1 00:00:07,394 --> 00:00:09,310 CHRISTINE BREINER: Welcome back to recitation. 2 00:00:09,310 --> 00:00:11,850 In this video, I'd like us to work on the following problem. 3 00:00:11,850 --> 00:00:14,910 We're going to let F be the vector field that's defined 4 00:00:14,910 --> 00:00:19,720 by r to the n, times the quantity x*i plus y*j. 5 00:00:19,720 --> 00:00:24,080 And r in this case is x squared plus y squared to the 1/2, 6 00:00:24,080 --> 00:00:25,480 as it usually is. 7 00:00:25,480 --> 00:00:27,530 The square root of x squared plus y squared. 8 00:00:27,530 --> 00:00:29,238 And then I'd like us to do the following. 9 00:00:29,238 --> 00:00:31,130 Use extended Green's theorem to show 10 00:00:31,130 --> 00:00:34,240 that F is conservative for all integers n. 11 00:00:34,240 --> 00:00:35,870 And then find a potential function. 12 00:00:35,870 --> 00:00:37,150 So there are two parts. 13 00:00:37,150 --> 00:00:40,110 The first part is that you want to show that F is conservative. 14 00:00:40,110 --> 00:00:41,890 And then once you know it's conservative, 15 00:00:41,890 --> 00:00:43,840 you can find a potential function. 16 00:00:43,840 --> 00:00:46,440 So why don't you take a little while to work on that. 17 00:00:46,440 --> 00:00:48,720 And then when you're feeling good about your answer, 18 00:00:48,720 --> 00:00:50,970 bring the video back up, and I'll show you what I did. 19 00:00:58,960 --> 00:01:00,440 OK, welcome back. 20 00:01:00,440 --> 00:01:02,300 So again, what was the point of this video? 21 00:01:02,300 --> 00:01:03,430 We want to do two things. 22 00:01:03,430 --> 00:01:04,763 We want to work on two problems. 23 00:01:04,763 --> 00:01:07,100 The first is to show that this vector field F 24 00:01:07,100 --> 00:01:08,750 I've given you is conservative. 25 00:01:08,750 --> 00:01:11,260 And then we want to find a potential function. 26 00:01:11,260 --> 00:01:12,790 And we want to be able to show it's 27 00:01:12,790 --> 00:01:14,360 conservative for all integers n. 28 00:01:14,360 --> 00:01:17,909 And what I want to point out is that for certain integer values 29 00:01:17,909 --> 00:01:19,700 of n, we're going to run into some problems 30 00:01:19,700 --> 00:01:22,220 with differentiability at the origin. 31 00:01:22,220 --> 00:01:22,720 OK? 32 00:01:22,720 --> 00:01:25,290 So we're going to try and deal with all of it at once, 33 00:01:25,290 --> 00:01:27,450 and simultaneously deal with all of the integers, 34 00:01:27,450 --> 00:01:32,750 by allowing ourselves to show that F is conservative 35 00:01:32,750 --> 00:01:36,130 even if we don't include the origin in our region. 36 00:01:36,130 --> 00:01:37,370 OK. 37 00:01:37,370 --> 00:01:41,010 So I want to point out a few things first. 38 00:01:41,010 --> 00:01:43,440 And the first thing I want to point out is if we denote F 39 00:01:43,440 --> 00:01:48,370 as we usually do in two dimensions as [M, N], 40 00:01:48,370 --> 00:01:52,720 then the curl of F is going to be N sub x minus M sub y. 41 00:01:52,720 --> 00:01:54,200 OK. 42 00:01:54,200 --> 00:01:56,390 I actually calculated these earlier, 43 00:01:56,390 --> 00:02:02,220 but I want to point out that M sub y is actually equal to n 44 00:02:02,220 --> 00:02:05,114 times r to the n minus 2, times x*y. 45 00:02:05,114 --> 00:02:06,780 Let me make sure I wrote that correctly. 46 00:02:06,780 --> 00:02:07,550 Yes. 47 00:02:07,550 --> 00:02:12,560 But that is also exactly equal to N sub x. 48 00:02:12,560 --> 00:02:14,150 And so what does that give us? 49 00:02:14,150 --> 00:02:17,010 Since N sub x minus M sub y is the curl of F, 50 00:02:17,010 --> 00:02:22,030 when we have this vector [M, N], we know that the curl of F is 51 00:02:22,030 --> 00:02:28,530 equal to 0 by this work. 52 00:02:28,530 --> 00:02:31,990 OK, now if our vector field was defined 53 00:02:31,990 --> 00:02:34,190 on a simply-connected region, then that's 54 00:02:34,190 --> 00:02:37,021 enough to show that F is conservative. 55 00:02:37,021 --> 00:02:37,520 OK? 56 00:02:37,520 --> 00:02:39,750 We just use Green's theorem right away. 57 00:02:39,750 --> 00:02:40,250 Right? 58 00:02:40,250 --> 00:02:42,195 But the problem is that we are not necessarily 59 00:02:42,195 --> 00:02:44,070 on a simply-connected region because we could 60 00:02:44,070 --> 00:02:45,994 have problems at the origin. 61 00:02:45,994 --> 00:02:47,410 And so I'm going to deal with this 62 00:02:47,410 --> 00:02:48,990 in a slightly different way. 63 00:02:48,990 --> 00:02:51,500 To show that F is conservative, what do I want to show? 64 00:02:51,500 --> 00:02:53,710 I want to show that when I take the line 65 00:02:53,710 --> 00:02:59,830 integral F dot dr over any closed loop that I get 0. 66 00:02:59,830 --> 00:03:01,590 That's ultimately what I'm trying to show. 67 00:03:01,590 --> 00:03:04,221 So there are fundamentally two types of curves 68 00:03:04,221 --> 00:03:05,220 that I'm concerned with. 69 00:03:05,220 --> 00:03:07,390 Two closed curves in R^2 that I'm concerned with, 70 00:03:07,390 --> 00:03:10,460 and I'm going to draw a picture of those two types of curves. 71 00:03:10,460 --> 00:03:17,110 So in R^2, I'm going to have curves that miss the origin-- 72 00:03:17,110 --> 00:03:21,110 some curve like this, which I'll call C_1. 73 00:03:21,110 --> 00:03:24,070 And then I'm going to have curves 74 00:03:24,070 --> 00:03:30,020 that go around the origin, and I'll call this C_2. 75 00:03:30,020 --> 00:03:31,020 OK? 76 00:03:31,020 --> 00:03:32,880 Fundamentally, there's a difference 77 00:03:32,880 --> 00:03:34,410 between this curve and this curve, 78 00:03:34,410 --> 00:03:38,210 because this curve contains the region where F is defined 79 00:03:38,210 --> 00:03:40,940 and differentiable, right? 80 00:03:40,940 --> 00:03:44,440 Every point on the interior of this curve, F 81 00:03:44,440 --> 00:03:47,510 is defined and differentiable and therefore, I 82 00:03:47,510 --> 00:03:50,610 can apply regular old Green's theorem here. 83 00:03:50,610 --> 00:03:51,300 OK? 84 00:03:51,300 --> 00:03:55,000 So I know by Green's theorem, the integral over the closed 85 00:03:55,000 --> 00:03:59,570 curve C_1 of F dot dr is equal to 0, 86 00:03:59,570 --> 00:04:02,970 and that's simply because the curl of F is equal to 0. 87 00:04:02,970 --> 00:04:03,470 Right? 88 00:04:03,470 --> 00:04:05,500 We can immediately use Green's theorem 89 00:04:05,500 --> 00:04:09,260 because we know that the integral over this loop C_1 90 00:04:09,260 --> 00:04:14,020 is equal to the integral over this region of the curl of F. 91 00:04:14,020 --> 00:04:15,260 That's just Green's theorem. 92 00:04:15,260 --> 00:04:17,900 So I can apply Green's theorem here. 93 00:04:17,900 --> 00:04:20,460 Now the problem here is I can't apply Green's theorem 94 00:04:20,460 --> 00:04:23,010 because this origin is a trouble spot. 95 00:04:23,010 --> 00:04:23,792 Right? 96 00:04:23,792 --> 00:04:25,500 I'm not necessarily differentiable there, 97 00:04:25,500 --> 00:04:27,660 so I have to be a little more careful. 98 00:04:27,660 --> 00:04:29,295 OK, and so what I do is I'm going 99 00:04:29,295 --> 00:04:32,950 to explain why, immediately, I can get the integral over C2 100 00:04:32,950 --> 00:04:35,170 is actually also 0. 101 00:04:35,170 --> 00:04:37,060 And what I'm going to do is I'm actually 102 00:04:37,060 --> 00:04:41,230 going to draw, hopefully, a circle that contains C_2. 103 00:04:41,230 --> 00:04:42,775 So I'm going to draw a circle. 104 00:04:42,775 --> 00:04:45,290 It's a lot of curves, but this is 105 00:04:45,290 --> 00:04:47,292 supposed to look like a circle. 106 00:04:47,292 --> 00:04:48,000 Sorry about that. 107 00:04:48,000 --> 00:04:50,990 It's a little big on the low side, but it's a circle. 108 00:04:50,990 --> 00:04:51,490 OK. 109 00:04:51,490 --> 00:04:53,350 This is a circle. 110 00:04:53,350 --> 00:04:56,010 And I'm going to call this C_3. 111 00:04:56,010 --> 00:05:00,380 Now, I can tell you right away that the integral 112 00:05:00,380 --> 00:05:05,220 over the curve C_3 of F dot dr is equal to 0, 113 00:05:05,220 --> 00:05:07,770 and let me explain why. 114 00:05:07,770 --> 00:05:08,550 OK? 115 00:05:08,550 --> 00:05:13,680 F is a normal vector field relative to a circle. 116 00:05:13,680 --> 00:05:14,900 Let's look at this again. 117 00:05:14,900 --> 00:05:16,790 It's radial, and that's why we know this. 118 00:05:16,790 --> 00:05:19,300 F is a radial vector field. 119 00:05:19,300 --> 00:05:22,880 It's really the vector field [x, y] times a scalar, 120 00:05:22,880 --> 00:05:24,550 depending on the radius. 121 00:05:24,550 --> 00:05:27,520 So if I look at this picture right here, 122 00:05:27,520 --> 00:05:31,090 then F is going to-- let me draw it 123 00:05:31,090 --> 00:05:34,890 in color-- F is going to, at any given point, 124 00:05:34,890 --> 00:05:36,850 be in the radial direction. 125 00:05:36,850 --> 00:05:40,040 But that is exactly normal to the tangent direction 126 00:05:40,040 --> 00:05:41,210 of this curve. 127 00:05:41,210 --> 00:05:43,589 So this is the direction F points, 128 00:05:43,589 --> 00:05:45,130 and this is the direction the tangent 129 00:05:45,130 --> 00:05:47,940 vector points to the curve. 130 00:05:47,940 --> 00:05:50,980 But remember, F dot dr is the same as F dotted 131 00:05:50,980 --> 00:05:53,670 with the tangent vector ds. 132 00:05:53,670 --> 00:05:54,170 OK? 133 00:05:54,170 --> 00:05:55,896 And so that is why for this circle, 134 00:05:55,896 --> 00:05:58,940 it's immediately obvious that F dot dr is equal to 0. 135 00:05:58,940 --> 00:06:01,170 Because at any given point on this circle, 136 00:06:01,170 --> 00:06:03,260 I'm taking a vector field, I'm dotting it 137 00:06:03,260 --> 00:06:05,810 with a vector field that's orthogonal to it, so I get 0, 138 00:06:05,810 --> 00:06:08,171 and when I integrate 0 I get 0. 139 00:06:08,171 --> 00:06:08,670 OK? 140 00:06:08,670 --> 00:06:10,332 So that's why this is 0. 141 00:06:10,332 --> 00:06:12,540 And now where the extended version of Green's theorem 142 00:06:12,540 --> 00:06:18,655 comes in, is the fact that, if I look in this region, 143 00:06:18,655 --> 00:06:21,460 F is defined and differentiable. 144 00:06:21,460 --> 00:06:21,960 Right? 145 00:06:21,960 --> 00:06:24,234 F is defined and differentiable in this entire region 146 00:06:24,234 --> 00:06:25,150 that I've just shaded. 147 00:06:25,150 --> 00:06:29,650 Which is the region between my circle and my curve C_2. 148 00:06:29,650 --> 00:06:33,486 And what that tells me is that because this one is 0-- when 149 00:06:33,486 --> 00:06:34,860 I integrate along this curve it's 150 00:06:34,860 --> 00:06:38,671 0-- the integral along this curve also has to be 0, right? 151 00:06:38,671 --> 00:06:40,420 That's what you actually have seen already 152 00:06:40,420 --> 00:06:42,140 when you talked about the extended version of Green's 153 00:06:42,140 --> 00:06:42,970 theorem. 154 00:06:42,970 --> 00:06:45,580 You can compare the integral along this curve 155 00:06:45,580 --> 00:06:48,580 to the integral along this curve because in the region 156 00:06:48,580 --> 00:06:51,220 between them, F is everywhere defined and differentiable. 157 00:06:51,220 --> 00:06:53,320 So you can apply Green's theorem there. 158 00:06:53,320 --> 00:06:55,100 It just now has two boundary components, 159 00:06:55,100 --> 00:06:56,780 instead of in this case where it just 160 00:06:56,780 --> 00:06:58,920 has one boundary component. 161 00:06:58,920 --> 00:07:02,770 And so since the integral on this curve is 0, 162 00:07:02,770 --> 00:07:06,770 and the curl of F is 0, and F is defined and differentiable 163 00:07:06,770 --> 00:07:09,030 everywhere in this region, that tells you 164 00:07:09,030 --> 00:07:12,804 that the integral on the curve C_2 is also 0. 165 00:07:15,470 --> 00:07:18,470 Let me say that one more time, OK? 166 00:07:18,470 --> 00:07:20,470 I'm going to label it in blue so you can see it. 167 00:07:20,470 --> 00:07:23,750 I'm going to call this region that's shaded R. 168 00:07:23,750 --> 00:07:27,620 So Green's theorem says that the double integral 169 00:07:27,620 --> 00:07:32,500 in R of the curl of F is equal to the integral 170 00:07:32,500 --> 00:07:34,040 around this curve. 171 00:07:34,040 --> 00:07:36,580 And then I come in and I go around this direction 172 00:07:36,580 --> 00:07:40,560 and I come back out, and that gives me 173 00:07:40,560 --> 00:07:45,730 the entire integral of the curl of F on this region. 174 00:07:45,730 --> 00:07:46,230 Right? 175 00:07:46,230 --> 00:07:49,450 The curl of F is 0 everywhere in this region, 176 00:07:49,450 --> 00:07:51,280 so that integral is 0. 177 00:07:51,280 --> 00:07:55,060 And so the sum of the integral on C_3 minus the integral 178 00:07:55,060 --> 00:07:56,760 on C_2 has to be 0. 179 00:07:56,760 --> 00:07:58,230 Since this one is 0, that one is 0. 180 00:07:58,230 --> 00:07:59,354 So you've seen this before. 181 00:07:59,354 --> 00:08:04,840 I just want to remind you about where that's coming from. 182 00:08:04,840 --> 00:08:05,780 All right. 183 00:08:05,780 --> 00:08:08,020 So now we have to do one other thing, 184 00:08:08,020 --> 00:08:11,455 and that's we have to find a potential function. 185 00:08:11,455 --> 00:08:13,830 OK, so let's talk about how to find a potential function. 186 00:08:18,400 --> 00:08:21,920 I'm going to do this by one of the methods we saw in lecture. 187 00:08:24,510 --> 00:08:25,730 have-- 188 00:08:25,730 --> 00:08:29,280 I'm in R^2, and I'm going to start at a certain point 189 00:08:29,280 --> 00:08:33,860 and I'm going to integrate up to (x_1, 190 00:08:33,860 --> 00:08:37,025 y_1) from this certain point. 191 00:08:37,025 --> 00:08:39,650 And then I'm going to figure out what the function is that way. 192 00:08:39,650 --> 00:08:41,191 So what I'm going to do-- again, I'll 193 00:08:41,191 --> 00:08:44,290 write it this way-- I'm going to figure out f of (x_1, y_1) 194 00:08:44,290 --> 00:08:49,060 by integrating along a certain curve, F dot dr. 195 00:08:49,060 --> 00:08:51,460 Now I can't do exactly what I did previously, 196 00:08:51,460 --> 00:08:54,820 because for certain values of n, I 197 00:08:54,820 --> 00:08:58,569 run into trouble with integrating F from the origin. 198 00:08:58,569 --> 00:09:00,610 So what I'm going to do is instead of integrating 199 00:09:00,610 --> 00:09:03,230 from the origin, I'm going to integrate 200 00:09:03,230 --> 00:09:05,940 from the point (1, 1). 201 00:09:05,940 --> 00:09:06,440 OK? 202 00:09:06,440 --> 00:09:09,420 So I'm going to start at the point 1 comma 1, 203 00:09:09,420 --> 00:09:11,890 and I'm going to integrate in the y-direction, 204 00:09:11,890 --> 00:09:14,320 and then I'm going to integrate in the x-direction. 205 00:09:14,320 --> 00:09:16,400 So this will be my first curve and this 206 00:09:16,400 --> 00:09:18,440 will be my second curve. 207 00:09:18,440 --> 00:09:22,425 And I will land at x_1 comma y_1. 208 00:09:22,425 --> 00:09:24,050 So again, this is one of the strategies 209 00:09:24,050 --> 00:09:25,630 we've seen previously. 210 00:09:25,630 --> 00:09:27,780 This is the idea that I'm going to integrate 211 00:09:27,780 --> 00:09:32,650 in the y-direction, from y equals 1 to y equals y_1. 212 00:09:32,650 --> 00:09:37,055 So this will be the point 1 comma y_1, so x is fixed there. 213 00:09:37,055 --> 00:09:40,340 And I'm going to integrate in the x-direction, 214 00:09:40,340 --> 00:09:45,070 from x equals 1 to x equals x_1, when y is equal to y_1. 215 00:09:45,070 --> 00:09:46,710 So let's break this down. 216 00:09:46,710 --> 00:09:50,800 And let me remind you, also, the integral along this curve C 217 00:09:50,800 --> 00:09:57,050 of F dot dr should be P*dx plus Q*dy. 218 00:09:57,050 --> 00:09:57,760 Right? 219 00:09:57,760 --> 00:10:02,270 And so I'm going to look at what P*dx is and what Q*dy is on C_1 220 00:10:02,270 --> 00:10:04,271 and on C_2. 221 00:10:04,271 --> 00:10:04,770 All right. 222 00:10:04,770 --> 00:10:05,820 So let's do that. 223 00:10:09,350 --> 00:10:13,240 OK, so I have to remind myself what P and Q actually 224 00:10:13,240 --> 00:10:14,830 are in order to do this. 225 00:10:14,830 --> 00:10:17,670 So let me write that down, because this will be helpful: 226 00:10:17,670 --> 00:10:19,960 [P, Q]. 227 00:10:19,960 --> 00:10:26,540 P is r to the n, x, and Q is r to the n, y. 228 00:10:26,540 --> 00:10:27,490 All right? 229 00:10:27,490 --> 00:10:30,760 So that's what we're dealing with here. 230 00:10:30,760 --> 00:10:32,580 I'm going to come back to this picture, 231 00:10:32,580 --> 00:10:33,730 and then I'm going to come back and forth 232 00:10:33,730 --> 00:10:35,120 a little bit at this point. 233 00:10:35,120 --> 00:10:41,000 So if I want to integrate P*dx plus Q*dy on the curve C_1, 234 00:10:41,000 --> 00:10:46,220 what I need to observe first is that x is fixed, so dx is 0. 235 00:10:46,220 --> 00:10:49,076 So I'm actually just going to integrate Q*dy. 236 00:10:49,076 --> 00:10:49,720 All right. 237 00:10:49,720 --> 00:10:52,420 So the first integral along C_1 is just 238 00:10:52,420 --> 00:10:54,090 a parameterization in y. 239 00:10:54,090 --> 00:10:57,080 So it's the integral from 0 to y_1 240 00:10:57,080 --> 00:11:04,860 of Q evaluated at x equal 1, and y going from 1 to y_1. 241 00:11:04,860 --> 00:11:06,090 SPEAKER 1: 1 to y_1. 242 00:11:06,090 --> 00:11:08,060 CHRISTINE BREINER: y going from 1 to y1. 243 00:11:08,060 --> 00:11:08,630 OK? 244 00:11:08,630 --> 00:11:09,129 Sorry. 245 00:11:09,129 --> 00:11:10,650 Yes. y going from 1 to y_1. 246 00:11:10,650 --> 00:11:11,470 Sorry about that. 247 00:11:11,470 --> 00:11:11,970 Right? 248 00:11:11,970 --> 00:11:13,540 I was avoiding the origin, so I'd better not 249 00:11:13,540 --> 00:11:15,206 put a 0 down there, because that's where 250 00:11:15,206 --> 00:11:16,570 I was running into problems. 251 00:11:16,570 --> 00:11:17,670 OK. 252 00:11:17,670 --> 00:11:20,780 So Q is r to the n, y. 253 00:11:20,780 --> 00:11:22,680 So I have to remember what r is. r 254 00:11:22,680 --> 00:11:28,960 is x squared plus y squared to the 1/2. 255 00:11:28,960 --> 00:11:34,985 So in this case, Q is-- x is 1, and then I square it 256 00:11:34,985 --> 00:11:37,880 and I get 1, and then I have y squared, and then 257 00:11:37,880 --> 00:11:40,690 to the n over 2-- so this is my r 258 00:11:40,690 --> 00:11:46,070 to the n part along the curve C_1-- and then I multiply by y, 259 00:11:46,070 --> 00:11:47,504 and then I take dy. 260 00:11:47,504 --> 00:11:48,920 So there are a lot of pieces here, 261 00:11:48,920 --> 00:11:51,253 so let me just make sure we understand what's happening. 262 00:11:51,253 --> 00:11:53,850 I am interested in this entire thing, 263 00:11:53,850 --> 00:11:58,110 P*dx plus Q*dy along the curve C_1. 264 00:11:58,110 --> 00:12:02,140 dx is 0 along that curve. x is 1. 265 00:12:02,140 --> 00:12:04,601 And y is going from 1 to y_1. 266 00:12:04,601 --> 00:12:06,850 So if I come back over here, I see I'm only interested 267 00:12:06,850 --> 00:12:08,340 in the Q*dy part. 268 00:12:08,340 --> 00:12:10,740 y is going from 1 to y1. 269 00:12:10,740 --> 00:12:17,230 And then this is r to the n, when x is 1 and y is y. 270 00:12:17,230 --> 00:12:18,340 And this is the y part. 271 00:12:18,340 --> 00:12:22,000 So this is exactly Q*dy on the curve C_1. 272 00:12:22,000 --> 00:12:24,094 Now let's look at what happens on the curve C_2. 273 00:12:24,094 --> 00:12:25,510 So if I come back over here again, 274 00:12:25,510 --> 00:12:29,870 I want to have P*dx plus Q*dy on the curve C_2. 275 00:12:29,870 --> 00:12:34,160 Notice y is fixed at y1 there, so dy is 0. 276 00:12:34,160 --> 00:12:36,760 And so I'm only interested in the P*dx part. 277 00:12:36,760 --> 00:12:38,650 Everything is going to be in terms of x. 278 00:12:38,650 --> 00:12:40,920 And let's see if we can do the same kind of thing. 279 00:12:40,920 --> 00:12:44,370 I'm going to be integrating from 1 to x_1. 280 00:12:44,370 --> 00:12:48,590 Now r is going to be of the form x plus y_1 squared, 281 00:12:48,590 --> 00:12:50,360 to the n over 2. 282 00:12:50,360 --> 00:12:56,450 And then-- P has an x here and not a y-- times x dx. 283 00:12:56,450 --> 00:12:59,690 So again, P is r to the n times x, 284 00:12:59,690 --> 00:13:03,790 so this is r to the n times x exactly on the curve C_2. 285 00:13:03,790 --> 00:13:06,240 Because on C_2, y is fixed at y_1, 286 00:13:06,240 --> 00:13:08,750 so that's why I actually substituted in a y_1 here. 287 00:13:08,750 --> 00:13:11,790 It's the same reason I substituted in a 1 288 00:13:11,790 --> 00:13:16,130 here for x, because x was fixed at 1 on the curve C_1. 289 00:13:16,130 --> 00:13:20,027 So now I have to integrate these two things. 290 00:13:20,027 --> 00:13:22,360 I'm going to just write down what you get in both cases, 291 00:13:22,360 --> 00:13:24,350 because it's really single-variable calculus 292 00:13:24,350 --> 00:13:27,120 at this point in both cases. 293 00:13:27,120 --> 00:13:29,317 The easiest way to do this, probably, in my mind, 294 00:13:29,317 --> 00:13:30,400 is to do a u-substitution. 295 00:13:33,340 --> 00:13:35,070 Oops, I made a mistake. 296 00:13:35,070 --> 00:13:36,320 This should be an x squared. 297 00:13:36,320 --> 00:13:36,990 I apologize. 298 00:13:36,990 --> 00:13:38,290 This should be an x squared, because this is 299 00:13:38,290 --> 00:13:39,720 supposed to be a radius, right? 300 00:13:39,720 --> 00:13:42,990 It's x squared plus whatever y is squared, to the n over 2. 301 00:13:42,990 --> 00:13:45,682 So if you didn't see the squared here, and you got nervous, 302 00:13:45,682 --> 00:13:46,390 you were correct. 303 00:13:46,390 --> 00:13:47,696 There should be a squared here. 304 00:13:47,696 --> 00:13:49,320 So anyway, I'm going to go back to what 305 00:13:49,320 --> 00:13:50,319 I was saying previously. 306 00:13:50,319 --> 00:13:56,020 To integrate these things, the easiest thing to do 307 00:13:56,020 --> 00:13:57,850 is to take what is inside the parentheses 308 00:13:57,850 --> 00:14:00,880 and set it equal to u, and then do a u-substitution from there. 309 00:14:00,880 --> 00:14:03,076 So again, I'm not going to actually do that for you, 310 00:14:03,076 --> 00:14:04,700 but I'm going to tell you what you get. 311 00:14:04,700 --> 00:14:06,980 Now, there are two different situations. 312 00:14:06,980 --> 00:14:09,550 And the situations follow when n is 313 00:14:09,550 --> 00:14:14,290 any integer except negative 2, and then when n is negative 2. 314 00:14:14,290 --> 00:14:16,280 And the reason is because when n is negative 2, 315 00:14:16,280 --> 00:14:18,550 this exponent is a minus 1. 316 00:14:18,550 --> 00:14:21,560 So when you integrate, you end up with a natural log. 317 00:14:21,560 --> 00:14:24,100 So let me just point out the two things 318 00:14:24,100 --> 00:14:26,110 that you get in each case, and then we'll 319 00:14:26,110 --> 00:14:28,501 evaluate and see what the solutions are in each case. 320 00:14:28,501 --> 00:14:30,000 So I'm just going to, at this point, 321 00:14:30,000 --> 00:14:32,220 write down what I got, because this is 322 00:14:32,220 --> 00:14:33,920 your single-variable calculus. 323 00:14:33,920 --> 00:14:40,850 OK, so what I got when n was not equal to minus 2, 324 00:14:40,850 --> 00:14:42,490 you get the following thing. 325 00:14:42,490 --> 00:14:50,030 You get 1 plus y squared, evaluated at n plus 2, 326 00:14:50,030 --> 00:14:53,420 over 2, over n plus 2. 327 00:14:53,420 --> 00:14:56,480 And this is evaluated from 1 to y_1. 328 00:14:56,480 --> 00:14:58,570 And then this one you get a similar thing there, 329 00:14:58,570 --> 00:15:00,250 but now the y_1 is fixed here. 330 00:15:00,250 --> 00:15:05,860 So you get an x squared plus y_1 squared, to the n plus 2, 331 00:15:05,860 --> 00:15:11,840 over 2, over n plus 2, evaluated from 1 to x_1. 332 00:15:11,840 --> 00:15:13,300 So here, the y_1 is fixed and it's 333 00:15:13,300 --> 00:15:15,210 the x-values that are changing, and here 334 00:15:15,210 --> 00:15:16,730 the y-values are changing. 335 00:15:16,730 --> 00:15:19,260 So when n is not equal to 2, I get exactly this quantity 336 00:15:19,260 --> 00:15:21,020 when I integrate these two terms. 337 00:15:21,020 --> 00:15:23,280 And so now, let's see what happens. 338 00:15:23,280 --> 00:15:23,960 OK? 339 00:15:23,960 --> 00:15:26,170 Exactly what happens is the following. 340 00:15:26,170 --> 00:15:29,760 Notice that when I put in y_1 here, I get a 1 plus y_1 341 00:15:29,760 --> 00:15:32,590 squared, to the n plus 2 over 2, over n plus 2. 342 00:15:32,590 --> 00:15:33,139 Right? 343 00:15:33,139 --> 00:15:35,680 I'm not going to write it down, because I'm going to show you 344 00:15:35,680 --> 00:15:37,270 it gets killed off immediately. 345 00:15:37,270 --> 00:15:39,000 Where does it get killed off? 346 00:15:39,000 --> 00:15:42,960 It gets killed off when I evaluate this one at 1. 347 00:15:42,960 --> 00:15:43,460 OK? 348 00:15:43,460 --> 00:15:47,090 So the upper bound here is the same as the lower bound here. 349 00:15:47,090 --> 00:15:49,200 When I put in a 1 here, I get 1 plus y_1 350 00:15:49,200 --> 00:15:52,940 squared to the n plus 2 over 2 over n plus 2. 351 00:15:52,940 --> 00:15:54,450 It's a lot of n's and 2's. 352 00:15:54,450 --> 00:15:57,740 But the point is that when I evaluate this one at y_1 353 00:15:57,740 --> 00:16:00,440 and I evaluate this one at 1, I get exactly the same thing, 354 00:16:00,440 --> 00:16:05,004 but the signs are opposite and so they subtract off. 355 00:16:05,004 --> 00:16:07,420 In the final answer, I'm not going to see this upper bound 356 00:16:07,420 --> 00:16:08,920 and I'm not going to see this lower bound, 357 00:16:08,920 --> 00:16:10,503 because they're going to subtract off. 358 00:16:10,503 --> 00:16:12,870 And what I'm actually left with is just two terms. 359 00:16:12,870 --> 00:16:15,120 And those two terms I'm going to write up here. 360 00:16:15,120 --> 00:16:18,810 Those two terms are going to be x_1 squared 361 00:16:18,810 --> 00:16:25,650 plus y_1 squared to the n plus 2, over 2, over n plus 2. 362 00:16:25,650 --> 00:16:29,730 Minus, 1 plus 1-- which is just 2-- to the n plus 2, 363 00:16:29,730 --> 00:16:32,450 over 2, over n plus 2. 364 00:16:32,450 --> 00:16:33,570 What it this really? 365 00:16:33,570 --> 00:16:39,880 This is just r to the n plus 2, over n plus 2, plus a constant. 366 00:16:39,880 --> 00:16:42,425 Because this is just a constant for any n. 367 00:16:42,425 --> 00:16:46,456 And notice n is not equal to minus 2-- negative 2. 368 00:16:46,456 --> 00:16:49,080 That was the place we were going to run into trouble otherwise. 369 00:16:49,080 --> 00:16:50,970 And so when n is not equal to negative 2-- 370 00:16:50,970 --> 00:16:52,480 when you do all the integration-- 371 00:16:52,480 --> 00:16:55,890 you should arrive at this as your potential function. 372 00:16:55,890 --> 00:16:57,000 OK? 373 00:16:57,000 --> 00:16:59,002 And again, what I did was I evaluated 374 00:16:59,002 --> 00:17:00,835 to make it simpler on ourselves so we didn't 375 00:17:00,835 --> 00:17:02,140 have to write everything out. 376 00:17:02,140 --> 00:17:05,370 I noticed that if I evaluate this at the two bounds, 377 00:17:05,370 --> 00:17:06,870 and evaluate this at the two bounds, 378 00:17:06,870 --> 00:17:09,940 and I add them together, that the evaluation here 379 00:17:09,940 --> 00:17:13,720 plus the evaluation here are the same numerically but opposite 380 00:17:13,720 --> 00:17:16,240 in sign, and so they subtract off. 381 00:17:16,240 --> 00:17:19,300 And then I just have to evaluate at this one and this one. 382 00:17:19,300 --> 00:17:26,000 So that's n not equal to negative 2. 383 00:17:26,000 --> 00:17:28,790 Now let's do the n equal to negative 2 case. 384 00:17:28,790 --> 00:17:31,730 OK, so now I'm integrating this exact same thing 385 00:17:31,730 --> 00:17:34,090 in the n equal to negative 2 case. 386 00:17:34,090 --> 00:17:37,080 And I'll just write down again what I get by the substitution. 387 00:17:37,080 --> 00:17:39,810 And what I get is natural log of 1 388 00:17:39,810 --> 00:17:45,740 plus y squared, over 2, evaluated from 1 to y_1. 389 00:17:45,740 --> 00:17:52,420 Plus, natural log of x squared plus y_1 squared, over 2, 390 00:17:52,420 --> 00:17:54,140 evaluated from 1 to x_1. 391 00:17:54,140 --> 00:17:55,640 Let me make sure I have that right. 392 00:17:55,640 --> 00:17:56,550 Yes. 393 00:17:56,550 --> 00:17:58,220 And the same kind of thing is going 394 00:17:58,220 --> 00:18:01,800 to happen that happened before, in terms of when I put y_1 395 00:18:01,800 --> 00:18:07,287 in here, and I put 1 in here, I get the same thing 396 00:18:07,287 --> 00:18:08,370 but with an opposite sign. 397 00:18:08,370 --> 00:18:09,980 Here it's a positive. 398 00:18:09,980 --> 00:18:12,540 It's natural log 1 plus y_1 squared over 2. 399 00:18:12,540 --> 00:18:15,630 And here it's natural log 1 plus y_1 squared over 2, 400 00:18:15,630 --> 00:18:18,240 but because it's the lower bound, it's a negative sign. 401 00:18:18,240 --> 00:18:22,080 So whatever I get here and what I get here subtract off. 402 00:18:22,080 --> 00:18:23,750 And then in the end, I wind up getting 403 00:18:23,750 --> 00:18:25,850 just the following two terms. 404 00:18:25,850 --> 00:18:32,520 I get x_1 squared plus y_1 squared over 2, 405 00:18:32,520 --> 00:18:36,500 minus natural log of 2 over 2. 406 00:18:36,500 --> 00:18:39,580 So this term comes from evaluating this at x_1. 407 00:18:39,580 --> 00:18:42,690 And this term comes from evaluating this one 408 00:18:42,690 --> 00:18:44,260 at y equal 1. 409 00:18:44,260 --> 00:18:45,900 And if you notice, what is this? 410 00:18:45,900 --> 00:18:50,110 This is exactly natural log of r plus a constant. 411 00:18:50,110 --> 00:18:53,830 So let me step to the other side so we can see it clearly. 412 00:18:53,830 --> 00:18:57,110 So this is natural log of r squared, 413 00:18:57,110 --> 00:19:00,650 but by log rules, that's really 2 times natural log of r, 414 00:19:00,650 --> 00:19:04,500 so it divides by 2 and I'm just left with natural log of r, 415 00:19:04,500 --> 00:19:06,220 and this is just a constant. 416 00:19:06,220 --> 00:19:09,910 And so my potential function in that case 417 00:19:09,910 --> 00:19:12,777 is exactly natural log of r plus a constant. 418 00:19:12,777 --> 00:19:14,235 All right, this was a long problem, 419 00:19:14,235 --> 00:19:16,320 so I'm just going to remind us where we came from 420 00:19:16,320 --> 00:19:17,100 and what we were doing. 421 00:19:17,100 --> 00:19:18,516 So let's go back to the beginning. 422 00:19:21,400 --> 00:19:23,960 So what we did initially, was we had this vector field F. 423 00:19:23,960 --> 00:19:28,782 It was a radial vector field. r to the n times x*i plus y*j. 424 00:19:28,782 --> 00:19:30,240 And we wanted to first show that it 425 00:19:30,240 --> 00:19:33,110 was conservative for any integer value of n, 426 00:19:33,110 --> 00:19:35,350 and then to find its potential function. 427 00:19:35,350 --> 00:19:36,980 And obviously we do it in that order, 428 00:19:36,980 --> 00:19:38,355 because if it's not conservative, 429 00:19:38,355 --> 00:19:41,880 we're not going to find a potential function. 430 00:19:41,880 --> 00:19:45,740 In this case, what I observed first was that the curl of F 431 00:19:45,740 --> 00:19:47,860 was 0. 432 00:19:47,860 --> 00:19:52,660 And so in places where I had a closed curve that 433 00:19:52,660 --> 00:19:54,544 didn't contain the origin, I knew 434 00:19:54,544 --> 00:19:56,460 that the integral all around that closed curve 435 00:19:56,460 --> 00:19:59,230 was 0 just by Green's theorem. 436 00:19:59,230 --> 00:20:02,400 But if I had a closed curve that contained the origin, because F 437 00:20:02,400 --> 00:20:04,950 is not differentiable for all the n-values there, 438 00:20:04,950 --> 00:20:06,790 I have to be a little careful. 439 00:20:06,790 --> 00:20:08,550 It's actually even 0, right? 440 00:20:08,550 --> 00:20:12,020 When x is 0 and y is 0, I'm going to get something 0 there. 441 00:20:12,020 --> 00:20:16,360 So I need to figure out a way to determine the line 442 00:20:16,360 --> 00:20:17,810 integral on C_2. 443 00:20:17,810 --> 00:20:18,310 Right? 444 00:20:18,310 --> 00:20:19,184 And that was my goal. 445 00:20:19,184 --> 00:20:20,760 For any C_2 that contains the origin, 446 00:20:20,760 --> 00:20:23,100 how do I figure out F dot dr. 447 00:20:23,100 --> 00:20:26,530 And so I just compared it to what 448 00:20:26,530 --> 00:20:29,260 I get when I take F dot dr around a circle. 449 00:20:29,260 --> 00:20:32,050 Because I know that I can always find a circle bigger, 450 00:20:32,050 --> 00:20:34,540 and then I can say I've got this region here-- in 451 00:20:34,540 --> 00:20:36,480 between-- on which F is defined everywhere, 452 00:20:36,480 --> 00:20:39,950 so I can apply Green's theorem to that inside region. 453 00:20:39,950 --> 00:20:43,940 And I know that the curl of F on the inside region is 0, 454 00:20:43,940 --> 00:20:48,391 and so the integral on C_2 and C_3 is going to agree, right? 455 00:20:48,391 --> 00:20:49,890 Because the integral on C_3 I showed 456 00:20:49,890 --> 00:20:52,240 was 0 just geometrically. 457 00:20:52,240 --> 00:20:56,096 And then the integral on C_2 then has to be 0. 458 00:20:56,096 --> 00:20:56,595 All right? 459 00:20:56,595 --> 00:21:00,010 And so that was just when you were using the extended version 460 00:21:00,010 --> 00:21:01,460 of Green's theorem. 461 00:21:01,460 --> 00:21:05,740 And then to find a potential function, we came over here. 462 00:21:05,740 --> 00:21:07,952 And we had to avoid the origin because 463 00:21:07,952 --> 00:21:09,910 of the differentiability problem at the origin. 464 00:21:09,910 --> 00:21:12,140 So we started-- instead of where we usually start, 465 00:21:12,140 --> 00:21:15,630 which is from (0, 0)-- we started from the point (1, 1). 466 00:21:15,630 --> 00:21:18,050 And we just determined the potential function 467 00:21:18,050 --> 00:21:22,460 going from the point (1, 1) to the point (x_1, y_1) 468 00:21:22,460 --> 00:21:25,039 along a curve that went straight up, so x was fixed, 469 00:21:25,039 --> 00:21:27,080 and then along the curve that went straight over, 470 00:21:27,080 --> 00:21:28,640 so y was fixed. 471 00:21:28,640 --> 00:21:30,925 And so then we were able to break up this thing where 472 00:21:30,925 --> 00:21:35,370 I'm integrating over C P*dx plus Q*dy into two separate pieces, 473 00:21:35,370 --> 00:21:38,300 and each of them was fairly simple to write down. 474 00:21:38,300 --> 00:21:41,930 So let's look at what they were. 475 00:21:41,930 --> 00:21:45,180 This first one was where we were moving up. 476 00:21:45,180 --> 00:21:48,440 And there was no dx. x was just fixed at 1. 477 00:21:48,440 --> 00:21:51,180 And y was going from 1 to y_1. 478 00:21:51,180 --> 00:21:51,980 Right? 479 00:21:51,980 --> 00:21:54,162 And so x is fixed at 1, so I put a 1 there. 480 00:21:54,162 --> 00:21:55,370 And y is going from 1 to y_1. 481 00:21:55,370 --> 00:21:59,240 So I evaluate Q*dy on that curve. 482 00:21:59,240 --> 00:22:01,740 And then the next one was P*dx on the curve where I'm moving 483 00:22:01,740 --> 00:22:03,620 straight across. 484 00:22:03,620 --> 00:22:06,970 Right? dy is 0 there, so I just pick up the P*dx. 485 00:22:06,970 --> 00:22:10,035 And my y-value was fixed at y_1, and x 486 00:22:10,035 --> 00:22:12,320 was varying from 1 to x_1. 487 00:22:12,320 --> 00:22:15,260 And so then I just had to be a little bit careful. 488 00:22:15,260 --> 00:22:17,110 I didn't show you exactly how you integrate, 489 00:22:17,110 --> 00:22:20,630 but using a substitution trick-- single-variable calculus-- 490 00:22:20,630 --> 00:22:22,970 shouldn't be too bad for you at this point. 491 00:22:22,970 --> 00:22:25,770 We distinguished between when n was not equal to negative 2 492 00:22:25,770 --> 00:22:27,460 and when n was equal to negative 2. 493 00:22:27,460 --> 00:22:30,420 In the case n not equal to negative 2, 494 00:22:30,420 --> 00:22:34,167 we determined the integral, we simplified, 495 00:22:34,167 --> 00:22:36,750 and we got to a place where the potential function was exactly 496 00:22:36,750 --> 00:22:42,160 equal to r to the n plus 2 over n plus 2, plus some constant. 497 00:22:42,160 --> 00:22:45,952 Then in the case where n was equal to negative 2, 498 00:22:45,952 --> 00:22:48,410 when you do the substitution, you get a different integral. 499 00:22:48,410 --> 00:22:50,430 And in that case, you get the natural log. 500 00:22:50,430 --> 00:22:52,630 And so again, we just had the natural log. 501 00:22:52,630 --> 00:22:54,237 We have these different functions. 502 00:22:54,237 --> 00:22:56,820 We're evaluating the natural log of these different functions. 503 00:22:56,820 --> 00:22:58,080 We have the bounds. 504 00:22:58,080 --> 00:23:00,850 We simplify everything, and we get exactly to the place 505 00:23:00,850 --> 00:23:03,650 where you have natural log of r plus a constant. 506 00:23:03,650 --> 00:23:06,070 And so we found our potential function in the case n 507 00:23:06,070 --> 00:23:10,190 is equal to negative 2, and then any other n-value. 508 00:23:10,190 --> 00:23:12,160 So, a very long problem. 509 00:23:12,160 --> 00:23:13,700 I hope you got something out of it. 510 00:23:13,700 --> 00:23:16,160 And this is where I will stop.