1 00:00:00,000 --> 00:00:07,670 2 00:00:07,670 --> 00:00:09,370 CHRISTINE BREINER: Welcome back to recitation. 3 00:00:09,370 --> 00:00:11,050 In this video, I'd like us to consider 4 00:00:11,050 --> 00:00:12,580 the following problem. 5 00:00:12,580 --> 00:00:16,270 The first part is I'd like to know for what values of b is 6 00:00:16,270 --> 00:00:18,700 this vector field F conservative. 7 00:00:18,700 --> 00:00:24,520 And F is defined as y i plus the quantity x plus byz j, 8 00:00:24,520 --> 00:00:26,930 plus the quantity y squared plus 1, k. 9 00:00:26,930 --> 00:00:29,270 So as you can see, the only thing we're allowed to 10 00:00:29,270 --> 00:00:32,800 manipulate in this problem is b. b will be some real number. 11 00:00:32,800 --> 00:00:38,390 And I want to know what real numbers can I put in for b so 12 00:00:38,390 --> 00:00:40,890 that this vector field is conservative. 13 00:00:40,890 --> 00:00:43,480 The second part of this problem is for each b-value 14 00:00:43,480 --> 00:00:47,680 you determined from one, find a potential function. 15 00:00:47,680 --> 00:00:52,200 So fix the b-value for one of the ones that is acceptable, 16 00:00:52,200 --> 00:00:55,990 based on number one, and then find the potential function. 17 00:00:55,990 --> 00:01:00,420 And then the third part says that you should explain why F 18 00:01:00,420 --> 00:01:03,530 dot dr is exact, and this is obviously for the b values 19 00:01:03,530 --> 00:01:04,990 determined from one. 20 00:01:04,990 --> 00:01:07,940 So the second and third part are once 21 00:01:07,940 --> 00:01:09,100 you know the b-values. 22 00:01:09,100 --> 00:01:11,330 And you're only going to use those b-values that make F 23 00:01:11,330 --> 00:01:13,970 conservative, because that's the place where we can talk 24 00:01:13,970 --> 00:01:16,670 about finding a potential function, and where we can 25 00:01:16,670 --> 00:01:18,650 talk about F dot dr being exact are 26 00:01:18,650 --> 00:01:20,560 exactly those values. 27 00:01:20,560 --> 00:01:21,250 OK. 28 00:01:21,250 --> 00:01:23,650 So why don't you pause the video, work on these three 29 00:01:23,650 --> 00:01:26,270 problems, and then when you're feeling good about them, bring 30 00:01:26,270 --> 00:01:27,720 the video back up and I'll show you what I did. 31 00:01:27,720 --> 00:01:36,300 32 00:01:36,300 --> 00:01:37,560 OK, welcome back. 33 00:01:37,560 --> 00:01:40,530 Again, we're interested in doing three things with this 34 00:01:40,530 --> 00:01:41,210 vector field. 35 00:01:41,210 --> 00:01:43,960 The first thing we want to do is to find the values of b 36 00:01:43,960 --> 00:01:45,880 that make this vector field conservative. 37 00:01:45,880 --> 00:01:47,920 So I will start with that part. 38 00:01:47,920 --> 00:01:50,270 And as we know from the lecture, the thing I 39 00:01:50,270 --> 00:01:53,470 ultimately need to do is I need to find the curl of F. 40 00:01:53,470 --> 00:01:56,390 OK, so the curl of F is going to measure how far F is from 41 00:01:56,390 --> 00:01:57,190 being conservative. 42 00:01:57,190 --> 00:02:00,580 So if the curl of F is 0, I'm going to have F being 43 00:02:00,580 --> 00:02:01,190 conservative. 44 00:02:01,190 --> 00:02:04,560 So that's really what I'm interested in doing first. 45 00:02:04,560 --> 00:02:09,320 So I'm actually just going to rewrite what the curl of F 46 00:02:09,320 --> 00:02:10,570 actually is. 47 00:02:10,570 --> 00:02:11,985 So I'm going to let F-- 48 00:02:11,985 --> 00:02:17,670 I'm going to denote in our usual way by P, Q, R-- 49 00:02:17,670 --> 00:02:26,710 in this case, be specifically y comma x plus byz comma y 50 00:02:26,710 --> 00:02:29,600 squared plus 1. 51 00:02:29,600 --> 00:02:34,940 OK, that's my P, Q, R. So P is y, Q is x plus byz, and R is y 52 00:02:34,940 --> 00:02:36,160 squared plus 1. 53 00:02:36,160 --> 00:02:37,460 And now the curl of F-- 54 00:02:37,460 --> 00:02:43,250 55 00:02:43,250 --> 00:02:46,560 which was found in the lecture, so I'm not going to 56 00:02:46,560 --> 00:02:48,710 show you again how to find it, I'm just going to write the 57 00:02:48,710 --> 00:02:52,880 formula for it-- is exactly the following vector. 58 00:02:52,880 --> 00:02:56,220 It's the derivative of the R component with respect to y, 59 00:02:56,220 --> 00:02:58,490 minus the Q component with respect to z. 60 00:02:58,490 --> 00:03:00,610 That's the i-value. 61 00:03:00,610 --> 00:03:08,650 The j-value is P sub z minus R sub x, j. 62 00:03:08,650 --> 00:03:15,990 The k-value is Q sub x minus P sub y, k. 63 00:03:15,990 --> 00:03:17,930 OK, so there are three components here. 64 00:03:17,930 --> 00:03:21,520 And let's just start figuring out with these values are, and 65 00:03:21,520 --> 00:03:25,090 then we'll see what kind of restrictions we have on b. 66 00:03:25,090 --> 00:03:25,940 So let's start doing this. 67 00:03:25,940 --> 00:03:30,900 So again, this is P, this is Q, and this is R. So R sub y 68 00:03:30,900 --> 00:03:33,390 is the derivative of this component with respect to y. 69 00:03:33,390 --> 00:03:36,230 That's just 2y. 70 00:03:36,230 --> 00:03:39,320 Q sub z is the derivative of this with respect to z. 71 00:03:39,320 --> 00:03:41,030 Well, this is 0, and this is by. 72 00:03:41,030 --> 00:03:44,560 73 00:03:44,560 --> 00:03:45,190 OK. 74 00:03:45,190 --> 00:03:48,660 Let's look at the rest first. P sub z: the derivative of 75 00:03:48,660 --> 00:03:51,535 this with respect to z is 0. 76 00:03:51,535 --> 00:03:56,782 The derivative of R with respect to x is 0. 77 00:03:56,782 --> 00:03:59,720 So that doesn't have any b's in it at all. 78 00:03:59,720 --> 00:04:01,580 And then Q sub x minus P sub y. 79 00:04:01,580 --> 00:04:02,440 Q is the middle one. 80 00:04:02,440 --> 00:04:03,970 Q sub x is 1. 81 00:04:03,970 --> 00:04:07,240 82 00:04:07,240 --> 00:04:08,050 P is the first one. 83 00:04:08,050 --> 00:04:09,300 P sub y is 1. 84 00:04:09,300 --> 00:04:12,130 85 00:04:12,130 --> 00:04:13,135 OK, so what do we get here? 86 00:04:13,135 --> 00:04:16,000 I should have written equals there, maybe. 87 00:04:16,000 --> 00:04:18,970 OK, so the j component is 0. 88 00:04:18,970 --> 00:04:20,910 And the k component is 0. 89 00:04:20,910 --> 00:04:28,140 So all I'm left with is 2y minus by, i. 90 00:04:28,140 --> 00:04:30,590 And if I want F to be conservative, this quantity 91 00:04:30,590 --> 00:04:33,360 has to be 0, so I see there's only one b-value that's going 92 00:04:33,360 --> 00:04:36,420 to work, and that is b is equal to 2. 93 00:04:36,420 --> 00:04:40,190 94 00:04:40,190 --> 00:04:40,820 OK? 95 00:04:40,820 --> 00:04:44,280 So I know in part one, the answer to the question is just 96 00:04:44,280 --> 00:04:47,280 for b equals 2 is F conservative. 97 00:04:47,280 --> 00:04:48,960 That was, maybe, poorly phrased. 98 00:04:48,960 --> 00:04:51,830 F is conservative only when b is 2. 99 00:04:51,830 --> 00:04:52,080 OK. 100 00:04:52,080 --> 00:04:55,960 And that's because the curl of F is 0 only when b is 2. 101 00:04:55,960 --> 00:04:56,310 All right. 102 00:04:56,310 --> 00:04:58,300 So now we can move on to the second part. 103 00:04:58,300 --> 00:05:01,230 And the second part is for this particular value of b, 104 00:05:01,230 --> 00:05:02,980 find a potential function. 105 00:05:02,980 --> 00:05:05,500 And our strategy for that is going to be one of the methods 106 00:05:05,500 --> 00:05:06,720 from lecture. 107 00:05:06,720 --> 00:05:08,370 And it's going to be the method from lecture that in 108 00:05:08,370 --> 00:05:11,820 three dimensions is much easier than the other. 109 00:05:11,820 --> 00:05:14,240 So the one method in lecture that's easy in three 110 00:05:14,240 --> 00:05:17,600 dimensions is where you start at the origin, and you 111 00:05:17,600 --> 00:05:20,680 integrate F dot dr along a curve that's 112 00:05:20,680 --> 00:05:21,950 made up of line segments. 113 00:05:21,950 --> 00:05:25,290 So this strategy I've done before in two dimensions in 114 00:05:25,290 --> 00:05:27,170 one of the problems in recitation. 115 00:05:27,170 --> 00:05:29,070 Now, we'll see it in three dimensions. 116 00:05:29,070 --> 00:05:31,360 So what we're going to do is we're going to integrate along 117 00:05:31,360 --> 00:05:37,240 a certain curve F dot dr. And this curve is going to go from 118 00:05:37,240 --> 00:05:41,500 the origin to (x1, y1, z1). 119 00:05:41,500 --> 00:05:50,210 And that will give us f of (x1, y1, z1). 120 00:05:50,210 --> 00:05:50,450 OK. 121 00:05:50,450 --> 00:05:53,360 So this is a sort of general strategy, and now we'll talk 122 00:05:53,360 --> 00:05:54,880 about it specifically. 123 00:05:54,880 --> 00:05:58,090 This will actually be f of (x1, y1, z1) plus a constant, 124 00:05:58,090 --> 00:06:00,470 but we'll deal with that part right at the end. 125 00:06:00,470 --> 00:06:03,400 OK, so C in this case is going to be made up of three curves. 126 00:06:03,400 --> 00:06:07,630 And I'm going to draw them in a picture and then we're going 127 00:06:07,630 --> 00:06:09,720 to describe them. 128 00:06:09,720 --> 00:06:12,380 So I'm going to start at (0, 0, 0). 129 00:06:12,380 --> 00:06:20,700 My first curve will go out to x1 comma 0, 0, and that's 130 00:06:20,700 --> 00:06:22,280 going to be the curve C1. 131 00:06:22,280 --> 00:06:23,060 Oops. 132 00:06:23,060 --> 00:06:24,560 I want that to go the other way. 133 00:06:24,560 --> 00:06:25,590 That way. 134 00:06:25,590 --> 00:06:29,300 OK, C1 is going to go from the origin to x1 comma 0, 0. 135 00:06:29,300 --> 00:06:32,920 So the y- and z-values are going to be 0 and 0 all the 136 00:06:32,920 --> 00:06:35,050 way along, and the x-value is going to change. 137 00:06:35,050 --> 00:06:35,660 My next one-- 138 00:06:35,660 --> 00:06:37,190 I'm going to make it long so I can have 139 00:06:37,190 --> 00:06:39,550 enough room to write-- 140 00:06:39,550 --> 00:06:41,900 that's going to be my C2. 141 00:06:41,900 --> 00:06:49,180 And that's going to be x1 comma y1 comma 0. 142 00:06:49,180 --> 00:06:52,020 So in the end, what I've done is I've taken my x-value, I've 143 00:06:52,020 --> 00:06:54,120 kept it fixed all the way along here, but I'm varying 144 00:06:54,120 --> 00:06:56,210 the y-value out to y1. 145 00:06:56,210 --> 00:06:59,730 And then the last one is going to go straight up. 146 00:06:59,730 --> 00:07:01,920 Right there. 147 00:07:01,920 --> 00:07:05,500 And so it's going to be with the x-value and y-value fixed. 148 00:07:05,500 --> 00:07:10,020 And at the end, I will be at x1, y1, z1. 149 00:07:10,020 --> 00:07:11,270 And this is C3. 150 00:07:11,270 --> 00:07:13,230 151 00:07:13,230 --> 00:07:15,590 And this one, I'm going to move in this direction. 152 00:07:15,590 --> 00:07:17,000 Those are my three curves. 153 00:07:17,000 --> 00:07:20,550 And I want to point out that in order to understand how to 154 00:07:20,550 --> 00:07:23,540 simplify this problem, I'm going to have to remind myself 155 00:07:23,540 --> 00:07:25,210 what F dot dr is. 156 00:07:25,210 --> 00:07:25,530 OK. 157 00:07:25,530 --> 00:07:37,310 So F dot dr is P dx plus Q dy plus R dz. 158 00:07:37,310 --> 00:07:38,120 Right? 159 00:07:38,120 --> 00:07:39,550 That's what F dot dr is. 160 00:07:39,550 --> 00:07:42,110 And so what I'm interested in, I'm going to integrate each of 161 00:07:42,110 --> 00:07:44,630 these things along C1, C2, C3. 162 00:07:44,630 --> 00:07:48,280 But let's notice what happens along certain 163 00:07:48,280 --> 00:07:51,500 numbers of these curves. 164 00:07:51,500 --> 00:07:52,660 If we come back over here. 165 00:07:52,660 --> 00:07:56,220 On C1, y is fixed and z as fixed. 166 00:07:56,220 --> 00:07:58,630 So dy and dz are both 0. 167 00:07:58,630 --> 00:08:02,740 So on C1, I only have to integrate P. OK, so I'm going 168 00:08:02,740 --> 00:08:04,240 to keep track of that. 169 00:08:04,240 --> 00:08:08,340 On C1, which C1 is parametrized by x-- 170 00:08:08,340 --> 00:08:08,860 0 to x-- 171 00:08:08,860 --> 00:08:14,380 I only need to worry about the P. P of x, 0, 0, dx. 172 00:08:14,380 --> 00:08:17,200 This is my C1 component, and there's nothing else there, 173 00:08:17,200 --> 00:08:19,890 because these two are both 0. 174 00:08:19,890 --> 00:08:20,180 Right? 175 00:08:20,180 --> 00:08:24,470 Now let's consider what happens on C2. 176 00:08:24,470 --> 00:08:25,930 If I look here. 177 00:08:25,930 --> 00:08:29,810 On C2, the x-value is fixed and the z-value is fixed. x is 178 00:08:29,810 --> 00:08:32,340 fixed at x1 and z is fixed at 0. 179 00:08:32,340 --> 00:08:36,080 And so dx and dz are both 0, because x 180 00:08:36,080 --> 00:08:37,190 and z are not changing. 181 00:08:37,190 --> 00:08:39,560 So there's only a dy component. 182 00:08:39,560 --> 00:08:43,810 So on C2, which is parametrized in y-- 183 00:08:43,810 --> 00:08:45,120 from 0 to y1-- 184 00:08:45,120 --> 00:08:52,300 I'm only interested in Q at x1 comma y comma 0, dy. 185 00:08:52,300 --> 00:08:57,730 Again, this component is 0 on C2 because dx is 0. 186 00:08:57,730 --> 00:09:00,770 And this component is 0 on C2 because dz is 0. 187 00:09:00,770 --> 00:09:01,900 And this component-- 188 00:09:01,900 --> 00:09:03,200 I'm evaluating it-- 189 00:09:03,200 --> 00:09:07,640 x is fixed at x1, z is fixed at 0, and the y is varying 190 00:09:07,640 --> 00:09:09,070 from 0 to y1. 191 00:09:09,070 --> 00:09:11,360 And then there's one more component, and I'm going to 192 00:09:11,360 --> 00:09:14,250 write it below, and then we'll do the rest over here. 193 00:09:14,250 --> 00:09:16,850 And the third component is the C3 component. 194 00:09:16,850 --> 00:09:18,540 Now, not surprisingly-- 195 00:09:18,540 --> 00:09:20,200 if I come back over here-- 196 00:09:20,200 --> 00:09:23,380 because x and y are fixed all along the C3 component, the 197 00:09:23,380 --> 00:09:25,130 only thing that's changing is z. 198 00:09:25,130 --> 00:09:28,050 So dx and dy are 0, so I'm only worried 199 00:09:28,050 --> 00:09:29,610 about the dz part. 200 00:09:29,610 --> 00:09:29,840 OK. 201 00:09:29,840 --> 00:09:33,910 So again, as happened before, I only had P in the first one 202 00:09:33,910 --> 00:09:35,940 and Q in the second one, and now I have R only 203 00:09:35,940 --> 00:09:37,000 in the third one. 204 00:09:37,000 --> 00:09:39,840 And it's parametrized in z, from 0 to z1. 205 00:09:39,840 --> 00:09:41,730 That's what z varies over. 206 00:09:41,730 --> 00:09:48,830 And it's going to be R at x1 comma y1 comma z dz. 207 00:09:48,830 --> 00:09:52,580 Because the x's are fixed at x1, the y is fixed at y1, but 208 00:09:52,580 --> 00:09:54,415 z is varying from 0 to z1. 209 00:09:54,415 --> 00:09:55,020 All right. 210 00:09:55,020 --> 00:09:57,800 So I have these three parts, and now I just have to fill 211 00:09:57,800 --> 00:10:01,750 them in with the vector field that I have. I want to find 212 00:10:01,750 --> 00:10:06,090 what P is at (x, 0, 0), what Q is at (x1, y1, 0), and what R 213 00:10:06,090 --> 00:10:07,870 is at (x1, y1, z). 214 00:10:07,870 --> 00:10:09,080 And then integrate. 215 00:10:09,080 --> 00:10:11,360 So I have two steps left. 216 00:10:11,360 --> 00:10:13,770 One is plugging in and one is evaluating. 217 00:10:13,770 --> 00:10:17,940 So let me remind us what P, Q, and R actually are, and then 218 00:10:17,940 --> 00:10:20,950 we'll see what we get. 219 00:10:20,950 --> 00:10:22,260 Let me write it again. 220 00:10:22,260 --> 00:10:23,630 Maybe here. 221 00:10:23,630 --> 00:10:31,870 P, Q, R was equal to y comma x plus 2yz-- 222 00:10:31,870 --> 00:10:34,860 I'll put it here so you don't have to look and I don't have 223 00:10:34,860 --> 00:10:38,410 to look-- and then y squared plus 1. 224 00:10:38,410 --> 00:10:39,270 OK. 225 00:10:39,270 --> 00:10:44,380 So P at x comma 0 comma 0. 226 00:10:44,380 --> 00:10:48,450 Well, if I plug in 0 for y, P is 0. 227 00:10:48,450 --> 00:10:53,160 So P at (x, 0, 0) is equal to 0. 228 00:10:53,160 --> 00:10:55,250 So I get nothing to integrate in the first part. 229 00:10:55,250 --> 00:10:56,340 That's nice. 230 00:10:56,340 --> 00:10:57,040 OK. 231 00:10:57,040 --> 00:11:02,320 Now, what is Q at x1 comma y comma 0? 232 00:11:02,320 --> 00:11:04,600 Well, that would be an x1 here. 233 00:11:04,600 --> 00:11:06,550 0 for y makes this term go away. 234 00:11:06,550 --> 00:11:09,130 So it's just equal to x1. 235 00:11:09,130 --> 00:11:09,930 Right? 236 00:11:09,930 --> 00:11:18,660 And then R at x1 comma y1 comma z is y1 squared plus 1. 237 00:11:18,660 --> 00:11:21,480 238 00:11:21,480 --> 00:11:23,790 So now I'm going to substitute these in to what I'm 239 00:11:23,790 --> 00:11:24,170 integrating. 240 00:11:24,170 --> 00:11:27,192 So in the first one, there's nothing there. 241 00:11:27,192 --> 00:11:28,820 Let me just write it right here. 242 00:11:28,820 --> 00:11:30,375 The Q is going to be the integral from 243 00:11:30,375 --> 00:11:34,150 0 to y1 of x1 dy. 244 00:11:34,150 --> 00:11:39,610 And the R part is going to be the integral from 0 to z1 of 245 00:11:39,610 --> 00:11:43,090 y1 squared plus 1 dz. 246 00:11:43,090 --> 00:11:46,060 OK, so the P part was disappeared. 247 00:11:46,060 --> 00:11:48,850 This is the Q part evaluated where I 248 00:11:48,850 --> 00:11:50,580 needed it to be evaluated. 249 00:11:50,580 --> 00:11:52,310 It's just x1 dy. 250 00:11:52,310 --> 00:11:56,030 And the R part evaluated at (x1, y1, z) is just y1 251 00:11:56,030 --> 00:11:57,130 squared plus 1. 252 00:11:57,130 --> 00:11:59,500 And so I integrate that in z. 253 00:11:59,500 --> 00:12:02,990 So if I integrate this in y, all I get is x1 y 254 00:12:02,990 --> 00:12:04,720 evaluated 0 and y1. 255 00:12:04,720 --> 00:12:08,070 So here I just get an x1 y1. 256 00:12:08,070 --> 00:12:09,510 Right? 257 00:12:09,510 --> 00:12:12,800 And then here, if I integrate this in z, I just get a z-- 258 00:12:12,800 --> 00:12:14,600 and so I evaluate that at z1 and 0-- 259 00:12:14,600 --> 00:12:18,890 I just get z1 times y1 squared plus 1. 260 00:12:18,890 --> 00:12:19,400 OK. 261 00:12:19,400 --> 00:12:22,330 So this is actually my potential function. 262 00:12:22,330 --> 00:12:24,190 And so let me write it formally. 263 00:12:24,190 --> 00:12:25,890 I should actually say, this is my final answer. 264 00:12:25,890 --> 00:12:26,760 Right? 265 00:12:26,760 --> 00:12:27,920 I was integrating. 266 00:12:27,920 --> 00:12:29,730 This is actually what I get. 267 00:12:29,730 --> 00:12:31,720 And so what I was trying to find-- if you remember. 268 00:12:31,720 --> 00:12:34,190 I'm going to come back here and just mention it again. 269 00:12:34,190 --> 00:12:36,550 What I was doing was I was integrating along a curve, F 270 00:12:36,550 --> 00:12:39,210 dot dr, to give me f of (x1, y1, z1). 271 00:12:39,210 --> 00:12:40,010 Right? 272 00:12:40,010 --> 00:12:41,530 So now I've found it. 273 00:12:41,530 --> 00:12:43,940 The only thing I said is we also have to allow for there 274 00:12:43,940 --> 00:12:45,170 to be a constant. 275 00:12:45,170 --> 00:12:45,440 OK. 276 00:12:45,440 --> 00:12:50,000 So the potential function is actually exactly this function 277 00:12:50,000 --> 00:12:52,990 plus a constant. 278 00:12:52,990 --> 00:12:53,950 OK. 279 00:12:53,950 --> 00:12:56,310 So this is f of (x1, y1, z1). 280 00:12:56,310 --> 00:12:58,160 And since I don't have much room above, I'll 281 00:12:58,160 --> 00:12:59,410 just write it below. 282 00:12:59,410 --> 00:13:05,330 283 00:13:05,330 --> 00:13:09,730 So that's my potential function for this vector 284 00:13:09,730 --> 00:13:12,920 field, capital F, when it is conservative. 285 00:13:12,920 --> 00:13:15,290 So when b is equal to 2. 286 00:13:15,290 --> 00:13:17,770 OK, and there was one last part to this question. 287 00:13:17,770 --> 00:13:17,970 Right? 288 00:13:17,970 --> 00:13:20,180 So if we come all the way back over, we're 289 00:13:20,180 --> 00:13:22,790 reminded of one last part. 290 00:13:22,790 --> 00:13:26,160 It was explain why F dot dr is exact for the b values 291 00:13:26,160 --> 00:13:27,790 determined from number one. 292 00:13:27,790 --> 00:13:30,080 And the reason is exactly because of 293 00:13:30,080 --> 00:13:32,410 the following thing. 294 00:13:32,410 --> 00:13:36,360 F is conservative based on the fact that b is 2. 295 00:13:36,360 --> 00:13:42,530 And so when we talk about when F dot dr is exact, the 296 00:13:42,530 --> 00:13:47,960 simplest case is capital F is conservative, and I'm on a 297 00:13:47,960 --> 00:13:49,810 simply connected domain. 298 00:13:49,810 --> 00:13:50,440 OK. 299 00:13:50,440 --> 00:13:56,250 And if you notice, capital F is defined for all values x, 300 00:13:56,250 --> 00:13:59,900 y, z, and is differentiable for all values x, y, z. 301 00:13:59,900 --> 00:14:02,370 So F is defined and differentiable 302 00:14:02,370 --> 00:14:04,550 everywhere on R3. 303 00:14:04,550 --> 00:14:06,040 R3 is simply connected. 304 00:14:06,040 --> 00:14:08,610 So we have a conservative vector field on a simply 305 00:14:08,610 --> 00:14:09,650 connected region. 306 00:14:09,650 --> 00:14:10,700 And that's what it means. 307 00:14:10,700 --> 00:14:14,100 That's one way that we have of knowing F dot dr is exact. 308 00:14:14,100 --> 00:14:16,060 And so that actually answers the 309 00:14:16,060 --> 00:14:17,910 third part of the question. 310 00:14:17,910 --> 00:14:20,010 So again, let me just remind you what we did. 311 00:14:20,010 --> 00:14:24,420 We started with a vector field F, we found values for b that 312 00:14:24,420 --> 00:14:27,310 made that vector field conservative, and then we used 313 00:14:27,310 --> 00:14:30,090 one of the techniques in class to find a potential function 314 00:14:30,090 --> 00:14:31,950 for that value of b. 315 00:14:31,950 --> 00:14:34,580 So there were a number of steps involved, but 316 00:14:34,580 --> 00:14:37,420 ultimately, again, it's the same type of problem you've 317 00:14:37,420 --> 00:14:40,710 seen before when F was a vector field in two 318 00:14:40,710 --> 00:14:41,750 dimensions. 319 00:14:41,750 --> 00:14:45,250 So it shouldn't be feeling too different from some of the 320 00:14:45,250 --> 00:14:47,120 stuff you saw earlier. 321 00:14:47,120 --> 00:14:49,520 OK, I think that's where I'll stop. 322 00:14:49,520 --> 00:14:50,624