1 00:00:01,000 --> 00:00:03,000 2 00:00:01,000 --> 00:00:03,000 The following content is provided under a Creative 3 00:00:03,000 --> 00:00:05,000 Commons license. Your support will help MIT 4 00:00:05,000 --> 00:00:08,000 OpenCourseWare continue to offer high quality educational 5 00:00:08,000 --> 00:00:13,000 resources for free. To make a donation or to view 6 00:00:13,000 --> 00:00:18,000 additional materials from hundreds of MIT courses, 7 00:00:18,000 --> 00:00:23,000 visit MIT OpenCourseWare at ocw.mit.edu. 8 00:00:23,000 --> 00:00:29,000 Let me just tell you first about the list of topics. 9 00:00:29,000 --> 00:00:35,000 Basically, the list of topics is simple. 10 00:00:35,000 --> 00:00:42,000 It is everything. I mean, everything we have seen 11 00:00:42,000 --> 00:00:48,000 so far is on the exam. But let me just remind you of 12 00:00:48,000 --> 00:00:52,000 the main topics that we have seen. 13 00:00:52,000 --> 00:00:58,000 First of all, we learned about vectors, 14 00:00:58,000 --> 00:01:03,000 how to use them, and dot-product. 15 00:01:03,000 --> 00:01:07,000 At this point, you probably should know that 16 00:01:07,000 --> 00:01:13,000 the dot-product of two vectors is obtained by summing products 17 00:01:13,000 --> 00:01:19,000 of components. And geometrically it is the 18 00:01:19,000 --> 00:01:29,000 length of A times the length of B times the cosine of the angle 19 00:01:29,000 --> 00:01:34,000 between them. And, in particular, 20 00:01:34,000 --> 00:01:41,000 we can use dot-product to measure angles by solving for 21 00:01:41,000 --> 00:01:49,000 cosine theta in this equality. And most importantly to detect 22 00:01:49,000 --> 00:01:54,000 whether two vectors are perpendicular to each other. 23 00:01:54,000 --> 00:02:00,000 Two vectors are perpendicular when their dot-product is zero. 24 00:02:00,000 --> 00:02:07,000 Any questions about that? No. 25 00:02:07,000 --> 00:02:15,000 Is everyone reasonably happy with dot-product by now? 26 00:02:15,000 --> 00:02:19,000 I see a stunned silence. Nobody happy with dot-product 27 00:02:19,000 --> 00:02:28,000 so far? OK. 28 00:02:28,000 --> 00:02:36,000 If you want to look at Practice 1A, a good example of a typical 29 00:02:36,000 --> 00:02:41,000 problem with dot-product would be problem 1. 30 00:02:41,000 --> 00:02:43,000 Let's see. We are going to go over the 31 00:02:43,000 --> 00:02:46,000 practice exam when I am done writing this list of topics. 32 00:02:46,000 --> 00:02:49,000 I think probably we actually will skip this problem because I 33 00:02:49,000 --> 00:02:51,000 think most of you know how to do it. 34 00:02:51,000 --> 00:02:57,000 And if not then you should run for help from me or from your 35 00:02:57,000 --> 00:03:02,000 recitation instructor to figure out how to do it. 36 00:03:02,000 --> 00:03:10,000 The second topic that we saw was cross-product. 37 00:03:10,000 --> 00:03:19,000 When you have two vectors in space, you can just form that 38 00:03:19,000 --> 00:03:25,000 cross-product by computing its determinant. 39 00:03:25,000 --> 00:03:31,000 So, implicitly, we should also know about 40 00:03:31,000 --> 00:03:35,000 determinants. By that I mean two by two and 41 00:03:35,000 --> 00:03:38,000 three by three. Don't worry about larger ones, 42 00:03:38,000 --> 00:03:43,000 even if you are interested, they won't be on the test. 43 00:03:43,000 --> 00:03:49,000 And applications of cross-product, 44 00:03:49,000 --> 00:03:55,000 for example, finding the area of a triangle 45 00:03:55,000 --> 00:04:04,000 or a parallelogram in space. If you have a triangle in space 46 00:04:04,000 --> 00:04:10,000 with sides A and B then its area is one-half of the length of A 47 00:04:10,000 --> 00:04:14,000 cross B. Because the length of A cross B 48 00:04:14,000 --> 00:04:19,000 is length A, length B sine theta, which is the same as the 49 00:04:19,000 --> 00:04:24,000 area of the parallelogram formed by these two vectors. 50 00:04:24,000 --> 00:04:28,000 And the other application of cross-product is to find a 51 00:04:28,000 --> 00:04:32,000 vector that's perpendicular to two given vectors. 52 00:04:32,000 --> 00:04:36,000 In particular, to find the vector that is 53 00:04:36,000 --> 00:04:42,000 normal to a plane and then find the equation of a plane. 54 00:04:42,000 --> 00:04:57,000 Another application is finding the normal vector to a plane and 55 00:04:57,000 --> 00:05:08,000 using that finding the equation of a plane. 56 00:05:08,000 --> 00:05:16,000 Basically, remember, to find the equation of a 57 00:05:16,000 --> 00:05:23,000 plane, ax by cz = d, what you need is the normal 58 00:05:23,000 --> 00:05:29,000 vector to the plane. And the components of the 59 00:05:29,000 --> 00:05:32,000 normal vector are exactly the coefficients that go into this. 60 00:05:32,000 --> 00:05:38,000 And we have seen an argument for why that happens to be the 61 00:05:38,000 --> 00:05:40,000 case. To find a normal vector to a 62 00:05:40,000 --> 00:05:44,000 plane typically what we will do is take two vectors that lie in 63 00:05:44,000 --> 00:05:47,000 the plane and will take their cross-product. 64 00:05:47,000 --> 00:05:54,000 And the cross-product will automatically be perpendicular 65 00:05:54,000 --> 00:05:59,000 to both of them. We are going to see an example 66 00:05:59,000 --> 00:06:04,000 of that when we look at problem 5 in practice 1A. 67 00:06:04,000 --> 00:06:12,000 I think we will try to do that one. 68 00:06:12,000 --> 00:06:17,000 Another application, well, we will just mention it 69 00:06:17,000 --> 00:06:21,000 as a topic that goes along with this one. 70 00:06:21,000 --> 00:06:34,000 We have seen also about equations of lines and how to 71 00:06:34,000 --> 00:06:43,000 find where a line intersects a plane. 72 00:06:43,000 --> 00:06:46,000 Just to refresh your memories, the equation of a line, 73 00:06:46,000 --> 00:06:50,000 well, we will be looking at parametric equations. 74 00:06:50,000 --> 00:06:54,000 To know the parametric equation of a line, we need to know a 75 00:06:54,000 --> 00:06:59,000 point on the line and we need to know a vector that is parallel 76 00:06:59,000 --> 00:07:03,000 to the line. And, if we know a point on the 77 00:07:03,000 --> 00:07:06,000 line and a vector along the line, 78 00:07:06,000 --> 00:07:11,000 then we can express the parametric equations for the 79 00:07:11,000 --> 00:07:16,000 motion of a point that is moving on the line. 80 00:07:16,000 --> 00:07:19,000 Actually, starting at point, at time zero, 81 00:07:19,000 --> 00:07:21,000 and moving with velocity v. 82 00:07:21,000 --> 00:07:38,000 83 00:07:38,000 --> 00:07:44,000 To put things in symbolic form, you will get a position of that 84 00:07:44,000 --> 00:07:48,000 point by starting with a position of time zero and adding 85 00:07:48,000 --> 00:07:53,000 t times the vector v. It gives you x, 86 00:07:53,000 --> 00:08:00,000 y and z in terms of t. And that is how we represent 87 00:08:00,000 --> 00:08:06,000 lines. We will look at problem 5 in a 88 00:08:06,000 --> 00:08:15,000 bit, but any general questions about these topics? 89 00:08:15,000 --> 00:08:20,000 No. Do you have a question? 90 00:08:20,000 --> 00:08:26,000 Do we have to know Taylor series? 91 00:08:26,000 --> 00:08:34,000 That is a good question. No, not on the exam. 92 00:08:34,000 --> 00:08:36,000 [APPLAUSE] Taylor series are something you 93 00:08:36,000 --> 00:08:38,000 should be aware of, generally speaking. 94 00:08:38,000 --> 00:08:40,000 It will be useful for you in real life, 95 00:08:40,000 --> 00:08:44,000 probably not when you go to the supermarket, 96 00:08:44,000 --> 00:08:48,000 but if you solve engineering problems you will need Taylor 97 00:08:48,000 --> 00:08:52,000 series. It would be good not to forget 98 00:08:52,000 --> 00:08:58,000 them entirely, but on the 18.02 exams they 99 00:08:58,000 --> 00:09:03,000 probably won't be there. Let me continue with more 100 00:09:03,000 --> 00:09:06,000 topics. And then we can see if you can 101 00:09:06,000 --> 00:09:11,000 think of other topics that should or should not be on the 102 00:09:11,000 --> 00:09:22,000 exam. Third topics would be matrices, 103 00:09:22,000 --> 00:09:34,000 linear systems, inverting matrices. 104 00:09:34,000 --> 00:09:37,000 I know that most of you that have calculators that can invert 105 00:09:37,000 --> 00:09:40,000 matrices, but still you are expected at this point to know 106 00:09:40,000 --> 00:09:42,000 how to do it by hand. If you have looked at the 107 00:09:42,000 --> 00:09:44,000 practice tests, both of them have a problem 108 00:09:44,000 --> 00:09:47,000 that asks you to invert a matrix or at least do part of it. 109 00:09:47,000 --> 00:09:53,000 And so it is very likely that tomorrow there will be a problem 110 00:09:53,000 --> 00:09:57,000 like that as well. In general, when a kind of 111 00:09:57,000 --> 00:10:00,000 problem is on both practice tests it's a good indication 112 00:10:00,000 --> 00:10:04,000 that it might be there also on the actual exam. 113 00:10:04,000 --> 00:10:08,000 Unfortunately, not with the same matrix so you 114 00:10:08,000 --> 00:10:12,000 cannot learn the answer by heart. 115 00:10:12,000 --> 00:10:17,000 Another thing that we have learned about, 116 00:10:17,000 --> 00:10:24,000 well, I should say this is going to be problem 3 on the 117 00:10:24,000 --> 00:10:29,000 test and will on the practice test. 118 00:10:29,000 --> 00:10:33,000 On the actual test, too, I think, 119 00:10:33,000 --> 00:10:37,000 actually. Anyway, we will come back to it 120 00:10:37,000 --> 00:10:39,000 later. A couple of things that you 121 00:10:39,000 --> 00:10:42,000 should remember. If you have a system of the 122 00:10:42,000 --> 00:10:46,000 form AX equals B then there are two cases. 123 00:10:46,000 --> 00:10:51,000 If a determinant of A is not zero then that means you can 124 00:10:51,000 --> 00:10:56,000 compute the inverse matrix and you can just solve by taking A 125 00:10:56,000 --> 00:11:02,000 inverse times B. And the other case is when the 126 00:11:02,000 --> 00:11:11,000 determinant of A is zero, and there is either no solution 127 00:11:11,000 --> 00:11:18,000 or there is infinitely many solutions. 128 00:11:18,000 --> 00:11:20,000 In particular, if you know that there is a 129 00:11:20,000 --> 00:11:21,000 solution, for example, 130 00:11:21,000 --> 00:11:24,000 if B is zero there always is an obvious solution, 131 00:11:24,000 --> 00:11:28,000 X equals zero, then you will actually have 132 00:11:28,000 --> 00:11:30,000 infinitely many. In general, we don't really 133 00:11:30,000 --> 00:11:33,000 know how to tell whether it is no solution or infinitely many. 134 00:11:33,000 --> 00:11:43,000 135 00:11:43,000 --> 00:11:52,000 Questions about that? Yes? 136 00:11:52,000 --> 00:11:58,000 Will we have to know how to rotate vectors and so on? 137 00:11:58,000 --> 00:12:01,000 Not in general, but you might still want to 138 00:12:01,000 --> 00:12:04,000 remember how to rotate a vector in a plane by 90 degrees because 139 00:12:04,000 --> 00:12:07,000 that has been useful when we have done problems about 140 00:12:07,000 --> 00:12:10,000 parametric equations, which is what I am coming to 141 00:12:10,000 --> 00:12:12,000 next. What we have seen about 142 00:12:12,000 --> 00:12:15,000 rotation matrices, that was the homework part B 143 00:12:15,000 --> 00:12:17,000 problem, you are not supposed to 144 00:12:17,000 --> 00:12:21,000 remember by heart everything that was in part B of your 145 00:12:21,000 --> 00:12:24,000 homework. It is a good idea to have some 146 00:12:24,000 --> 00:12:27,000 vague knowledge because it is useful culture, 147 00:12:27,000 --> 00:12:28,000 I would say, useful background for later in 148 00:12:28,000 --> 00:12:33,000 your lives, but I won't ask you by heart to 149 00:12:33,000 --> 00:12:40,000 know what is the formation for a rotation matrix. 150 00:12:40,000 --> 00:12:49,000 And then we come to, last by not least, 151 00:12:49,000 --> 00:13:00,000 the problem of finding parametric equations. 152 00:13:00,000 --> 00:13:03,000 And, in particular, possibly by decomposing the 153 00:13:03,000 --> 00:13:06,000 position vector into a sum of simpler vectors. 154 00:13:06,000 --> 00:13:13,000 You have seen quite an evil exam of that on the last problem 155 00:13:13,000 --> 00:13:19,000 set with this picture that maybe by now you have had some 156 00:13:19,000 --> 00:13:23,000 nightmares about. Anyway, the one on the exam 157 00:13:23,000 --> 00:13:27,000 will certainly be easier than that. 158 00:13:27,000 --> 00:13:36,000 But, as you have seen -- I mean, you should know, 159 00:13:36,000 --> 00:13:40,000 basically, how to analyze a motion that is 160 00:13:40,000 --> 00:13:44,000 being described to you and express it in terms of vectors 161 00:13:44,000 --> 00:13:48,000 and then figure out what the parametric equation will be. 162 00:13:48,000 --> 00:13:52,000 Now, again, it won't be as complicated on the exam as the 163 00:13:52,000 --> 00:13:57,000 one in the problem set. But there are a couple of those 164 00:13:57,000 --> 00:14:01,000 on the practice exam, so that gives you an idea of 165 00:14:01,000 --> 00:14:04,000 what is realistically expected of you. 166 00:14:04,000 --> 00:14:09,000 And now once we have parametric equations for motion, 167 00:14:09,000 --> 00:14:13,000 so that means when we know how to find the position vector as a 168 00:14:13,000 --> 00:14:15,000 function of a parameter maybe of time, 169 00:14:15,000 --> 00:14:24,000 then we have seen also about velocity and acceleration, 170 00:14:24,000 --> 00:14:28,000 which the vector is obtained by taking the first and second 171 00:14:28,000 --> 00:14:31,000 derivatives of a position vector. 172 00:14:31,000 --> 00:14:41,000 And so one topic that I will add in there as well is somehow 173 00:14:41,000 --> 00:14:50,000 how to prove things about motions by differentiating 174 00:14:50,000 --> 00:14:55,000 vector identities. One example of that, 175 00:14:55,000 --> 00:14:58,000 for example, is when we try to look at 176 00:14:58,000 --> 00:15:03,000 Kepler's law in class last time. We look at Kepler's second law 177 00:15:03,000 --> 00:15:06,000 of planetary motion, and we reduced it to a 178 00:15:06,000 --> 00:15:10,000 calculation about a derivative of the cross-product R cross v. 179 00:15:10,000 --> 00:15:14,000 Now, on the exam you don't need to know the details of Kepler's 180 00:15:14,000 --> 00:15:16,000 law, but you need to be able to 181 00:15:16,000 --> 00:15:22,000 manipulate vector quantities a bit in the way that we did. 182 00:15:22,000 --> 00:15:27,000 And so on practice exam 1A, you actually have a variety of 183 00:15:27,000 --> 00:15:30,000 problems on this topics because you have problems two, 184 00:15:30,000 --> 00:15:36,000 four and six, all about parametric motions. 185 00:15:36,000 --> 00:15:42,000 Probably tomorrow there will not be three distinct problems 186 00:15:42,000 --> 00:15:48,000 about parametric motions, but maybe a couple of them. 187 00:15:48,000 --> 00:15:51,000 I think that is basically the list of topics. 188 00:15:51,000 --> 00:15:54,000 Anybody spot something that I have forgotten to put on the 189 00:15:54,000 --> 00:15:58,000 exam or questions about something that should or should 190 00:15:58,000 --> 00:16:03,000 not be there? You go first. 191 00:16:03,000 --> 00:16:11,000 Yeah? How about parametrizing weird 192 00:16:11,000 --> 00:16:18,000 trigonometric functions? I am not sure what you mean by 193 00:16:18,000 --> 00:16:21,000 that. Well, parametric curves, 194 00:16:21,000 --> 00:16:25,000 you need to know how to parameterize motions, 195 00:16:25,000 --> 00:16:30,000 and that involves a little bit of trigonometrics. 196 00:16:30,000 --> 00:16:33,000 When we have seen these problems about rotating wheels, 197 00:16:33,000 --> 00:16:34,000 say the cycloid, for example, 198 00:16:34,000 --> 00:16:38,000 and so on there is a bit of cosine and sine and so on. 199 00:16:38,000 --> 00:16:41,000 I think not much more on that. You won't need obscure 200 00:16:41,000 --> 00:16:48,000 trigonometric identities. You're next. 201 00:16:48,000 --> 00:16:51,000 Any proofs on the exam or just like problems? 202 00:16:51,000 --> 00:16:55,000 Well, a problem can ask you to show things. 203 00:16:55,000 --> 00:16:58,000 It is not going to be a complicated proof. 204 00:16:58,000 --> 00:17:00,000 The proofs are going to be fairly easy. 205 00:17:00,000 --> 00:17:04,000 If you look at practice 1A, the last problem does have a 206 00:17:04,000 --> 00:17:08,000 little bit of proof. 6B says that show that blah, 207 00:17:08,000 --> 00:17:11,000 blah, blah. But, as you will see, 208 00:17:11,000 --> 00:17:13,000 it is not a difficult kind of proof. 209 00:17:13,000 --> 00:17:24,000 So, about the same. Yes? 210 00:17:24,000 --> 00:17:29,000 Are there equations of 3D shapes that we should know at 211 00:17:29,000 --> 00:17:32,000 this point? We should know definitely a lot 212 00:17:32,000 --> 00:17:34,000 about the equations of planes on lines. 213 00:17:34,000 --> 00:17:37,000 And you should probably know that a sphere centered at the 214 00:17:37,000 --> 00:17:40,000 origin is the set of points where distance to the center is 215 00:17:40,000 --> 00:17:43,000 equal to the radius of the sphere. 216 00:17:43,000 --> 00:17:45,000 We don't need more at this point. 217 00:17:45,000 --> 00:17:48,000 As the semester goes on, we will start seeing cones and 218 00:17:48,000 --> 00:17:52,000 things like that. But at this point planes, lines. 219 00:17:52,000 --> 00:17:58,000 And maybe you need to know about circles and spheres, 220 00:17:58,000 --> 00:18:03,000 but nothing beyond that. More questions? 221 00:18:03,000 --> 00:18:11,000 Yes? If there is a formula that you 222 00:18:11,000 --> 00:18:14,000 have proved on the homework then, yes, you can assume it on 223 00:18:14,000 --> 00:18:17,000 the test. Maybe you want to write on your 224 00:18:17,000 --> 00:18:20,000 test that this is a formula you have seen in homework just so 225 00:18:20,000 --> 00:18:24,000 that we know that you remember it from homework and not from 226 00:18:24,000 --> 00:18:27,000 looking over your neighbor's shoulder or whatever. 227 00:18:27,000 --> 00:18:31,000 Yes, it is OK to use things that you know general-speaking. 228 00:18:31,000 --> 00:18:33,000 That being said, for example, 229 00:18:33,000 --> 00:18:37,000 probably there will be a linear system to solve. 230 00:18:37,000 --> 00:18:40,000 It will say on the exam you are supposed to solve that using 231 00:18:40,000 --> 00:18:43,000 matrices, not by elimination. There are things like that. 232 00:18:43,000 --> 00:18:47,000 If a problem says solve by using vector methods, 233 00:18:47,000 --> 00:18:51,000 things like that, then try to use at least a 234 00:18:51,000 --> 00:18:54,000 vector somewhere. But, in general, 235 00:18:54,000 --> 00:18:58,000 you are allowed to use things that you know. 236 00:18:58,000 --> 00:19:07,000 Yes? Will we need to go from 237 00:19:07,000 --> 00:19:09,000 parametric equations to xy equations? 238 00:19:09,000 --> 00:19:16,000 Well, let's say only if it is very easy. 239 00:19:16,000 --> 00:19:19,000 If I give you a parametric curve, sin t, 240 00:19:19,000 --> 00:19:25,000 sin t, then you should be able to observe that it is on the 241 00:19:25,000 --> 00:19:29,000 line y equals x, not beyond that. 242 00:19:29,000 --> 00:19:37,000 Yes? Do we have to use -- Yes. 243 00:19:37,000 --> 00:19:40,000 I don't know if you will have to use it, but certainly you 244 00:19:40,000 --> 00:19:43,000 should know a little bit about the unit tangent vector. 245 00:19:43,000 --> 00:19:51,000 Just remember the main thing to know that the unit tangent 246 00:19:51,000 --> 00:19:57,000 vector is velocity divided by the speed. 247 00:19:57,000 --> 00:20:04,000 I mean there is not much more to it when you think about it. 248 00:20:04,000 --> 00:20:11,000 Yes? Kepler's law, 249 00:20:11,000 --> 00:20:13,000 well, you are allowed to use it if it helps you, 250 00:20:13,000 --> 00:20:16,000 if you find a way to squeeze it in. 251 00:20:16,000 --> 00:20:20,000 You don't have to know Kepler's law in detail. 252 00:20:20,000 --> 00:20:22,000 You just have to know how to reproduce the general steps. 253 00:20:22,000 --> 00:20:28,000 If I tell you R cross v is constant, you might be expected 254 00:20:28,000 --> 00:20:33,000 to know what to do with that. I would say -- Basically, 255 00:20:33,000 --> 00:20:36,000 you don't need to know Kepler's law. 256 00:20:36,000 --> 00:20:38,000 You need to know the kind of stuff that we saw when we 257 00:20:38,000 --> 00:20:41,000 derived it such as how to take the derivative of a dot-product 258 00:20:41,000 --> 00:20:52,000 or a cross-product. That is basically the answer. 259 00:20:52,000 --> 00:20:55,000 I don't see any questions anymore. 260 00:20:55,000 --> 00:21:03,000 Oh, you are raising your hand. Yes. 261 00:21:03,000 --> 00:21:06,000 How to calculate the distance between two lines and the 262 00:21:06,000 --> 00:21:08,000 distance between two planes? Well, you have seen, 263 00:21:08,000 --> 00:21:10,000 probably recently, that it is quite painful to do 264 00:21:10,000 --> 00:21:13,000 in general. And, no, I don't think that 265 00:21:13,000 --> 00:21:16,000 will be on the exam by itself. You need to know how to compute 266 00:21:16,000 --> 00:21:19,000 the distance between two points. That certainly you need to know. 267 00:21:19,000 --> 00:21:24,000 And also maybe how to find the compliment of a vector in a 268 00:21:24,000 --> 00:21:28,000 certain direction. And that is about it, 269 00:21:28,000 --> 00:21:33,000 I would say. I mean the more you know about 270 00:21:33,000 --> 00:21:38,000 things the better. Things that come up on part Bs 271 00:21:38,000 --> 00:21:43,000 of the problem sets are interesting things, 272 00:21:43,000 --> 00:21:48,000 but they are usually not needed on the exams. 273 00:21:48,000 --> 00:21:53,000 If you have more questions then you are not raising your hand 274 00:21:53,000 --> 00:21:55,000 high enough for me to see it. OK. 275 00:21:55,000 --> 00:22:00,000 Let's try to do a bit of this practice exam 1A. 276 00:22:00,000 --> 00:22:03,000 Hopefully, everybody has it. If you don't have it, 277 00:22:03,000 --> 00:22:07,000 hopefully your neighbor has it. If you don't have it and your 278 00:22:07,000 --> 00:22:09,000 neighbor doesn't have it then please raise your hand. 279 00:22:09,000 --> 00:22:10,000 I have a couple. 280 00:22:10,000 --> 00:22:57,000 281 00:22:57,000 --> 00:23:02,000 If you neighbor has it then just follow with them for now. 282 00:23:02,000 --> 00:23:05,000 I think there are a few people behind you over there. 283 00:23:05,000 --> 00:23:06,000 I will stop handing them out now. 284 00:23:06,000 --> 00:23:11,000 If you really need one, it is on the website, 285 00:23:11,000 --> 00:23:15,000 it will be here at the end of class. 286 00:23:15,000 --> 00:23:23,000 Let's see. Well, I think we are going to 287 00:23:23,000 --> 00:23:27,000 just skip problems 1 and 2 because they are pretty 288 00:23:27,000 --> 00:23:32,000 straightforward and I hope that you know how to do them. 289 00:23:32,000 --> 00:23:37,000 I mean I don't know. Let's see. 290 00:23:37,000 --> 00:23:42,000 How many of you have no problem with problem 1? 291 00:23:42,000 --> 00:23:46,000 How many of you have trouble with problem 1? 292 00:23:46,000 --> 00:23:50,000 OK. How many of you haven't raised 293 00:23:50,000 --> 00:23:52,000 your hands? OK. 294 00:23:52,000 --> 00:23:56,000 How many of you have trouble with problem 2? 295 00:23:56,000 --> 00:23:58,000 OK. Well, if you have questions 296 00:23:58,000 --> 00:24:01,000 about those, maybe you should just come see me at the end 297 00:24:01,000 --> 00:24:04,000 because that is probably more efficient that way. 298 00:24:04,000 --> 00:24:11,000 I am going to start right away with problem 3, 299 00:24:11,000 --> 00:24:18,000 actually. Problem 3 says we have a matrix 300 00:24:18,000 --> 00:24:25,000 given to us |1 3 2; 2 0 - 1; 1 1 0|. 301 00:24:25,000 --> 00:24:30,000 And it tells us determinant of A is 2 and inverse equals 302 00:24:30,000 --> 00:24:34,000 something, but we are missing two values A and B and we are 303 00:24:34,000 --> 00:24:41,000 supposed to find them. That means we need to do the 304 00:24:41,000 --> 00:24:51,000 steps of the algorithm to find the inverse of A. 305 00:24:51,000 --> 00:25:00,000 We are told that A inverse is one-half of |1 ... 306 00:25:00,000 --> 00:25:06,000 ...; - 1 - 2 5; 2 2 - 6|. 307 00:25:06,000 --> 00:25:10,000 And here there are two unknown values. 308 00:25:10,000 --> 00:25:15,000 Remember, to invert a matrix, first we compute the minors. 309 00:25:15,000 --> 00:25:17,000 Then we flip some signs to get the cofactors. 310 00:25:17,000 --> 00:25:19,000 Then we transpose. And, finally, 311 00:25:19,000 --> 00:25:23,000 we divide by the determinant. Let's try to be smart about 312 00:25:23,000 --> 00:25:26,000 this. Do we need to compute all nine 313 00:25:26,000 --> 00:25:27,000 minors? No. 314 00:25:27,000 --> 00:25:30,000 We only need to compute two of them, right? 315 00:25:30,000 --> 00:25:34,000 Which minors do we need to compute? 316 00:25:34,000 --> 00:25:39,000 Here and here or here and here? Yeah, that looks better. 317 00:25:39,000 --> 00:25:43,000 Because, remember, we need to transpose things so 318 00:25:43,000 --> 00:25:48,000 these two guys will end up here. I claim we should compute these 319 00:25:48,000 --> 00:25:51,000 two minors. And we will see if that is good 320 00:25:51,000 --> 00:25:52,000 enough. If you start doing others and 321 00:25:52,000 --> 00:25:56,000 you find that they don't end up in the right place then just do 322 00:25:56,000 --> 00:25:58,000 more, but you don't need to spend 323 00:25:58,000 --> 00:26:00,000 your time computing all nine of them. 324 00:26:00,000 --> 00:26:03,000 If you are worried about not doing it right then, 325 00:26:03,000 --> 00:26:06,000 of course, you can maybe compute one or two more to just 326 00:26:06,000 --> 00:26:11,000 double-check your answers. But let us just do those that 327 00:26:11,000 --> 00:26:16,000 we think are needed. The matrix of minors. 328 00:26:16,000 --> 00:26:22,000 The one that goes in the middle position is obtained by deleting 329 00:26:22,000 --> 00:26:26,000 this row and that column, and we are left with a 330 00:26:26,000 --> 00:26:32,000 determinant |3 2;1 0|, 3 times 0 minus 1 times 2 331 00:26:32,000 --> 00:26:40,000 should be - 2 should be - 2. Then the one in the lower left 332 00:26:40,000 --> 00:26:45,000 corner, we delete the last row and the first column, 333 00:26:45,000 --> 00:26:47,000 we are left with |3 2; 0 - 1|. 334 00:26:47,000 --> 00:26:50,000 3 times (- 1) is negative 3 minus 0. 335 00:26:50,000 --> 00:26:58,000 We are still left with negative three. 336 00:26:58,000 --> 00:27:08,000 Is that step clear for everyone? Then we need to go to cofactors. 337 00:27:08,000 --> 00:27:10,000 That means we need to change signs. 338 00:27:10,000 --> 00:27:24,000 The rule is -- We change signs in basically these four places. 339 00:27:24,000 --> 00:27:32,000 That means we will be left with positive 2 and negative 3. 340 00:27:32,000 --> 00:27:44,000 Then we take the transpose. That means the first column 341 00:27:44,000 --> 00:27:49,000 will copy into the first row, so this guy we still don't 342 00:27:49,000 --> 00:27:52,000 know, but here we will have two and 343 00:27:52,000 --> 00:27:59,000 here we will have minus three. Finally, we have to divide by 344 00:27:59,000 --> 00:28:05,000 the determinant of A. And here we are actually told 345 00:28:05,000 --> 00:28:08,000 that the determinant of A is two. 346 00:28:08,000 --> 00:28:12,000 So we will divide by two. But there is only one-half here 347 00:28:12,000 --> 00:28:18,000 so actually it is done for us. The values that we will put up 348 00:28:18,000 --> 00:28:22,000 there are going to be 2 and negative 3. 349 00:28:22,000 --> 00:28:30,000 350 00:28:30,000 --> 00:28:38,000 Now let's see how we use that to solve a linear system. 351 00:28:38,000 --> 00:28:46,000 If we have to solve a linear system, Ax equals B, 352 00:28:46,000 --> 00:28:50,000 well, if the matrix is invertible, its determinant is 353 00:28:50,000 --> 00:28:54,000 not zero, so we can certainly write x 354 00:28:54,000 --> 00:29:01,000 equals A inverse B. So we have to multiply, 355 00:29:01,000 --> 00:29:09,000 that is one-half | 1 2 - 3; - 1 - 2 5; 2 2 - 6|. 356 00:29:09,000 --> 00:29:14,000 Times B [ 1, - 2,1]. 357 00:29:14,000 --> 00:29:17,000 Remember, to do a matrix multiplication you take the rows 358 00:29:17,000 --> 00:29:21,000 in here, the columns in here and you do dot-products. 359 00:29:21,000 --> 00:29:25,000 The first entry will be one times one plus two times minus 360 00:29:25,000 --> 00:29:30,000 two plus minus three times one, one minus four minus three 361 00:29:30,000 --> 00:29:35,000 should be negative six, except I still have, 362 00:29:35,000 --> 00:29:43,000 of course, a one-half in front. Then minus one plus four plus 363 00:29:43,000 --> 00:29:50,000 five should be 8. Two minus four minus six should 364 00:29:50,000 --> 00:29:57,000 be -8. That will simplify to [- 3,4, 365 00:29:57,000 --> 00:30:04,000 - 5]. Any questions about that? 366 00:30:04,000 --> 00:30:08,000 OK. Now we come to part C which is 367 00:30:08,000 --> 00:30:13,000 the harder part of this problem. It says let's take this matrix 368 00:30:13,000 --> 00:30:17,000 A and let's replace the two in the upper right corner by some 369 00:30:17,000 --> 00:30:25,000 other number C. That means we will look at 1 3 370 00:30:25,000 --> 00:30:34,000 C; 2 0 - 1; 1 1 0|. And let's call that M. 371 00:30:34,000 --> 00:30:40,000 And it first asks you to find the value of C for which this 372 00:30:40,000 --> 00:30:48,000 matrix is not invertible. M is not invertible exactly 373 00:30:48,000 --> 00:30:53,000 when the determinant of M is zero. 374 00:30:53,000 --> 00:31:05,000 Let's compute the determinant. Well, we should do one times 375 00:31:05,000 --> 00:31:12,000 that smaller determinant, which is zero minus negative 376 00:31:12,000 --> 00:31:16,000 one, which is 1 times 1 minus three 377 00:31:16,000 --> 00:31:23,000 times that determinant, which is zero plus one is 1. 378 00:31:23,000 --> 00:31:28,000 And then we have plus C times the lower left determinant which 379 00:31:28,000 --> 00:31:31,000 is two times one minus zero is 2. 380 00:31:31,000 --> 00:31:39,000 That gives us one minus three is - 2 2C. 381 00:31:39,000 --> 00:31:47,000 That is zero when C equals 1. For C equals 1, 382 00:31:47,000 --> 00:31:53,000 this matrix is not invertible. For other values it is 383 00:31:53,000 --> 00:31:58,000 invertible. It goes on to say let's look at 384 00:31:58,000 --> 00:32:04,000 this value of C and let's look at the system Mx equals zero. 385 00:32:04,000 --> 00:32:14,000 I am going to put value one in there. 386 00:32:14,000 --> 00:32:20,000 Now, if we look at Mx equals zero, well, this has either no 387 00:32:20,000 --> 00:32:24,000 solution or infinitely many solutions. 388 00:32:24,000 --> 00:32:26,000 But here there is an obvious solution. 389 00:32:26,000 --> 00:32:30,000 Namely x equals zero is a solution. 390 00:32:30,000 --> 00:32:35,000 Maybe let me rewrite it more geometrically. 391 00:32:35,000 --> 00:32:44,000 X 3 y z = 0. 2x - z = 0. 392 00:32:44,000 --> 00:32:52,000 And x y = 0. You see we have an obvious 393 00:32:52,000 --> 00:32:54,000 solution, (0,0, 0). 394 00:32:54,000 --> 00:32:57,000 But we have more solutions. How do we find more solutions? 395 00:32:57,000 --> 00:33:01,000 Well, (x, y, z) is a solution if it is in 396 00:33:01,000 --> 00:33:06,000 all three of these planes. That is a way to think about it. 397 00:33:06,000 --> 00:33:13,000 Probably we are actually in this situation where, 398 00:33:13,000 --> 00:33:20,000 in fact, we have three planes that are all passing through the 399 00:33:20,000 --> 00:33:25,000 origin and all parallel to the same line. 400 00:33:25,000 --> 00:33:28,000 And so that would be the line of solutions. 401 00:33:28,000 --> 00:33:32,000 To find it actually we can think of this as follows. 402 00:33:32,000 --> 00:33:37,000 The first observation is that actually in this situation we 403 00:33:37,000 --> 00:33:41,000 don't need all three equations. The fact that the system has 404 00:33:41,000 --> 00:33:45,000 infinitely many solutions means that actually one of the 405 00:33:45,000 --> 00:33:49,000 equations is redundant. If you look at it long enough 406 00:33:49,000 --> 00:33:51,000 you will see, for example, 407 00:33:51,000 --> 00:33:55,000 if you multiply three times this equation and you subtract 408 00:33:55,000 --> 00:33:58,000 that one then you will get the first equation. 409 00:33:58,000 --> 00:34:05,000 Three times (x y) - (2x - z) will be x 3y z. 410 00:34:05,000 --> 00:34:08,000 Now, we don't actually need to see that to solve a problem. 411 00:34:08,000 --> 00:34:10,000 I am just showing you that is what happens when you have a 412 00:34:10,000 --> 00:34:14,000 matrix with determinant zero. One of the equations is somehow 413 00:34:14,000 --> 00:34:19,000 a duplicate of the others. We don't actually need to 414 00:34:19,000 --> 00:34:24,000 figure out how exactly. What that means is really we 415 00:34:24,000 --> 00:34:28,000 want to solve, let's say start with two of the 416 00:34:28,000 --> 00:34:33,000 equations. To find the solution we can 417 00:34:33,000 --> 00:34:41,000 observe that the first equation says actually that 00:34:45,000 y, z> dot-product with 419 00:34:45,000 --> 00:34:46,000 420 00:34:46,000 --> 00:34:51,000 =0. And the second equation says 421 00:34:51,000 --> 00:34:52,000 422 00:34:52,000 --> 00:34:57,000 dot-product with <2,0,- 1> is zero. 423 00:34:57,000 --> 00:35:03,000 And the third equation, if we really want to keep it, 424 00:35:03,000 --> 00:35:08,000 says we should be also having this. 425 00:35:08,000 --> 00:35:11,000 Now, these equations now written like this, 426 00:35:11,000 --> 00:35:15,000 they are just saying we want an x, y, z that is perpendicular to 427 00:35:15,000 --> 00:35:19,000 these vectors. Let's forget this one and let's 428 00:35:19,000 --> 00:35:23,000 just look at these two. They are saying we want a 429 00:35:23,000 --> 00:35:27,000 vector that is perpendicular to these two given vectors. 430 00:35:27,000 --> 00:35:35,000 How do we find that? We do the cross-product. 431 00:35:35,000 --> 00:35:43,000 To find x, y, z perpendicular to <1,3, 432 00:35:43,000 --> 00:35:51,000 1> and <2,0, - 1>, we take the 433 00:35:51,000 --> 00:35:56,000 cross-product. And that will give us 434 00:35:56,000 --> 00:35:58,000 something. Well, let me just give you the 435 00:35:58,000 --> 00:36:00,000 answer. I am sure you know how to do 436 00:36:00,000 --> 00:36:07,000 cross-products by now. I don't have the answer here, 437 00:36:07,000 --> 00:36:18,000 so I guess I have to do it. That should be <- 3, 438 00:36:18,000 --> 00:36:26,000 probably positive 3, and then - 6>. 439 00:36:26,000 --> 00:36:29,000 That is the solution. And any multiple of that is a 440 00:36:29,000 --> 00:36:31,000 solution. If you like to neatly simplify 441 00:36:31,000 --> 00:36:34,000 them you could say negative one, one, negative two. 442 00:36:34,000 --> 00:36:37,000 If you like larger numbers you can multiply that by a million. 443 00:36:37,000 --> 00:36:50,000 That is also a solution. Any questions about that? 444 00:36:50,000 --> 00:36:56,000 Yes? That is correct. 445 00:36:56,000 --> 00:37:00,000 If you pick these two guys instead, you will get the same 446 00:37:00,000 --> 00:37:02,000 solution. Well, up to a multiple. 447 00:37:02,000 --> 00:37:06,000 It could be if you do the cross-product of these two guys 448 00:37:06,000 --> 00:37:10,000 you actually get something that is a multiple -- Actually, 449 00:37:10,000 --> 00:37:14,000 I think if you do the cross-product of the first and 450 00:37:14,000 --> 00:37:17,000 third one you will get actually minus one, one, 451 00:37:17,000 --> 00:37:20,000 minus two, the smaller one. But it doesn't matter. 452 00:37:20,000 --> 00:37:22,000 I mean it is really in the same direction. 453 00:37:22,000 --> 00:37:27,000 This is all because a plane has actually normal vectors of all 454 00:37:27,000 --> 00:37:32,000 sizes. Yes? 455 00:37:32,000 --> 00:37:35,000 I don't think so because -- An important thing to remember 456 00:37:35,000 --> 00:37:38,000 about cross-product is we compute for minors, 457 00:37:38,000 --> 00:37:40,000 but then we put a minus sign on the second component. 458 00:37:40,000 --> 00:37:44,000 The coefficient of j in here, the second component, 459 00:37:44,000 --> 00:37:47,000 you do one times minus two times one. 460 00:37:47,000 --> 00:37:51,000 That is negative three indeed. But then you actually change 461 00:37:51,000 --> 00:37:54,000 that to a positive three. Yes? 462 00:37:54,000 --> 00:38:14,000 463 00:38:14,000 --> 00:38:21,000 Well, we don't have parametric equations here. 464 00:38:21,000 --> 00:38:25,000 Oh, solving by elimination. Well, if it says that you have 465 00:38:25,000 --> 00:38:29,000 to use vector methods then you should use vector methods. 466 00:38:29,000 --> 00:38:32,000 If it says you should use vectors and matrices then you 467 00:38:32,000 --> 00:38:41,000 are expected to do it that way. Yes? 468 00:38:41,000 --> 00:38:43,000 It depends what the problem is asking. 469 00:38:43,000 --> 00:38:45,000 The question is, is it enough to find the 470 00:38:45,000 --> 00:38:47,000 components of a vector or do we have to find the equation of a 471 00:38:47,000 --> 00:38:50,000 line? Here it says find one solution 472 00:38:50,000 --> 00:38:55,000 using vector operations. We have found one solution. 473 00:38:55,000 --> 00:38:58,000 If you wanted to find the line then it would all the things 474 00:38:58,000 --> 00:39:04,000 that are proportional to this. It would be maybe minus 3t, 475 00:39:04,000 --> 00:39:09,000 3t minus 6t, all the multiples of that 476 00:39:09,000 --> 00:39:12,000 vector. We do because (0,0, 477 00:39:12,000 --> 00:39:17,000 0) is an obvious solution. Maybe I should write that on 478 00:39:17,000 --> 00:39:19,000 the board. You had another question? 479 00:39:19,000 --> 00:39:28,000 480 00:39:28,000 --> 00:39:31,000 Not quite. Let me re-explain first how we 481 00:39:31,000 --> 00:39:35,000 get all the solutions and why I did that cross-product. 482 00:39:35,000 --> 00:40:08,000 483 00:40:08,000 --> 00:40:09,000 First of all, why did I take that 484 00:40:09,000 --> 00:40:12,000 cross-product again? I took that cross-product 485 00:40:12,000 --> 00:40:17,000 because I looked at my three equations and I observed that my 486 00:40:17,000 --> 00:40:21,000 three equations can be reformulated in terms of these 487 00:40:21,000 --> 00:40:25,000 dot-products saying that x, y, z is actually perpendicular 488 00:40:25,000 --> 00:40:29,000 these guys and these guys have normal vectors to the planes. 489 00:40:29,000 --> 00:40:33,000 Remember, to be in all three planes it has to be 490 00:40:33,000 --> 00:40:36,000 perpendicular to the normal vectors. 491 00:40:36,000 --> 00:40:40,000 That is how we got here. And now, if we want something 492 00:40:40,000 --> 00:40:43,000 that is perpendicular to a bunch of given vectors, 493 00:40:43,000 --> 00:40:45,000 well, to be perpendicular to two vectors, 494 00:40:45,000 --> 00:40:48,000 an easy way to find one is to take that cross-product. 495 00:40:48,000 --> 00:40:51,000 And, if you take any two of them, you will get something 496 00:40:51,000 --> 00:40:54,000 that is the same up to scaling. Now, what it means 497 00:40:54,000 --> 00:41:00,000 geometrically is that when we have our three planes and they 498 00:41:00,000 --> 00:41:06,000 all actually contain the same line -- And we know that is 499 00:41:06,000 --> 00:41:11,000 actually the smae case because they all pass through the 500 00:41:11,000 --> 00:41:16,000 origin. They pass through the origin 501 00:41:16,000 --> 00:41:20,000 because the constant terms are just zero. 502 00:41:20,000 --> 00:41:26,000 What happens is that the normal vectors to these planes are, 503 00:41:26,000 --> 00:41:29,000 in fact, all perpendicular to that line. 504 00:41:29,000 --> 00:41:40,000 The normal vectors -- Say this line is vertical. 505 00:41:40,000 --> 00:41:45,000 The normal vectors are all going to be horizontal. 506 00:41:45,000 --> 00:41:49,000 Well, it is kind of hard to draw. 507 00:41:49,000 --> 00:41:53,000 By taking the cross-product between two normal vectors we 508 00:41:53,000 --> 00:42:01,000 found this direction. Now, to find actually all the 509 00:42:01,000 --> 00:42:06,000 solutions. What we know so far is that we 510 00:42:06,000 --> 00:42:11,000 have this direction <-3 3 - 6>. 511 00:42:11,000 --> 00:42:14,000 That is going to be parallel to the line of intersections. 512 00:42:14,000 --> 00:42:19,000 Let me do it here, for example, 513 00:42:19,000 --> 00:42:29,000 . Now we have one particular 514 00:42:29,000 --> 00:42:34,000 solution. 0,0, 0. 515 00:42:34,000 --> 00:42:39,000 Actually, we have found another one, too, which is <- 3,3, 516 00:42:39,000 --> 00:42:46,000 - 6>. Anyway, if a line of solutions 517 00:42:46,000 --> 00:42:55,000 -- -- has parametric equation x = - 3t, y = 3t, 518 00:42:55,000 --> 00:43:04,000 z = - 6t, anything proportional to that. 519 00:43:04,000 --> 00:43:07,000 That is how we would find all the solutions if we wanted them. 520 00:43:07,000 --> 00:43:18,000 521 00:43:18,000 --> 00:43:22,000 It is almost time. I think I need to jump ahead to 522 00:43:22,000 --> 00:43:24,000 other problems. Let's see. 523 00:43:24,000 --> 00:43:29,000 I think problem 4 you can probably find for yourselves. 524 00:43:29,000 --> 00:43:32,000 It is a reasonably straightforward parametric 525 00:43:32,000 --> 00:43:35,000 equation problem. You just have to find the 526 00:43:35,000 --> 00:43:39,000 coordinates of point P. And for that it is a very 527 00:43:39,000 --> 00:43:41,000 simple trick. Problem 5. 528 00:43:41,000 --> 00:43:43,000 Find the area of a spaced triangle. 529 00:43:43,000 --> 00:43:47,000 It sounds like a cross-product. Find the equation of a plane 530 00:43:47,000 --> 00:43:50,000 also sounds like a cross-product. 531 00:43:50,000 --> 00:43:53,000 And find the intersection of this plane with a line means we 532 00:43:53,000 --> 00:43:56,000 find first the parametric equation of the line and then we 533 00:43:56,000 --> 00:43:59,000 plug that into the equation of the plane to get where they 534 00:43:59,000 --> 00:44:02,000 intersect. Does that sound reasonable? 535 00:44:02,000 --> 00:44:06,000 Who is disparate about problem 5? 536 00:44:06,000 --> 00:44:15,000 OK. Let me repeat problem 5. First part we need to find the 537 00:44:15,000 --> 00:44:19,000 area of a triangle. And the way to do that is to 538 00:44:19,000 --> 00:44:22,000 just do one-half the length of a cross-product. 539 00:44:22,000 --> 00:44:32,000 If we have three points, P0, P1, P2 then maybe we can 540 00:44:32,000 --> 00:44:38,000 form vectors P0P1 and P0P2. And, if we take that 541 00:44:38,000 --> 00:44:42,000 cross-product and take the length of that and divide by 542 00:44:42,000 --> 00:44:46,000 two, that will give us the area of a triangle. 543 00:44:46,000 --> 00:44:52,000 Here it turns out that this guy is , 544 00:44:52,000 --> 00:44:59,000 if I look at the solutions, so you will end up with square 545 00:44:59,000 --> 00:45:06,000 root of 6 over 2. The second is asking you for 546 00:45:06,000 --> 00:45:13,000 the equation of a plane containing these three points. 547 00:45:13,000 --> 00:45:23,000 Well, first of all, we know that a normal vector to 548 00:45:23,000 --> 00:45:34,000 the plane is going to be given by this cross-product again. 549 00:45:34,000 --> 00:45:40,000 That means that the equation of plane will be of a form x plus y 550 00:45:40,000 --> 00:45:45,000 plus 2z equals something. If a coefficient is here it 551 00:45:45,000 --> 00:45:49,000 comes from the normal vector. And to find what goes in the 552 00:45:49,000 --> 00:45:51,000 right-hand side, we just plug in any of the 553 00:45:51,000 --> 00:45:55,000 points. If you plug in P0, 554 00:45:55,000 --> 00:46:03,000 which is (2,1, 0) then two plus one seems like 555 00:46:03,000 --> 00:46:06,000 it is 3. And, if you want to 556 00:46:06,000 --> 00:46:08,000 double-check your answer, you can take P1 and P2 and 557 00:46:08,000 --> 00:46:15,000 check that you also get three. It is a good way to check your 558 00:46:15,000 --> 00:46:18,000 answer. Then the third part. 559 00:46:18,000 --> 00:46:25,000 We have a line parallel to the vector v equals one, 560 00:46:25,000 --> 00:46:31,000 one, one through the point S, which is (- 1,0, 561 00:46:31,000 --> 00:46:34,000 0). That means you can find its 562 00:46:34,000 --> 00:46:37,000 parametric equation. X will start at - 1, 563 00:46:37,000 --> 00:46:41,000 increases at rate 1. Y starts at zero, 564 00:46:41,000 --> 00:46:44,000 increases at rate one. Z starts at zero, 565 00:46:44,000 --> 00:46:48,000 increases at rate one. You plug these into the plane 566 00:46:48,000 --> 00:46:52,000 equation, and that will tell you where they intersect. 567 00:46:52,000 --> 00:46:58,000 Is that clear? And now, in the last one 568 00:46:58,000 --> 00:47:03,000 minute, on that side I have one minute, 569 00:47:03,000 --> 00:47:08,000 let me just say very quickly -- Well, 570 00:47:08,000 --> 00:47:10,000 do you want to hear about problem 6 anyway very quickly? 571 00:47:10,000 --> 00:47:20,000 Yeah. OK. Problem 6 is one of these like 572 00:47:20,000 --> 00:47:24,000 vector calculations. It says we have a position 573 00:47:24,000 --> 00:47:28,000 vector R. And it asks you how do we find 574 00:47:28,000 --> 00:47:32,000 the derivative of R dot R? Well, remember we have a 575 00:47:32,000 --> 00:47:34,000 product rule for taking the derivative. 576 00:47:34,000 --> 00:47:37,000 UV prime is U prime V plus UV prime. 577 00:47:37,000 --> 00:47:44,000 It also applies for dot-product. That is dR by dt dot R plus R 578 00:47:44,000 --> 00:47:49,000 dot dR by dt. And these are both the same 579 00:47:49,000 --> 00:47:52,000 thing. You get two R dot dR/dt, 580 00:47:52,000 --> 00:47:57,000 but dR/dt is v for velocity vector. 581 00:47:57,000 --> 00:48:00,000 Hopefully you have seen things like that. 582 00:48:00,000 --> 00:48:04,000 Now, it says show that if R has constant length then they are 583 00:48:04,000 --> 00:48:10,000 perpendicular. All you need to write basically 584 00:48:10,000 --> 00:48:15,000 is we assume length R is constant. 585 00:48:15,000 --> 00:48:18,000 That is what it says, R has constant length. 586 00:48:18,000 --> 00:48:21,000 Well, how do we get to, say, something we probably want 587 00:48:21,000 --> 00:48:26,000 to reduce to that? Well, if R is constant in 588 00:48:26,000 --> 00:48:32,000 length then R dot R is also constant. 589 00:48:32,000 --> 00:48:39,000 And so that means d by dt of R dot R is zero. 590 00:48:39,000 --> 00:48:41,000 That is what it means to be constant. 591 00:48:41,000 --> 00:48:46,000 And so that means R dot v is zero. 592 00:48:46,000 --> 00:48:52,000 That means R is perpendicular to v. 593 00:48:52,000 --> 00:48:58,000 That is a proof. It is not a scary proof. 594 00:48:58,000 --> 00:49:04,000 And then the last question of the exam says let's continue to 595 00:49:04,000 --> 00:49:10,000 assume that R has constant length, and let's try to find R 596 00:49:10,000 --> 00:49:13,000 dot v. If there is acceleration then 597 00:49:13,000 --> 00:49:17,000 probably we should bring it in somewhere, maybe by taking a 598 00:49:17,000 --> 00:49:21,000 derivative of something. If we know that R dot v equals 599 00:49:21,000 --> 00:49:24,000 zero, let's take the derivative of that. 600 00:49:24,000 --> 00:49:32,000 That is still zero. But now, using the product 601 00:49:32,000 --> 00:49:40,000 rule, dR/dt is v dot v plus R dot dv/dt is going to be zero. 602 00:49:40,000 --> 00:49:43,000 That means that you are asked about R dot A. 603 00:49:43,000 --> 00:49:48,000 Well, that is equal to minus V dot V. 604 00:49:48,000 --> 00:49:50,000 And that is it. 605 00:49:50,000 --> 00:49:55,000