1 00:00:00,040 --> 00:00:02,460 The following content is provided under a Creative 2 00:00:02,460 --> 00:00:03,870 Commons license. 3 00:00:03,870 --> 00:00:06,320 Your support will help MIT OpenCourseWare 4 00:00:06,320 --> 00:00:10,560 continue to offer high quality educational resources for free. 5 00:00:10,560 --> 00:00:13,300 To make a donation or view additional materials 6 00:00:13,300 --> 00:00:17,210 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,210 --> 00:00:19,650 at ocw.mit.edu. 8 00:00:33,930 --> 00:00:35,140 PROFESSOR: Hi. 9 00:00:35,140 --> 00:00:37,780 Today's lecture starts off with what 10 00:00:37,780 --> 00:00:41,260 probably is the single most important topic 11 00:00:41,260 --> 00:00:44,100 in the entire course, not just to date, 12 00:00:44,100 --> 00:00:45,920 but in the entire course. 13 00:00:45,920 --> 00:00:49,720 And it's a rather sneaky thing in the sense that 14 00:00:49,720 --> 00:00:51,640 takes a while to grow into it. 15 00:00:51,640 --> 00:00:55,390 Things have been going so smoothly so far, that perhaps 16 00:00:55,390 --> 00:00:58,250 this point is going to be a little bit subtle to make, 17 00:00:58,250 --> 00:01:00,770 and we'll try to lead into it gradually. 18 00:01:00,770 --> 00:01:05,620 There are probably ten different ways to introduce this topic. 19 00:01:05,620 --> 00:01:08,150 And whichever one you pick, I think, 20 00:01:08,150 --> 00:01:10,680 any one of the other nine would've been better. 21 00:01:10,680 --> 00:01:15,510 But without any further ado, let me talk about today's lesson 22 00:01:15,510 --> 00:01:19,770 in terms of something called the directional derivative 23 00:01:19,770 --> 00:01:22,360 and indicate, in a manner of speaking, 24 00:01:22,360 --> 00:01:25,590 that when we deal with functions of several variables, 25 00:01:25,590 --> 00:01:28,020 especially in the case that n equals 2, 26 00:01:28,020 --> 00:01:32,150 there is a very obvious geometrical interpretation, one 27 00:01:32,150 --> 00:01:35,380 that we made some use of in the previous lecture, 28 00:01:35,380 --> 00:01:37,010 but which I hope to make more use 29 00:01:37,010 --> 00:01:39,430 of in this particular lecture. 30 00:01:39,430 --> 00:01:42,300 Let's take a look and see what the situation is. 31 00:01:42,300 --> 00:01:45,120 Let's suppose I'm given that w is a function of the two 32 00:01:45,120 --> 00:01:49,300 independent variables x and y, and I have that w is therefore 33 00:01:49,300 --> 00:01:51,260 some function of x and y. 34 00:01:51,260 --> 00:01:54,310 And I can now talk about the graph of w, which 35 00:01:54,310 --> 00:01:57,870 we assume is some surface. 36 00:01:57,870 --> 00:02:00,900 And I've drawn it in this particular way. 37 00:02:00,900 --> 00:02:04,960 I take some point a comma b in the xy-plane 38 00:02:04,960 --> 00:02:07,170 on which f is defined. 39 00:02:07,170 --> 00:02:11,820 And at that particular point, I go to the corresponding point 40 00:02:11,820 --> 00:02:13,960 on the surface, which I call P_0, 41 00:02:13,960 --> 00:02:15,530 which has coordinates what? 42 00:02:15,530 --> 00:02:20,820 a, b, and c, where c is simply f of a comma b. 43 00:02:20,820 --> 00:02:23,850 That, of course, is how you graph a function 44 00:02:23,850 --> 00:02:25,910 of two independent variables. 45 00:02:25,910 --> 00:02:30,710 The w coordinate, so to speak, is simply the functional value 46 00:02:30,710 --> 00:02:34,050 of the x- and y-coordinates. 47 00:02:34,050 --> 00:02:38,140 Now, what we did last time was we essentially said, lookit, 48 00:02:38,140 --> 00:02:42,470 if we were to slice this surface by a plane which 49 00:02:42,470 --> 00:02:47,240 was either parallel to the wy-plane or to the wx-plane, 50 00:02:47,240 --> 00:02:49,980 we get special curves of intersection, 51 00:02:49,980 --> 00:02:54,810 and we use this to indicate the idea of partial derivatives. 52 00:02:54,810 --> 00:02:56,997 Now, without too much imagination, 53 00:02:56,997 --> 00:02:58,580 I think it should be very easy for you 54 00:02:58,580 --> 00:03:03,390 to visualize an arbitrary surface over your heads. 55 00:03:03,390 --> 00:03:06,230 You're standing in a particular point on the floor. 56 00:03:06,230 --> 00:03:10,290 At that point on the floor, you visualize a coordinate axis, 57 00:03:10,290 --> 00:03:12,670 which we'll call the x- and y-axis. 58 00:03:12,670 --> 00:03:15,970 Now, obviously, you can move out from the point at which you're 59 00:03:15,970 --> 00:03:18,940 standing along the x-axis. 60 00:03:18,940 --> 00:03:22,260 You can move out in the direction along the y-axis. 61 00:03:22,260 --> 00:03:24,350 You can visualize that how rapidly 62 00:03:24,350 --> 00:03:28,190 the height above your head is changing as you move out 63 00:03:28,190 --> 00:03:30,420 certainly will depend on whether you're moving 64 00:03:30,420 --> 00:03:33,170 along the x-axis or the y-axis. 65 00:03:33,170 --> 00:03:35,840 But the key point-- at least, the key point 66 00:03:35,840 --> 00:03:38,150 from the point of view of today's lecture-- 67 00:03:38,150 --> 00:03:41,070 is the fact that why were you restricted to either 68 00:03:41,070 --> 00:03:42,370 of these two directions? 69 00:03:42,370 --> 00:03:44,620 Why couldn't you have moved in any one 70 00:03:44,620 --> 00:03:47,020 of infinitely many different directions, 71 00:03:47,020 --> 00:03:50,030 namely from the point at which you're originating, 72 00:03:50,030 --> 00:03:53,040 you can move in any direction at all, because we're assuming, 73 00:03:53,040 --> 00:03:55,200 at least, that the floor on which you're standing 74 00:03:55,200 --> 00:03:56,440 is continuous. 75 00:03:56,440 --> 00:03:57,310 It's unbroken. 76 00:03:57,310 --> 00:03:59,230 You can move in these directions. 77 00:03:59,230 --> 00:04:02,610 And the idea that you get can be shown pictorially 78 00:04:02,610 --> 00:04:04,900 by saying, lookit. 79 00:04:04,900 --> 00:04:07,570 Let's assume that we're at the point a comma b. 80 00:04:07,570 --> 00:04:11,250 We have this surface over our head just as we did before. 81 00:04:11,250 --> 00:04:15,620 But now, instead of picking a direction either parallel 82 00:04:15,620 --> 00:04:19,459 to the wx-plane or to the wy-plane, 83 00:04:19,459 --> 00:04:24,230 let's pick some arbitrary direction s in which to move. 84 00:04:24,230 --> 00:04:26,620 And now, what I'm saying is if I now 85 00:04:26,620 --> 00:04:32,470 take the plane that passes through this direction 86 00:04:32,470 --> 00:04:37,370 s perpendicular to the xy-plane, that plane will also 87 00:04:37,370 --> 00:04:41,580 intersect this surface, passing through the point P_0. 88 00:04:41,580 --> 00:04:43,790 I get a curve of intersection, and I 89 00:04:43,790 --> 00:04:47,242 can talk about the slope of that particular curve. 90 00:04:47,242 --> 00:04:48,700 See, ultimately, this is what we're 91 00:04:48,700 --> 00:04:49,783 going to be talking about. 92 00:04:49,783 --> 00:04:52,180 Now, what is the slope of that particular curve? 93 00:04:52,180 --> 00:04:54,500 Well it's a derivative. 94 00:04:54,500 --> 00:04:58,060 It's a derivative of w with respect to s, 95 00:04:58,060 --> 00:05:01,090 where all of a sudden we've now chosen the s direction 96 00:05:01,090 --> 00:05:03,120 over here to be what we're going to take 97 00:05:03,120 --> 00:05:05,110 our derivative with respect to. 98 00:05:05,110 --> 00:05:07,720 In other words, it seems to make sense 99 00:05:07,720 --> 00:05:12,890 to talk about the derivative of f in the direction of s, 100 00:05:12,890 --> 00:05:15,320 evaluated at the point a comma b. 101 00:05:15,320 --> 00:05:18,260 In other words, just like we talked about f sub x of a comma 102 00:05:18,260 --> 00:05:21,160 b and f sub y of a comma b, why can't we 103 00:05:21,160 --> 00:05:26,220 talk about the derivative of f evaluated at a comma b, 104 00:05:26,220 --> 00:05:28,050 in the direction of s? 105 00:05:28,050 --> 00:05:31,620 And if we want to correlate this with the notation in the text, 106 00:05:31,620 --> 00:05:35,420 observe that we call that dw/ds. 107 00:05:35,420 --> 00:05:39,440 By the way, notice-- not the partial of w with respect to s, 108 00:05:39,440 --> 00:05:43,570 but the derivative of w with respect to s. 109 00:05:43,570 --> 00:05:46,780 Remember, the partials were essentially 110 00:05:46,780 --> 00:05:51,190 defined in terms of holding either x or y constant. 111 00:05:51,190 --> 00:05:55,820 Notice that in our general directional derivative idea, 112 00:05:55,820 --> 00:05:59,470 neither x nor y is being held constant. 113 00:05:59,470 --> 00:06:04,910 Notice that x and y are varying as we move along s. 114 00:06:04,910 --> 00:06:08,730 Notice, also by the way, that once x and y are restricted 115 00:06:08,730 --> 00:06:12,930 to move along s, they are no longer independent variables. 116 00:06:12,930 --> 00:06:14,650 It's a rather interesting point. 117 00:06:14,650 --> 00:06:17,410 We talk about w being a function of the two 118 00:06:17,410 --> 00:06:19,990 independent variables x and y, but as soon 119 00:06:19,990 --> 00:06:23,710 as we pick that direction s in the xy-plane, 120 00:06:23,710 --> 00:06:26,709 x and y have to be very specially related-- namely, 121 00:06:26,709 --> 00:06:28,750 according to the equation of a straight line that 122 00:06:28,750 --> 00:06:32,450 determines s-- for that point to be on the line. 123 00:06:32,450 --> 00:06:35,016 And that's why we talk about dw/ds. 124 00:06:35,016 --> 00:06:38,040 w is a function of a single variable 125 00:06:38,040 --> 00:06:41,110 once you restrict yourself to the particular direction s. 126 00:06:41,110 --> 00:06:43,970 At any rate, the question that we want to come to grips with 127 00:06:43,970 --> 00:06:46,950 is, how do you find dw/ds? 128 00:06:46,950 --> 00:06:50,640 Well, here again is the beauty of our logical structure. 129 00:06:50,640 --> 00:06:53,890 By definition-- this is the same definition we knew early 130 00:06:53,890 --> 00:06:55,605 in part one of our course. 131 00:06:55,605 --> 00:06:59,890 dw/ds, by definition, is the limit as delta 132 00:06:59,890 --> 00:07:05,450 s approaches 0, delta w divided by delta s. 133 00:07:05,450 --> 00:07:08,790 Now, the thing is, what's very difficult to compute 134 00:07:08,790 --> 00:07:10,420 in real life is delta w. 135 00:07:10,420 --> 00:07:13,465 After all, this w, being f of x, y, 136 00:07:13,465 --> 00:07:17,030 f can be a very, very complicated surface. 137 00:07:17,030 --> 00:07:20,360 And to actually find the true change in w-- well, heck. 138 00:07:20,360 --> 00:07:25,060 We already saw this in part one when we compared this delta y 139 00:07:25,060 --> 00:07:26,540 with delta y tan. 140 00:07:26,540 --> 00:07:30,800 To actually find the change in y was a much more difficult job 141 00:07:30,800 --> 00:07:33,470 than to find the change in y to the tangent line. 142 00:07:33,470 --> 00:07:35,490 What we're saying here is, lookit. 143 00:07:35,490 --> 00:07:39,340 We already know how to find the change in w, 144 00:07:39,340 --> 00:07:42,080 not to the surface, but to the plane which 145 00:07:42,080 --> 00:07:45,650 is tangent to the surface at our point P sub 0. 146 00:07:45,650 --> 00:07:47,820 In other words, we have already discussed 147 00:07:47,820 --> 00:07:51,100 that delta w tan is the partial of f 148 00:07:51,100 --> 00:07:54,830 with respect to x evaluated at a comma b times delta x, 149 00:07:54,830 --> 00:07:58,270 plus the partial of f with respect to y at the point 150 00:07:58,270 --> 00:08:01,530 a comma b times delta y. 151 00:08:01,530 --> 00:08:03,780 And now, what we say is-- and I've 152 00:08:03,780 --> 00:08:06,260 written this to accentuate it, because I 153 00:08:06,260 --> 00:08:08,010 have to talk very strongly about this. 154 00:08:08,010 --> 00:08:11,960 It's a point that, if I don't make, most of you, at least, 155 00:08:11,960 --> 00:08:14,680 will allow me to slip over this and not even notice 156 00:08:14,680 --> 00:08:16,940 that I've missed something very, very crucial. 157 00:08:16,940 --> 00:08:21,780 But let me assume that I can approximate delta w by delta w 158 00:08:21,780 --> 00:08:22,280 tan. 159 00:08:22,280 --> 00:08:26,850 In other words, let's suppose that delta w 160 00:08:26,850 --> 00:08:30,440 tan is a reasonable approximation for delta w. 161 00:08:30,440 --> 00:08:33,549 You say, well, what's such a big assumption about that? 162 00:08:33,549 --> 00:08:35,309 And I'm going to save that for the very 163 00:08:35,309 --> 00:08:38,650 last part of the lecture, because my belief is 164 00:08:38,650 --> 00:08:41,950 that the subtlety is so great that I would like to leave that 165 00:08:41,950 --> 00:08:45,630 for the very end, and go through as if the subtlety didn't exist 166 00:08:45,630 --> 00:08:47,430 so that you get the computational feeling 167 00:08:47,430 --> 00:08:48,790 as to what's going on here. 168 00:08:48,790 --> 00:08:50,660 But here's the interesting point. 169 00:08:50,660 --> 00:08:55,590 Notice that delta w is a change in w as you move from the point 170 00:08:55,590 --> 00:08:58,280 a comma b to some other point in the plane. 171 00:08:58,280 --> 00:08:59,880 It's a change in height. 172 00:08:59,880 --> 00:09:02,080 Now, obviously, that change in w is 173 00:09:02,080 --> 00:09:04,980 going to depend very strongly on what direction 174 00:09:04,980 --> 00:09:06,060 you're moving in. 175 00:09:06,060 --> 00:09:09,780 On the other hand, how was delta w tan computed? 176 00:09:09,780 --> 00:09:17,140 Delta w tan was computed just by knowing 177 00:09:17,140 --> 00:09:19,200 two special directional derivatives known 178 00:09:19,200 --> 00:09:21,300 as the partial derivatives. 179 00:09:21,300 --> 00:09:23,820 You see, notice that to get delta w 180 00:09:23,820 --> 00:09:27,310 tan, I have made the assumption that all I have to know 181 00:09:27,310 --> 00:09:29,440 is what's happening in the x direction 182 00:09:29,440 --> 00:09:32,240 and what's happening in the y direction, everything 183 00:09:32,240 --> 00:09:36,010 else being determined from the function evaluated at a comma 184 00:09:36,010 --> 00:09:36,750 b. 185 00:09:36,750 --> 00:09:39,720 So this is really a very strong assumption, 186 00:09:39,720 --> 00:09:44,800 that delta w is determined pretty closely by delta w tan. 187 00:09:44,800 --> 00:09:48,130 And it turns out to be almost universally true 188 00:09:48,130 --> 00:09:50,990 and we're going to save that part, as I say, for the end. 189 00:09:50,990 --> 00:09:53,110 But for now, let's suppose that we 190 00:09:53,110 --> 00:09:57,530 are allowed to replace delta w by delta w tan. 191 00:09:57,530 --> 00:10:02,550 If we do that, and we go back to our definition for dw/ds-- 192 00:10:02,550 --> 00:10:05,350 namely, the limit as delta s approaches 0, 193 00:10:05,350 --> 00:10:08,450 delta w divided by delta s-- we now 194 00:10:08,450 --> 00:10:14,130 replace delta w by delta w tan and divide through by delta 195 00:10:14,130 --> 00:10:17,480 s to find delta w over delta s. 196 00:10:17,480 --> 00:10:22,090 We have, assuming that this is a legal substitution, 197 00:10:22,090 --> 00:10:27,160 that delta w over delta s is the partial of f with respect to x, 198 00:10:27,160 --> 00:10:31,450 evaluated at a comma b, times delta x divided by delta 199 00:10:31,450 --> 00:10:34,250 s plus the partial of f with respect to y 200 00:10:34,250 --> 00:10:39,180 evaluated at a comma b times the change in y 201 00:10:39,180 --> 00:10:42,990 divided by the change in s-- delta y divided by delta s. 202 00:10:42,990 --> 00:10:46,450 Now, remember, in the s direction, x and y 203 00:10:46,450 --> 00:10:48,330 are not independent variables. 204 00:10:48,330 --> 00:10:50,560 In fact, how are they related? 205 00:10:50,560 --> 00:10:53,960 Let's isolate our little diagram here so 206 00:10:53,960 --> 00:10:56,760 that we see in what direction we're moving here. 207 00:10:56,760 --> 00:10:58,780 We're starting at the point a comma b. 208 00:10:58,780 --> 00:11:00,860 And by the way, to get the proper orientation 209 00:11:00,860 --> 00:11:04,290 here, as I've drawn the xy-plane here, 210 00:11:04,290 --> 00:11:08,150 imagine the surface coming out from the blackboard. 211 00:11:08,150 --> 00:11:12,000 In other words, the height is really being measured away 212 00:11:12,000 --> 00:11:13,330 from the blackboard here. 213 00:11:13,330 --> 00:11:14,850 That's where my surface is. 214 00:11:14,850 --> 00:11:16,920 I move in the direction of s. 215 00:11:16,920 --> 00:11:18,630 Here is a delta x. 216 00:11:18,630 --> 00:11:20,762 Here is a delta y. 217 00:11:20,762 --> 00:11:23,740 s has a constant direction, a constant slope. 218 00:11:23,740 --> 00:11:25,030 It's a straight line. 219 00:11:25,030 --> 00:11:28,940 Call the angle that it makes with the positive x-axis phi. 220 00:11:28,940 --> 00:11:32,950 Notice that no matter how big delta x and delta y are, 221 00:11:32,950 --> 00:11:36,280 they are related by similar triangles to what? 222 00:11:36,280 --> 00:11:40,720 The delta x divided by delta s will always be cosine phi, 223 00:11:40,720 --> 00:11:45,300 and delta y divided by delta s will always be sine phi. 224 00:11:45,300 --> 00:11:48,890 And therefore, if I replace delta x divided by delta 225 00:11:48,890 --> 00:11:53,640 s by cosine phi, delta y divided by delta s by sine phi, 226 00:11:53,640 --> 00:11:56,050 I obtain that-- what? 227 00:11:56,050 --> 00:12:01,110 Delta w divided by delta s is equal to the partial 228 00:12:01,110 --> 00:12:04,730 of f with respect to x evaluated at point a comma b 229 00:12:04,730 --> 00:12:08,460 times the cosine of phi, plus the partial of f with respect 230 00:12:08,460 --> 00:12:12,060 to y evaluated at a comma b times the sine of phi, 231 00:12:12,060 --> 00:12:13,720 where phi is the what? 232 00:12:13,720 --> 00:12:18,444 The angle that the direction s makes with the positive x-axis. 233 00:12:18,444 --> 00:12:20,860 At any rate, notice, by the way, that these things are all 234 00:12:20,860 --> 00:12:25,710 constants once s is chosen, once the point a comma b is fixed. 235 00:12:25,710 --> 00:12:29,720 Therefore, when I pass to the limit, nothing really changes. 236 00:12:29,720 --> 00:12:31,590 In other words, this thing that I'm 237 00:12:31,590 --> 00:12:34,610 calling the directional derivative of f 238 00:12:34,610 --> 00:12:38,870 in the direction of s evaluated at a comma b 239 00:12:38,870 --> 00:12:40,350 turns out to be what? 240 00:12:40,350 --> 00:12:43,950 The partial of f with respect to x at the point a comma b times 241 00:12:43,950 --> 00:12:47,970 cosine phi, plus the partial of f with respect to y 242 00:12:47,970 --> 00:12:51,220 evaluated at a comma b times the sine of phi. 243 00:12:51,220 --> 00:12:55,130 And what I'd like you to notice is that in these two terms, 244 00:12:55,130 --> 00:12:58,760 one factor of each term is determined solely 245 00:12:58,760 --> 00:13:00,750 by the point a comma b. 246 00:13:00,750 --> 00:13:04,580 In other words, notice that these two factors here 247 00:13:04,580 --> 00:13:06,880 are just partial derivatives and have nothing 248 00:13:06,880 --> 00:13:07,830 to do with direction. 249 00:13:07,830 --> 00:13:10,630 They're determined solely by the choice of the function 250 00:13:10,630 --> 00:13:13,340 f and the point a comma b. 251 00:13:13,340 --> 00:13:17,330 On the other hand, notice that these two factors, cosine phi 252 00:13:17,330 --> 00:13:20,300 and sine phi, have nothing to do with f 253 00:13:20,300 --> 00:13:23,270 and have to do only with the direction 254 00:13:23,270 --> 00:13:27,000 s itself, which, again, should make intuitive sense to you. 255 00:13:27,000 --> 00:13:29,470 That if you're asking, in terms of the surface 256 00:13:29,470 --> 00:13:32,560 being up over your head, for a directional derivative, 257 00:13:32,560 --> 00:13:35,070 obviously, once the surface is given, 258 00:13:35,070 --> 00:13:36,460 the directional derivative should 259 00:13:36,460 --> 00:13:40,880 depend on two things: one, what point you start at, 260 00:13:40,880 --> 00:13:44,950 and the other, what direction you move in once you've started 261 00:13:44,950 --> 00:13:46,340 at that particular point. 262 00:13:46,340 --> 00:13:47,972 And what direction you move in has 263 00:13:47,972 --> 00:13:49,680 nothing to do with what the surface looks 264 00:13:49,680 --> 00:13:51,410 like over your head. 265 00:13:51,410 --> 00:13:55,600 At any rate, what we now do is invoke our dot product 266 00:13:55,600 --> 00:13:57,990 notation-- in other words, this little trick 267 00:13:57,990 --> 00:14:00,420 that we talked about when we learned dot products. 268 00:14:00,420 --> 00:14:06,400 This particular sum can be written very suggestively 269 00:14:06,400 --> 00:14:09,070 as the dot product of two vectors. 270 00:14:09,070 --> 00:14:11,550 Namely, I will write this as what? 271 00:14:11,550 --> 00:14:13,710 The vector whose components are f 272 00:14:13,710 --> 00:14:18,210 sub x and f sub y dotted with the vector whose components are 273 00:14:18,210 --> 00:14:21,080 cosine phi and sine phi, because remember, 274 00:14:21,080 --> 00:14:24,230 when you dot two vectors in Cartesian coordinates, 275 00:14:24,230 --> 00:14:27,670 you multiply them coefficient by coefficient. 276 00:14:27,670 --> 00:14:31,650 To make a long story short, dw/ds is what? 277 00:14:31,650 --> 00:14:34,960 It's this vector, whose i component 278 00:14:34,960 --> 00:14:39,040 is f sub x evaluated at a comma b, whose j complement is 279 00:14:39,040 --> 00:14:41,810 f sub y evaluated at a comma b. 280 00:14:41,810 --> 00:14:45,450 I call that vector g of a comma b. 281 00:14:45,450 --> 00:14:48,010 I'm going to give that a special name a little bit later. 282 00:14:48,010 --> 00:14:50,590 But for now, it's very important to notice 283 00:14:50,590 --> 00:14:52,460 that this is not a number. 284 00:14:52,460 --> 00:14:54,990 It's an ordered pair of numbers. 285 00:14:54,990 --> 00:14:57,134 In other words, it's a vector, and if you 286 00:14:57,134 --> 00:14:59,300 want to think of this as a vector, what we're saying 287 00:14:59,300 --> 00:15:00,000 is what? 288 00:15:00,000 --> 00:15:02,790 Think of the vector whose i component 289 00:15:02,790 --> 00:15:06,130 is the partial of f with respect to x evaluated at a comma 290 00:15:06,130 --> 00:15:11,370 b, whose j complement is f sub y evaluated at a comma b. 291 00:15:11,370 --> 00:15:13,450 Notice that these are numbers, because a 292 00:15:13,450 --> 00:15:15,530 and b are fixed constants here. 293 00:15:15,530 --> 00:15:20,210 Therefore, this vector, g of a comma b, is a constant vector, 294 00:15:20,210 --> 00:15:22,300 and it's in the xy-plane. 295 00:15:22,300 --> 00:15:24,000 It's a 2-tuple. 296 00:15:24,000 --> 00:15:27,490 On the other hand, the other vector, (cosine phi, sine phi), 297 00:15:27,490 --> 00:15:31,830 hopefully, you recognize by now is nothing more than the unit 298 00:15:31,830 --> 00:15:34,250 vector in the direction of s. 299 00:15:34,250 --> 00:15:36,760 You see, this is the unit vector in the direction of s. 300 00:15:36,760 --> 00:15:42,530 So if I now use my abbreviation, dw/ds evaluated at a comma 301 00:15:42,530 --> 00:15:47,120 b-- in other words, the directional derivative of w 302 00:15:47,120 --> 00:15:51,200 in the direction of s evaluated at a comma b-- is just 303 00:15:51,200 --> 00:15:55,580 my vector g of a comma b dotted with the unit vector 304 00:15:55,580 --> 00:15:57,110 in the direction of s. 305 00:15:57,110 --> 00:15:58,880 Now, observe have two things. 306 00:15:58,880 --> 00:16:01,920 First of all, when you dot, this is a constant. 307 00:16:01,920 --> 00:16:04,780 I can't change this once a and b are given. 308 00:16:04,780 --> 00:16:06,510 This is fixed. 309 00:16:06,510 --> 00:16:09,220 So all I can vary is u sub s. 310 00:16:09,220 --> 00:16:12,950 But u sub s is a unit vector, so the only way I can vary u 311 00:16:12,950 --> 00:16:15,240 sub s is to change its direction. 312 00:16:15,240 --> 00:16:19,010 Notice that for two vectors of constant magnitude, 313 00:16:19,010 --> 00:16:21,270 their dot product is maximum when 314 00:16:21,270 --> 00:16:23,210 the two vectors are parallel. 315 00:16:23,210 --> 00:16:26,610 In other words, this will be as big as possible 316 00:16:26,610 --> 00:16:32,310 when u sub s is chosen to be in the direction of my vector g. 317 00:16:32,310 --> 00:16:35,960 In other words, dw/ds at a comma b 318 00:16:35,960 --> 00:16:40,810 is maximum in the direction of g of a comma b. 319 00:16:40,810 --> 00:16:42,600 That's the first thing to observe. 320 00:16:42,600 --> 00:16:45,690 The second thing is that when they are parallel, 321 00:16:45,690 --> 00:16:48,200 the cosine of the angle between them is 1. 322 00:16:48,200 --> 00:16:50,760 So the magnitude of this vector will just 323 00:16:50,760 --> 00:16:53,180 be the product of these two magnitudes. 324 00:16:53,180 --> 00:16:56,730 But the magnitude of u sub s, u sub s being a unit vector, 325 00:16:56,730 --> 00:16:57,740 is 1. 326 00:16:57,740 --> 00:17:00,720 Therefore, the maximum magnitude not only 327 00:17:00,720 --> 00:17:05,490 occurs in the direction of g, but it is also numerically 328 00:17:05,490 --> 00:17:09,220 equal to the magnitude of g. 329 00:17:09,220 --> 00:17:13,660 In other words, the maximum value of dw/ds 330 00:17:13,660 --> 00:17:17,140 evaluated at a comma b not only occurs 331 00:17:17,140 --> 00:17:19,119 in the direction of the vector g, 332 00:17:19,119 --> 00:17:23,910 but that maximum magnitude is the magnitude of g evaluated 333 00:17:23,910 --> 00:17:25,400 at a comma b. 334 00:17:25,400 --> 00:17:30,600 For that reason, g of a comma b is given a very important name. 335 00:17:30,600 --> 00:17:34,790 And I decided to hold off on the name until as late as possible 336 00:17:34,790 --> 00:17:36,820 so that the name wouldn't frighten you. 337 00:17:36,820 --> 00:17:39,380 But the name is the gradient vector. 338 00:17:39,380 --> 00:17:42,280 In other words, the vector g of a comma b 339 00:17:42,280 --> 00:17:45,990 is called the gradient of f at a comma b. 340 00:17:45,990 --> 00:17:48,640 And it's usually written in this notation. 341 00:17:48,640 --> 00:17:51,250 An upside down delta. 342 00:17:51,250 --> 00:17:54,200 It's called del, usually, with an arrow over it, 343 00:17:54,200 --> 00:17:56,570 or in boldface print in the text. 344 00:17:56,570 --> 00:17:58,990 And it's written this way, and it's read what? 345 00:17:58,990 --> 00:18:02,750 The gradient of f evaluated at a comma b. 346 00:18:02,750 --> 00:18:06,060 What is the gradient of f evaluated at a comma b? 347 00:18:06,060 --> 00:18:09,660 It's the vector, which gives you the hint as to how 348 00:18:09,660 --> 00:18:12,620 to compute the directional derivative in any direction 349 00:18:12,620 --> 00:18:13,650 that you wish. 350 00:18:13,650 --> 00:18:16,280 Namely, in terms of the gradient vector, 351 00:18:16,280 --> 00:18:20,420 the directional derivative of f in the direction of s, 352 00:18:20,420 --> 00:18:23,410 evaluated at a, b, is the gradient 353 00:18:23,410 --> 00:18:28,260 of f evaluated at a comma b dotted with the unit vector 354 00:18:28,260 --> 00:18:29,520 in the direction of s. 355 00:18:29,520 --> 00:18:31,830 By the way, you may recall that when 356 00:18:31,830 --> 00:18:35,180 you dot a vector with a unit vector, 357 00:18:35,180 --> 00:18:39,310 you get the projection of that vector in the direction 358 00:18:39,310 --> 00:18:40,420 of the unit vector. 359 00:18:40,420 --> 00:18:42,990 In other words, the directional derivative-- another way 360 00:18:42,990 --> 00:18:45,960 of looking at this physically is nothing more 361 00:18:45,960 --> 00:18:49,840 than the projection of the gradient vector 362 00:18:49,840 --> 00:18:53,700 onto the given direction in which you're moving. 363 00:18:53,700 --> 00:18:57,390 And the important point is that this particular definition 364 00:18:57,390 --> 00:19:00,570 does not depend on our coordinate system. 365 00:19:00,570 --> 00:19:04,380 What is interesting is that, in Cartesian coordinates, 366 00:19:04,380 --> 00:19:07,970 there is a very simple way of computing the gradient vector. 367 00:19:07,970 --> 00:19:11,030 Namely, the i component of the gradient vector 368 00:19:11,030 --> 00:19:12,900 is just the partial of f with respect 369 00:19:12,900 --> 00:19:16,010 to x and the j component is just the partial 370 00:19:16,010 --> 00:19:17,700 of f with respect to y. 371 00:19:17,700 --> 00:19:20,840 But that was a very special case, because, you see, 372 00:19:20,840 --> 00:19:26,430 i and x happen to have the same direction, as do y and j. 373 00:19:26,430 --> 00:19:30,450 For arbitrary coordinate systems, this need not be true. 374 00:19:30,450 --> 00:19:33,032 And I'm going to drill you on that in the exercises. 375 00:19:33,032 --> 00:19:34,490 But in other words, what I'm saying 376 00:19:34,490 --> 00:19:36,640 is remember the gradient vector in terms 377 00:19:36,640 --> 00:19:38,890 of a maximum directional derivative. 378 00:19:38,890 --> 00:19:42,310 Don't memorize it as a formula, because if you do, 379 00:19:42,310 --> 00:19:43,700 you're going to get in trouble. 380 00:19:43,700 --> 00:19:46,830 For example, if I were to give my surface 381 00:19:46,830 --> 00:19:51,060 in polar coordinates, say w of some function of r and theta, 382 00:19:51,060 --> 00:19:53,060 then it turns out-- and there's an exercise 383 00:19:53,060 --> 00:19:55,750 on this in the notes-- that the gradient of f 384 00:19:55,750 --> 00:20:01,550 is the partial of w with respect to r times u sub r plus-- 385 00:20:01,550 --> 00:20:03,440 and here's the big difference-- 1 386 00:20:03,440 --> 00:20:08,040 over r times the partial of w with respect to theta times 387 00:20:08,040 --> 00:20:09,140 u sub theta. 388 00:20:09,140 --> 00:20:11,020 In other words, the gradient vector 389 00:20:11,020 --> 00:20:14,580 is not the partial of w with respect to r times 390 00:20:14,580 --> 00:20:18,870 u sub r plus the partial of w with respect to theta times 391 00:20:18,870 --> 00:20:19,580 u sub theta. 392 00:20:19,580 --> 00:20:22,660 In other words, you don't just mechanically differentiate 393 00:20:22,660 --> 00:20:24,230 with respect to these variables. 394 00:20:24,230 --> 00:20:27,500 And the key reason that you can't do this-- well, lookit. 395 00:20:27,500 --> 00:20:29,290 Let's just look at this little diagram. 396 00:20:29,290 --> 00:20:31,830 And I think the whole idea will become very clear. 397 00:20:31,830 --> 00:20:34,000 Remember that in polar coordinates 398 00:20:34,000 --> 00:20:36,280 r is denoted this way. 399 00:20:36,280 --> 00:20:39,730 u sub theta is at right angles to r. 400 00:20:39,730 --> 00:20:44,920 Notice that u sub theta is not in the direction of theta. 401 00:20:44,920 --> 00:20:46,920 Notice that the direction of theta 402 00:20:46,920 --> 00:20:50,390 is sort of the tangent to this circle of radius 403 00:20:50,390 --> 00:20:51,750 r at this point. 404 00:20:51,750 --> 00:20:54,580 If I call this increment d theta, 405 00:20:54,580 --> 00:20:57,062 notice that this arc length is r d 406 00:20:57,062 --> 00:21:00,250 theta, so the vector in the direction of u sub theta 407 00:21:00,250 --> 00:21:03,480 is r d theta, not d theta. 408 00:21:03,480 --> 00:21:05,890 I don't know if you noticed that, but coming back up 409 00:21:05,890 --> 00:21:09,240 here for a moment, notice that this was OK here, 410 00:21:09,240 --> 00:21:13,190 because r was in the same direction as u sub r. 411 00:21:13,190 --> 00:21:18,010 Notice, however, that it's r d theta which is in the u sub 412 00:21:18,010 --> 00:21:19,770 theta direction. 413 00:21:19,770 --> 00:21:22,630 Again, I leave most of these details for the notes. 414 00:21:22,630 --> 00:21:25,460 But I feel that if I don't say these things to you, 415 00:21:25,460 --> 00:21:28,430 it becomes very easy to miss these points when 416 00:21:28,430 --> 00:21:30,660 we talk about them or write about them, 417 00:21:30,660 --> 00:21:33,530 but somehow I hope that by you hearing me say this, 418 00:21:33,530 --> 00:21:35,390 you will be keyed in when you come 419 00:21:35,390 --> 00:21:39,460 to these concepts in the unit that we're studying. 420 00:21:39,460 --> 00:21:42,230 But I think the best way to augment what we're doing 421 00:21:42,230 --> 00:21:45,100 is by means of a specific example. 422 00:21:45,100 --> 00:21:49,890 Let us suppose that we're given the surface w equals f of x, y, 423 00:21:49,890 --> 00:21:53,250 where f of x, y is x to the fifth plus x cubed y 424 00:21:53,250 --> 00:21:54,870 plus y to the fifth. 425 00:21:54,870 --> 00:21:57,290 And we want to compute the directional derivative 426 00:21:57,290 --> 00:22:01,540 of f at the point 1 comma 1 in the direction-- 427 00:22:01,540 --> 00:22:06,940 let's call it s sub 1, where s sub 1 is the direction that 428 00:22:06,940 --> 00:22:11,212 goes from the point 1 comma 1 to the point 4 comma 5. 429 00:22:11,212 --> 00:22:13,170 Now, what we're saying is-- and I guess, maybe, 430 00:22:13,170 --> 00:22:17,570 if we look at these two diagrams concurrently, 431 00:22:17,570 --> 00:22:19,180 maybe this'll be easier to see. 432 00:22:19,180 --> 00:22:21,600 Here we are at the point 1 comma 1. 433 00:22:21,600 --> 00:22:26,650 We want to see how fast the slope over our head-- the w 434 00:22:26,650 --> 00:22:30,160 value-- is changing in the direction of s_1, where 435 00:22:30,160 --> 00:22:32,950 s_1 is chosen to be what? 436 00:22:32,950 --> 00:22:35,540 We're moving from the point 1 comma 1 437 00:22:35,540 --> 00:22:39,290 in the xy-plane to the point 4 comma 5. 438 00:22:39,290 --> 00:22:41,100 See, we're moving in this direction 439 00:22:41,100 --> 00:22:45,450 and we want to see how fast w is changing over our heads, which 440 00:22:45,450 --> 00:22:49,120 geometrically means you draw this plane, 441 00:22:49,120 --> 00:22:53,840 intersect it with the particular surface here. 442 00:22:53,840 --> 00:22:59,090 And this point, P_0-- what we really want geometrically 443 00:22:59,090 --> 00:22:59,780 is what? 444 00:22:59,780 --> 00:23:03,030 The slope of the line tangent to this curve 445 00:23:03,030 --> 00:23:08,440 in the w, s_1 plane tangent to this curve at the point P_0. 446 00:23:08,440 --> 00:23:11,420 And my claim is that this can be done very, very 447 00:23:11,420 --> 00:23:13,870 easy from a mechanical point of view 448 00:23:13,870 --> 00:23:16,540 now that we have our gradient vector behind us. 449 00:23:16,540 --> 00:23:19,570 Namely, what we do is, given what 450 00:23:19,570 --> 00:23:25,430 w looks like as a function of x and y, we take the partial of w 451 00:23:25,430 --> 00:23:28,600 with respect to both x and y, which, hopefully, 452 00:23:28,600 --> 00:23:30,620 you can all do quite mechanically now 453 00:23:30,620 --> 00:23:32,570 based on our last unit's work. 454 00:23:32,570 --> 00:23:35,360 We differentiate, first holding y constant, 455 00:23:35,360 --> 00:23:37,310 then holding x constant. 456 00:23:37,310 --> 00:23:39,060 At any rate, we obtain what? 457 00:23:39,060 --> 00:23:41,800 That the partial of w with respect to x 458 00:23:41,800 --> 00:23:45,290 is 5 x to the fourth plus 3 x squared y. 459 00:23:45,290 --> 00:23:47,880 And so if we compute that at the point 1 460 00:23:47,880 --> 00:23:50,210 comma 1, when x and y are both 1, 461 00:23:50,210 --> 00:23:52,830 this simply turns out to be 8. 462 00:23:52,830 --> 00:23:56,840 In a similar way, the partial of w with respect to y 463 00:23:56,840 --> 00:23:59,710 is x cubed plus 6 y to the fifth. 464 00:23:59,710 --> 00:24:02,810 So if we compute that at the point 1 comma 1, 465 00:24:02,810 --> 00:24:04,710 that turns out to be 7. 466 00:24:04,710 --> 00:24:06,610 In other words, then, by definition 467 00:24:06,610 --> 00:24:10,900 of our gradient, which is the partial of f with respect 468 00:24:10,900 --> 00:24:14,200 to x evaluated at 1 comma 1 times i, 469 00:24:14,200 --> 00:24:16,420 plus the partial of f with respect 470 00:24:16,420 --> 00:24:19,580 to y evaluated at 1 comma 1 times j, 471 00:24:19,580 --> 00:24:24,640 the gradient of f at 1 comma 1 is just 8i plus 7j. 472 00:24:24,640 --> 00:24:26,875 Very easy to write down mechanically when you're 473 00:24:26,875 --> 00:24:28,650 using Cartesian coordinates. 474 00:24:28,650 --> 00:24:32,510 Now, let me make a brief aside, an interruption here. 475 00:24:32,510 --> 00:24:36,600 The idea is to emphasize what the gradient vector means. 476 00:24:36,600 --> 00:24:41,030 What this tells me is that if I were to leave the point 1 477 00:24:41,030 --> 00:24:46,820 comma 1 in the direction of the vector 8i plus 4j-- 478 00:24:46,820 --> 00:24:49,710 if I were to leave in that direction, 479 00:24:49,710 --> 00:24:51,230 that would be the direction in which 480 00:24:51,230 --> 00:24:53,610 the directional derivative would be maximum. 481 00:24:53,610 --> 00:24:56,900 And moreover, that maximum directional derivative 482 00:24:56,900 --> 00:25:00,260 would just be the magnitude of this gradient vector. 483 00:25:00,260 --> 00:25:02,510 The magnitude of that gradient vector 484 00:25:02,510 --> 00:25:06,870 is just the square root of 8 squared plus 7 squared, which 485 00:25:06,870 --> 00:25:09,020 is the square root of 113. 486 00:25:09,020 --> 00:25:10,970 In other words, what this tells us 487 00:25:10,970 --> 00:25:14,550 is that the maximum directional derivative, leaving the point 1 488 00:25:14,550 --> 00:25:17,900 comma 1, is the square root of 113, 489 00:25:17,900 --> 00:25:21,630 and it occurs in a direction 8i plus 7j 490 00:25:21,630 --> 00:25:23,890 as you leave the point 1 comma 1. 491 00:25:23,890 --> 00:25:25,770 At any rate, getting back to the main stream 492 00:25:25,770 --> 00:25:29,250 of the problem, what we want is a directional derivative 493 00:25:29,250 --> 00:25:31,950 in the direction of s_1. 494 00:25:31,950 --> 00:25:32,810 That means what? 495 00:25:32,810 --> 00:25:36,140 We take our gradient vector, which is 8 comma 7m 496 00:25:36,140 --> 00:25:39,860 and dot that with the unit vector in the direction of s_1. 497 00:25:39,860 --> 00:25:42,430 You may recall from this diagram here 498 00:25:42,430 --> 00:25:46,320 that the vector in the direction of s_1 499 00:25:46,320 --> 00:25:50,940 has its i component equal to 3, its j complement equal to 4. 500 00:25:50,940 --> 00:25:53,980 This makes this a 3, 4, 5 right triangle. 501 00:25:53,980 --> 00:25:56,270 So the unit vector in this direction 502 00:25:56,270 --> 00:26:00,610 has as its components 3/5 and 4/5. 503 00:26:00,610 --> 00:26:04,200 In other words, the directional derivative of f at the point 1 504 00:26:04,200 --> 00:26:09,960 comma 1 in the given direction s_1 is just the gradient dotted 505 00:26:09,960 --> 00:26:14,170 with the unit vector 3/5*i plus 4/5*j. 506 00:26:14,170 --> 00:26:16,810 Just mechanically carrying out this operation 507 00:26:16,810 --> 00:26:19,210 leads to 52 over 5. 508 00:26:19,210 --> 00:26:23,160 And by the way, this had better turn out 509 00:26:23,160 --> 00:26:29,430 to be less than this, because this is what? 510 00:26:29,430 --> 00:26:32,740 The maximum value that the directional derivative 511 00:26:32,740 --> 00:26:33,270 can have. 512 00:26:33,270 --> 00:26:35,850 In other words, if we haven't made a mistake here, 513 00:26:35,850 --> 00:26:37,390 one of the checkpoints is what? 514 00:26:37,390 --> 00:26:41,450 That the vector can't project to be any longer 515 00:26:41,450 --> 00:26:43,920 than what it really is in this. 516 00:26:43,920 --> 00:26:48,630 It can't be more than the gradient vector. 517 00:26:48,630 --> 00:26:51,860 But at any rate, let's now conclude the lecture 518 00:26:51,860 --> 00:26:54,750 by coming to the part which is probably 519 00:26:54,750 --> 00:26:57,260 the hardest thing that we're going to encounter 520 00:26:57,260 --> 00:26:58,220 in the whole course. 521 00:26:58,220 --> 00:27:00,190 In a way, I feel a little bit like a man 522 00:27:00,190 --> 00:27:01,890 who fell off the Empire State Building. 523 00:27:01,890 --> 00:27:03,740 And when he went past the 40th floor, 524 00:27:03,740 --> 00:27:05,198 somebody said, "How are you doing?" 525 00:27:05,198 --> 00:27:06,650 And he said, "So far so good." 526 00:27:06,650 --> 00:27:10,010 And that is, we've taken some tremendous liberties here. 527 00:27:10,010 --> 00:27:12,110 And the biggest liberty that we've taken-- 528 00:27:12,110 --> 00:27:13,680 and it's not just a liberty. 529 00:27:13,680 --> 00:27:16,660 It's the kind of a liberty that to solve 530 00:27:16,660 --> 00:27:19,430 involves the foundations of our entire course. 531 00:27:19,430 --> 00:27:23,050 We are now at the grassroots of what 532 00:27:23,050 --> 00:27:25,990 at least the calculus of functions of several variables 533 00:27:25,990 --> 00:27:27,000 is all about. 534 00:27:27,000 --> 00:27:28,240 And that's this trouble spot. 535 00:27:28,240 --> 00:27:31,110 First of all, does delta w tan exist? 536 00:27:31,110 --> 00:27:32,690 That's the first question. 537 00:27:32,690 --> 00:27:35,210 Namely, how do you know that there is a tangent plane? 538 00:27:35,210 --> 00:27:37,640 Just because the surface happens to be smooth 539 00:27:37,640 --> 00:27:41,730 when you cut it by a plane parallel to the wy-plane 540 00:27:41,730 --> 00:27:45,910 and smooth when you cut it by a plane parallel to the wx-plane, 541 00:27:45,910 --> 00:27:50,076 how do you that it's going to be smooth for any given direction? 542 00:27:50,076 --> 00:27:52,140 See, that's the first intellectual question 543 00:27:52,140 --> 00:27:54,270 that comes up in the reading assignment that 544 00:27:54,270 --> 00:27:57,060 has to be solved effectively. 545 00:27:57,060 --> 00:28:00,150 First of all, does delta w tan exist meaningfully? 546 00:28:00,150 --> 00:28:04,700 And secondly, if it does exist, how is it related to delta w? 547 00:28:04,700 --> 00:28:07,970 And now, we come to that key theorem, the proof of which 548 00:28:07,970 --> 00:28:09,200 is quite hairy. 549 00:28:09,200 --> 00:28:11,200 It's done in the text. 550 00:28:11,200 --> 00:28:15,700 It's also done as an optional exercise 551 00:28:15,700 --> 00:28:19,400 to help you generalize what's done in the text. 552 00:28:19,400 --> 00:28:21,510 And it's the counterpart of what happens 553 00:28:21,510 --> 00:28:25,250 with differentials in functions of a single real variable. 554 00:28:25,250 --> 00:28:28,300 But the key theorem, which I'll state here without proof, 555 00:28:28,300 --> 00:28:29,710 is simply this. 556 00:28:29,710 --> 00:28:32,710 Suppose that w is a function of x and y 557 00:28:32,710 --> 00:28:36,180 and that f sub x and f sub y both happen 558 00:28:36,180 --> 00:28:39,901 to exist in some neighborhood of the point a comma b. 559 00:28:39,901 --> 00:28:40,400 All right? 560 00:28:40,400 --> 00:28:41,960 So far so good. 561 00:28:41,960 --> 00:28:45,360 Now, here is they key additional hypothesis. 562 00:28:45,360 --> 00:28:49,570 Suppose also that f sub x and f sub y happen 563 00:28:49,570 --> 00:28:52,630 to be continuous at a comma b. 564 00:28:52,630 --> 00:28:55,660 My claim is that if this additional hypothesis is 565 00:28:55,660 --> 00:28:58,620 obeyed, the tangent plane will exist. 566 00:28:58,620 --> 00:29:01,940 In other words, it's not enough for the directional derivative 567 00:29:01,940 --> 00:29:04,380 to exist in the x and y directions 568 00:29:04,380 --> 00:29:07,000 in order to guarantee that the directional derivative will 569 00:29:07,000 --> 00:29:08,650 exist in every direction. 570 00:29:08,650 --> 00:29:12,340 But it is enough provided that these directional derivatives 571 00:29:12,340 --> 00:29:14,150 happen to be continuous. 572 00:29:14,150 --> 00:29:17,020 And by the way, if these conditions are met, 573 00:29:17,020 --> 00:29:21,760 we say that f is a continuously differentiable function 574 00:29:21,760 --> 00:29:22,790 of x and y. 575 00:29:22,790 --> 00:29:26,180 But I'll talk about that more next time, or in the notes, 576 00:29:26,180 --> 00:29:27,940 or in the exercises. 577 00:29:27,940 --> 00:29:30,550 We're going to make a big issue over this sooner or later. 578 00:29:30,550 --> 00:29:33,310 But for now what I do want to do is just 579 00:29:33,310 --> 00:29:35,670 end with what the key theorem is. 580 00:29:35,670 --> 00:29:37,680 The key theorem says, lookit. 581 00:29:37,680 --> 00:29:41,020 Just like with one variable, if these conditions are met, 582 00:29:41,020 --> 00:29:44,590 then there is a very reasonable approximation 583 00:29:44,590 --> 00:29:47,160 to delta w by delta w tan. 584 00:29:47,160 --> 00:29:49,920 Namely, what the theorem says is, in this case, 585 00:29:49,920 --> 00:29:53,300 delta w will be the partial of f with respect 586 00:29:53,300 --> 00:29:57,170 to x evaluated at a comma b times delta x 587 00:29:57,170 --> 00:29:59,390 plus the partial of f with respect to y 588 00:29:59,390 --> 00:30:03,100 evaluated at a comma b times delta y-- and notice, 589 00:30:03,100 --> 00:30:05,940 of course, that this is a thing that we've been calling delta y 590 00:30:05,940 --> 00:30:10,880 tan-- plus an arrow. 591 00:30:10,880 --> 00:30:15,680 And the arrow has the form k_1 delta x plus k_2 delta y, 592 00:30:15,680 --> 00:30:20,900 where k_1 and k_2 both approach 0 as delta x and delta y 593 00:30:20,900 --> 00:30:21,930 approach 0. 594 00:30:21,930 --> 00:30:24,090 And this is very, very crucial. 595 00:30:24,090 --> 00:30:25,910 It's not enough, as we're going to see 596 00:30:25,910 --> 00:30:29,570 in the very next lecture, that delta x and delta y approach 0. 597 00:30:29,570 --> 00:30:34,680 It's that these things go to 0 as delta x and delta y go to 0. 598 00:30:34,680 --> 00:30:38,620 Consequently, these terms go to 0 faster. 599 00:30:38,620 --> 00:30:42,030 They go to 0 as a second-order infinitesimal. 600 00:30:42,030 --> 00:30:44,400 And what this really says is, look, 601 00:30:44,400 --> 00:30:48,660 for very small values of delta x and delta y, even when you're 602 00:30:48,660 --> 00:30:53,310 dealing with 0 over 9 forms, if you pick a sufficiently 603 00:30:53,310 --> 00:30:56,550 small neighborhood of the point a comma b-- 604 00:30:56,550 --> 00:30:59,400 and that's the key point, a sufficiently small neighborhood 605 00:30:59,400 --> 00:31:02,590 of the point a comma b-- then delta w 606 00:31:02,590 --> 00:31:06,510 is approximately equal to delta w tan 607 00:31:06,510 --> 00:31:10,920 And by the way, this holds also in several variables. 608 00:31:10,920 --> 00:31:16,070 In other words, I picked the case n equals 2 here simply 609 00:31:16,070 --> 00:31:18,660 so that we can utilize the geometry. 610 00:31:18,660 --> 00:31:22,510 Namely, what we're saying is, in terms of the geometry, 611 00:31:22,510 --> 00:31:25,800 if f happens to be a continuously differentiable 612 00:31:25,800 --> 00:31:29,800 function of x and y, and we look at the surface w 613 00:31:29,800 --> 00:31:34,650 equals f of x, y above the point a comma b, what we're saying 614 00:31:34,650 --> 00:31:38,100 is that in a neighborhood of that point, 615 00:31:38,100 --> 00:31:40,710 there is-- well, first of all we're saying what? 616 00:31:40,710 --> 00:31:45,300 A tangent plane exists to the surface above that point 617 00:31:45,300 --> 00:31:48,400 and that in a neighborhood of that point of tangency, 618 00:31:48,400 --> 00:31:51,990 the tangent plane is an excellent approximation 619 00:31:51,990 --> 00:31:54,810 for the true change in w. 620 00:31:54,810 --> 00:31:57,450 Now, what happens is, if n is greater than 2, 621 00:31:57,450 --> 00:32:01,890 we can no longer use the geometric interpretation. 622 00:32:01,890 --> 00:32:05,130 But what is important is that the analytic proof never 623 00:32:05,130 --> 00:32:06,610 makes use of the picture. 624 00:32:06,610 --> 00:32:08,300 And the key point is-- and I'm going 625 00:32:08,300 --> 00:32:10,320 to exploit this in future lectures. 626 00:32:10,320 --> 00:32:15,720 The really key point is that the delta w tan never gets messy, 627 00:32:15,720 --> 00:32:18,300 that the variables-- delta x, delta y, 628 00:32:18,300 --> 00:32:21,880 et cetera-- all occur as linear terms. 629 00:32:21,880 --> 00:32:25,830 And this is why the so-called linear algebra subject 630 00:32:25,830 --> 00:32:29,500 becomes so important in the study of functions 631 00:32:29,500 --> 00:32:30,900 of several variables. 632 00:32:30,900 --> 00:32:34,160 At any rate, I think this is enough for one lesson. 633 00:32:34,160 --> 00:32:36,800 And in our next lesson, what we will do 634 00:32:36,800 --> 00:32:39,980 is show how using this key theorem 635 00:32:39,980 --> 00:32:43,340 has its analog in something called the chain rule, 636 00:32:43,340 --> 00:32:46,290 just as it did in the case of part one 637 00:32:46,290 --> 00:32:49,730 when we studied functions of a single independent variable. 638 00:32:49,730 --> 00:32:52,070 At any rate, then, until next time. 639 00:32:52,070 --> 00:32:54,600 Good bye. 640 00:32:54,600 --> 00:32:56,980 Funding for the publication of this video 641 00:32:56,980 --> 00:33:01,850 was provided by the Gabrielle and Paul Rosenbaum Foundation. 642 00:33:01,850 --> 00:33:06,020 Help OCW continue to provide free and open access to MIT 643 00:33:06,020 --> 00:33:13,730 courses by making a donation at ocw.mit.edu/donate.