1 00:00:07,454 --> 00:00:09,370 CHRISTINE BREINER: Welcome back to recitation. 2 00:00:09,370 --> 00:00:12,580 In this video, I'd like us to consider the following problem. 3 00:00:12,580 --> 00:00:16,000 The first part is I'd like to know, for what values of b 4 00:00:16,000 --> 00:00:18,700 is this vector field F conservative? 5 00:00:18,700 --> 00:00:24,800 And F is defined as y*i plus the quantity x plus b*y*z j plus 6 00:00:24,800 --> 00:00:26,930 the quantity y squared plus 1, k. 7 00:00:26,930 --> 00:00:28,729 So as you can see, the only thing 8 00:00:28,729 --> 00:00:30,520 we're allowed to manipulate in this problem 9 00:00:30,520 --> 00:00:32,800 is b. b will be some real number. 10 00:00:32,800 --> 00:00:38,390 And I want to know, what real numbers can I put in for b so 11 00:00:38,390 --> 00:00:40,890 that this vector field is conservative. 12 00:00:40,890 --> 00:00:42,680 The second part of this problem is 13 00:00:42,680 --> 00:00:46,180 for each b-value you determined from one, 14 00:00:46,180 --> 00:00:47,680 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 --> 00:00:59,880 And then the third part says that you should explain 18 00:00:59,880 --> 00:01:02,750 why F dot dr is exact, and this is obviously 19 00:01:02,750 --> 00:01:04,990 for the b-values determined from one. 20 00:01:04,990 --> 00:01:09,000 So the second and third part are once you know the b-values. 21 00:01:09,000 --> 00:01:11,000 And you're only going to use those b-values that 22 00:01:11,000 --> 00:01:13,770 make F conservative, because that's the place where we can 23 00:01:13,770 --> 00:01:15,760 talk about finding a potential function, 24 00:01:15,760 --> 00:01:18,540 and that's where we can talk about F dot dr being exact, 25 00:01:18,540 --> 00:01:20,560 are exactly those values. 26 00:01:20,560 --> 00:01:21,250 OK. 27 00:01:21,250 --> 00:01:23,650 So why don't you pause the video, work on these three 28 00:01:23,650 --> 00:01:26,070 problems, and then when you're feeling good about them, 29 00:01:26,070 --> 00:01:28,153 bring the video back up, I'll show you what I did. 30 00:01:36,300 --> 00:01:37,560 OK, welcome back. 31 00:01:37,560 --> 00:01:40,711 Again, we're interested in doing three things with this vector 32 00:01:40,711 --> 00:01:41,210 field. 33 00:01:41,210 --> 00:01:43,290 The first thing we want to do is to find 34 00:01:43,290 --> 00:01:45,880 the values of b that make this vector field conservative. 35 00:01:45,880 --> 00:01:47,920 So I will start with that part. 36 00:01:47,920 --> 00:01:50,170 And as we know from the lecture, the thing 37 00:01:50,170 --> 00:01:53,750 I ultimately need to do is I need to find the curl of F. OK, 38 00:01:53,750 --> 00:01:55,880 so the curl of F is going to measure how far F 39 00:01:55,880 --> 00:01:57,190 is from being conservative. 40 00:01:57,190 --> 00:01:59,380 So if the curl of F is 0, I'm going 41 00:01:59,380 --> 00:02:01,190 to have F being conservative. 42 00:02:01,190 --> 00:02:04,560 So that's really what I'm interested in doing first. 43 00:02:04,560 --> 00:02:09,320 So I'm actually just going to rewrite what the curl of F 44 00:02:09,320 --> 00:02:10,570 actually is. 45 00:02:10,570 --> 00:02:14,210 So I'm going to let F-- I'm going to denote in our usual 46 00:02:14,210 --> 00:02:19,410 way by P, Q, R. And in this case, 47 00:02:19,410 --> 00:02:27,130 that's specifically y comma x plus b*y*z comma y squared plus 48 00:02:27,130 --> 00:02:29,600 1. 49 00:02:29,600 --> 00:02:33,670 OK, that's my P, Q, R. So P is y, Q is x plus b*y*z, 50 00:02:33,670 --> 00:02:36,160 and R is y squared plus 1. 51 00:02:36,160 --> 00:02:45,580 And now the curl of F-- which was found in the lecture, 52 00:02:45,580 --> 00:02:48,020 so I'm not going to show you again how to find it, 53 00:02:48,020 --> 00:02:50,410 I'm just going to write the formula for it-- is exactly 54 00:02:50,410 --> 00:02:52,880 the following vector. 55 00:02:52,880 --> 00:02:56,220 It's the derivative of the R-th component with respect to y, 56 00:02:56,220 --> 00:02:58,490 minus the Q-th component with respect to z. 57 00:02:58,490 --> 00:03:00,610 That's the i-value. 58 00:03:00,610 --> 00:03:08,650 The j-value is P sub z minus R sub x, j. 59 00:03:08,650 --> 00:03:15,990 The k-th value is Q sub x minus P sub y, k. 60 00:03:15,990 --> 00:03:17,930 OK, so there are three components here. 61 00:03:17,930 --> 00:03:21,410 And let's just start figuring out what these values are, 62 00:03:21,410 --> 00:03:24,857 and then we'll see what kind of restrictions we have on b. 63 00:03:24,857 --> 00:03:25,940 So let's start doing this. 64 00:03:25,940 --> 00:03:28,420 So again, this is P, this is Q, and this 65 00:03:28,420 --> 00:03:32,020 is R. So R sub y is the derivative of this component 66 00:03:32,020 --> 00:03:33,390 with respect to y. 67 00:03:33,390 --> 00:03:36,230 That's just 2y. 68 00:03:36,230 --> 00:03:39,320 Q sub z is the derivative of this with respect to z. 69 00:03:39,320 --> 00:03:41,030 Well, this is 0, and this is b*y. 70 00:03:44,560 --> 00:03:45,190 OK. 71 00:03:45,190 --> 00:03:47,410 Let's look at the rest first. 72 00:03:47,410 --> 00:03:51,535 P sub z: the derivative of this with respect to z is 0. 73 00:03:51,535 --> 00:03:56,782 The derivative of R with respect to x is 0. 74 00:03:56,782 --> 00:03:59,720 So that doesn't have any b's in it at all. 75 00:03:59,720 --> 00:04:01,580 And then Q sub x minus P sub y. 76 00:04:01,580 --> 00:04:02,440 Q is the middle one. 77 00:04:02,440 --> 00:04:03,970 Q sub x is 1. 78 00:04:07,240 --> 00:04:08,050 P is the first one. 79 00:04:08,050 --> 00:04:09,030 P sub y is 1. 80 00:04:12,010 --> 00:04:13,135 OK, so what do we get here? 81 00:04:13,135 --> 00:04:16,000 I should have written equals there, maybe. 82 00:04:16,000 --> 00:04:18,970 OK, so the j-th component is 0. 83 00:04:18,970 --> 00:04:20,910 And the k-th component is 0. 84 00:04:20,910 --> 00:04:28,140 So all I'm left with is 2y minus b*y, i. 85 00:04:28,140 --> 00:04:29,960 And if I want F to be conservative, 86 00:04:29,960 --> 00:04:31,880 this quantity has to be 0, so I see 87 00:04:31,880 --> 00:04:34,070 there's only one b-value that's going to work, 88 00:04:34,070 --> 00:04:36,420 and that is b is equal to 2. 89 00:04:40,190 --> 00:04:40,820 OK? 90 00:04:40,820 --> 00:04:43,740 So I know in part one, the answer to the question 91 00:04:43,740 --> 00:04:47,280 is just for b equals 2, is F conservative. 92 00:04:47,280 --> 00:04:48,960 That was, maybe, poorly phrased. 93 00:04:48,960 --> 00:04:51,580 F is conservative only when b is 2. 94 00:04:51,580 --> 00:04:52,080 OK. 95 00:04:52,080 --> 00:04:55,810 And that's because the curl of F is 0 only when b is 2. 96 00:04:55,810 --> 00:04:56,310 All right. 97 00:04:56,310 --> 00:04:58,300 So now we can move on to the second part. 98 00:04:58,300 --> 00:05:01,230 And the second part is for this particular value of b, 99 00:05:01,230 --> 00:05:02,980 find a potential function. 100 00:05:02,980 --> 00:05:04,610 And our strategy for that is going 101 00:05:04,610 --> 00:05:06,720 to be one of the methods from lecture. 102 00:05:06,720 --> 00:05:08,761 And it's going to be the method from lecture that 103 00:05:08,761 --> 00:05:11,820 in three dimensions is much easier than the other. 104 00:05:11,820 --> 00:05:15,140 So the one method in lecture that's easy in three dimensions 105 00:05:15,140 --> 00:05:17,220 is where you start at the origin, 106 00:05:17,220 --> 00:05:20,680 and you integrate F dot dr along a curve that's 107 00:05:20,680 --> 00:05:21,950 made up of line segments. 108 00:05:21,950 --> 00:05:25,130 So this strategy I've done before in two dimensions, 109 00:05:25,130 --> 00:05:27,170 in one of the problems in recitation. 110 00:05:27,170 --> 00:05:29,044 Now, we'll see it in three dimensions. 111 00:05:29,044 --> 00:05:30,460 So what we're going to do is we're 112 00:05:30,460 --> 00:05:34,930 going to integrate along a certain curve F dot dr. 113 00:05:34,930 --> 00:05:38,630 And this curve is going to go from the origin to 114 00:05:38,630 --> 00:05:41,500 (x_1, y_1, z_1). 115 00:05:41,500 --> 00:05:49,950 And that will give us f of (x_1, y_1, z_1). 116 00:05:49,950 --> 00:05:50,450 OK. 117 00:05:50,450 --> 00:05:52,600 So this is a sort of general strategy, 118 00:05:52,600 --> 00:05:54,880 and now we'll talk about it specifically. 119 00:05:54,880 --> 00:05:58,090 This will actually be f of (x_1, y_1, z_1) plus a constant, 120 00:05:58,090 --> 00:06:00,470 but we'll deal with that part right at the end. 121 00:06:00,470 --> 00:06:03,400 OK, so C in this case is going to be made up of three curves. 122 00:06:03,400 --> 00:06:06,970 And I'm going to draw them, in a picture, 123 00:06:06,970 --> 00:06:09,720 and then we're going to describe them. 124 00:06:09,720 --> 00:06:12,380 So I'm going to start at (0, 0, 0). 125 00:06:12,380 --> 00:06:20,310 My first curve will go out to x_1 comma 0, 0, 126 00:06:20,310 --> 00:06:22,280 and that's going to be the curve C_1. 127 00:06:22,280 --> 00:06:23,060 Oops. 128 00:06:23,060 --> 00:06:24,560 I want that to go the other way. 129 00:06:24,560 --> 00:06:25,590 That way. 130 00:06:25,590 --> 00:06:29,300 OK, C_1 is going to go from the origin to x_1 comma 0, 0. 131 00:06:29,300 --> 00:06:31,860 So the y- and z-values are going to be 132 00:06:31,860 --> 00:06:35,016 0 and 0 all the way along, and the x-value is going to change. 133 00:06:35,016 --> 00:06:36,640 My next one-- I'm going to make it long 134 00:06:36,640 --> 00:06:39,770 so I can have enough room to write-- that's 135 00:06:39,770 --> 00:06:41,900 going to be my C_2. 136 00:06:41,900 --> 00:06:49,180 And that's going to be x_1 comma y_1 comma 0. 137 00:06:49,180 --> 00:06:51,830 So in the end, what I've done is I've taken my x-value, 138 00:06:51,830 --> 00:06:53,580 I've kept it fixed all the way along here, 139 00:06:53,580 --> 00:06:56,210 but I'm varying the y-value out to y_1. 140 00:06:56,210 --> 00:06:59,730 And then the last one is going to go straight up. 141 00:06:59,730 --> 00:07:01,920 Right there. 142 00:07:01,920 --> 00:07:05,500 And so it's going to be with the x-value and y-value fixed. 143 00:07:05,500 --> 00:07:10,020 And at the end, I will be at x_1, y_1, z_1. 144 00:07:10,020 --> 00:07:11,110 And this is C3. 145 00:07:11,110 --> 00:07:13,790 So those are my three curves-- And this one, 146 00:07:13,790 --> 00:07:15,590 I'm going to move in this direction. 147 00:07:15,590 --> 00:07:17,000 Those are my three curves. 148 00:07:17,000 --> 00:07:19,450 And I want to point out that in order 149 00:07:19,450 --> 00:07:22,190 to understand how to simplify this problem, 150 00:07:22,190 --> 00:07:25,030 I'm going to have to remind myself what F dot dr is. 151 00:07:25,030 --> 00:07:25,530 OK. 152 00:07:25,530 --> 00:07:37,310 So F dot dr is P*dx plus Q*dy plus R*dz. 153 00:07:37,310 --> 00:07:38,120 Right? 154 00:07:38,120 --> 00:07:39,550 That's what F dot dr is. 155 00:07:39,550 --> 00:07:41,100 And so what I'm interested in, I'm 156 00:07:41,100 --> 00:07:44,630 going to integrate each of these things along C_1, C_2, C_3. 157 00:07:44,630 --> 00:07:48,700 But let's notice what happens along certain numbers 158 00:07:48,700 --> 00:07:51,500 of these curves. 159 00:07:51,500 --> 00:07:52,660 If we come back over here. 160 00:07:52,660 --> 00:07:56,220 On C_1, y is fixed and z as fixed. 161 00:07:56,220 --> 00:07:58,630 So dy and dz are both 0. 162 00:07:58,630 --> 00:08:02,490 So on C1, I only have to integrate P. OK, 163 00:08:02,490 --> 00:08:04,240 so I'm going to keep track of that. 164 00:08:04,240 --> 00:08:06,770 On C_1-- which, C_1 is parametrized 165 00:08:06,770 --> 00:08:10,420 by x, 0 to x-- I only need to worry 166 00:08:10,420 --> 00:08:14,380 about the P. P of x, 0, 0, dx. 167 00:08:14,380 --> 00:08:17,200 This is my C_1 component, and there's nothing else there, 168 00:08:17,200 --> 00:08:19,680 because these two are both 0. 169 00:08:19,680 --> 00:08:20,180 Right? 170 00:08:20,180 --> 00:08:24,470 Now let's consider what happens on C_2. 171 00:08:24,470 --> 00:08:25,930 If I look here. 172 00:08:25,930 --> 00:08:29,400 On C_2, the x-value is fixed and the z-value is fixed. 173 00:08:29,400 --> 00:08:32,340 x is fixed at x_1 and z is fixed at 0. 174 00:08:32,340 --> 00:08:36,590 And so dx and dz are both 0, because x and z are not 175 00:08:36,590 --> 00:08:37,190 changing. 176 00:08:37,190 --> 00:08:39,560 So there's only a dy component. 177 00:08:39,560 --> 00:08:45,120 So on C_2, which is parametrized in y-- from 0 to y_1-- 178 00:08:45,120 --> 00:08:52,300 I'm only interested in Q at x_1 comma y comma 0, dy. 179 00:08:52,300 --> 00:08:57,730 Again, this component is 0 on C_2 because dx is 0. 180 00:08:57,730 --> 00:09:00,770 And this component is 0 on C_2 because dz is 0. 181 00:09:00,770 --> 00:09:03,200 And this component-- I'm evaluating it-- 182 00:09:03,200 --> 00:09:06,440 x is fixed at x_1, z is fixed at 0, 183 00:09:06,440 --> 00:09:09,070 and the y is varying from 0 to y1. 184 00:09:09,070 --> 00:09:10,932 And then there's one more component, 185 00:09:10,932 --> 00:09:12,640 and I'm going to write it below, and then 186 00:09:12,640 --> 00:09:14,250 we'll do the rest over here. 187 00:09:14,250 --> 00:09:16,850 And the third component is the C_3 component. 188 00:09:16,850 --> 00:09:20,200 Now, not surprisingly-- if I come back over here-- 189 00:09:20,200 --> 00:09:23,290 because x and y are fixed all along the C3 component, 190 00:09:23,290 --> 00:09:25,130 the only thing that's changing is z. 191 00:09:25,130 --> 00:09:29,340 So dx and dy are 0, so I'm only worried about the dz part. 192 00:09:29,340 --> 00:09:29,840 OK. 193 00:09:29,840 --> 00:09:32,852 So again, as happened before, I only 194 00:09:32,852 --> 00:09:34,810 had P in the first one and Q in the second one, 195 00:09:34,810 --> 00:09:37,000 and now I have R, only, in the third one. 196 00:09:37,000 --> 00:09:39,840 And it's parametrized in z, from 0 to z_1. 197 00:09:39,840 --> 00:09:41,730 That's what z varies over. 198 00:09:41,730 --> 00:09:48,830 And it's going to be R at x_1 comma y_1 comma z dz. 199 00:09:48,830 --> 00:09:52,390 Because the x's are fixed at x_1, the y is fixed at y_1, 200 00:09:52,390 --> 00:09:54,415 but z is varying from 0 to z_1. 201 00:09:54,415 --> 00:09:55,020 All right. 202 00:09:55,020 --> 00:09:57,050 So I have these three parts, and now 203 00:09:57,050 --> 00:10:00,830 I just have to fill them in with the vector field that I have. 204 00:10:00,830 --> 00:10:03,390 I want to find what P is at (x, 0, 0), 205 00:10:03,390 --> 00:10:07,870 what Q is at (x_1, y_1, 0), and what R is at (x_1, y_1, z). 206 00:10:07,870 --> 00:10:09,080 And then integrate. 207 00:10:09,080 --> 00:10:11,360 So I have two steps left. 208 00:10:11,360 --> 00:10:13,770 One is plugging in and one is evaluating. 209 00:10:13,770 --> 00:10:17,940 So let me remind us what P, Q, and R actually are, and then 210 00:10:17,940 --> 00:10:20,950 we'll see what we get. 211 00:10:20,950 --> 00:10:22,260 Let me write it again. 212 00:10:22,260 --> 00:10:23,630 Maybe here. 213 00:10:23,630 --> 00:10:31,870 [P, Q, R] was equal to y comma x plus 2y*z-- 214 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 215 00:10:34,860 --> 00:10:38,410 to look-- and then y squared plus 1. 216 00:10:38,410 --> 00:10:39,270 OK. 217 00:10:39,270 --> 00:10:44,380 So P at x comma 0 comma 0. 218 00:10:44,380 --> 00:10:48,450 Well, if I plug in 0 for y, P is 0. 219 00:10:48,450 --> 00:10:53,160 So P at (x, 0, 0) is equal to 0. 220 00:10:53,160 --> 00:10:55,250 So I get nothing to integrate in the first part. 221 00:10:55,250 --> 00:10:56,340 That's nice. 222 00:10:56,340 --> 00:10:57,040 OK. 223 00:10:57,040 --> 00:11:02,320 Now, what is Q at x_1 comma y comma 0? 224 00:11:02,320 --> 00:11:04,600 Well, that would be an x_1 here. 225 00:11:04,600 --> 00:11:06,550 0 for y makes this term go away. 226 00:11:06,550 --> 00:11:09,130 So it's just equal to x_1. 227 00:11:09,130 --> 00:11:09,930 Right? 228 00:11:09,930 --> 00:11:18,660 And then R at x_1 comma y_1 comma z is y_1 squared plus 1. 229 00:11:21,480 --> 00:11:23,087 So now I'm going to substitute these 230 00:11:23,087 --> 00:11:24,170 into what I'm integrating. 231 00:11:24,170 --> 00:11:27,192 So in the first one, there's nothing there. 232 00:11:27,192 --> 00:11:28,820 Let me just write it right here. 233 00:11:28,820 --> 00:11:30,330 The Q is going to be the integral 234 00:11:30,330 --> 00:11:34,150 from 0 to y_1 of x_1 dy. 235 00:11:34,150 --> 00:11:39,160 And the R part is going to be the integral from 0 to z_1 236 00:11:39,160 --> 00:11:43,090 of y_1 squared plus 1 dz. 237 00:11:43,090 --> 00:11:46,060 OK, so the P part was disappeared. 238 00:11:46,060 --> 00:11:47,780 This is the Q part evaluated where 239 00:11:47,780 --> 00:11:50,580 I needed it to be evaluated. 240 00:11:50,580 --> 00:11:52,310 It's just x_1 dy. 241 00:11:52,310 --> 00:11:55,200 And the R part evaluated at (x_1, y_1, z) 242 00:11:55,200 --> 00:11:57,130 is just y_1 squared plus 1. 243 00:11:57,130 --> 00:11:59,500 And so I integrate that in z. 244 00:11:59,500 --> 00:12:03,910 So if I integrate this in y, all I get is x_1*y evaluated 0 245 00:12:03,910 --> 00:12:04,720 and y1. 246 00:12:04,720 --> 00:12:08,070 So here I just get an x_1*y_1. 247 00:12:08,070 --> 00:12:09,510 Right? 248 00:12:09,510 --> 00:12:12,800 And then here, if I integrate this in z, I just get a z-- 249 00:12:12,800 --> 00:12:14,600 and so I evaluate that at z_1 and 0-- 250 00:12:14,600 --> 00:12:18,890 I just get z_1 times y1 squared plus 1. 251 00:12:18,890 --> 00:12:19,400 OK. 252 00:12:19,400 --> 00:12:22,330 So this is actually my potential function. 253 00:12:22,330 --> 00:12:23,932 And so let me write it formally. 254 00:12:23,932 --> 00:12:25,890 I should actually say, this is my final answer. 255 00:12:25,890 --> 00:12:26,760 Right? 256 00:12:26,760 --> 00:12:27,920 I was integrating. 257 00:12:27,920 --> 00:12:29,730 This is actually what I get. 258 00:12:29,730 --> 00:12:31,180 And so what I was trying to find, 259 00:12:31,180 --> 00:12:33,429 if you remember-- I'm going to come back here and just 260 00:12:33,429 --> 00:12:34,190 mention it again. 261 00:12:34,190 --> 00:12:36,740 What I was doing was I was integrating along a curve F dot 262 00:12:36,740 --> 00:12:39,210 dr, to give me f of (x_1, y_1, z_1). 263 00:12:39,210 --> 00:12:40,010 Right? 264 00:12:40,010 --> 00:12:41,530 So now I've found it. 265 00:12:41,530 --> 00:12:43,650 The only thing I said is we also have to allow 266 00:12:43,650 --> 00:12:44,941 for there to be a constant. 267 00:12:44,941 --> 00:12:45,440 OK. 268 00:12:45,440 --> 00:12:50,000 So the potential function is actually exactly this function 269 00:12:50,000 --> 00:12:52,990 plus a constant. 270 00:12:52,990 --> 00:12:53,950 OK. 271 00:12:53,950 --> 00:12:56,280 So this is f of (x_1, y_1, z_1). 272 00:12:56,280 --> 00:12:57,905 And since I don't have much room above, 273 00:12:57,905 --> 00:12:59,373 I'll just write it below. 274 00:13:03,450 --> 00:13:05,330 This is f of x_1, y_1, z_1. 275 00:13:05,330 --> 00:13:10,810 So that's my potential function for this vector field, capital 276 00:13:10,810 --> 00:13:12,920 F, when it is conservative. 277 00:13:12,920 --> 00:13:15,290 So when b is equal to 2. 278 00:13:15,290 --> 00:13:17,470 OK, and there was one last part to this question. 279 00:13:17,470 --> 00:13:17,970 Right? 280 00:13:17,970 --> 00:13:19,920 So if we come all the way back over, 281 00:13:19,920 --> 00:13:22,790 we're reminded of one last part. 282 00:13:22,790 --> 00:13:26,160 It was explain why F dot dr is exact for the b values 283 00:13:26,160 --> 00:13:27,790 determined from number one. 284 00:13:27,790 --> 00:13:32,410 And the reason is exactly because of the following thing. 285 00:13:32,410 --> 00:13:36,360 F is conservative based on the fact that b is 2. 286 00:13:36,360 --> 00:13:42,240 And so when we talk about when F dot dr is exact, 287 00:13:42,240 --> 00:13:47,080 the simplest case is capital F is conservative, 288 00:13:47,080 --> 00:13:49,810 and I'm on a simply connected domain. 289 00:13:49,810 --> 00:13:50,440 OK. 290 00:13:50,440 --> 00:13:57,320 And if you notice, capital F is defined for all values x, y, z, 291 00:13:57,320 --> 00:13:59,900 and is differentiable for all values of x, y, z. 292 00:13:59,900 --> 00:14:03,250 So F is defined and differentiable everywhere 293 00:14:03,250 --> 00:14:04,550 on R^3. 294 00:14:04,550 --> 00:14:06,040 R^3 is simply connected. 295 00:14:06,040 --> 00:14:08,120 So we have a conservative vector field 296 00:14:08,120 --> 00:14:09,650 on a simply connected region. 297 00:14:09,650 --> 00:14:10,700 And that's what it means. 298 00:14:10,700 --> 00:14:14,100 That's one way that we have of knowing F dot dr is exact. 299 00:14:14,100 --> 00:14:17,910 And so that actually answers the third part of the question. 300 00:14:17,910 --> 00:14:20,010 So again, let me just remind you what we did. 301 00:14:20,010 --> 00:14:22,640 We started with a vector field F, 302 00:14:22,640 --> 00:14:26,320 we found values for b that made that vector field conservative, 303 00:14:26,320 --> 00:14:27,915 and then we used one of the techniques 304 00:14:27,915 --> 00:14:31,950 in class to find a potential function for that value of b. 305 00:14:31,950 --> 00:14:34,350 So there were a number of steps involved, 306 00:14:34,350 --> 00:14:37,240 but ultimately, again, it's the same type of problem 307 00:14:37,240 --> 00:14:41,750 you've seen before, when F was a vector field in two dimensions. 308 00:14:41,750 --> 00:14:45,110 So it shouldn't be feeling too different from some 309 00:14:45,110 --> 00:14:47,120 of the stuff you saw earlier. 310 00:14:47,120 --> 00:14:49,520 OK, I think that's where I'll stop.