as

97
00:05:30,250 --> 00:05:34,820
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 --> 00:05:46,540
and so on the right hand side
you have a 1, and your

102
00:05:46,540 --> 00:05:48,930
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 --> 00:05:56,520
And so the boundary of R is an
ellipse described by this

105
00:05:56,520 --> 00:06:01,070
equation, but the boundary
of R is actually just C.

106
00:06:01,070 --> 00:06:05,670
So C we now know is exactly the
curve that is carved out

107
00:06:05,670 --> 00:06:08,840
by this equation on the plane,
on the xy plane.

108
00:06:08,840 --> 00:06:11,930
That's an ellipse.

109
00:06:11,930 --> 00:06:15,700
So just to remind you what
we were trying to do.

110
00:06:15,700 --> 00:06:18,470
If you come back here, we were
trying to figure out a way to

111
00:06:18,470 --> 00:06:23,020
minimize the certain integral
over a path.

112
00:06:23,020 --> 00:06:26,730
And what we did was instead of
trying to look at a bunch of

113
00:06:26,730 --> 00:06:29,900
paths and figure out what would
minimize that, we tried

114
00:06:29,900 --> 00:06:31,570
to see if Green's theorem
would help us.

115
00:06:31,570 --> 00:06:34,440
So Green's theorem allowed us to
take something that was an

116
00:06:34,440 --> 00:06:35,820
integral over a path
and change it to an

117
00:06:35,820 --> 00:06:37,790
integral over a region.

118
00:06:37,790 --> 00:06:40,720
And then when we look at what
we ended up with, we realize

119
00:06:40,720 --> 00:06:45,080
that we could make this as small
as the most minimum if

120
00:06:45,080 --> 00:06:48,310
we let this be on the region
where it was negative

121
00:06:48,310 --> 00:06:49,360
everywhere.

122
00:06:49,360 --> 00:06:51,540
So we were looking for a region
where this quantity was

123
00:06:51,540 --> 00:06:55,250
everywhere negative on the
inside and 0 on the boundary.

124
00:06:55,250 --> 00:06:56,950
And that's exactly
what we did.

125
00:06:56,950 --> 00:07:00,500
And then we see that we get to a
point where the boundary has

126
00:07:00,500 --> 00:07:03,340
this equation, x squared plus
4y squared equals 4.

127
00:07:03,340 --> 00:07:06,310
We see then the boundary's an
ellipse, and C is indeed the

128
00:07:06,310 --> 00:07:07,660
boundary of the region.

129
00:07:07,660 --> 00:07:10,220
So we see that C is the ellipse

130
00:07:10,220 --> 00:07:12,420
described by this equation.

131
00:07:12,420 --> 00:07:14,190
So that's where I'll stop.

132
00:07:14,190 --> 00:07:14,434