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PROFESSOR: Hi.

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00:00:58,860 --> 00:01:04,610
Today we're going to
conclude our study of vectors

10
00:01:04,610 --> 00:01:07,650
as applied to
motion in the plane.

11
00:01:07,650 --> 00:01:09,880
Now recall that for
the last two units,

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00:01:09,880 --> 00:01:12,120
we were discussing
polar coordinates.

13
00:01:12,120 --> 00:01:14,360
So today what we
would like to do

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00:01:14,360 --> 00:01:20,400
is investigate what velocity
and acceleration vectors would

15
00:01:20,400 --> 00:01:25,360
have looked like, had we elected
to pick representative vectors

16
00:01:25,360 --> 00:01:27,750
in terms of polar coordinates.

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00:01:27,750 --> 00:01:30,420
Now what do we mean by
representative vectors?

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00:01:30,420 --> 00:01:33,910
We mean, of course,
analogs of i and j,

19
00:01:33,910 --> 00:01:37,920
just like T and N in
tangential-normal components

20
00:01:37,920 --> 00:01:41,400
were parallels to i and j.

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00:01:41,400 --> 00:01:44,140
Let's take a look and
see what that means here.

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00:01:44,140 --> 00:01:47,750
We call today's lecture
"Vectors in Polar Coordinates."

23
00:01:47,750 --> 00:01:50,250
The idea is that
we're given a curve

24
00:01:50,250 --> 00:01:53,430
C, which we have for
some reason or other

25
00:01:53,430 --> 00:01:57,330
elected to express in
terms of polar coordinates.

26
00:01:57,330 --> 00:02:02,440
The polar equation of the
curve C is r equals f of theta.

27
00:02:02,440 --> 00:02:05,250
A typical point on
the curve C would

28
00:02:05,250 --> 00:02:10,360
be called r comma theta say,
where theta was the angle made

29
00:02:10,360 --> 00:02:13,680
by the horizontal and
the radius vector,

30
00:02:13,680 --> 00:02:17,830
and r was the distance from
the origin to the point.

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00:02:17,830 --> 00:02:20,120
Now what we're saying
is, that in terms

32
00:02:20,120 --> 00:02:23,630
of polar coordinates,
a very natural vector

33
00:02:23,630 --> 00:02:25,910
to pick-- especially
if we think later

34
00:02:25,910 --> 00:02:27,570
in terms of the
simple force fields

35
00:02:27,570 --> 00:02:30,160
that we've talked
about earlier, the idea

36
00:02:30,160 --> 00:02:34,370
that there may be a force
that has its line of action

37
00:02:34,370 --> 00:02:36,810
from the origin to the
point on the curve--

38
00:02:36,810 --> 00:02:40,060
a very natural vector
to choose is the vector

39
00:02:40,060 --> 00:02:42,920
that we elect to call u sub r.

40
00:02:42,920 --> 00:02:45,900
And what that vector
is, apparently,

41
00:02:45,900 --> 00:02:51,480
is the vector 1 unit
long having the direction

42
00:02:51,480 --> 00:02:56,640
of the radius vector R. So
here is u sub r over here.

43
00:02:56,640 --> 00:03:00,070
If we think of u sub r
as playing the role of i,

44
00:03:00,070 --> 00:03:02,550
then the vector which
plays the role of j

45
00:03:02,550 --> 00:03:07,700
should be a positive
90-degree rotation of u sub r.

46
00:03:07,700 --> 00:03:12,330
And we elect to call that vector
u sub theta, using the theta

47
00:03:12,330 --> 00:03:14,360
here more to indicate
the fact that we're

48
00:03:14,360 --> 00:03:17,550
using polar coordinates
than to indicate anything

49
00:03:17,550 --> 00:03:19,550
about the angle theta itself.

50
00:03:19,550 --> 00:03:22,600
In other words, notice that
u sub theta, by definition,

51
00:03:22,600 --> 00:03:26,350
is just a positive 90-degree
rotation of u sub r, where

52
00:03:26,350 --> 00:03:31,834
u sub r is a unit vector in the
direction of the radius vector.

53
00:03:31,834 --> 00:03:33,500
Now if we want to see
this in terms of i

54
00:03:33,500 --> 00:03:35,630
and j components,
what we're saying

55
00:03:35,630 --> 00:03:41,210
is that u sub r is a unit vector
whose i component is what?

56
00:03:41,210 --> 00:03:43,390
Since this angle
here is also theta,

57
00:03:43,390 --> 00:03:47,470
its i component is cosine
theta, and its j component

58
00:03:47,470 --> 00:03:48,700
is sine theta

59
00:03:48,700 --> 00:03:52,550
So u sub r is cosine
theta i plus sine theta j.

60
00:03:52,550 --> 00:03:55,980
u sub theta-- and I'm going
to do a little twist here

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00:03:55,980 --> 00:03:58,470
that I didn't do in
the T and N components,

62
00:03:58,470 --> 00:04:00,379
just to show you
another approach.

63
00:04:00,379 --> 00:04:02,420
Rather than to start with
derivatives or anything

64
00:04:02,420 --> 00:04:05,750
like this, notice that what
we know about u sub theta

65
00:04:05,750 --> 00:04:08,260
is that it's
obtained from u sub r

66
00:04:08,260 --> 00:04:11,530
by a positive 90-degree
rotation of theta.

67
00:04:11,530 --> 00:04:16,910
So that means if I replace theta
in the expression for u sub r,

68
00:04:16,910 --> 00:04:21,200
by theta plus 90 degrees, that
should give me u sub theta.

69
00:04:21,200 --> 00:04:24,700
If I now remember my
trigonometric identities,

70
00:04:24,700 --> 00:04:28,210
that tells me that u sub
theta is minus sine theta

71
00:04:28,210 --> 00:04:31,220
i plus cosine theta j.

72
00:04:31,220 --> 00:04:34,580
By the way, if we now
look at this expression

73
00:04:34,580 --> 00:04:37,600
and compare it with the
expression for u sub r,

74
00:04:37,600 --> 00:04:42,750
we see at once that u sub theta
is the derivative of u sub

75
00:04:42,750 --> 00:04:44,710
r with respect to theta.

76
00:04:44,710 --> 00:04:47,970
And notice, as we said
before, that part of this

77
00:04:47,970 --> 00:04:50,550
should have been
known to us by now.

78
00:04:50,550 --> 00:04:53,930
Namely, since u sub
r varies with theta

79
00:04:53,930 --> 00:04:56,490
but it has a constant
magnitude, we

80
00:04:56,490 --> 00:05:00,030
know that the derivative of
u sub r with respect to theta

81
00:05:00,030 --> 00:05:03,980
has to be perpendicular
to u sub r.

82
00:05:03,980 --> 00:05:06,400
In other words, we knew
that u sub theta had

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00:05:06,400 --> 00:05:09,275
to be either plus or minus
the derivative of u sub

84
00:05:09,275 --> 00:05:11,040
r with respect to
theta, but now we

85
00:05:11,040 --> 00:05:13,230
have a direct way
of showing this.

86
00:05:13,230 --> 00:05:16,090
And by the way, going
one step further,

87
00:05:16,090 --> 00:05:20,350
if we now differentiate u sub
theta with respect to theta,

88
00:05:20,350 --> 00:05:21,360
we get what?

89
00:05:21,360 --> 00:05:25,330
Minus cosine theta i
minus sine theta j,

90
00:05:25,330 --> 00:05:28,450
which is just minus u sub r.

91
00:05:28,450 --> 00:05:30,430
In other words, if
you differentiate

92
00:05:30,430 --> 00:05:33,010
u sub r with respect
to theta once,

93
00:05:33,010 --> 00:05:36,790
you get u sub theta,
just as we should.

94
00:05:36,790 --> 00:05:41,300
If you differentiate a second
time, you get minus u sub r.

95
00:05:41,300 --> 00:05:44,750
And therefore, it appears
that the operation

96
00:05:44,750 --> 00:05:48,160
of differentiating with
respect to theta rotates

97
00:05:48,160 --> 00:05:51,300
u sub r by 90 degrees.

98
00:05:51,300 --> 00:05:52,970
By the way, I should
mention that I

99
00:05:52,970 --> 00:05:54,880
could have made all
of these remarks

100
00:05:54,880 --> 00:05:58,537
when we were studying tangential
and normal components.

101
00:05:58,537 --> 00:06:00,370
In other words-- I just
wrote this out here,

102
00:06:00,370 --> 00:06:04,230
but this was true with T
and N. But the point was,

103
00:06:04,230 --> 00:06:07,200
we never had to
differentiate N with respect

104
00:06:07,200 --> 00:06:10,410
to t to find the
acceleration vector a.

105
00:06:10,410 --> 00:06:13,780
In other words, recall
that the key step in using

106
00:06:13,780 --> 00:06:16,760
tangential and normal
vectors-- and I'll mention this

107
00:06:16,760 --> 00:06:18,370
in a little bit
more detail later--

108
00:06:18,370 --> 00:06:21,430
was that the velocity
vector was simply

109
00:06:21,430 --> 00:06:26,790
ds/dt times the unit tangent
vector T. The coefficient of N

110
00:06:26,790 --> 00:06:28,670
was 0.

111
00:06:28,670 --> 00:06:32,780
In polar coordinates,
notice that v in general

112
00:06:32,780 --> 00:06:37,390
will have both a u sub r
and a u sub theta component.

113
00:06:37,390 --> 00:06:41,260
Therefore, to compute a, I have
to differentiate v with respect

114
00:06:41,260 --> 00:06:42,210
to t.

115
00:06:42,210 --> 00:06:44,030
That means, among
other things, I'm

116
00:06:44,030 --> 00:06:46,350
going to have to take
the derivative of u sub

117
00:06:46,350 --> 00:06:48,090
theta with respect to t.

118
00:06:48,090 --> 00:06:50,330
By the chain rule,
that's going to be

119
00:06:50,330 --> 00:06:52,150
the same as taking the
derivative of u sub

120
00:06:52,150 --> 00:06:55,820
theta with respect to
theta times d theta / dt.

121
00:06:55,820 --> 00:06:57,880
But the important point
is, is that someplace

122
00:06:57,880 --> 00:07:00,040
along the line in
studying kinematics

123
00:07:00,040 --> 00:07:02,150
and polar coordinates,
I am going

124
00:07:02,150 --> 00:07:06,070
to have to differentiate u sub
theta with respect to theta.

125
00:07:06,070 --> 00:07:07,950
And just to show you
again very, very quickly

126
00:07:07,950 --> 00:07:10,090
what I mean by
this, all I'm saying

127
00:07:10,090 --> 00:07:14,230
is we already know in kinematics
that the velocity vector is

128
00:07:14,230 --> 00:07:16,430
always tangential to the curve.

129
00:07:16,430 --> 00:07:18,040
Notice in this
particular diagram,

130
00:07:18,040 --> 00:07:22,090
for example, that if you look
at u sub r and u sub theta,

131
00:07:22,090 --> 00:07:24,760
if you think of a vector
whose direction is

132
00:07:24,760 --> 00:07:27,570
tangent to the curve at
this particular point,

133
00:07:27,570 --> 00:07:29,580
that vector will
have, in general,

134
00:07:29,580 --> 00:07:33,200
both a u sub theta and
a u sub r component.

135
00:07:33,200 --> 00:07:35,380
In fact, in this
particular diagram,

136
00:07:35,380 --> 00:07:39,595
I shouldn't say "in general,"
it will have u sub r and a u sub

137
00:07:39,595 --> 00:07:40,960
theta component.

138
00:07:40,960 --> 00:07:41,460
OK.

139
00:07:41,460 --> 00:07:47,210
So far so good, but now I want
to make one little caution,

140
00:07:47,210 --> 00:07:50,540
a caution which is not at all
self-evident, at least to me,

141
00:07:50,540 --> 00:07:52,850
and which gave me
great difficulty myself

142
00:07:52,850 --> 00:07:54,560
when I was a student.

143
00:07:54,560 --> 00:07:59,130
And that is, my feeling was
that u sub r was simply the unit

144
00:07:59,130 --> 00:08:00,785
vector in the direction of r.

145
00:08:00,785 --> 00:08:02,630
In fact, I said that
earlier, that u sub

146
00:08:02,630 --> 00:08:06,410
r was the unit vector in the
direction of the radius vector

147
00:08:06,410 --> 00:08:09,990
R. And this is one of the
reasons why even though

148
00:08:09,990 --> 00:08:12,240
Professor Thomas in the
textbook doesn't make such

149
00:08:12,240 --> 00:08:18,120
an issue over this, why I am
such a bug on using the phrase

150
00:08:18,120 --> 00:08:20,610
"sense" as well as direction.

151
00:08:20,610 --> 00:08:25,210
And that is, my claim
is that the unit vector

152
00:08:25,210 --> 00:08:30,140
u sub r need not be the
radius vector R divided

153
00:08:30,140 --> 00:08:33,980
by the magnitude of R. It'll
have the same direction,

154
00:08:33,980 --> 00:08:36,220
but watch what
happens with sense.

155
00:08:36,220 --> 00:08:38,409
Instead of talking about
this thing abstractly,

156
00:08:38,409 --> 00:08:40,460
let me give you a
concrete example.

157
00:08:40,460 --> 00:08:43,309
Let's take the curve
which in polar coordinates

158
00:08:43,309 --> 00:08:46,400
has the equation r
equals cosine theta.

159
00:08:46,400 --> 00:08:47,540
OK?

160
00:08:47,540 --> 00:08:50,490
As you recall, this would be
this particular circle here.

161
00:08:50,490 --> 00:08:53,490
Now let's take theta
to be 120 degrees.

162
00:08:53,490 --> 00:08:57,350
If I use our definition for u
sub r, which is cosine theta

163
00:08:57,350 --> 00:09:00,700
i plus sine theta
j, and replace theta

164
00:09:00,700 --> 00:09:04,650
by 120 degrees, what I
get is, is that u sub r

165
00:09:04,650 --> 00:09:09,180
is cos 120 degrees i
plus sine 120 degrees j.

166
00:09:09,180 --> 00:09:12,080
Remember that the cosine
of 120 is minus 1/2

167
00:09:12,080 --> 00:09:16,610
and the sine of 120 is plus
1/2 the square root of 3.

168
00:09:16,610 --> 00:09:21,800
u sub r turns out to
be minus 1/2 i plus 1/2

169
00:09:21,800 --> 00:09:23,890
the square root of 3 j.

170
00:09:23,890 --> 00:09:28,490
On the other hand, my claim is
that when theta is 120 degrees,

171
00:09:28,490 --> 00:09:30,900
what point are we at the curve?

172
00:09:30,900 --> 00:09:32,840
You see, if I take
theta to be 120

173
00:09:32,840 --> 00:09:39,190
degrees-- notice that
when theta is 120 degrees,

174
00:09:39,190 --> 00:09:40,790
r is negative 1/2.

175
00:09:43,550 --> 00:09:47,330
And that therefore I'm
at the point P_0 here.

176
00:09:47,330 --> 00:09:51,440
Recall that by definition,
R, the radius vector,

177
00:09:51,440 --> 00:09:53,980
is measured from the
origin to the point.

178
00:09:53,980 --> 00:09:57,650
In other words, according
to our previous definition,

179
00:09:57,650 --> 00:10:00,090
it's this vector,
which would be called

180
00:10:00,090 --> 00:10:03,660
R. But our definition
says that it's

181
00:10:03,660 --> 00:10:07,320
this vector, which is u sub r.

182
00:10:07,320 --> 00:10:09,890
And in fact, if you
just check the figures

183
00:10:09,890 --> 00:10:11,940
that we've obtained
over here, notice

184
00:10:11,940 --> 00:10:16,230
that u sub r has its i
component equal to minus 1/2,

185
00:10:16,230 --> 00:10:20,710
its j component being plus
1/2 the square root of 3.

186
00:10:20,710 --> 00:10:22,980
Therefore x is negative.

187
00:10:22,980 --> 00:10:24,480
y is positive.

188
00:10:24,480 --> 00:10:26,960
But if x is negative
and y is positive,

189
00:10:26,960 --> 00:10:30,790
you're in the second quadrant,
not the fourth quadrant.

190
00:10:30,790 --> 00:10:31,290
You see?

191
00:10:31,290 --> 00:10:34,970
In other words, u sub r is
almost the radius vector.

192
00:10:34,970 --> 00:10:39,380
In fact, it would have been,
if in polar coordinates,

193
00:10:39,380 --> 00:10:41,360
little r happened
to be positive.

194
00:10:41,360 --> 00:10:44,760
In fact, let me summarize
that in a different way.

195
00:10:44,760 --> 00:10:47,980
Let's assume that we have a
curve C whose polar equation

196
00:10:47,980 --> 00:10:49,970
is r equals f of theta.

197
00:10:49,970 --> 00:10:51,540
Then the idea is this.

198
00:10:51,540 --> 00:10:55,280
If f of theta happens to
be at least as big as 0,

199
00:10:55,280 --> 00:10:59,090
then u sub r is equal
to the radius vector

200
00:10:59,090 --> 00:11:01,400
R divided by its magnitude.

201
00:11:01,400 --> 00:11:04,740
In other words, u sub r will
be in the same direction

202
00:11:04,740 --> 00:11:06,800
as the radius vector.

203
00:11:06,800 --> 00:11:09,860
And by the way, recall this is
just another way of saying r.

204
00:11:09,860 --> 00:11:12,260
What we're saying is
again, if we had never

205
00:11:12,260 --> 00:11:15,210
let little r in polar
coordinates be negative,

206
00:11:15,210 --> 00:11:17,400
no problem would have occurred.

207
00:11:17,400 --> 00:11:19,180
But we do let little
r be negative.

208
00:11:19,180 --> 00:11:21,740
So we have to be a mite careful.

209
00:11:21,740 --> 00:11:23,770
The careful part comes in where?

210
00:11:23,770 --> 00:11:25,770
If r happens to be negative.

211
00:11:25,770 --> 00:11:29,800
In which case, u sub r
has the opposite sense

212
00:11:29,800 --> 00:11:32,320
of the radius vector
capital R, which

213
00:11:32,320 --> 00:11:34,050
we saw in the previous example.

214
00:11:34,050 --> 00:11:36,260
In other words, in
this case, u sub r

215
00:11:36,260 --> 00:11:40,100
is the negative of
the radius vector

216
00:11:40,100 --> 00:11:42,220
divided by its magnitude.

217
00:11:42,220 --> 00:11:43,400
In what case is that?

218
00:11:43,400 --> 00:11:45,550
If r happens to be negative.

219
00:11:45,550 --> 00:11:50,150
The important point to notice,
however, that in either case,

220
00:11:50,150 --> 00:11:57,220
the radius vector R is equal to
the polar coordinate r times u

221
00:11:57,220 --> 00:11:58,000
sub r.

222
00:11:58,000 --> 00:12:00,590
In other words, if r
happens to be positive,

223
00:12:00,590 --> 00:12:02,810
these two vectors
have the same sense.

224
00:12:02,810 --> 00:12:06,030
If r happens to be
negative, these two vectors

225
00:12:06,030 --> 00:12:07,600
have the opposite sense.

226
00:12:07,600 --> 00:12:10,120
In other words, in either
case, this expression

227
00:12:10,120 --> 00:12:11,350
is always correct.

228
00:12:11,350 --> 00:12:13,320
But the important
thing to notice

229
00:12:13,320 --> 00:12:18,180
is that the sense of u sub r
is determined by theta, not

230
00:12:18,180 --> 00:12:21,150
by r, not by f of theta.

231
00:12:21,150 --> 00:12:22,530
OK?

232
00:12:22,530 --> 00:12:26,120
At any rate, once we have this
particular recipe established,

233
00:12:26,120 --> 00:12:30,230
we can now go ahead and
study motion in the plane.

234
00:12:30,230 --> 00:12:32,910
Namely, notice that
our radius vector

235
00:12:32,910 --> 00:12:38,020
R is now given by the polar
coordinate r times u sub r.

236
00:12:38,020 --> 00:12:40,200
Or I guess I should
say here that I'm

237
00:12:40,200 --> 00:12:42,720
assuming that the
equation of motion

238
00:12:42,720 --> 00:12:46,160
is given by r is some
function of theta.

239
00:12:46,160 --> 00:12:48,960
That's the r that
I'm using in here.

240
00:12:48,960 --> 00:12:51,760
At any rate, what is
the velocity vector?

241
00:12:51,760 --> 00:12:55,470
By definition, it's just the
derivative of the radius vector

242
00:12:55,470 --> 00:12:57,120
with respect to time.

243
00:12:57,120 --> 00:13:01,060
That's just d/dt
of r times u sub r.

244
00:13:01,060 --> 00:13:04,210
Now keep in mind,
that r and u sub r

245
00:13:04,210 --> 00:13:06,470
are both functions
of time-- namely,

246
00:13:06,470 --> 00:13:09,240
the distance of the
particle from the origin

247
00:13:09,240 --> 00:13:11,760
as well as the direction
of the line of action

248
00:13:11,760 --> 00:13:14,450
that joins the
particle to the origin

249
00:13:14,450 --> 00:13:16,550
will, in general,
depend on time.

250
00:13:16,550 --> 00:13:19,280
Consequently, I must use
the product rule here.

251
00:13:19,280 --> 00:13:22,220
I already know that I can use
the product rule for vector

252
00:13:22,220 --> 00:13:25,860
and/or scalar functions and
any combination thereof.

253
00:13:25,860 --> 00:13:29,170
So I just differentiate this
thing with respect to time.

254
00:13:29,170 --> 00:13:30,060
I get what?

255
00:13:30,060 --> 00:13:32,730
This is dr/dt-- in other
words, the derivative

256
00:13:32,730 --> 00:13:36,680
of the first times the
second, which is u sub r, OK?

257
00:13:36,680 --> 00:13:39,680
Plus the first
times the derivative

258
00:13:39,680 --> 00:13:41,980
of the second, which
is the derivative of u

259
00:13:41,980 --> 00:13:44,540
sub r with respect to time.

260
00:13:44,540 --> 00:13:46,570
Now keep in mind,
again, there is

261
00:13:46,570 --> 00:13:48,550
nothing wrong with this recipe.

262
00:13:48,550 --> 00:13:52,280
But what I would like is to have
the velocity expressed in u sub

263
00:13:52,280 --> 00:13:54,490
r and u sub theta components.

264
00:13:54,490 --> 00:13:58,380
So far, I have it expressed
in terms of u sub r component

265
00:13:58,380 --> 00:14:01,280
and a d(u sub r)/dt component.

266
00:14:01,280 --> 00:14:04,970
But now, here again is why the
chain rule is so important.

267
00:14:04,970 --> 00:14:10,380
Keep in mind that I already know
that if this expression here

268
00:14:10,380 --> 00:14:13,840
had been the derivative
of u sub r with respect

269
00:14:13,840 --> 00:14:18,120
to theta instead of
with respect to t,

270
00:14:18,120 --> 00:14:21,120
what would this
expression have been?

271
00:14:21,120 --> 00:14:22,990
It would have been u sub theta.

272
00:14:22,990 --> 00:14:26,060
We solved that earlier
in the lecture.

273
00:14:26,060 --> 00:14:27,640
Well, here's what we do.

274
00:14:27,640 --> 00:14:29,380
We say, OK, these
aren't the same.

275
00:14:29,380 --> 00:14:32,040
But let's cross this out.

276
00:14:32,040 --> 00:14:33,594
Let's use the chain rule.

277
00:14:33,594 --> 00:14:35,135
And by the chain
rule, the derivative

278
00:14:35,135 --> 00:14:38,970
of u sub r with respect to t
is the same as the derivative

279
00:14:38,970 --> 00:14:44,310
of u sub r with respect to
theta-- times d theta / dt.

280
00:14:44,310 --> 00:14:47,640
And rewriting this so
that it becomes legible,

281
00:14:47,640 --> 00:14:53,560
we have that the velocity vector
is dr/dt times u sub r plus r d

282
00:14:53,560 --> 00:14:56,690
theta / dt times u sub theta.

283
00:14:56,690 --> 00:15:02,610
And again notice that as
far as u sub r and u sub

284
00:15:02,610 --> 00:15:06,080
theta are concerned, even if
I did not notice my subtlety--

285
00:15:06,080 --> 00:15:09,120
and by the way, I'm going to
leave this for the exercises--

286
00:15:09,120 --> 00:15:13,560
but even if I didn't notice the
subtlety that u sub r need not

287
00:15:13,560 --> 00:15:16,960
have the same sense as the
radius vector r-- notice

288
00:15:16,960 --> 00:15:20,060
that if I did not try to
draw this thing to scale,

289
00:15:20,060 --> 00:15:23,770
I can still get the-- I
shouldn't have said the scale--

290
00:15:23,770 --> 00:15:27,500
but if I didn't try to
graph the answer here given

291
00:15:27,500 --> 00:15:31,090
r as a function of theta and
theta is a function of t,

292
00:15:31,090 --> 00:15:34,400
notice that dr/dt
and r d theta / dt

293
00:15:34,400 --> 00:15:38,570
are well-defined arithmetically
with no possible chance

294
00:15:38,570 --> 00:15:40,440
of making a geometrical mistake.

295
00:15:40,440 --> 00:15:42,680
The place that you can
make the biggest mistake

296
00:15:42,680 --> 00:15:45,210
is if you automatically
think that u sub r must

297
00:15:45,210 --> 00:15:48,410
have the same sense as
capital R. But as I say,

298
00:15:48,410 --> 00:15:50,380
we'll leave any
additional discussion

299
00:15:50,380 --> 00:15:52,720
of that for the exercises.

300
00:15:52,720 --> 00:15:55,100
I should also point
out that when I first

301
00:15:55,100 --> 00:15:57,570
learned this recipe
myself, it turned out

302
00:15:57,570 --> 00:16:00,580
that we were ahead of--
the physics class was ahead

303
00:16:00,580 --> 00:16:01,780
of the math class.

304
00:16:01,780 --> 00:16:03,890
And we learned this thing
in the physics class

305
00:16:03,890 --> 00:16:05,460
almost intuitively.

306
00:16:05,460 --> 00:16:07,900
In other words, as
a geometric aside,

307
00:16:07,900 --> 00:16:10,550
notice that if I'm
given the curve

308
00:16:10,550 --> 00:16:13,410
and say s indicates the
direction of increasing arc

309
00:16:13,410 --> 00:16:20,800
length here, what I could do is
think of a little differential

310
00:16:20,800 --> 00:16:21,850
region here.

311
00:16:21,850 --> 00:16:24,480
Namely, here's my
radius vector r,

312
00:16:24,480 --> 00:16:28,550
and here's my velocity vector
in the direction of u sub r.

313
00:16:28,550 --> 00:16:31,080
Then I take a little
increment of angle d theta,

314
00:16:31,080 --> 00:16:35,210
and I now think of v sub
theta, which is at right angles

315
00:16:35,210 --> 00:16:40,560
to v sub r, as being tangent
to the circle I would have

316
00:16:40,560 --> 00:16:44,290
obtained if I had imagined
that this particular point--

317
00:16:44,290 --> 00:16:48,140
the particle was being viewed
with respect to the circle

318
00:16:48,140 --> 00:16:49,800
rather than to the curve itself.

319
00:16:49,800 --> 00:16:52,260
To make a long story
short, what I'm driving at

320
00:16:52,260 --> 00:16:54,810
is that physically,
it's very easy

321
00:16:54,810 --> 00:16:58,305
to justify that the
magnitude of the u sub r

322
00:16:58,305 --> 00:17:02,220
component of the velocity
is the magnitude of dr/dt--

323
00:17:02,220 --> 00:17:05,440
how fast the radius vector
is changing instantaneously.

324
00:17:05,440 --> 00:17:09,510
On the other hand, notice that
for the u sub theta component,

325
00:17:09,510 --> 00:17:14,460
this arc length is given in
differential form by r d theta.

326
00:17:14,460 --> 00:17:17,940
If I divide the arc length
by the time, which is dt,

327
00:17:17,940 --> 00:17:23,380
I get r d theta divided by dt,
which leads to r d theta / dt,

328
00:17:23,380 --> 00:17:27,470
which is the same expression
that we got analytically.

329
00:17:27,470 --> 00:17:29,620
But the point that I
want to bring out here

330
00:17:29,620 --> 00:17:33,720
is that our derivation required
no geometrical physical

331
00:17:33,720 --> 00:17:34,640
insight.

332
00:17:34,640 --> 00:17:36,560
And the reason that I
want to bring this out

333
00:17:36,560 --> 00:17:39,950
is I followed this argument
fine in my elementary physics

334
00:17:39,950 --> 00:17:40,970
course.

335
00:17:40,970 --> 00:17:43,930
The place I got hung up is
that the instructor then went

336
00:17:43,930 --> 00:17:47,290
into a fantastic hand-waving
type of demonstration

337
00:17:47,290 --> 00:17:49,530
and showed us how the
acceleration looked

338
00:17:49,530 --> 00:17:52,620
in terms of u sub r and
u sub theta components.

339
00:17:52,620 --> 00:17:54,720
And actually, that
was a blessing for me,

340
00:17:54,720 --> 00:17:56,730
because it was that
day that I decided

341
00:17:56,730 --> 00:17:58,960
to become a math major
rather than a physics

342
00:17:58,960 --> 00:18:02,089
major, which was a blessing both
for me and society, I guess.

343
00:18:02,089 --> 00:18:03,630
But the thing that
I want to show you

344
00:18:03,630 --> 00:18:06,810
is that the beauty of
our mathematical approach

345
00:18:06,810 --> 00:18:10,460
is that we can now obtain
a, the acceleration vector,

346
00:18:10,460 --> 00:18:13,370
from the velocity vector
without having to know

347
00:18:13,370 --> 00:18:14,950
any great physical insight.

348
00:18:14,950 --> 00:18:18,790
In fact, we have to know no
physical insight to do this.

349
00:18:18,790 --> 00:18:23,520
Namely, by definition, a is
the derivative of the velocity

350
00:18:23,520 --> 00:18:25,330
vector with respect to time.

351
00:18:25,330 --> 00:18:27,170
We also have seen
that the velocity

352
00:18:27,170 --> 00:18:31,050
vector is the expression
that I have here in brackets.

353
00:18:31,050 --> 00:18:33,910
So I have to differentiate
that with respect to time.

354
00:18:33,910 --> 00:18:38,190
Notice that this is the sum
of two terms, one of which

355
00:18:38,190 --> 00:18:41,100
is a product of two factors,
and the other of which

356
00:18:41,100 --> 00:18:43,170
is a product of three factors.

357
00:18:43,170 --> 00:18:46,900
And by the way, among other
things to review here,

358
00:18:46,900 --> 00:18:49,380
this is the first
time in this course

359
00:18:49,380 --> 00:18:53,040
that we have actually
had to use the product

360
00:18:53,040 --> 00:18:59,630
rule for a function consisting
of three variable factors.

361
00:18:59,630 --> 00:19:01,810
Even though we discussed
this in part one,

362
00:19:01,810 --> 00:19:04,650
here is a case where in
a real-life situation,

363
00:19:04,650 --> 00:19:09,550
what we need is the derivative
rule for a product of three

364
00:19:09,550 --> 00:19:10,110
functions.

365
00:19:10,110 --> 00:19:14,370
At any rate, this is done
in great detail in the text.

366
00:19:14,370 --> 00:19:16,200
I do it more in the notes.

367
00:19:16,200 --> 00:19:18,360
So I'm just going to
hit the highlights here.

368
00:19:18,360 --> 00:19:22,840
The point is I now differentiate
this sum term by term.

369
00:19:22,840 --> 00:19:25,530
Namely, to differentiate
this, I take

370
00:19:25,530 --> 00:19:30,210
the derivative of the
first term times the second

371
00:19:30,210 --> 00:19:35,030
plus the first term times
the derivative of the second.

372
00:19:35,030 --> 00:19:37,410
Now to differentiate
this term, I

373
00:19:37,410 --> 00:19:40,940
have to differentiate a
product of three factors.

374
00:19:40,940 --> 00:19:43,780
And recall-- and by the way,
as I told you in part one,

375
00:19:43,780 --> 00:19:47,280
whenever I say "recall," that
means if you don't recall,

376
00:19:47,280 --> 00:19:50,180
it's my polite way of
saying, "look it up."

377
00:19:50,180 --> 00:19:53,870
But to differentiate a
product of three functions,

378
00:19:53,870 --> 00:19:56,710
we write the product
down three times,

379
00:19:56,710 --> 00:19:59,910
and each time differentiate
a different factor.

380
00:19:59,910 --> 00:20:01,880
For example, the
first time we'll

381
00:20:01,880 --> 00:20:05,600
differentiate r with respect
to t, which is dr/dt.

382
00:20:05,600 --> 00:20:07,270
The second time,
we'll differentiate

383
00:20:07,270 --> 00:20:09,750
d theta / dt with
respect to t, which

384
00:20:09,750 --> 00:20:12,380
is the second derivative
of theta with respect to t.

385
00:20:12,380 --> 00:20:14,080
And the third time,
we'll differentiate

386
00:20:14,080 --> 00:20:17,850
u sub theta with respect to t,
which is the derivative of u

387
00:20:17,850 --> 00:20:19,540
sub theta with respect to t.

388
00:20:19,540 --> 00:20:22,260
And summarizing that,
what do I have here?

389
00:20:22,260 --> 00:20:28,180
I have dr/dt d theta / dt times
u sub theta plus r d^2 theta /

390
00:20:28,180 --> 00:20:32,900
dt squared times u sub theta
plus r d theta / dt times

391
00:20:32,900 --> 00:20:36,570
the derivative of u sub
theta with respect to t.

392
00:20:36,570 --> 00:20:40,010
Now the point is that if I
look at these five terms,

393
00:20:40,010 --> 00:20:43,610
some of them are in nice form.

394
00:20:43,610 --> 00:20:45,720
Namely, here's a u sub r term.

395
00:20:45,720 --> 00:20:47,470
Here's a u sub theta term.

396
00:20:47,470 --> 00:20:49,170
Here's a u sub theta term.

397
00:20:49,170 --> 00:20:52,090
But these terms are
sort of mongrelized.

398
00:20:52,090 --> 00:20:56,960
Namely, what I have to do here
is utilize the chain rule.

399
00:20:56,960 --> 00:20:58,800
And remember that the
derivative of u sub

400
00:20:58,800 --> 00:21:02,650
r with respect to theta
would have been u sub theta.

401
00:21:02,650 --> 00:21:06,310
The derivative of u sub
theta with respect to theta

402
00:21:06,310 --> 00:21:08,240
would have been minus u sub r.

403
00:21:08,240 --> 00:21:09,860
So by the chain
rule-- you see what

404
00:21:09,860 --> 00:21:13,400
I'm going to do is, I'll
replace each of these terms

405
00:21:13,400 --> 00:21:16,170
by their chain rule expression.

406
00:21:16,170 --> 00:21:17,864
Then I'll collect terms.

407
00:21:17,864 --> 00:21:19,280
And the reason I'm
going over this

408
00:21:19,280 --> 00:21:24,060
fairly rapidly is that it is
a problem of sheer mechanics.

409
00:21:24,060 --> 00:21:27,560
But the punch line is that
if I now collect my terms,

410
00:21:27,560 --> 00:21:32,080
the acceleration vector has as
its u sub r component d^2 r /

411
00:21:32,080 --> 00:21:36,310
dt squared minus r d
theta / dt squared.

412
00:21:36,310 --> 00:21:41,975
And the u sub theta component is
r d^2 theta / dt squared plus 2

413
00:21:41,975 --> 00:21:44,329
dr/dt d theta / dt.

414
00:21:44,329 --> 00:21:45,870
And the beautiful
part, from my point

415
00:21:45,870 --> 00:21:47,600
of view about all
of this, is if I

416
00:21:47,600 --> 00:21:50,240
don't understand
any physics at all,

417
00:21:50,240 --> 00:21:52,470
this particular result is valid.

418
00:21:52,470 --> 00:21:54,720
It's mathematically
self-contained.

419
00:21:54,720 --> 00:21:57,420
Now certainly there
is no harm in a man

420
00:21:57,420 --> 00:22:00,700
who understands physics well
enough to say, look at it.

421
00:22:00,700 --> 00:22:04,740
This is the acceleration in
the radius direction alone.

422
00:22:04,740 --> 00:22:08,180
And this is some kind of a
correction factor proportional

423
00:22:08,180 --> 00:22:11,070
to the square of the
angular velocity, see,

424
00:22:11,070 --> 00:22:13,990
d theta / dt being angular
velocity and what have you.

425
00:22:13,990 --> 00:22:16,170
And go through this
particular thing.

426
00:22:16,170 --> 00:22:17,910
I'm saying fine,
if you can do that.

427
00:22:17,910 --> 00:22:19,120
But notice the beauty.

428
00:22:19,120 --> 00:22:22,310
This complicated
expression gives us

429
00:22:22,310 --> 00:22:25,560
the acceleration vector in
terms of u sub r and u sub

430
00:22:25,560 --> 00:22:27,570
theta with no hand-waving.

431
00:22:27,570 --> 00:22:29,950
It's mathematically
self-contained.

432
00:22:29,950 --> 00:22:32,230
And by the way, keep
in mind that one

433
00:22:32,230 --> 00:22:35,880
of the reasons that we study
polar coordinate motion

434
00:22:35,880 --> 00:22:37,780
is the fact that,
in many cases, we

435
00:22:37,780 --> 00:22:42,200
are going to be dealing
with a central force field.

436
00:22:42,200 --> 00:22:50,910
And the interesting thing
is that in a central--

437
00:22:50,910 --> 00:22:56,350
I'll just abbreviate this--
in a central force situation,

438
00:22:56,350 --> 00:22:57,620
this expression is 0.

439
00:22:57,620 --> 00:22:59,360
See, central force means what?

440
00:22:59,360 --> 00:23:01,810
That the force is in
the radial direction.

441
00:23:01,810 --> 00:23:03,340
That means all of
the acceleration--

442
00:23:03,340 --> 00:23:05,100
if you're using
Newtonian physics,

443
00:23:05,100 --> 00:23:08,090
F equals ma-- all
the acceleration is

444
00:23:08,090 --> 00:23:09,790
in the direction of u sub r.

445
00:23:09,790 --> 00:23:13,150
Therefore, the component in
the direction of u sub theta

446
00:23:13,150 --> 00:23:14,300
must be 0.

447
00:23:14,300 --> 00:23:16,820
So this fairly
complicated expression--

448
00:23:16,820 --> 00:23:22,310
r d^2 theta / dt squared plus
2 dr/dt d theta / dt equals 0

449
00:23:22,310 --> 00:23:27,060
becomes the fundamental equation
for central force field motion.

450
00:23:27,060 --> 00:23:30,400
But again, we'll talk about
that more in the exercises.

451
00:23:30,400 --> 00:23:32,850
What I wanted to do now
was to make what I think

452
00:23:32,850 --> 00:23:35,130
is a very important summary.

453
00:23:35,130 --> 00:23:38,180
And that is that when we're
studying the position vector

454
00:23:38,180 --> 00:23:44,950
R, and the velocity vector v,
and the acceleration vector a,

455
00:23:44,950 --> 00:23:48,890
that none of these depend
on the coordinate system.

456
00:23:48,890 --> 00:23:51,610
It's only their
components that do.

457
00:23:51,610 --> 00:23:54,150
In other words, at
the expense of having

458
00:23:54,150 --> 00:23:58,140
a fairly jumbled figure which I
rationalize here-- it is small,

459
00:23:58,140 --> 00:24:00,290
but I think it is
clear from context.

460
00:24:00,290 --> 00:24:03,260
What I'm saying is, let's
suppose I have a curve C,

461
00:24:03,260 --> 00:24:07,410
and some point P_0
on this curve C.

462
00:24:07,410 --> 00:24:10,530
I can draw in the pair
of orthogonal vectors

463
00:24:10,530 --> 00:24:12,200
i and j in the plane.

464
00:24:12,200 --> 00:24:15,490
I can draw in the pair
of orthogonal vectors

465
00:24:15,490 --> 00:24:17,510
u sub r-- "orthogonal"
means perpendicular,

466
00:24:17,510 --> 00:24:21,090
if we haven't said that before--
u sub r and u sub theta.

467
00:24:21,090 --> 00:24:23,690
I can draw those in.

468
00:24:23,690 --> 00:24:28,650
And I can draw in
T and N. Now all

469
00:24:28,650 --> 00:24:32,130
I know is that if I have
the velocity vector v,

470
00:24:32,130 --> 00:24:34,560
it must be tangential
to the curve.

471
00:24:34,560 --> 00:24:37,390
Hopefully by this time, we
realize that the acceleration

472
00:24:37,390 --> 00:24:39,930
vector has no such restriction.

473
00:24:39,930 --> 00:24:43,710
Let's just draw in a v and an a,
call these the velocity vectors

474
00:24:43,710 --> 00:24:45,690
and the acceleration vectors.

475
00:24:45,690 --> 00:24:50,190
The point is that v and a are
determined by the motion--

476
00:24:50,190 --> 00:24:52,180
not by the coordinate system.

477
00:24:52,180 --> 00:24:54,660
In other words, when we're
talking about the velocity

478
00:24:54,660 --> 00:24:58,850
of this particle at the point
P:0, its velocity is the same,

479
00:24:58,850 --> 00:25:01,230
no matter what coordinate
system we're using.

480
00:25:01,230 --> 00:25:03,480
It just happens that
if we're dealing

481
00:25:03,480 --> 00:25:06,870
with Cartesian coordinates,
the velocity vector

482
00:25:06,870 --> 00:25:10,600
is dx/dt i plus dy/dt j.

483
00:25:10,600 --> 00:25:13,060
In other words, it's this
particular combination

484
00:25:13,060 --> 00:25:14,310
of i and j.

485
00:25:14,310 --> 00:25:16,920
If we're using T
and N components,

486
00:25:16,920 --> 00:25:21,080
the particular combination
of T and N is what?

487
00:25:21,080 --> 00:25:28,020
ds/dt times the unit
tangent vector plus 0 N.

488
00:25:28,020 --> 00:25:31,620
And if we happen to be
using polar coordinates,

489
00:25:31,620 --> 00:25:35,586
the expression is dr/dt * u
sub r plus r * d theta / dt

490
00:25:35,586 --> 00:25:37,140
* u sub theta.

491
00:25:37,140 --> 00:25:39,095
But let me circle
these, because it's

492
00:25:39,095 --> 00:25:41,070
the same velocity in each case.

493
00:25:41,070 --> 00:25:44,860
We use this when horizontal and
vertical motion are important.

494
00:25:44,860 --> 00:25:48,960
We use this when we're
interested in motion

495
00:25:48,960 --> 00:25:50,250
along the curve.

496
00:25:50,250 --> 00:25:52,960
And we use this primarily
in central force fields.

497
00:25:52,960 --> 00:25:54,200
But it makes no difference.

498
00:25:54,200 --> 00:25:55,860
It's the same velocity vector.

499
00:25:55,860 --> 00:26:00,580
And in a similar way, it's also
the same acceleration vector,

500
00:26:00,580 --> 00:26:02,940
whichever system
you happen to use.

501
00:26:02,940 --> 00:26:05,880
Namely, if we use
Cartesian coordinates,

502
00:26:05,880 --> 00:26:08,260
the acceleration vector
is the second derivative

503
00:26:08,260 --> 00:26:15,280
of x with respect to t times i
plus the second derivative of y

504
00:26:15,280 --> 00:26:18,570
with respect to t times j.

505
00:26:18,570 --> 00:26:22,550
That same vector, if we express
it in T and N components,

506
00:26:22,550 --> 00:26:26,200
is d^2 s / dt squared
times T, plus kappa,

507
00:26:26,200 --> 00:26:32,500
the curvature number, times
ds/dt squared times N.

508
00:26:32,500 --> 00:26:35,770
And if we express it in
terms of polar coordinates,

509
00:26:35,770 --> 00:26:38,430
as we just saw earlier
in our lecture,

510
00:26:38,430 --> 00:26:41,330
this is the expression
that we get.

511
00:26:41,330 --> 00:26:45,880
In other words then,
this summarizes our study

512
00:26:45,880 --> 00:26:52,460
of motion in the plane
using either Cartesian

513
00:26:52,460 --> 00:26:58,260
or polar or tangential
and normal components.

514
00:26:58,260 --> 00:27:01,250
You see, the point is
that we pick whichever

515
00:27:01,250 --> 00:27:06,250
coordinate system happens to be
of the greatest interest to us,

516
00:27:06,250 --> 00:27:08,160
the greatest value to us.

517
00:27:08,160 --> 00:27:10,540
We make the coordinate
system our slave,

518
00:27:10,540 --> 00:27:12,230
rather than the
other way around,

519
00:27:12,230 --> 00:27:15,370
and tackle the problem from
that particular point of view.

520
00:27:15,370 --> 00:27:20,270
At any rate, that ends this
phase of our particular course.

521
00:27:20,270 --> 00:27:22,510
And in the next
phase of our course,

522
00:27:22,510 --> 00:27:26,150
we get to probably what is
the most fundamental building

523
00:27:26,150 --> 00:27:28,410
block of the entire course.

524
00:27:28,410 --> 00:27:30,430
We get to that
particular topic which,

525
00:27:30,430 --> 00:27:34,400
by and large, most courses in
functions of several variables

526
00:27:34,400 --> 00:27:35,450
begin with.

527
00:27:35,450 --> 00:27:39,930
But we'll talk about that
more the next time we meet.

528
00:27:39,930 --> 00:27:43,750
And until that time, goodbye.

529
00:27:43,750 --> 00:27:46,130
Funding for the
publication of this video

530
00:27:46,130 --> 00:27:51,000
was provided by the Gabriella
and Paul Rosenbaum Foundation.

531
00:27:51,000 --> 00:27:55,180
Help OCW continue to provide
free and open access to MIT

532
00:27:55,180 --> 00:27:59,575
courses by making a donation
at ocw.mit.edu/donate.