1 00:00:00,000 --> 00:00:08,402 CHRISTINE BREINER: Welcome back to recitation. 2 00:00:08,402 --> 00:00:12,460 In this video I'd like us to work on the following problem. 3 00:00:12,460 --> 00:00:16,010 What values of b will make this vector field F a gradient 4 00:00:16,010 --> 00:00:21,020 field where F is determined by e to the x plus y times x plus 5 00:00:21,020 --> 00:00:23,480 bi plus xj? 6 00:00:23,480 --> 00:00:26,030 So the e to the x plus y is in both the i 7 00:00:26,030 --> 00:00:28,640 component and the j component. 8 00:00:28,640 --> 00:00:30,820 And then once you've determined what values b will 9 00:00:30,820 --> 00:00:32,970 make that a gradient field-- 10 00:00:32,970 --> 00:00:35,800 for this b, or I should've said these b's-- 11 00:00:35,800 --> 00:00:37,700 find a potential function f using both 12 00:00:37,700 --> 00:00:39,590 methods from the lecture. 13 00:00:39,590 --> 00:00:42,910 So why don't you pause the video, work on this, and then 14 00:00:42,910 --> 00:00:45,960 when you are ready to look at how I do it bring 15 00:00:45,960 --> 00:00:47,210 the video back up. 16 00:00:47,210 --> 00:00:55,440 17 00:00:55,440 --> 00:00:55,700 OK. 18 00:00:55,700 --> 00:00:56,680 Welcome back. 19 00:00:56,680 --> 00:00:59,410 So I'm going to start off working on the first part of 20 00:00:59,410 --> 00:01:01,920 this problem, which is to find the values of b that will make 21 00:01:01,920 --> 00:01:04,310 this vector field F a gradient field. 22 00:01:04,310 --> 00:01:07,170 And to clarify things for myself, I'm going to write 23 00:01:07,170 --> 00:01:12,370 down what M and what N are based on F. So just to have it 24 00:01:12,370 --> 00:01:23,110 clear, M is equal to e to the x plus y times x plus b and N 25 00:01:23,110 --> 00:01:27,170 is equal to x times e to the x plus y. 26 00:01:27,170 --> 00:01:29,360 So those are our values for M and N. 27 00:01:29,360 --> 00:01:32,070 And now if I want f to be a gradient field, what I have to 28 00:01:32,070 --> 00:01:36,670 do is I have to have M sub y equal N sub x. 29 00:01:36,670 --> 00:01:39,530 So I'm going to determine M sub y and I'm going to 30 00:01:39,530 --> 00:01:42,750 determine N sub x and I'm going to compare them and see 31 00:01:42,750 --> 00:01:44,130 what value of b I get. 32 00:01:44,130 --> 00:01:48,280 So M sub y, fairly straightforward because this 33 00:01:48,280 --> 00:01:50,390 is a constant in y. 34 00:01:50,390 --> 00:01:53,380 And the derivative of this in terms of y is just this back. 35 00:01:53,380 --> 00:01:53,570 Right? 36 00:01:53,570 --> 00:01:56,820 It's an exponential function with the value that it has in 37 00:01:56,820 --> 00:01:57,620 y is linear. 38 00:01:57,620 --> 00:01:59,120 So you get exactly that thing back. 39 00:01:59,120 --> 00:02:04,450 So it actually is just either the x plus y times x plus b. 40 00:02:04,450 --> 00:02:07,850 So the derivative of M sub y is just itself. 41 00:02:07,850 --> 00:02:09,220 The derivative of M with respect to y. 42 00:02:09,220 --> 00:02:09,510 Sorry. 43 00:02:09,510 --> 00:02:12,140 Not the derivative of M sub y. 44 00:02:12,140 --> 00:02:12,760 OK. 45 00:02:12,760 --> 00:02:13,680 That's an x. 46 00:02:13,680 --> 00:02:15,930 Let me just rewrite that. 47 00:02:15,930 --> 00:02:19,940 OK, now N sub x is going to have two parts. 48 00:02:19,940 --> 00:02:24,720 N sub x, the derivative with respect to x of this is 1. 49 00:02:24,720 --> 00:02:26,540 And so I have an e to the x plus y. 50 00:02:26,540 --> 00:02:29,120 51 00:02:29,120 --> 00:02:31,750 And the derivative with respect to the x of e to the x 52 00:02:31,750 --> 00:02:34,310 plus y is just e to the plus y. 53 00:02:34,310 --> 00:02:37,300 For the same reason as the derivative with respect to y 54 00:02:37,300 --> 00:02:38,400 was the same. 55 00:02:38,400 --> 00:02:42,690 So then I'm just going to get a plus x e to the x plus y. 56 00:02:42,690 --> 00:02:46,440 So that means if I factor that out, I get an e to the x plus 57 00:02:46,440 --> 00:02:49,360 y times 1 plus x. 58 00:02:49,360 --> 00:02:52,540 And we see that if F is going to be a gradient field then b 59 00:02:52,540 --> 00:02:53,700 has to equal 1. 60 00:02:53,700 --> 00:02:55,930 Because it can only have one value, and so b 61 00:02:55,930 --> 00:02:58,830 has to equal 1. 62 00:02:58,830 --> 00:03:03,840 To get N sub x to equal M sub y, b has to equal 1. 63 00:03:03,840 --> 00:03:07,050 So now what I'm going to do is I'm going to erase that b, put 64 00:03:07,050 --> 00:03:11,510 in a 1, so that the rest of my calculations refer to that. 65 00:03:11,510 --> 00:03:15,830 So now the second part said for this b find a potential 66 00:03:15,830 --> 00:03:18,530 function f using both methods from the lecture. 67 00:03:18,530 --> 00:03:21,370 So we're going to go through both methods and hopefully we 68 00:03:21,370 --> 00:03:23,220 get the same answer both times. 69 00:03:23,220 --> 00:03:24,890 So let me come back here. 70 00:03:24,890 --> 00:03:28,220 71 00:03:28,220 --> 00:03:31,830 The first method is where I'm integrating along a curve from 72 00:03:31,830 --> 00:03:34,460 (0, 0) to (x1, y1). 73 00:03:34,460 --> 00:03:37,440 So I'm going to do it in the following way. 74 00:03:37,440 --> 00:03:39,620 I'm going to let C1-- 75 00:03:39,620 --> 00:03:41,446 so here's (0, 0)-- 76 00:03:41,446 --> 00:03:45,250 I'm going to let C1 be the curve from (0, 77 00:03:45,250 --> 00:03:48,110 0) up to (0, y1). 78 00:03:48,110 --> 00:03:50,010 And then C2 be the curve-- 79 00:03:50,010 --> 00:03:51,240 so it's parameterized in that direction-- 80 00:03:51,240 --> 00:03:58,600 C2 be the curve from (0, y1) to (x1, y1). 81 00:03:58,600 --> 00:04:00,460 OK? 82 00:04:00,460 --> 00:04:03,580 So that's what I'm going to do and I'm going to let C equal 83 00:04:03,580 --> 00:04:05,190 the curve C1 plus C2. 84 00:04:05,190 --> 00:04:07,160 So I'm going to have C be the full curve. 85 00:04:07,160 --> 00:04:11,770 And what I'm interested in doing is finding f of (x1, 86 00:04:11,770 --> 00:04:21,720 y1), which will just equal the integral along C of f dot dr. 87 00:04:21,720 --> 00:04:23,670 So now we need to figure out some important things 88 00:04:23,670 --> 00:04:25,900 about C1 and C2. 89 00:04:25,900 --> 00:04:28,620 What's happening on C1 and what's happening on C2. 90 00:04:28,620 --> 00:04:30,870 And the first thing I want to point out-- actually, before I 91 00:04:30,870 --> 00:04:32,690 do that, let me remind you that this is going to be the 92 00:04:32,690 --> 00:04:39,260 integral on C of M dx plus N dy. 93 00:04:39,260 --> 00:04:41,010 So this will be helpful to refer back to. 94 00:04:41,010 --> 00:04:44,830 That's really what we're also doing here. 95 00:04:44,830 --> 00:04:47,390 So on C1, what do I notice? 96 00:04:47,390 --> 00:04:54,050 On C1, x is 0 and dx is 0. 97 00:04:54,050 --> 00:04:56,690 And y goes between 0 and y1. 98 00:04:56,690 --> 00:04:59,280 99 00:04:59,280 --> 00:05:05,740 And then on C2, y is equal to y1. 100 00:05:05,740 --> 00:05:12,770 So dy is equal to 0 and x is going between 0 and x1. 101 00:05:12,770 --> 00:05:16,220 So those are the values that are important to me. 102 00:05:16,220 --> 00:05:19,950 So if you notice from this fact and this fact, we see 103 00:05:19,950 --> 00:05:24,550 that if we look at the integral just along C1, 104 00:05:24,550 --> 00:05:26,840 there's going to be no M dx term. 105 00:05:26,840 --> 00:05:31,190 And if we look at the integral along C2, there's going to be 106 00:05:31,190 --> 00:05:33,580 no dy term because of that. 107 00:05:33,580 --> 00:05:36,690 So let me write down the terms that do exist, and we'll see 108 00:05:36,690 --> 00:05:38,170 some other things drop out also. 109 00:05:38,170 --> 00:05:40,990 110 00:05:40,990 --> 00:05:45,510 If I look along first just C1, I'm only going to get-- 111 00:05:45,510 --> 00:05:48,980 I said the dy term, which-- 112 00:05:48,980 --> 00:05:50,470 let me just make sure-- dx is 0. 113 00:05:50,470 --> 00:05:52,890 I'm only going to get the dy term, which is-- 114 00:05:52,890 --> 00:05:54,100 well, x is 0 there. 115 00:05:54,100 --> 00:05:58,310 So I'm going to get 0 times e to the 0 plus y. 116 00:05:58,310 --> 00:05:59,270 dy. 117 00:05:59,270 --> 00:06:00,930 From 0 to y1. 118 00:06:00,930 --> 00:06:03,110 Well that's nice and easy to calculate, thank goodness. 119 00:06:03,110 --> 00:06:04,620 That's just 0. 120 00:06:04,620 --> 00:06:08,440 So all I have to do for this one is just leave it at 0. 121 00:06:08,440 --> 00:06:10,690 That's everything that happens along C1. 122 00:06:10,690 --> 00:06:12,220 That's what I'm interested in. 123 00:06:12,220 --> 00:06:13,840 I just get 0 there. 124 00:06:13,840 --> 00:06:17,940 And if I integrate along C2, as I mentioned, dy is 0. 125 00:06:17,940 --> 00:06:21,490 So we don't have any component with N dy. 126 00:06:21,490 --> 00:06:25,530 We just have the component M dx that we're integrating. 127 00:06:25,530 --> 00:06:29,805 OK, so if I integrate along C2, I just have M dx and M is 128 00:06:29,805 --> 00:06:32,610 e to the x plus y times x plus 1. 129 00:06:32,610 --> 00:06:34,660 And y is fixed at y1. 130 00:06:34,660 --> 00:06:40,620 So it's e to the x plus y1 times x plus 1 dx. 131 00:06:40,620 --> 00:06:44,540 And I'm going from 0 to x1. 132 00:06:44,540 --> 00:06:46,210 I'm going to make sure I didn't make any mistakes. 133 00:06:46,210 --> 00:06:48,130 I'm going to check my work here. 134 00:06:48,130 --> 00:06:49,520 Yes, I'm looking good. 135 00:06:49,520 --> 00:06:50,020 OK. 136 00:06:50,020 --> 00:06:51,550 So this one is 0. 137 00:06:51,550 --> 00:06:54,800 So all I have to do is find an antiderivative of this. 138 00:06:54,800 --> 00:06:58,830 And the term-- if I multiply through, I see that here I get 139 00:06:58,830 --> 00:07:01,610 exactly the same thing when I look for an antiderivative. 140 00:07:01,610 --> 00:07:04,960 And here I get, I believe, two terms when I look for an 141 00:07:04,960 --> 00:07:05,420 antiderivative. 142 00:07:05,420 --> 00:07:08,180 But I'm going to get some cancellation. 143 00:07:08,180 --> 00:07:11,600 And ultimately, when I'm all done I'm going to get this. 144 00:07:11,600 --> 00:07:16,660 xe to the x plus y1 evaluated at 0 and x1. 145 00:07:16,660 --> 00:07:18,050 You could do this. 146 00:07:18,050 --> 00:07:19,900 This is really now a single variable problem. 147 00:07:19,900 --> 00:07:23,210 So I'm not going to work out all the details, but you might 148 00:07:23,210 --> 00:07:25,930 want to do an integration by parts on that first part of 149 00:07:25,930 --> 00:07:26,740 it, if that helps. 150 00:07:26,740 --> 00:07:29,260 Or an integration by parts on the whole thing. 151 00:07:29,260 --> 00:07:31,820 That would also do the trick. 152 00:07:31,820 --> 00:07:32,790 So what do I get here? 153 00:07:32,790 --> 00:07:37,800 Then I get x1 e to the x1 plus y1. 154 00:07:37,800 --> 00:07:40,800 And then when I put in 0 for x here, I get 0, so that's it. 155 00:07:40,800 --> 00:07:47,430 So this, plus possibly a constant, is equal to my f. 156 00:07:47,430 --> 00:07:52,280 So I see that in general I get f of xy is equal to xe to the 157 00:07:52,280 --> 00:07:55,080 x plus y plus a constant. 158 00:07:55,080 --> 00:07:58,060 So that's what I get in the first method. 159 00:07:58,060 --> 00:08:01,594 So now let's use the second method. 160 00:08:01,594 --> 00:08:05,770 So I should say f of (x1, y1). 161 00:08:05,770 --> 00:08:08,900 In the second method, what I do is M, remember, is 162 00:08:08,900 --> 00:08:10,550 equal to f sub x. 163 00:08:10,550 --> 00:08:13,890 164 00:08:13,890 --> 00:08:23,020 So f sub x is equal to M which is equal to e to the x plus y 165 00:08:23,020 --> 00:08:26,350 times x plus 1. 166 00:08:26,350 --> 00:08:28,440 So if I want to find an antiderivative-- 167 00:08:28,440 --> 00:08:32,230 if I want to find f, I should take an antiderivative, right? 168 00:08:32,230 --> 00:08:33,930 With respect to x. 169 00:08:33,930 --> 00:08:35,820 And so notice I already did that, actually. 170 00:08:35,820 --> 00:08:38,450 If I just put this as y, I already did that here. 171 00:08:38,450 --> 00:08:41,851 And so I should get something that looks like this. 172 00:08:41,851 --> 00:08:46,490 xe to the x plus y plus possibly a function that only 173 00:08:46,490 --> 00:08:47,880 depends on y. 174 00:08:47,880 --> 00:08:50,400 And the reason is when I take a derivative with respect to x 175 00:08:50,400 --> 00:08:52,500 of this, obviously this would be 0. 176 00:08:52,500 --> 00:08:53,680 So it doesn't show up over here. 177 00:08:53,680 --> 00:08:57,850 So this, we make sure that-- 178 00:08:57,850 --> 00:08:59,660 oh, that I shouldn't write equals. 179 00:08:59,660 --> 00:09:00,610 Sorry. 180 00:09:00,610 --> 00:09:02,260 That I shouldn't write equals. 181 00:09:02,260 --> 00:09:03,110 OK? 182 00:09:03,110 --> 00:09:06,030 This would imply that this is equal to f. 183 00:09:06,030 --> 00:09:07,310 Sorry about that. 184 00:09:07,310 --> 00:09:09,790 f sub x was equal to M was equal to this. 185 00:09:09,790 --> 00:09:10,560 That implies-- 186 00:09:10,560 --> 00:09:12,530 when I take an antiderivative of an x-- 187 00:09:12,530 --> 00:09:16,520 that xe to the x plus y plus g of y is equal to f. 188 00:09:16,520 --> 00:09:17,670 So I apologize. 189 00:09:17,670 --> 00:09:19,740 That wouldn't have been an equals because obviously those 190 00:09:19,740 --> 00:09:20,860 two things are not equal. 191 00:09:20,860 --> 00:09:22,720 That would imply, I think-- 192 00:09:22,720 --> 00:09:25,130 yeah, that would imply something very bad 193 00:09:25,130 --> 00:09:26,530 mathematically. 194 00:09:26,530 --> 00:09:29,330 So make sure you understand I put an equals sign where I 195 00:09:29,330 --> 00:09:32,600 should not have. This is actually a derivative of that. 196 00:09:32,600 --> 00:09:35,790 So this is antiderivative of this. 197 00:09:35,790 --> 00:09:38,690 So now I have a candidate for f. 198 00:09:38,690 --> 00:09:40,680 And so now I'm going to take the derivative of that. 199 00:09:40,680 --> 00:09:44,700 And what's the derivative of that with respect to y? 200 00:09:44,700 --> 00:09:51,920 So f sub y based on this is going to be equal to xe to the 201 00:09:51,920 --> 00:09:55,650 x plus y plus g prime of y. 202 00:09:55,650 --> 00:10:00,060 So the prime here indicates it's in a derivative in y. 203 00:10:00,060 --> 00:10:07,470 And now that f sub y should also equal N. And N equals xe 204 00:10:07,470 --> 00:10:09,820 to the x plus y. 205 00:10:09,820 --> 00:10:11,270 So what do I get here? 206 00:10:11,270 --> 00:10:14,930 I see xe to the x plus y has to equal xe to the x plus y 207 00:10:14,930 --> 00:10:17,050 plus g prime of y. 208 00:10:17,050 --> 00:10:21,300 Which means g prime of y is equal to 0. 209 00:10:21,300 --> 00:10:24,460 Which means when I take an antiderivative of that I just 210 00:10:24,460 --> 00:10:25,710 get a constant. 211 00:10:25,710 --> 00:10:27,620 212 00:10:27,620 --> 00:10:29,590 That means g of y was a constant. 213 00:10:29,590 --> 00:10:37,390 So that implies that this boxed expression right here is 214 00:10:37,390 --> 00:10:41,350 f of xy if g of y is just a constant. 215 00:10:41,350 --> 00:10:43,640 So let me go through that logic one more time. 216 00:10:43,640 --> 00:10:45,190 I had f sub x. 217 00:10:45,190 --> 00:10:48,360 I took an antiderivative to get f but it involved a 218 00:10:48,360 --> 00:10:50,960 constant in x that was a function of y. 219 00:10:50,960 --> 00:10:54,650 I take a derivative of that in y. 220 00:10:54,650 --> 00:10:57,890 I compare that to what I know the derivative is in y. 221 00:10:57,890 --> 00:10:59,770 That gives me that this is 0. 222 00:10:59,770 --> 00:11:03,090 So its antiderivative, which is g of y, is just a constant. 223 00:11:03,090 --> 00:11:08,420 And so all together this implies that f of xy is equal 224 00:11:08,420 --> 00:11:12,970 to xe to the x plus y plus a constant. 225 00:11:12,970 --> 00:11:15,060 Which is exactly what I got before. 226 00:11:15,060 --> 00:11:18,410 Fortunately, I got two answers that are the same. 227 00:11:18,410 --> 00:11:19,700 So that's it. 228 00:11:19,700 --> 00:11:21,550 I'll stop there.