1 00:00:01 --> 00:00:03 The following content is provided under a Creative 2 00:00:03 --> 00:00:05 Commons license. Your support will help MIT 3 00:00:05 --> 00:00:08 OpenCourseWare continue to offer high quality educational 4 00:00:08 --> 00:00:13 resources for free. To make a donation or to view 5 00:00:13 --> 00:00:18 additional materials from hundreds of MIT courses, 6 00:00:18 --> 00:00:23 visit MIT OpenCourseWare at ocw.mit.edu. 7 00:00:24 --> 00:00:28 The other thing is the last lecture before the break ended a 8 00:00:28 --> 00:00:31 bit abruptly because I ran out of time, 9 00:00:31 --> 00:00:34 so just to summarize what the main point was, 10 00:00:34 --> 00:00:37 I mean probably you figured this out if you looked at the 11 00:00:37 --> 00:00:39 notes. It is not that important, 12 00:00:39 --> 00:00:43 but anyway. I just wanted to remind you, 13 00:00:43 --> 00:00:48 just to clarify what happened at the end, we got the diffusion 14 00:00:48 --> 00:00:52 equation from two bits of information. 15 00:00:52 --> 00:00:55 I mean the unknown, in this partial differential 16 00:00:55 --> 00:00:59 equation, is a function that we call u that corresponds to the 17 00:00:59 --> 00:01:04 concentration of some substance. And we used a vector field that 18 00:01:04 --> 00:01:12 represents the flow of whatever the substance is whose diffusion 19 00:01:12 --> 00:01:17 we are studying. And so we got two bits of 20 00:01:17 --> 00:01:25 information, one that came from physics that said the flow goes 21 00:01:25 --> 00:01:31 from high concentration to smaller concentration. 22 00:01:31 --> 00:01:37 And that told us that the flow is proportional to a negative 23 00:01:37 --> 00:01:42 gradient of a concentration. And the second piece of 24 00:01:42 --> 00:01:46 information that we got was from the divergence theorem, 25 00:01:46 --> 00:01:50 and that was the one I spent time trying to explain. 26 00:01:50 --> 00:01:57 And that one told us that the divergence of F is actually 27 00:01:57 --> 00:02:02 negative partial u over partial t. 28 00:02:02 --> 00:02:07 When you combine these two relations together that is how 29 00:02:07 --> 00:02:11 you get the diffusion equation. Sorry, I should say this is not 30 00:02:11 --> 00:02:12 the statement of a divergence theorem. 31 00:02:12 --> 00:02:17 This is something that would derive from it with quite a few 32 00:02:17 --> 00:02:21 steps involved. And so what we got out of that 33 00:02:21 --> 00:02:27 is a diffusion equation because we end up getting that partial u 34 00:02:27 --> 00:02:33 over partial t is minus div F, which is therefore positive k 35 00:02:33 --> 00:02:39 times divergence of grad u, which is what we denoted by del 36 00:02:39 --> 00:02:42 square u, the Laplacian. 37 00:02:42 --> 00:02:59 So, that is how we got the diffusion equation. 38 00:02:59 --> 00:03:04 Anyway, I will let you have a look at the notes that were 39 00:03:04 --> 00:03:08 handed out in case you really want to see more. 40 00:03:08 --> 00:03:16 I just wanted to give the missing part of the last 41 00:03:16 --> 00:03:19 lecture. Let me just switch gears 42 00:03:19 --> 00:03:24 completely and switch to today's topic, which is line integrals 43 00:03:24 --> 00:03:27 and work in 3D. That is going to look a lot 44 00:03:27 --> 00:03:29 like what we did in the plane, except, of course, 45 00:03:29 --> 00:03:32 there is a z coordinate. You will see it doesn't change 46 00:03:32 --> 00:03:35 things much when it comes to computing a line integral. 47 00:03:35 --> 00:03:39 It changes things quite a bit, however, when it comes to 48 00:03:39 --> 00:03:42 testing whether a field is a gradient field. 49 00:03:42 --> 00:03:45 That is why we have to be more careful. 50 00:03:45 --> 00:03:57 Let's start right away with line integrals in space. 51 00:03:57 --> 00:04:08 Let's say that we have a vector field F with components P, 52 00:04:08 --> 00:04:12 Q and R. We should think of it maybe as 53 00:04:12 --> 00:04:18 representing a force. And let's say that we have a 54 00:04:18 --> 00:04:24 curve C in space. Then the work done by the field 55 00:04:24 --> 00:04:29 will be the line integral along C of F dot dr. 56 00:04:29 --> 00:04:34 That is a familiar formula. And what we do with that 57 00:04:34 --> 00:04:37 formula is also familiar, except now, of course, 58 00:04:37 --> 00:04:44 we have a z coordinate. We are going to think of vector 59 00:04:44 --> 00:04:50 dr as a space vector with components dx, 60 00:04:50 --> 00:04:56 dy and dz. When we do the dot product of F 61 00:04:56 --> 00:05:04 with dr that will tell us that we have to integrate Pdx Qdy 62 00:05:04 --> 00:05:07 Rdz. But it is still a line integral 63 00:05:07 --> 00:05:12 so it is still going to turn into a single integral when you 64 00:05:12 --> 00:05:16 plug in the correct values. So the method will be exactly 65 00:05:16 --> 00:05:19 the same as in the plane, namely we will find some way to 66 00:05:19 --> 00:05:22 parameterize our curve, x plus x, y z in terms of a 67 00:05:22 --> 00:05:30 single variable, and then we will integrate with 68 00:05:30 --> 00:05:44 respect to that variable. The way that we evaluate is by 69 00:05:44 --> 00:05:56 parameterizing C and express x, y, z, dx, dy, 70 00:05:56 --> 00:06:07 dz in terms of the parameter. Let's do an example just to 71 00:06:07 --> 00:06:09 convince you that you actually know how to do this, 72 00:06:09 --> 00:06:12 or at least you should know how to do this. 73 00:06:12 --> 00:06:21 Let's say that I give you the vector field with components yz, 74 00:06:21 --> 00:06:27 xz and xy. And let's say that we have a 75 00:06:27 --> 00:06:33 curve given by x equals t^3, y equals t^2, 76 00:06:33 --> 00:06:40 z equals t for t going from zero to one. 77 00:06:40 --> 00:06:45 The way we will set up the line integral for the work done will 78 00:06:45 --> 00:06:49 be -- Well, sorry. Before we actually set up the 79 00:06:49 --> 00:06:51 line integral, we need to know how we will 80 00:06:51 --> 00:06:54 express everything in terms of t and dt. 81 00:06:54 --> 00:06:56 x, y and z, in terms of t, are given here. 82 00:06:56 --> 00:07:01 We just need to do also dx, dy and dz. 83 00:07:01 --> 00:07:05 By differentiating you get dx is 3t^2 dt. 84 00:07:05 --> 00:07:14 That is the derivative of t^3, dy will be 2t dt and dz will 85 00:07:14 --> 00:07:23 just be dt. And we will evaluate the line 86 00:07:23 --> 00:07:36 integral for work. That will be the integral of yz 87 00:07:36 --> 00:07:48 dx xz dy xy dz, which will become -- yz is t^3 88 00:07:48 --> 00:08:04 times dx is 3t^2 dt plus xz is t^4 times dy is 2t dt plus xy is 89 00:08:04 --> 00:08:08 t^5 dt. That just becomes the integral 90 00:08:08 --> 00:08:11 from, well, I guess t goes from zero to one, actually. 91 00:08:11 --> 00:08:14 And we are integrating three plus two plus one. 92 00:08:14 --> 00:08:22 That is 6t^5 dt which, I am sure you know, 93 00:08:22 --> 00:08:30 integrates to t^6, so we will just get one. 94 00:08:30 --> 00:08:33 It is the same method as usual. And if you are being given a 95 00:08:33 --> 00:08:36 geometric description of a curve then, of course, 96 00:08:36 --> 00:08:39 you will have to decide for yourself what the best parameter 97 00:08:39 --> 00:08:43 will be. It might be some time parameter 98 00:08:43 --> 00:08:45 t like here. It might be one of the 99 00:08:45 --> 00:08:49 coordinates. Here we could have used z as 100 00:08:49 --> 00:08:53 our parameter because, in fact, this curve is x equals 101 00:08:53 --> 00:08:57 x3 and y equals z2. And we could also have used 102 00:08:57 --> 00:08:59 maybe some angle. Well, not here, 103 00:08:59 --> 00:09:04 but if we had been moving on a circle or something like that. 104 00:09:04 --> 00:09:11 Any questions so far? No. 105 00:09:11 --> 00:09:26 OK. Well, 106 00:09:26 --> 00:09:31 because we can do a bit more practice, 107 00:09:31 --> 00:09:40 let's do another one where we do the same vector field F but 108 00:09:40 --> 00:09:49 our curve C will be going from the origin to the point (1,0, 109 00:09:49 --> 00:09:54 0) along the x-axis. Let's call that C1. 110 00:09:54 --> 00:10:02 Then to (1,1, 0). Let's call that C2 by moving 111 00:10:02 --> 00:10:08 parallel to the y-axis. And then up to (1,1, 112 00:10:08 --> 00:10:14 1) parallel to the z-axis, let's call that C3. 113 00:10:14 --> 00:10:19 I am sure that some of you at least are suspecting what I am 114 00:10:19 --> 00:10:23 getting at here, but let's not spoil it for 115 00:10:23 --> 00:10:27 those who don't see it yet. OK. 116 00:10:27 --> 00:10:33 If we want to compute the line integral along this guy then we 117 00:10:33 --> 00:10:38 have to break it into a sum of three terms. 118 00:10:38 --> 00:10:42 Well, maybe I should call that C', not C, because that is not 119 00:10:42 --> 00:10:48 the same C anymore. I want to do the sum of the 120 00:10:48 --> 00:10:55 line integrals along C1, C2 and C3. 121 00:10:55 --> 00:11:13 And, well, if I look at C1 and C2 they take place inside the x, 122 00:11:13 --> 00:11:19 y plane. In fact, you know that z will 123 00:11:19 --> 00:11:25 be zero and dz will also be zero on both of these. 124 00:11:25 --> 00:11:30 And if you just look at the formula for line integral in the 125 00:11:30 --> 00:11:34 role of yz dx plus xz dy plus xy dz, 126 00:11:34 --> 00:11:37 well, it looks like if you plug z equals zero and dz equals zero 127 00:11:37 --> 00:11:47 you will just get zero. These are actually very fast. 128 00:11:47 --> 00:12:00 Let me write it. This is going to be zero, 129 00:12:00 --> 00:12:04 this is going to be zero, this is going to zero, 130 00:12:04 --> 00:12:07 and we will get zero. Now, if we do C3, 131 00:12:07 --> 00:12:11 well, there we might have to do some calculation, 132 00:12:11 --> 00:12:18 but it won't be all that bad. C3, well, x and y are both 133 00:12:18 --> 00:12:22 equal to one. And, of course, 134 00:12:22 --> 00:12:26 because they are constant that means dx is zero and dy is zero. 135 00:12:26 --> 00:12:35 On the other hand, z varies from zero to one. 136 00:12:35 --> 00:12:42 If I look at the line integral on C3 -- The first two terms, 137 00:12:42 --> 00:12:48 yz, dx and xz dy go away because the dx and dy are zero, 138 00:12:48 --> 00:12:55 so I am just left with xy dz. But because x and y are one it 139 00:12:55 --> 00:13:01 is just the integral of dz from zero to one, and that will just 140 00:13:01 --> 00:13:07 end up being one. If add these numbers together, 141 00:13:07 --> 00:13:13 zero plus zero plus one, I get one again. 142 00:13:13 --> 00:13:15 And, of course, it is not a coincidence because 143 00:13:15 --> 00:13:18 this vector field is a gradient field. 144 00:13:18 --> 00:13:20 I am sure some of you have already figured out what it is 145 00:13:20 --> 00:13:24 the gradient of. Otherwise, we will figure it 146 00:13:24 --> 00:13:28 out together. And so that is why we get the 147 00:13:28 --> 00:13:33 same answer for these two paths going both from the origin to 148 00:13:33 --> 00:13:36 (1,1, 1). Maybe I should point out, 149 00:13:36 --> 00:13:40 to make it clear, that if you plug t equals zero 150 00:13:40 --> 00:13:42 in up there you will get (0,0, 0). 151 00:13:42 --> 00:13:45 If you plug t equals one, you will get (1,1, 152 00:13:45 --> 00:13:56 1). In fact -- F that we have here 153 00:13:56 --> 00:14:11 happens to be conservative. And, if you plug the two curves 154 00:14:11 --> 00:14:20 together -- Well, I am not really sure if I know 155 00:14:20 --> 00:14:28 how to plot this correctly. It is not exactly how it looks. 156 00:14:28 --> 00:14:33 Whatever. The first curve C goes from the 157 00:14:33 --> 00:14:39 origin to this point, and so does C ', 158 00:14:39 --> 00:14:45 just in a slightly more roundabout way. 159 00:14:45 --> 00:14:51 They both go from the origin to (1,1, 1). 160 00:14:51 --> 00:15:00 It is not a surprise that you will get the same answer for 161 00:15:00 --> 00:15:05 both line integrals. And how do we see that? 162 00:15:05 --> 00:15:11 Well, actually here it is not very hard to find a function 163 00:15:11 --> 00:15:15 whose gradient is this vector field. 164 00:15:15 --> 00:15:21 Namely, the gradient of x, y, z looks like it should be 165 00:15:21 --> 00:15:26 exactly what we want. If you take partial of this 166 00:15:26 --> 00:15:29 with respect to x, you will get yz, 167 00:15:29 --> 00:15:33 then with respect to y, xz, and with respect to z, 168 00:15:33 --> 00:15:35 xy. And so, in fact, 169 00:15:35 --> 00:15:38 what was the easier way to compute these line integrals was 170 00:15:38 --> 00:15:41 to use the fundamental theorem of calculus. 171 00:15:41 --> 00:15:45 Once we have this remark, we don't need to compute these 172 00:15:45 --> 00:15:56 line integrals anymore. We can just use the fundamental 173 00:15:56 --> 00:16:08 theorem. If we know this fundamental 174 00:16:08 --> 00:16:20 theorem -- -- for line integrals, 175 00:16:20 --> 00:16:27 that tells us that the line integral of a gradient field is 176 00:16:27 --> 00:16:33 equal to the value of the potential at the final point 177 00:16:33 --> 00:16:40 minus the value of the potential at the starting point. 178 00:16:40 --> 00:16:43 And that, of course, only applies if you have a 179 00:16:43 --> 00:16:46 potential. So, in particular, 180 00:16:46 --> 00:16:52 only if you have a conservative field, a gradient field. 181 00:16:52 --> 00:16:58 Here, in our example, we have to look at, 182 00:16:58 --> 00:17:06 let's call little f of x, y, z that potential xyz, 183 00:17:06 --> 00:17:12 then we take f(1,1, 1) - f(0,0, 0). 184 00:17:12 --> 00:17:17 And that indeed is one minus zero which is one. 185 00:17:17 --> 00:17:24 Everything is consistent. All this stuff so far works 186 00:17:24 --> 00:17:32 exactly as in the plane. Any questions? 187 00:17:32 --> 00:17:35 No. OK. 188 00:17:35 --> 00:17:40 Let's try to see where things do get a little bit different. 189 00:17:40 --> 00:17:44 And the first such place is when we try to test whether a 190 00:17:44 --> 00:17:47 vector field is a gradient field. 191 00:17:47 --> 00:17:53 Remember when we had a vector field in the plane, 192 00:17:53 --> 00:17:56 to know whether it was a gradient of a function of two 193 00:17:56 --> 00:17:59 variables we just had to check one condition, 194 00:17:59 --> 00:18:04 N sub x equals M sub y. Now we actually have three 195 00:18:04 --> 00:18:06 different conditions to check, and that means, 196 00:18:06 --> 00:18:07 of course, more work. 197 00:18:07 --> 00:18:23 198 00:18:23 --> 00:18:34 OK. So what is our test for 199 00:18:34 --> 00:18:44 gradient fields? We want to know whether a given 200 00:18:44 --> 00:18:50 vector field with components P, Q and R can be written as f sub 201 00:18:50 --> 00:18:55 x, f sub y and f sub z for a same function F. 202 00:18:55 --> 00:18:59 And for that to possibly happen, well, 203 00:18:59 --> 00:19:03 we need certainly some relations between P, 204 00:19:03 --> 00:19:05 Q and R. And, as before, 205 00:19:05 --> 00:19:11 this comes from the fact that the mixed second derivatives are 206 00:19:11 --> 00:19:16 the same, no matter in which order you take them. 207 00:19:16 --> 00:19:21 If that is the case then I can compute f sub xy, 208 00:19:21 --> 00:19:28 which is the same as f sub yx in two different ways. 209 00:19:28 --> 00:19:35 F sub xy should be P sub y. F sub yx, well, 210 00:19:35 --> 00:19:39 since f sub y is Q, that should be Q sub x. 211 00:19:39 --> 00:19:43 That is a part of a criterion that we already had when we had 212 00:19:43 --> 00:19:46 only two variables. But now, of course, 213 00:19:46 --> 00:19:50 we need to do the same thing when we look at x and z or y and 214 00:19:50 --> 00:19:53 z. That gives us two more 215 00:19:53 --> 00:19:58 conditions. P sub z is f sub xz, 216 00:19:58 --> 00:20:10 which is the same as f sub zx, so it should be the same as R 217 00:20:10 --> 00:20:14 sub x. Finally, Q sub z, 218 00:20:14 --> 00:20:22 which is f sub yz, equals f sub zy equals R sub y. 219 00:20:22 --> 00:20:33 We have three conditions, so our criterion -- Vector 220 00:20:33 --> 00:20:41 field F equals . 221 00:20:41 --> 00:20:46 And here, to be completely truthful, I have to say defined 222 00:20:46 --> 00:20:50 in a simply connected region. Otherwise, we might have the 223 00:20:50 --> 00:20:53 same kind of strange things happening as before. 224 00:20:53 --> 00:21:03 Let's not worry too much about it. 225 00:21:03 --> 00:21:07 For accuracy we need our vector field to be defined in a simply 226 00:21:07 --> 00:21:10 connected region. And example is just if it is 227 00:21:10 --> 00:21:14 defined everywhere. If you don't have any evil 228 00:21:14 --> 00:21:21 eliminators then you can just go ahead and there is no problem. 229 00:21:21 --> 00:21:36 It is a gradient field. We need three conditions. 230 00:21:36 --> 00:21:43 Let's do it in order. P sub y equals Q sub x. 231 00:21:43 --> 00:21:52 And we have P sub z equals R sub x and Q sub z equals R sub 232 00:21:52 --> 00:21:55 y. How do you remember these three 233 00:21:55 --> 00:21:56 conditions? Well, it is pretty easy. 234 00:21:56 --> 00:22:01 You pick any two components, say the x and the z component, 235 00:22:01 --> 00:22:04 and you take the partial of the x component with respect to z, 236 00:22:04 --> 00:22:07 the partial of the z component with respect to x and you must 237 00:22:07 --> 00:22:12 make them equal. And the same with every pair of 238 00:22:12 --> 00:22:15 variables. In fact, if you had a function 239 00:22:15 --> 00:22:17 of many more variables the criterion would still look 240 00:22:17 --> 00:22:21 exactly like that. For every pair of components 241 00:22:21 --> 00:22:24 the mixed partials must be the same. 242 00:22:24 --> 00:22:27 But we are not going to go beyond three variables so you 243 00:22:27 --> 00:22:32 don't need to know that. This you need to know so let me 244 00:22:32 --> 00:22:33 box it. 245 00:22:33 --> 00:22:51 246 00:22:51 --> 00:23:02 That is pretty straightforward. Let's do an example just to see 247 00:23:02 --> 00:23:11 how it goes. By the way, we can also think 248 00:23:11 --> 00:23:19 of it in terms of differentials. Before I do the example, 249 00:23:19 --> 00:23:22 let me just say in a different language. 250 00:23:22 --> 00:23:29 If we have a differential given to us of a form Pdx Qdy Rdz is 251 00:23:29 --> 00:23:33 going to be an exact differential, 252 00:23:33 --> 00:23:40 which means it is equal to df for some function F exactly and 253 00:23:40 --> 00:23:43 of the same conditions. That is the same thing. 254 00:23:43 --> 00:23:51 Just in the language of differentials. 255 00:23:51 --> 00:23:57 The example that I promised. Of course, I could do again the 256 00:23:57 --> 00:24:00 same one over there and check that it satisfies the condition, 257 00:24:00 --> 00:24:02 but then it wouldn't be much fun. 258 00:24:02 --> 00:24:10 So let's do a better one. Actually, let's do it in a way 259 00:24:10 --> 00:24:18 that looks like an exam problem. Let's say for which a and b is 260 00:24:18 --> 00:24:29 a xy dx plus -- Oh, it is not going to fit here. 261 00:24:29 --> 00:24:39 But it will fit here. a xy dx ( x^2 z^3) dy (byz^2 - 262 00:24:39 --> 00:24:51 4z^3) dz, an exact differential. Or, if you don't like exact 263 00:24:51 --> 00:24:54 differentials, for which a and b is the 264 00:24:54 --> 00:24:58 corresponding vector field with i, j and k instead, 265 00:24:58 --> 00:25:02 a gradient field. Let's just apply the criterion. 266 00:25:02 --> 00:25:06 And, of course, you can guess that what will 267 00:25:06 --> 00:25:11 follow is figuring out how to find the potential when there is 268 00:25:11 --> 00:25:12 one. 269 00:25:12 --> 00:25:33 270 00:25:33 --> 00:25:40 Let's do it one by one. We want to compare P sub y with 271 00:25:40 --> 00:25:45 Q sub x, we want to compare P sub z with 272 00:25:45 --> 00:25:53 R sub x and we want to compare Q sub z with R sub y where we call 273 00:25:53 --> 00:25:57 P, Q and R these guys. Let's see. 274 00:25:57 --> 00:26:04 What is P sub y? That seems to be ax. 275 00:26:04 --> 00:26:09 What is Q sub x? 2x. 276 00:26:09 --> 00:26:12 Q is this one. Actually, let me write them 277 00:26:12 --> 00:26:16 down. Because otherwise I am going to 278 00:26:16 --> 00:26:21 get confused myself. This guy here, 279 00:26:21 --> 00:26:32 that is P, this guy here, that is Q and that guy here, 280 00:26:32 --> 00:26:38 that is R. This one tells us that a should 281 00:26:38 --> 00:26:43 be equal to two of the first product that you hold. 282 00:26:43 --> 00:26:49 OK. Let's look at P sub z. That is just zero. 283 00:26:49 --> 00:26:51 R sub x? Well, R doesn't have any x 284 00:26:51 --> 00:26:56 either so that is zero. This one is not a problem. 285 00:26:56 --> 00:27:02 Q sub z? Well, that seems to be 3z2. 286 00:27:02 --> 00:27:14 R sub y seems to be bz2, so b should be equal to three. 287 00:27:14 --> 00:27:19 We need to have a equals two and, this is an and, 288 00:27:19 --> 00:27:23 not or, b equals three for this to be exact. 289 00:27:23 --> 00:27:27 For those values of a and b, we can look for a potential 290 00:27:27 --> 00:27:30 using the method that we are going to see right now. 291 00:27:30 --> 00:27:33 For any other values of a and b we cannot. 292 00:27:33 --> 00:27:38 If we have to compute a line integral, we have to do it by 293 00:27:38 --> 00:27:42 finding a parameter and setting up everything. 294 00:27:42 --> 00:27:57 Any questions at this point? Yes? 295 00:27:57 --> 00:28:00 I see. Well, if I got the same answer, 296 00:28:00 --> 00:28:05 oh, did say bz^2 or 3bz^2? Well, 3bz^2, for example, 297 00:28:05 --> 00:28:09 I would need b to be zero because the only time that 3bz2 298 00:28:09 --> 00:28:13 equals bz2 as not just at one point but everywhere, 299 00:28:13 --> 00:28:15 I need them to be the same function of x, 300 00:28:15 --> 00:28:18 y, z. Well, if a coefficient of z2 is 301 00:28:18 --> 00:28:22 the same that would be give b equals 3b, that would give me b 302 00:28:22 --> 00:28:25 equals zero. If you got bz2 on both sides 303 00:28:25 --> 00:28:29 then it would mean for any value of b it works, 304 00:28:29 --> 00:28:35 and you wouldn't have to worry about what the value of b is. 305 00:28:35 --> 00:28:41 Any other questions? No. 306 00:28:41 --> 00:28:50 OK. Now, how do we find the 307 00:28:50 --> 00:28:58 potential? Well, there are two methods as 308 00:28:58 --> 00:28:59 before. One of them, 309 00:28:59 --> 00:29:02 I don't remember if it was the first one or the second one last 310 00:29:02 --> 00:29:04 time, but it really doesn't matter. 311 00:29:04 --> 00:29:10 One of them was just to say that the value of F at the 312 00:29:10 --> 00:29:15 point, let me call that x1, y1, z1, 313 00:29:15 --> 00:29:22 is equal to the line integral of my field along a well-chosen 314 00:29:22 --> 00:29:25 curve plus, of course, a constant, 315 00:29:25 --> 00:29:30 which is going to be the integration constant. 316 00:29:30 --> 00:29:39 And the kind of curve that I will take to do this calculation 317 00:29:39 --> 00:29:48 will just be my favorite curve going from the origin to the 318 00:29:48 --> 00:29:51 point x1, y1, z1. 319 00:29:51 --> 00:29:57 And so, typically the most common choice would be to go 320 00:29:57 --> 00:30:04 just first along the x-axis, then parallel to the y-axis and 321 00:30:04 --> 00:30:11 then parallel to the z-axis all the way to my point x1, 322 00:30:11 --> 00:30:14 y1, z1. I would just calculate three 323 00:30:14 --> 00:30:18 easy line integrals. Add them together and that 324 00:30:18 --> 00:30:21 would give me the value of my function. 325 00:30:21 --> 00:30:27 That method works exactly the same way as it did in two 326 00:30:27 --> 00:30:30 variables. Now, I seem to recall that you 327 00:30:30 --> 00:30:32 guys mostly preferred the other method. 328 00:30:32 --> 00:30:34 I am going to tell you about the other method as well, 329 00:30:34 --> 00:30:37 but I just want to point out this one actually doesn't become 330 00:30:37 --> 00:30:40 more complicated. The other one has actually more 331 00:30:40 --> 00:30:42 steps. I mean, of course, 332 00:30:42 --> 00:30:45 here there are also a bit more steps because you have three 333 00:30:45 --> 00:30:47 parts to your path instead of two. 334 00:30:47 --> 00:30:54 You have three line integrals to compute instead of two, 335 00:30:54 --> 00:30:59 but conceptually it remains exactly the same idea. 336 00:30:59 --> 00:31:07 I should say it works the same way as in 2D. 337 00:31:07 --> 00:31:17 Not much changes. Let's look at the other method 338 00:31:17 --> 00:31:24 using anti-derivatives. Remember we want to find a 339 00:31:24 --> 00:31:29 function little f whose partials are exactly the things we have 340 00:31:29 --> 00:31:31 been given. We want to solve, 341 00:31:31 --> 00:31:36 well, let me plug in the values of a and b that will work. 342 00:31:36 --> 00:31:47 We said a should be two, so f sub x should be 2xy, 343 00:31:47 --> 00:31:59 f sub y should be x2 plus z3, and f sub z should be 3yz^2 344 00:31:59 --> 00:32:04 minus 4z^3. We are going to look at them 345 00:32:04 --> 00:32:07 one at a time and get partial information on the function. 346 00:32:07 --> 00:32:11 And then we will compare with the others to get more 347 00:32:11 --> 00:32:15 information until we are completely done. 348 00:32:15 --> 00:32:25 The first thing we will do, we know that f sub x is 2xy. 349 00:32:25 --> 00:32:27 That should tell us something about f. 350 00:32:27 --> 00:32:31 Well, let's just integrate that with respect to x. 351 00:32:31 --> 00:32:36 Let me write integral dx next to that. 352 00:32:36 --> 00:32:41 That tells us that f should be, well, if we integrate that with 353 00:32:41 --> 00:32:44 respect to x, 2x integrates to x^2, 354 00:32:44 --> 00:32:47 so we should get x2y. Plus, of course, 355 00:32:47 --> 00:32:50 an integration constant. Now, what do we mean by 356 00:32:50 --> 00:32:53 integration constant. It means that for given values 357 00:32:53 --> 00:32:58 of y and z we will get a term that does not depend on x. 358 00:32:58 --> 00:33:02 It still depends on y and z. In fact, what we get is a 359 00:33:02 --> 00:33:07 function of y and z. See, if you took the derivative 360 00:33:07 --> 00:33:12 of this with respect to x you will get 2xy and this guy will 361 00:33:12 --> 00:33:16 go away because there is no x in it. 362 00:33:16 --> 00:33:19 That is the first step. Now we need to get some 363 00:33:19 --> 00:33:21 information on g. How do we do that? 364 00:33:21 --> 00:33:28 Well, we look at the other partials. 365 00:33:28 --> 00:33:35 F sub y, we want that to be x^2 z^3. 366 00:33:35 --> 00:33:41 But we have another way to find it, which is starting from this 367 00:33:41 --> 00:33:50 and differentiating. Let me try to use color for 368 00:33:50 --> 00:33:54 this. Now, if I take the partial of 369 00:33:54 --> 00:33:58 this with respect to y, I am going to get a different 370 00:33:58 --> 00:34:06 formula for f sub y. That will be x^2 plus g sub y. 371 00:34:06 --> 00:34:15 Well, if I compare these two expressions that tells me that g 372 00:34:15 --> 00:34:23 sub y should be z3. Now, if I have this I can 373 00:34:23 --> 00:34:32 integrate with respect to y. That will tell me that g is 374 00:34:32 --> 00:34:39 actually yz^3 plus an integration constant. 375 00:34:39 --> 00:34:42 That constant, again, does not depend on y, 376 00:34:42 --> 00:34:46 but it can still depend on z because we still have not said 377 00:34:46 --> 00:34:49 anything about partial with respect to z. 378 00:34:49 --> 00:34:54 In fact, that constant I will write as a function h of z. 379 00:34:54 --> 00:35:00 If I have this function of z and I take its partial with 380 00:35:00 --> 00:35:04 respect to y, I will still get z^3 no matter 381 00:35:04 --> 00:35:06 what h was. Now, how do I find h? 382 00:35:06 --> 00:35:09 Well, obviously, I have to look at f sub z. 383 00:35:09 --> 00:35:41 384 00:35:41 --> 00:35:47 F sub z. We know from the given vector 385 00:35:47 --> 00:35:53 field that we want it to be 3yz^2 minus 4z^3. 386 00:35:53 --> 00:35:56 In case you are wondering where that came from, 387 00:35:56 --> 00:36:01 that was R. But that is also obtained by 388 00:36:01 --> 00:36:09 differentiating with respect to z what we had so far. 389 00:36:09 --> 00:36:15 Sorry. What did we have so far? 390 00:36:15 --> 00:36:20 Well, we had f equals x^2y plus g. 391 00:36:20 --> 00:36:30 And we said g is actually yz^3 plus h of z. 392 00:36:30 --> 00:36:37 That is what we have so far. If we take the derivative of 393 00:36:37 --> 00:36:44 that with respect to z, we will get zero plus 3yz^2 394 00:36:44 --> 00:36:51 plus h prime of z, or dh dz as you want. 395 00:36:51 --> 00:36:59 Now, if we compare these two, we will get the derivative of 396 00:36:59 --> 00:37:03 h. It will tell us that h prime is 397 00:37:03 --> 00:37:08 negative for z3. That means that h is negative 398 00:37:08 --> 00:37:13 z^4 plus a constant. And this it is at last an 399 00:37:13 --> 00:37:17 actual constant. Because it does not depend on z 400 00:37:17 --> 00:37:21 and there is nothing else for it to depend on. 401 00:37:21 --> 00:37:33 Now we plug this into what we had before, and that will give 402 00:37:33 --> 00:37:44 us our function f. We get that f=x^2y yz^3 - z^4 403 00:37:44 --> 00:37:49 plus constant. If you just wanted to find one 404 00:37:49 --> 00:37:51 potential, you can just forget the constant. 405 00:37:51 --> 00:37:55 This guy was a potential. If you want all the potentials 406 00:37:55 --> 00:37:58 they differ by this constant. OK. 407 00:37:58 --> 00:38:01 Just to recap the method what did we do? 408 00:38:01 --> 00:38:04 We started with -- And, of course, you can do it in 409 00:38:04 --> 00:38:07 whichever order you prefer, but you have to still follow 410 00:38:07 --> 00:38:11 the systematic method. You start with f sub x and you 411 00:38:11 --> 00:38:13 integrate that with respect to x. 412 00:38:13 --> 00:38:19 That gives you f up to a function of y and z only. 413 00:38:19 --> 00:38:25 Now you compare f sub y as given to you by the vector field 414 00:38:25 --> 00:38:32 with the formula you get from this expression for f. 415 00:38:32 --> 00:38:36 And, of course, this one will involve g sub y. 416 00:38:36 --> 00:38:41 Out of this, you will get the value of g sub 417 00:38:41 --> 00:38:44 y. When you have g sub y that 418 00:38:44 --> 00:38:48 gives you g up to a function of z only. 419 00:38:48 --> 00:38:52 And so now you have f up to a function of z only. 420 00:38:52 --> 00:38:57 And what you will do is look at the derivative with respect to 421 00:38:57 --> 00:38:59 z, the one you want coming from 422 00:38:59 --> 00:39:02 the vector field and the one you have coming from this formula 423 00:39:02 --> 00:39:05 for f, match them and that will tell 424 00:39:05 --> 00:39:09 you h prime. You will get h and then you 425 00:39:09 --> 00:39:20 will get f. Any questions? 426 00:39:20 --> 00:39:25 Who still prefers this method? OK, still most of you. 427 00:39:25 --> 00:39:29 Who is thinking that maybe the other method was not so bad 428 00:39:29 --> 00:39:33 after all? OK. That is still a minority. 429 00:39:33 --> 00:39:36 You can choose whichever one you prefer. 430 00:39:36 --> 00:39:42 I would encourage you to get some practice by trying both on 431 00:39:42 --> 00:39:47 least a couple of examples just to make sure that you know how 432 00:39:47 --> 00:39:53 to do them both and then stick to whichever one you prefer. 433 00:39:53 --> 00:39:59 Any questions on that? No. I guess I already asked. 434 00:39:59 --> 00:40:07 Still no questions? OK. 435 00:40:07 --> 00:40:13 The next logical thing is going to be curl. 436 00:40:13 --> 00:40:17 And the theorem that is going to replace Green's theorem for 437 00:40:17 --> 00:40:21 work in this setting is going to be called Stokes' theorem. 438 00:40:21 --> 00:40:34 Let me start by telling you about curl in 3D. 439 00:40:34 --> 00:40:39 Here is the statement. The curl is just going to 440 00:40:39 --> 00:40:43 measure how much your vector field fails to be conservative. 441 00:40:43 --> 00:40:47 And, if you want to think about it in terms of motions, 442 00:40:47 --> 00:40:51 that also will measure the rotation part of the motion. 443 00:40:51 --> 00:40:55 Well, let me first give a definition. 444 00:40:55 --> 00:41:00 Let's say that my vector field has components P, 445 00:41:00 --> 00:41:06 Q and R. Then we define the curl of F to 446 00:41:06 --> 00:41:16 be R sub y minus Q sub z times i plus P sub z minus R sub x times 447 00:41:16 --> 00:41:21 j plus Q sub x minus P sub y times k. 448 00:41:21 --> 00:41:25 And of course nobody can remember this formula, 449 00:41:25 --> 00:41:28 so what is the structure of this formula? 450 00:41:28 --> 00:41:35 Well, you see, each of these guys is one of 451 00:41:35 --> 00:41:43 the things that have to be zero for our field to be 452 00:41:43 --> 00:41:52 conservative. If F is defined in a simply 453 00:41:52 --> 00:42:05 connected region then we have that F is conservative and is 454 00:42:05 --> 00:42:15 equivalent to if and only if curl F is zero. 455 00:42:15 --> 00:42:19 Now, an important difference between curl here and curl in 456 00:42:19 --> 00:42:23 the plane is that now the curl of a vector field is again a 457 00:42:23 --> 00:42:27 vector field. These expressions are functions 458 00:42:27 --> 00:42:31 of x, y, z and together you form a vector out of them. 459 00:42:31 --> 00:42:36 The curl of a vector field in space is actually a vector 460 00:42:36 --> 00:42:43 field, not a scalar function. I have delayed the inevitable. 461 00:42:43 --> 00:42:47 I have to really tell you how to remember this evil formula. 462 00:42:47 --> 00:42:55 The secret is that, in fact, you can think of this 463 00:42:55 --> 00:43:01 as del cross f. Maybe you have seen that in 464 00:43:01 --> 00:43:04 physics. This is really where this del 465 00:43:04 --> 00:43:08 notation becomes extremely useful, because that is 466 00:43:08 --> 00:43:13 basically the only way to remember the formula for curl. 467 00:43:13 --> 00:43:30 468 00:43:30 --> 00:43:34 Remember we introduced the dell operator. 469 00:43:34 --> 00:43:42 That was this symbolic vector operator in which the components 470 00:43:42 --> 00:43:47 are the partial derivative operators. 471 00:43:47 --> 00:43:59 We have seen that if you apply this to a scalar function then 472 00:43:59 --> 00:44:08 that will give you the gradient. And we have seen that if you do 473 00:44:08 --> 00:44:13 the dot product between dell and a vector field, 474 00:44:13 --> 00:44:19 maybe I should give it components P, 475 00:44:19 --> 00:44:24 Q and R, you will get partial P over 476 00:44:24 --> 00:44:30 partial x plus partial Q over partial y plus partial R over 477 00:44:30 --> 00:44:36 partial z, which is the divergence. 478 00:44:36 --> 00:44:47 And so now what is new is that if I try to do dell cross F, 479 00:44:47 --> 00:44:53 well, what is dell cross F? I have to set up a 480 00:44:53 --> 00:44:58 cross-product between this strange thing that is not really 481 00:44:58 --> 00:45:02 a vector. I mean, I cannot really think 482 00:45:02 --> 00:45:05 of partial over partial x as a number. 483 00:45:05 --> 00:45:10 And my vector field . 484 00:45:10 --> 00:45:14 See, that is really a completely perverted use of a 485 00:45:14 --> 00:45:18 determinant notation. Initially, determinants were 486 00:45:18 --> 00:45:21 just supposed to be you had a three by three table of numbers 487 00:45:21 --> 00:45:23 and you computed a number out of them. 488 00:45:23 --> 00:45:28 These guys are functions so they count as numbers, 489 00:45:28 --> 00:45:33 but these are vectors and these are partial derivatives. 490 00:45:33 --> 00:45:37 It doesn't really make much sense, except this notation. 491 00:45:37 --> 00:45:43 If you try to enter this into a calculator or computer, 492 00:45:43 --> 00:45:48 it will just yell back at you saying are you crazy. 493 00:45:48 --> 00:45:50 [LAUGHTER] We just use that as a notation 494 00:45:50 --> 00:45:53 to remember what is in there. Let's try and see how that 495 00:45:53 --> 00:45:55 works. The component of i in this 496 00:45:55 --> 00:45:58 cross-product, remember that is this smaller 497 00:45:58 --> 00:46:01 determinant, that smaller determinant is 498 00:46:01 --> 00:46:07 partial over partial y of R minus partial over partial z of 499 00:46:07 --> 00:46:10 Q, the coefficient of i. 500 00:46:10 --> 00:46:14 And that seems to be what I had over there. 501 00:46:14 --> 00:46:18 If not then I made a mistake. Minus the next determinant 502 00:46:18 --> 00:46:20 times z. Remember there is always a 503 00:46:20 --> 00:46:23 minus sign in front of a j component when you do a 504 00:46:23 --> 00:46:27 cross-product. The other one is partial over 505 00:46:27 --> 00:46:33 partial x R minus partial over partial z of P plus the 506 00:46:33 --> 00:46:39 component of z which is going to be partial over partial x Q 507 00:46:39 --> 00:46:46 minus partial over partial y P. And that is indeed going to be 508 00:46:46 --> 00:46:48 the curl of F. In practice, 509 00:46:48 --> 00:46:51 if you have to compute the curl of a vector field, 510 00:46:51 --> 00:46:53 you know, don't try to remember this formula. 511 00:46:53 --> 00:46:59 Just set up this cross-product with whatever formulas you have 512 00:46:59 --> 00:47:04 for the components of a field and then compute it. 513 00:47:04 --> 00:47:15 Don't bother to try to remember the general formula, 514 00:47:15 --> 00:47:25 just remember this. What is the geometric 515 00:47:25 --> 00:47:35 interpretation of curl, just to finish? 516 00:47:35 --> 00:47:48 In a way, I will say just curl measures the rotation component 517 00:47:48 --> 00:47:55 in a velocity field. An exercise that you can do, 518 00:47:55 --> 00:47:59 which is actually pretty easy to check, is say that we have a 519 00:47:59 --> 00:48:04 fluid that is just rotating about the x-axis uniformly. 520 00:48:04 --> 00:48:10 Your fluid is just rotating like that about the z-axis. 521 00:48:10 --> 00:48:15 If I take a rotation about the z-axis. 522 00:48:15 --> 00:48:28 That is given by a velocity field with components at angular 523 00:48:28 --> 00:48:35 velocity omega. That will be negative omega 524 00:48:35 --> 00:48:41 times y, then omega x and zero. And the curl of that you can 525 00:48:41 --> 00:48:45 compute, and you will find two omega times k. 526 00:48:45 --> 00:48:48 Concretely, this curl gives you the angular 527 00:48:48 --> 00:48:51 velocity of the rotation, well, with a factor two but 528 00:48:51 --> 00:48:53 that doesn't matter, and the axis of rotation, 529 00:48:53 --> 00:48:56 the direction of the axis of rotation. 530 00:48:56 --> 00:48:59 It tells you it is rotating about a vertical axis. 531 00:48:59 --> 00:49:01 And, in general, if you have a complicated 532 00:49:01 --> 00:49:03 motion some of it might be, you know, there is a 533 00:49:03 --> 00:49:05 translation. And then within that 534 00:49:05 --> 00:49:08 translation there is maybe expansion and rotation and 535 00:49:08 --> 00:49:11 sharing and everything. And the curl will compute how 536 00:49:11 --> 00:49:14 much rotation is taking place. It will tell you, 537 00:49:14 --> 00:49:16 say that you have a very small solid, 538 00:49:16 --> 00:49:19 I don't know like a ping pong ball in your flow, 539 00:49:19 --> 00:49:21 and it is just going with the flow, 540 00:49:21 --> 00:49:25 it tells you how it is going to start rotating. 541 00:49:25 --> 00:49:32 That is what curl measures. On Thursday we will see Stokes' 542 00:49:32 --> 00:49:37 theorem, which will be the last ingredient before the next exam. 543 00:49:37 --> 00:49:40 And then on Friday we will review stuff. 544 00:49:40 --> 00:49:41