1
00:00:00,000 --> 00:00:09,400
CHRISTINE BREINER: Welcome
back to recitation.

2
00:00:09,400 --> 00:00:13,100
In this video, I'd like us to do
the following two problems,

3
00:00:13,100 --> 00:00:16,660
both related to the same
position vector.

4
00:00:16,660 --> 00:00:20,690
So we're starting off with a
position vector defined as r

5
00:00:20,690 --> 00:00:25,720
of t is equal to 1 minus 2t
squared i plus t squared j

6
00:00:25,720 --> 00:00:28,870
plus negative 2 plus
2t squared k.

7
00:00:28,870 --> 00:00:31,350
So that's our position vector,
and I'd like us to do the

8
00:00:31,350 --> 00:00:32,760
following two things.

9
00:00:32,760 --> 00:00:35,740
And you'll notice this problem
is pretty much just a

10
00:00:35,740 --> 00:00:36,830
computational problem.

11
00:00:36,830 --> 00:00:40,290
We're going to make sure that we
know what these things I'm

12
00:00:40,290 --> 00:00:43,140
about to talk about are, how you
define them, and how you

13
00:00:43,140 --> 00:00:45,830
get from the position vector
to each of these things.

14
00:00:45,830 --> 00:00:50,260
So we want to compute the
velocity, the speed, the

15
00:00:50,260 --> 00:00:52,590
acceleration, and find
the unit tangent

16
00:00:52,590 --> 00:00:55,250
vector for r of t.

17
00:00:55,250 --> 00:00:57,860
And then, the second part, we
want to compute the arc length

18
00:00:57,860 --> 00:01:02,330
of the trajectory from t
equals 0 to t equals 2.

19
00:01:02,330 --> 00:01:04,030
So I'll give you a moment
to do that problem.

20
00:01:04,030 --> 00:01:06,070
Why don't you pause the video,
work on the problem.

21
00:01:06,070 --> 00:01:08,430
When you're ready to check your
work, bring the video

22
00:01:08,430 --> 00:01:09,860
back up and I'll show
you how I do it.

23
00:01:09,860 --> 00:01:18,930

24
00:01:18,930 --> 00:01:20,120
OK, welcome back.

25
00:01:20,120 --> 00:01:22,700
Well, hopefully, you felt
comfortable with answering

26
00:01:22,700 --> 00:01:23,850
these questions.

27
00:01:23,850 --> 00:01:26,330
So now I will answer them
and you can compare your

28
00:01:26,330 --> 00:01:27,200
answers with mine.

29
00:01:27,200 --> 00:01:31,180
So let me start off
with part a.

30
00:01:31,180 --> 00:01:33,840
Part a, the first thing
we're going to

31
00:01:33,840 --> 00:01:35,850
do is find the velocity.

32
00:01:35,850 --> 00:01:41,050
So velocity is really-- all
we need to do is take the

33
00:01:41,050 --> 00:01:43,640
derivative of the position
vector with respect to t.

34
00:01:43,640 --> 00:01:46,260
So I'm just going to
take r prime of t.

35
00:01:46,260 --> 00:01:49,880
And now I'm going to write it in
the shorthand notation that

36
00:01:49,880 --> 00:01:54,950
you've seen with the brackets
to denote that it's not a

37
00:01:54,950 --> 00:01:56,750
point, but it's a vector.

38
00:01:56,750 --> 00:02:00,250
So this is what you've seen
to denote a vector

39
00:02:00,250 --> 00:02:01,380
rather than a point.

40
00:02:01,380 --> 00:02:04,160
So the derivative with respect
to t of the first component is

41
00:02:04,160 --> 00:02:07,100
just negative 4t.

42
00:02:07,100 --> 00:02:09,760
The derivative with respect to
t of the second component is

43
00:02:09,760 --> 00:02:13,520
just 2t, because we had t
squared, so when we take its

44
00:02:13,520 --> 00:02:15,450
derivative, we just get 2t.

45
00:02:15,450 --> 00:02:19,490
And the third component was
negative two plus 2t squared,

46
00:02:19,490 --> 00:02:23,020
so when I take its derivative,
I get a 4t, so that is

47
00:02:23,020 --> 00:02:25,180
actually v of t, OK?

48
00:02:25,180 --> 00:02:27,710
And then the next thing I asked
you to do is determine

49
00:02:27,710 --> 00:02:31,710
the speed, and the speed, of
course, is just the length of

50
00:02:31,710 --> 00:02:33,540
the velocity vector, right?

51
00:02:33,540 --> 00:02:39,540
So we just need to find the
length of v. Now, to do that,

52
00:02:39,540 --> 00:02:41,770
to remind ourselves what we do
for that, we actually take the

53
00:02:41,770 --> 00:02:44,110
inner product of v with itself,
the dot product of v

54
00:02:44,110 --> 00:02:49,900
with itself, and then we take
the square root of that.

55
00:02:49,900 --> 00:02:52,070
So let's look at what the
dot product will be.

56
00:02:52,070 --> 00:02:56,040
Let me find the squared thing
first, and then I will take

57
00:02:56,040 --> 00:02:57,830
the square root.

58
00:02:57,830 --> 00:03:01,270
So v dotted with v, the first
component I'm going to have

59
00:03:01,270 --> 00:03:03,840
negative 4t quantity
squared, so that's

60
00:03:03,840 --> 00:03:06,760
going to be 16t squared.

61
00:03:06,760 --> 00:03:09,540
And then the second component
is going to be 2t quantity

62
00:03:09,540 --> 00:03:13,830
squared, so I'm going to
have plus 4t squared.

63
00:03:13,830 --> 00:03:15,920
And the third component is going
to be another-- it's

64
00:03:15,920 --> 00:03:18,130
going to be 4t quantity
squared, so I

65
00:03:18,130 --> 00:03:21,720
get another 16t squared.

66
00:03:21,720 --> 00:03:27,980
So when I add those up, I
believe I get 36t squared?

67
00:03:27,980 --> 00:03:29,230
Yes, good.

68
00:03:29,230 --> 00:03:31,500

69
00:03:31,500 --> 00:03:33,920
And so then, I just have to take
the square root of both

70
00:03:33,920 --> 00:03:38,050
sides to get what the speed
actually is instead of the

71
00:03:38,050 --> 00:03:39,300
square of the speed.

72
00:03:39,300 --> 00:03:41,730

73
00:03:41,730 --> 00:03:46,090
So I get 6t, OK?

74
00:03:46,090 --> 00:03:50,080
So that's the velocity;
that's the speed.

75
00:03:50,080 --> 00:03:52,280
Now I need to find the
acceleration and I need to

76
00:03:52,280 --> 00:03:54,100
find the unit tangent vector.

77
00:03:54,100 --> 00:03:55,250
OK, so let me see.

78
00:03:55,250 --> 00:03:58,040
I will come over here.

79
00:03:58,040 --> 00:04:01,190
Let me step off here and I will
find the acceleration and

80
00:04:01,190 --> 00:04:02,200
the unit tangent vector.

81
00:04:02,200 --> 00:04:04,950
So the acceleration, if you
remember, the acceleration is

82
00:04:04,950 --> 00:04:07,440
actually just the derivative
of the velocity

83
00:04:07,440 --> 00:04:08,380
with respect to t.

84
00:04:08,380 --> 00:04:12,020
So the acceleration is going
to be the derivative of

85
00:04:12,020 --> 00:04:14,250
negative 4t is just
negative 4.

86
00:04:14,250 --> 00:04:18,470
The derivative of 2t is just 2,
and the derivative of 4t is

87
00:04:18,470 --> 00:04:20,660
just 4, all with respect to t.

88
00:04:20,660 --> 00:04:21,750
OK.

89
00:04:21,750 --> 00:04:26,420
So the acceleration vector is
equal to negative 4 comma 2

90
00:04:26,420 --> 00:04:28,620
comma 4, so you see this
actually has constant

91
00:04:28,620 --> 00:04:30,040
acceleration.

92
00:04:30,040 --> 00:04:32,040
So at any point, your
acceleration is always this

93
00:04:32,040 --> 00:04:35,220
value, so it's not surprising
that our velocity is

94
00:04:35,220 --> 00:04:38,280
increasing, and actually, it's
increasing-- you'll notice,

95
00:04:38,280 --> 00:04:40,230
each of these components
is constant.

96
00:04:40,230 --> 00:04:43,140
The velocity, each of the
components is linear, and if

97
00:04:43,140 --> 00:04:46,490
we went back, we look at the
position vector, each of those

98
00:04:46,490 --> 00:04:47,950
components is quadratic.

99
00:04:47,950 --> 00:04:51,200
And this is exactly what you
expect from just your

100
00:04:51,200 --> 00:04:54,810
understanding of derivatives in
single-variable calculus.

101
00:04:54,810 --> 00:04:57,120
If you start off with a constant
and you find an

102
00:04:57,120 --> 00:04:58,970
antiderivative, it's going to
be linear, and you find

103
00:04:58,970 --> 00:05:02,710
another antiderivative, you're
going to have a quadratic, so

104
00:05:02,710 --> 00:05:04,680
we shouldn't be surprised
by any of that.

105
00:05:04,680 --> 00:05:07,650
Now we have one more thing to do
with Part a, and that is to

106
00:05:07,650 --> 00:05:08,770
find the unit tangent vector.

107
00:05:08,770 --> 00:05:11,320
And that's fairly easy, because
all we have to do is,

108
00:05:11,320 --> 00:05:16,770
if you notice, we have the
velocity vector and we have

109
00:05:16,770 --> 00:05:17,940
its length.

110
00:05:17,940 --> 00:05:19,770
And so to find the unit tangent
vector, all we have to

111
00:05:19,770 --> 00:05:22,900
do is take the velocity and
divide it by its length, and

112
00:05:22,900 --> 00:05:24,110
that will normalize it.

113
00:05:24,110 --> 00:05:28,140
That means that its length will
be one at that point,

114
00:05:28,140 --> 00:05:30,370
because you're taking a vector,
dividing by its

115
00:05:30,370 --> 00:05:32,810
length, so the length of the
new vector will have to be

116
00:05:32,810 --> 00:05:33,760
length one.

117
00:05:33,760 --> 00:05:36,670
So let me write that down.

118
00:05:36,670 --> 00:05:39,235
And actually, I guess the point
to remember here is that

119
00:05:39,235 --> 00:05:44,180
the velocity vector is tangent
to the path you're carving

120
00:05:44,180 --> 00:05:45,430
out, to the trajectory.

121
00:05:45,430 --> 00:05:50,020

122
00:05:50,020 --> 00:05:52,440
OK, so this is a vector.

123
00:05:52,440 --> 00:05:54,220
This is a scalar.

124
00:05:54,220 --> 00:05:59,220
So I'm going to take 1 over 6t,
and I'm going to multiply

125
00:05:59,220 --> 00:06:08,310
it by negative 4t, 2t, 4t, and
this gives me, when I do my

126
00:06:08,310 --> 00:06:12,110
division, looks like it gives
me a negative 2/3,-- right--

127
00:06:12,110 --> 00:06:15,400

128
00:06:15,400 --> 00:06:18,490
1/3, 2/3.

129
00:06:18,490 --> 00:06:21,880

130
00:06:21,880 --> 00:06:28,370
So that is the unit tangent
vector, OK?

131
00:06:28,370 --> 00:06:31,500
OK, and now, we have one more
point we want to make, and

132
00:06:31,500 --> 00:06:35,490
that is having to do with the
arc length of the trajectory.

133
00:06:35,490 --> 00:06:37,400
That was the second part of this
problem, was to find the

134
00:06:37,400 --> 00:06:41,320
arc length of the trajectory
from t equals 0 to t equals 2.

135
00:06:41,320 --> 00:06:44,770
So let me just draw
another line here.

136
00:06:44,770 --> 00:06:47,735
And what we want to do there
then is-- really what we want

137
00:06:47,735 --> 00:06:56,610
to do is we want to integrate
the speed, right?

138
00:06:56,610 --> 00:06:59,040
We want to integrate the
speed from 0 to 2.

139
00:06:59,040 --> 00:07:00,580
So this--

140
00:07:00,580 --> 00:07:01,860
let me come over here--

141
00:07:01,860 --> 00:07:07,520
this absolute v, you might have
also seen it written as

142
00:07:07,520 --> 00:07:11,060
ds dt, right?

143
00:07:11,060 --> 00:07:15,060
And so we want to integrate
this in dt--

144
00:07:15,060 --> 00:07:18,220
in t-- sorry-- from 0 to 2.

145
00:07:18,220 --> 00:07:21,180
And so we come over here.

146
00:07:21,180 --> 00:07:27,200
We want to integrate
from 0 to 2, 6t dt.

147
00:07:27,200 --> 00:07:28,140
That's fairly easy.

148
00:07:28,140 --> 00:07:33,890
That's going to be 6t squared
over 2, evaluated from 0 to 2.

149
00:07:33,890 --> 00:07:37,660
And so when I write that down,
I'm going to get 6 times 4

150
00:07:37,660 --> 00:07:40,840
divided by 2, 24, divided by
2, I just get 12, and the

151
00:07:40,840 --> 00:07:42,020
other term is 0.

152
00:07:42,020 --> 00:07:44,950
So the arc length is--

153
00:07:44,950 --> 00:07:50,310
of the trajectory from 0
to 2 is just 12 units.

154
00:07:50,310 --> 00:07:52,760
So this really was a purely
computational type of problem.

155
00:07:52,760 --> 00:07:55,510
All we were doing, if you come
back over here and you recall

156
00:07:55,510 --> 00:07:58,520
what we were trying to
do, is we started off

157
00:07:58,520 --> 00:08:00,050
with a position vector.

158
00:08:00,050 --> 00:08:02,310
We just did a lot
of computation.

159
00:08:02,310 --> 00:08:04,730
We found the velocity, the
speed, the acceleration, the

160
00:08:04,730 --> 00:08:07,940
unit tangent vector, and then we
just wanted to find the arc

161
00:08:07,940 --> 00:08:08,840
length of the trajectory.

162
00:08:08,840 --> 00:08:11,540
So this is all very
computational, but just to

163
00:08:11,540 --> 00:08:14,730
make sure we understood what
all the terms meant and how

164
00:08:14,730 --> 00:08:17,190
they were related
to one another.

165
00:08:17,190 --> 00:08:18,950
So I'll stop there.

166
00:08:18,950 --> 00:08:19,499