1
00:00:00,980 --> 00:00:01,560
So, I think

2
00:00:03,590 --> 00:00:06,040
hopefully everybody knows the distance
formula, if

3
00:00:06,040 --> 00:00:08,900
not I made it just on one slide.

4
00:00:08,900 --> 00:00:11,630
So, distance equals rate times time.

5
00:00:11,630 --> 00:00:15,800
This just means if I'm say driving at a
certain speed say

6
00:00:15,800 --> 00:00:19,219
30 miles per hour for three hours, I'm
going to go 90 miles.

7
00:00:20,780 --> 00:00:25,160
it's actually amazing how often this is
the correct answer to

8
00:00:25,160 --> 00:00:28,570
a question, or the corre, correct way to
solve a question.

9
00:00:28,570 --> 00:00:33,520
if you can recognize what you're going to
think of as distance, what you're going

10
00:00:33,520 --> 00:00:35,950
to think of as rate, and what you're going
to think of as time.

11
00:00:37,660 --> 00:00:41,430
This form of the distance formula really
only works

12
00:00:41,430 --> 00:00:45,210
when the rate is constant across the
entire time.

13
00:00:45,210 --> 00:00:48,020
But now that we can integrate, we're
going to be

14
00:00:48,020 --> 00:00:50,030
able to allow that to change a little bit.

15
00:00:56,640 --> 00:01:00,250
So, it provides intuition for a lot of
problems, and

16
00:01:00,250 --> 00:01:04,940
all that we're really doing is imagining
rate on one

17
00:01:04,940 --> 00:01:07,700
axis, time on another axis, and then the
distance you're

18
00:01:07,700 --> 00:01:11,850
going to cover is going to be equal to
this area.

19
00:01:11,850 --> 00:01:17,260
So that's just the definition of the area
of a rectangle is height times width.

20
00:01:22,058 --> 00:01:26,170
So let's look at something I call the, the
job problem, which

21
00:01:26,170 --> 00:01:28,280
seems to give a lot of people quite a lot
of trouble.

22
00:01:30,025 --> 00:01:36,060
so, I take my motivation quite often from
the Simpsons.

23
00:01:36,060 --> 00:01:40,170
So, I said if Lenny can do a job in 6
hours and Carl can do

24
00:01:40,170 --> 00:01:44,620
the same job in 4 hours, how long does the
job take if they work together?

25
00:01:45,950 --> 00:01:47,150
a lot of times when you ask

26
00:01:47,150 --> 00:01:49,940
people this question, they will say
something the first

27
00:01:49,940 --> 00:01:52,640
number that pops into their head is 5
hours.

28
00:01:54,050 --> 00:01:57,980
which seems to make sense, because it's
just the average of 4 and 6.

29
00:01:57,980 --> 00:02:00,790
But that would actually mean that by
helping

30
00:02:00,790 --> 00:02:04,750
out Carl, Lenny is making the job take
longer.

31
00:02:04,750 --> 00:02:07,890
So, you know, clearly that's not the right
way to think about the problem.

32
00:02:09,690 --> 00:02:12,280
And the right way to think about this
problem is

33
00:02:12,280 --> 00:02:14,089
that it's the rates that are additive.

34
00:02:16,510 --> 00:02:22,100
So, Lenny, if he can do a job in 6 hours,
he can do 1/6th of the job per hour.

35
00:02:23,490 --> 00:02:27,830
And Carl, can do 1 quarter of the job per
hour.

36
00:02:27,830 --> 00:02:31,818
So if they're working together, their rate
is going to be 1/6th

37
00:02:31,818 --> 00:02:35,520
plus 1/4th, and that ends up being 5/12ths
of the job per hour.

38
00:02:37,640 --> 00:02:40,350
And so, here, the job is what's going to
end up

39
00:02:40,350 --> 00:02:44,740
being the distance and the time is what
we're asking for.

40
00:02:44,740 --> 00:02:47,660
How long does it take them to do the job
if they're working together?

41
00:02:49,370 --> 00:02:53,180
So, solving for time, I just get distance
divided by rate.

42
00:02:53,180 --> 00:02:57,240
And then I've worked out what these
numbers are, we have one job.

43
00:02:58,490 --> 00:03:03,010
The rate that they're working is 5 over 12
jobs per hour.

44
00:03:04,850 --> 00:03:09,484
I can multiply the top and bottom of that
by 12 over 5 and get 12/5th.

45
00:03:09,484 --> 00:03:12,950
So the job is going to take 2 hours and 24
minutes if they work together.

46
00:03:14,110 --> 00:03:18,020
But we've still made the assumption, that
Lenny

47
00:03:18,020 --> 00:03:19,840
and Carl are working at a constant rate.

48
00:03:24,450 --> 00:03:29,420
So, suppose instead, that Lenny works at a
rate L

49
00:03:29,420 --> 00:03:34,444
of x and Carl works at a rate C of x, then
my

50
00:03:34,444 --> 00:03:40,070
condition that I started out with, I could
rewrite

51
00:03:40,070 --> 00:03:44,730
this in terms of definite integrals.
So I know

52
00:03:44,730 --> 00:03:50,850
that if I let Lenny work for 6 hours, he's
going to finish one job.

53
00:03:50,850 --> 00:03:56,231
So that means that the integral of L of x
from 0

54
00:03:56,231 --> 00:04:01,823
to 6 has to be equal to 1, and then oops,
now I've

55
00:04:01,823 --> 00:04:06,740
confused L's and my C's.
So I think this should be the C

56
00:04:09,450 --> 00:04:15,600
and the integral from 0 to 4 of c of x
should also be equal to 1.

57
00:04:15,600 --> 00:04:20,860
And that's just the way of mathematically
writing down the condition, or the

58
00:04:20,860 --> 00:04:27,410
statement that Lenny can do one job in 6
hours and Carl can do one job in 4 hours.

59
00:04:31,760 --> 00:04:39,000
And so now my job question is, can I find
a capital T,

60
00:04:39,000 --> 00:04:46,220
so that's a, a time, so that when I add
these rates together I get 1.

61
00:04:46,220 --> 00:04:46,720
And

62
00:04:49,120 --> 00:04:51,380
so now, this is not quite as nice

63
00:04:51,380 --> 00:04:53,320
a problem, because there are actually a
whole

64
00:04:53,320 --> 00:04:55,350
bunch of different rates that I could add

65
00:04:55,350 --> 00:04:58,050
together that would, that would give me
this condition.

66
00:04:58,050 --> 00:05:01,510
So for instance I could just say well, I
guess I can't.

67
00:05:01,510 --> 00:05:04,110
So I have some constraints to, so I am

68
00:05:04,110 --> 00:05:07,170
bound by the constraint that Carl works at
this rate.

69
00:05:07,170 --> 00:05:10,630
So the integral of that one has to be over
0 to

70
00:05:10,630 --> 00:05:14,490
4 to be 1 and the integral of L of x over

71
00:05:14,490 --> 00:05:23,090
0 to 6 also has to be 1.

72
00:05:23,090 --> 00:05:27,970
So I just invented some functions to, to
try and show, demonstrate this.

73
00:05:27,970 --> 00:05:31,120
So I chose 12 minus x divided by 54.

74
00:05:31,120 --> 00:05:34,030
And all this is going to, all this is
going to do,

75
00:05:36,580 --> 00:05:41,350
is when we're at time zero.
So when x is zero, I'm going to have 12

76
00:05:41,350 --> 00:05:47,050
minus zero divided by 54.
And at the end of the job, so when x

77
00:05:47,050 --> 00:05:51,870
is equal to 6, I'm going to have 12 minus
6, which is going to be 6 divided by 54.

78
00:05:51,870 --> 00:05:56,030
And we can forget about the bottom,
because that's the same.

79
00:05:56,030 --> 00:05:58,470
So at the beginning, we have 12 over
something

80
00:05:58,470 --> 00:06:00,420
and at the end, we have six over
something.

81
00:06:00,420 --> 00:06:01,410
Six is half of 12,

82
00:06:01,410 --> 00:06:06,900
so all this is saying is that Lenny is
working twice as hard

83
00:06:06,900 --> 00:06:09,360
at the beginning of the time period as he
is at the end.

84
00:06:09,360 --> 00:06:11,840
So he gets tired while he's working, so by
the

85
00:06:11,840 --> 00:06:14,310
end of the 6 hours, he's going only half
as fast.

86
00:06:14,310 --> 00:06:19,590
And then I just have to put this 54 down
here so that the, the condition,

87
00:06:19,590 --> 00:06:24,460
the constraint that he finishes the job in
6 hours is met.

88
00:06:24,460 --> 00:06:26,440
And then I'm going to do

89
00:06:26,440 --> 00:06:28,920
exactly the same thing for Carl.

90
00:06:28,920 --> 00:06:32,030
So initially he's working at 8 over 24, at
the

91
00:06:32,030 --> 00:06:35,790
end of his 4 hours, he's working at 4 over
24.

92
00:06:35,790 --> 00:06:38,990
So he works faster, so he's also getting
tired faster.

93
00:06:38,990 --> 00:06:39,490
And

94
00:06:41,510 --> 00:06:46,280
now, how long is it going to take them if
they, if they work together?

95
00:06:46,280 --> 00:06:51,000
So, would it be more or less than the 2
hours and 24 minutes?

96
00:06:56,820 --> 00:07:02,540
So basically what I want to do is find T,
so that when I integrate this

97
00:07:02,540 --> 00:07:08,290
function, L of x plus C of x from zero to
T, I get my one job finished.

98
00:07:15,320 --> 00:07:19,660
And so I'm just going to start by plugging
in the two functions.

99
00:07:19,660 --> 00:07:25,880
So L of x was 12 minus x over 54 and C of
x was 8 minus x over 24.

100
00:07:25,880 --> 00:07:28,808
And then I don't like having these numbers
in the denominator.

101
00:07:28,808 --> 00:07:34,270
So I'm going to multiply both sides of
this by 216, that's going to be

102
00:07:34,270 --> 00:07:40,255
4 times 54.
So the 4 times 50 is 200, 4 times 4 is 16,

103
00:07:40,255 --> 00:07:42,640
216.
Or it'll be 9 times 24.

104
00:07:42,640 --> 00:07:48,030
You can do that in your head pretty easily
too, because 10 times 24 would be 240,

105
00:07:48,030 --> 00:07:53,460
and that's 24 more than 216 so that's got
to be 9 times 24.

106
00:07:53,460 --> 00:07:53,960
And so

107
00:07:56,860 --> 00:07:59,000
that's going to get everything out of the
denominator.

108
00:07:59,000 --> 00:08:01,700
So, mostly the reason I wanted to do that,
it makes

109
00:08:01,700 --> 00:08:03,988
the lines a little skinnier so it all fits
on one slide.

110
00:08:03,988 --> 00:08:11,390
But it also means that I can now add these
together, so my integral is going to

111
00:08:11,390 --> 00:08:14,853
simplify itself a bit before I actually
have to compute an anti-derivative.

112
00:08:17,300 --> 00:08:19,040
And so what I end up having to compute

113
00:08:19,040 --> 00:08:24,470
the anti-derivative of is this function
here, 120 minus 13x.

114
00:08:24,470 --> 00:08:27,380
So I can just use my anti-power rule to
get that.

115
00:08:29,990 --> 00:08:36,720
And that's going to give me 120x minus 13
over 2 x squared from 0 to T.

116
00:08:36,720 --> 00:08:41,880
And now, even though the variable T is on
this side, I, I treat it exactly the

117
00:08:41,880 --> 00:08:47,020
same way, so I'm just going to evaluate my
anti deriv, oops, my anti derivative here.

118
00:08:47,020 --> 00:08:50,180
I evaluate it at T and then subtract it,
evaluate

119
00:08:50,180 --> 00:08:55,309
it at 0, so I get 120T minus 13 over

120
00:08:55,309 --> 00:08:56,249
2 T squared.

121
00:08:57,710 --> 00:09:03,156
And then luckily on this side, since I've
got that zero, when I plug in zero to the

122
00:09:03,156 --> 00:09:09,320
to, to this anti-derivative, I just get 0,
so that makes things a little bit simpler.

123
00:09:12,390 --> 00:09:17,200
So I end up with 13 T squared.
So I, I multiplied both sides of this by,

124
00:09:18,640 --> 00:09:24,440
by 2 as well, 13T squared minus 240T plus
216, oops, I

125
00:09:27,050 --> 00:09:28,990
think I forgot to multiply that bit by 2.
And

126
00:09:31,240 --> 00:09:34,480
then you can solve that by using the
quadratic formula.

127
00:09:34,480 --> 00:09:36,620
And that gives 57 minutes.

128
00:09:36,620 --> 00:09:38,950
Although I think it gives 57 minutes for
this one that

129
00:09:38,950 --> 00:09:41,000
I forgot to multiply by 2, so I'll have to
check

130
00:09:43,880 --> 00:09:44,070
that.

131
00:09:44,070 --> 00:09:47,480
And then another application where this is
going to

132
00:09:47,480 --> 00:09:51,490
turn of for those of you that are taking
461.

133
00:09:51,490 --> 00:09:52,770
in the next couple of weeks, we're
going to

134
00:09:52,770 --> 00:09:55,438
learn about something called a probability
density function.

135
00:09:55,438 --> 00:10:02,330
So, I'm going to denote a probability
density function for a random variable x.

136
00:10:02,330 --> 00:10:10,042
It says f sub X, and its argument is going
be little x, and I'm

137
00:10:10,042 --> 00:10:18,510
going to use it to assign probabilities to
a random variable being in an interval.

138
00:10:18,510 --> 00:10:20,790
So I'm reading this thing on the left hand
side.

139
00:10:20,790 --> 00:10:24,352
It's the probability that my random
variable is in the interval

140
00:10:24,352 --> 00:10:28,200
[a, b].
It's just going to be the integral of my

141
00:10:28,200 --> 00:10:37,220
probability density function over the
interval [a,b].

142
00:10:37,220 --> 00:10:40,160
The support of a random variable is just
going to be

143
00:10:40,160 --> 00:10:44,338
the, the set of points, so generally it's
an interval.

144
00:10:44,338 --> 00:10:50,380
a set of points S, where the probability
density function is greater

145
00:10:50,380 --> 00:10:56,618
than or is greater than zero, when x is in
S.

146
00:10:56,618 --> 00:11:02,535
And so, some of the profi-, properties of
a probability density function

147
00:11:02,535 --> 00:11:08,161
are, if f sub X is a probability, so I'm
going to eval, abbreviate that

148
00:11:08,161 --> 00:11:13,787
as pdf, then I have to have f of x is
greater than or equal to 0 for all real

149
00:11:13,787 --> 00:11:19,351
numbers.

150
00:11:19,351 --> 00:11:22,051
And when I integrate this, so sometimes
instead of

151
00:11:22,051 --> 00:11:24,271
instead of writing an interval as just
from a

152
00:11:24,271 --> 00:11:27,571
to b, I might want to write a set that
might be made up of more than one

153
00:11:27,571 --> 00:11:31,453
interval density function.

154
00:11:31,453 --> 00:11:34,693
And so, in that case, when I've defined S
to be a set.

155
00:11:34,693 --> 00:11:39,440
So, here S is the set of points where this
thing strictly greater than 0.

156
00:11:39,440 --> 00:11:41,290
I can write the integral sign, and I'll
just

157
00:11:41,290 --> 00:11:43,550
put the name of the set at the bottom.

158
00:11:45,248 --> 00:11:47,239
And the integal has to be equal to 1.

159
00:11:49,550 --> 00:11:52,750
So the exercise that'll show up quite
often is, can you

160
00:11:54,180 --> 00:11:59,000
prove that a certain function is actually
a probability density function?

161
00:11:59,000 --> 00:12:01,620
And so I'll go through one example of that
pretty quick.

162
00:12:01,620 --> 00:12:09,565
So the PDF of a beta 2-2 distribution, is
defined to be 6x

163
00:12:09,565 --> 00:12:15,450
times 1 minus x, when x is in the interval
[0, 1], and its defined to be 0 otherwise.

164
00:12:20,670 --> 00:12:24,250
So if I want to verify that f of x is

165
00:12:24,250 --> 00:12:26,950
greater than or equal to zero, that should
be pretty straightforward.

166
00:12:26,950 --> 00:12:33,240
If I have a number between zero and one,
then it's greater than or equal to zero.

167
00:12:33,240 --> 00:12:35,680
So 6x has to be greater than or equal to
zero.

168
00:12:36,680 --> 00:12:40,750
And one minus x, so the smallest this
could be

169
00:12:40,750 --> 00:12:43,760
is when x is equal to 1 and that's zero.

170
00:12:43,760 --> 00:12:45,950
So, it's pretty easy to verify the first
condition.

171
00:12:47,920 --> 00:12:51,129
The, the work is usually verifying that
the

172
00:12:51,129 --> 00:12:54,530
PDF is going to integrate is going to
integrate

173
00:12:57,370 --> 00:12:58,680
to one over it's support.

174
00:13:00,823 --> 00:13:08,550
So here going to have the, the integral,
so S is going to be 0 to 1.

175
00:13:08,550 --> 00:13:11,310
So, it doesn't actually include the
endpoints, but

176
00:13:11,310 --> 00:13:13,690
those end up not mattering for the
integral.

177
00:13:14,898 --> 00:13:17,380
So, I'm going to say this is just an
integral from

178
00:13:17,380 --> 00:13:21,920
0 to 1 of 6 x times 1 minus x.

179
00:13:21,920 --> 00:13:23,860
Oops, again that typo.

180
00:13:24,946 --> 00:13:29,950
dx, so this x, shouldn't be here.
Hopefully that's going to go away.

181
00:13:31,190 --> 00:13:31,190
[COUGH]

182
00:13:31,190 --> 00:13:31,690
And

183
00:13:34,120 --> 00:13:35,890
so, using some of the properties I've

184
00:13:38,280 --> 00:13:39,849
listed in the first set of slides.

185
00:13:41,170 --> 00:13:44,690
I have, I can think of x times 1 minus x
as

186
00:13:44,690 --> 00:13:48,310
my f of x, and so I'm just multiplying
that by a constant.

187
00:13:48,310 --> 00:13:51,650
So I can take that constant outside of my
integral sign to make things a

188
00:13:51,650 --> 00:13:54,130
little bit simpler, and then I'm just

189
00:13:54,130 --> 00:13:56,070
going to just go ahead and do the
multiplication.

190
00:13:56,070 --> 00:13:59,910
So I have x times 1 gives me x.

191
00:13:59,910 --> 00:14:03,020
And then x times x gives me the x squared
here.

192
00:14:07,360 --> 00:14:11,300
And then I can calculate the
anti-derivative of that, just using my

193
00:14:11,300 --> 00:14:17,120
anti power rule on each term.
So the first one came from 1 half x, 1

194
00:14:17,120 --> 00:14:23,439
half x squared, and the second term came
from 1/3rd x cubed.

195
00:14:23,439 --> 00:14:27,360
And now that I've written down my
anti-derivative here,

196
00:14:27,360 --> 00:14:29,720
I've moved my limits to the right hand
side.

197
00:14:29,720 --> 00:14:33,020
So it means this notation again, just
means evaluate

198
00:14:33,020 --> 00:14:36,430
my anti-derivative upper end point and
subtract

199
00:14:36,430 --> 00:14:38,970
the anti-derivative evaluated at the lower
end point.

200
00:14:42,550 --> 00:14:46,532
So, again, because I have an x in each
term, the bottom

201
00:14:46,532 --> 00:14:49,300
endpoint is just going to be zero, that
simplifies things quite a bit.

202
00:14:50,320 --> 00:14:52,931
So, I end up with just 1 half minus 1/3rd.

203
00:14:52,931 --> 00:15:01,162
Oops, and that's equal, so 1 half is, I
can write that as 3 over 6.

204
00:15:01,162 --> 00:15:07,302
1/3rd, I could think of as 2 over 6, so 3
over 6 minus 2 over 6 is 1 over 6.

205
00:15:07,302 --> 00:15:07,558
And I

206
00:15:07,558 --> 00:15:11,387
have this 6 out in front and that cancels
each other out.

207
00:15:11,387 --> 00:15:14,782
So, I verified that this function here, fx

208
00:15:14,782 --> 00:15:20,000
of x is a valid probability density
function.