1 00:00:01 --> 00:00:03 The following content is provided under a Creative 2 00:00:03 --> 00:00:05 Commons license. Your support will help MIT 3 00:00:05 --> 00:00:08 OpenCourseWare continue to offer high quality educational 4 00:00:08 --> 00:00:13 resources for free. To make a donation or to view 5 00:00:13 --> 00:00:18 additional materials from hundreds of MIT courses, 6 00:00:18 --> 00:00:23 visit MIT OpenCourseWare at ocw.mit.edu. 7 00:00:23 --> 00:00:35 OK, so we're going to continue looking at what happens when we 8 00:00:35 --> 00:00:45 have non-independent variables. So, I'm afraid we don't take 9 00:00:45 --> 00:00:50 deliveries during class time, sorry. 10 00:00:50 --> 00:01:00 Please take a seat, thanks. [LAUGHTER] 11 00:01:00 --> 00:01:05 [APPLAUSE] OK, so Jason, 12 00:01:05 --> 00:01:17 you please claim your package at the end of lecture. 13 00:01:17 --> 00:01:19 OK, so last time we saw how to use 14 00:01:19 --> 00:01:23 Lagrange multipliers to find the minimum or maximum of a function 15 00:01:23 --> 00:01:27 of several variables when the variables are not independent. 16 00:01:27 --> 00:01:29 And, today we're going to try to figure out more about 17 00:01:29 --> 00:01:33 relations between the variables, and how to handle functions 18 00:01:33 --> 00:01:36 that depend on several variables when they're related. 19 00:01:36 --> 00:01:40 So, just to give you an example, 20 00:01:40 --> 00:01:44 in physics, very often, you have functions that depend 21 00:01:44 --> 00:01:49 on pressure, volume, and temperature where pressure, 22 00:01:49 --> 00:01:52 volume, and temperature are actually 23 00:01:52 --> 00:01:55 not independent. But they are related, 24 00:01:55 --> 00:01:58 say, by PV=nRT. So, of course, 25 00:01:58 --> 00:02:01 then you can substitute and expressed a function in terms of 26 00:02:01 --> 00:02:04 two of them only, but very often it's convenient 27 00:02:04 --> 00:02:06 to keep all three. But then we have to figure out, 28 00:02:06 --> 00:02:11 what are the rates of change with respect to t, 29 00:02:11 --> 00:02:14 with respect to each other, the rate of change of f with 30 00:02:14 --> 00:02:16 respect to these variables, and so on. 31 00:02:16 --> 00:02:21 So, we have to figure out what we mean by partial derivatives 32 00:02:21 --> 00:02:24 again. So, 33 00:02:24 --> 00:02:31 OK, more generally, let's say just for the sake of 34 00:02:31 --> 00:02:33 notation, I'm going to think of a 35 00:02:33 --> 00:02:35 function of three variables, x, y, z, 36 00:02:35 --> 00:02:39 where the variables are related by some equation, 37 00:02:39 --> 00:02:44 but I will put in the form g of x, y, z equals some constant. 38 00:02:44 --> 00:02:48 OK, so that's the same kind of setup as we had last time, 39 00:02:48 --> 00:02:52 except now we are not just looking for minima and maxima. 40 00:02:52 --> 00:02:59 We are trying to understand partial derivatives. 41 00:02:59 --> 00:03:06 So, the first observation is that if x, y, 42 00:03:06 --> 00:03:09 and z are related, then that means, 43 00:03:09 --> 00:03:11 in principle, we could solve for one of them, 44 00:03:11 --> 00:03:15 and express it as a function of the two others. 45 00:03:15 --> 00:03:19 So, in particular, can we understand even without 46 00:03:19 --> 00:03:21 solving? Maybe we can not solve. 47 00:03:21 --> 00:03:27 Can we understand how the variables are related to each 48 00:03:27 --> 00:03:29 other? So, for example, 49 00:03:29 --> 00:03:33 z, you can think of z as a function of x and y. 50 00:03:33 --> 00:03:40 So, we can ask ourselves, what are the rates of change of 51 00:03:40 --> 00:03:44 z with respect to x, keeping y constant, 52 00:03:44 --> 00:03:49 or with respect to y keeping x constant? 53 00:03:49 --> 00:03:51 And, of course, if we can solve, 54 00:03:51 --> 00:03:53 that we know the formula for this. 55 00:03:53 --> 00:03:55 And then we can compute these guys. 56 00:03:55 --> 00:04:03 But, what if we can't solve? So, how do we find these things 57 00:04:03 --> 00:04:11 without solving? Well, so let's do an example. 58 00:04:11 --> 00:04:19 Let's say that my relation is x^2 yz z^3=8. 59 00:04:19 --> 00:04:24 And, let's say that I'm looking near the point (x, 60 00:04:24 --> 00:04:27 y, z) equals (2,3, 1). 61 00:04:27 --> 00:04:33 So, let me check 2^2 plus three times one plus 1^3 is indeed 62 00:04:33 --> 00:04:34 eight. OK, but now, 63 00:04:34 --> 00:04:38 if I change x and y a little bit, how does z change? 64 00:04:38 --> 00:04:41 Well, of course I could solve for z in here. 65 00:04:41 --> 00:04:43 It's a cubic equation. There is actually a formula. 66 00:04:43 --> 00:04:45 But that formula is quite complicated. 67 00:04:45 --> 00:04:47 We actually don't want to do that. 68 00:04:47 --> 00:04:58 There's an easier way. So, how can we do it? 69 00:04:58 --> 00:05:07 Well, let's look at the differential -- -- of this 70 00:05:07 --> 00:05:15 constraint quantity. OK, so if we called this g, 71 00:05:15 --> 00:05:21 let's look at dg. So, what's the differential of 72 00:05:21 --> 00:05:26 this? So, the differential of x^2 is 73 00:05:26 --> 00:05:32 2x dx plus, I think there's a zdy. 74 00:05:32 --> 00:05:38 There's a ydz, and there's also a 3z^2 dz. 75 00:05:38 --> 00:05:42 OK, you can get this either by implicit differentiation and the 76 00:05:42 --> 00:05:45 product rule, or you could get this just by 77 00:05:45 --> 00:05:46 putting here, here, 78 00:05:46 --> 00:05:51 and here the partial derivatives of this with respect 79 00:05:51 --> 00:05:56 to x, y, and z. OK, any questions about how I 80 00:05:56 --> 00:05:58 got this? No? 81 00:05:58 --> 00:06:03 OK. So, now, what do I do with this? 82 00:06:03 --> 00:06:07 Well, this represents, somehow, variations of g. 83 00:06:07 --> 00:06:12 But, well, I've set this thing equal to eight. 84 00:06:12 --> 00:06:16 And, eight is a constant. So, it doesn't change. 85 00:06:16 --> 00:06:26 So, in fact, well, we can set this to zero 86 00:06:26 --> 00:06:35 because, well, they call this g. 87 00:06:35 --> 00:06:39 Then, g equals eight is constant. 88 00:06:39 --> 00:06:43 That means we set dg equal to zero. 89 00:06:43 --> 00:06:53 OK, so, now let's just plug in some values at this point. 90 00:06:53 --> 00:06:58 That tells us, well, so if x equals two, 91 00:06:58 --> 00:07:07 that's 4dx plus z is one. So, dy plus y 3z^2 should be 92 00:07:07 --> 00:07:13 6dz equals zero. And now, this equation, 93 00:07:13 --> 00:07:17 here, tells us about a relation between the changes in x, 94 00:07:17 --> 00:07:21 y, and z near that point. It tells us how you change x 95 00:07:21 --> 00:07:25 and y, well, how z will change. Or, it tells you actually 96 00:07:25 --> 00:07:28 anything you might want to know about the relations between 97 00:07:28 --> 00:07:30 these variables so, for example, 98 00:07:30 --> 00:07:34 you can move dz to that side, and then express dz in terms of 99 00:07:34 --> 00:07:37 dx and dy. Or, you can move dy to that 100 00:07:37 --> 00:07:41 side and express dy in terms of dx and dz, and so on. 101 00:07:41 --> 00:07:46 It tells you at the level of the derivatives how each of the 102 00:07:46 --> 00:07:49 variables depends on the two others. 103 00:07:49 --> 00:07:57 OK, so, just to clarify this: if we want to view z as a 104 00:07:57 --> 00:08:03 function of x and y, then what we will do is we will 105 00:08:03 --> 00:08:06 just move the dz's to the other side, 106 00:08:06 --> 00:08:15 and it will tell us dz equals minus one over six times 4dx 107 00:08:15 --> 00:08:20 plus dy. And, so that should tell you 108 00:08:20 --> 00:08:26 that partial z over partial x is minus four over six. 109 00:08:26 --> 00:08:35 Well, that's minus two thirds, and partial z over partial y is 110 00:08:35 --> 00:08:42 going to be minus one sixth. OK, another way to think about 111 00:08:42 --> 00:08:47 this: when we compute partial z over partial x, 112 00:08:47 --> 00:08:51 that means that actually we keep y constant. 113 00:08:51 --> 00:08:55 OK, let me actually add some subtitles here. 114 00:08:55 --> 00:09:00 So, here that means we keep y constant. 115 00:09:00 --> 00:09:05 And so, if we keep y constant, another way to think about it 116 00:09:05 --> 00:09:10 is we set dy to zero. We set dy equals zero. 117 00:09:10 --> 00:09:14 So if we do that, we get dx equals negative four 118 00:09:14 --> 00:09:17 sixths dx. That tells us the rate of 119 00:09:17 --> 00:09:24 change of z with respect to x. Here, we set x constant. 120 00:09:24 --> 00:09:29 So, that means we set dx equal to zero. 121 00:09:29 --> 00:09:32 And, if we set dx equal to zero, then we have dz equals 122 00:09:32 --> 00:09:36 negative one sixth of dy. That tells us the rate of 123 00:09:36 --> 00:09:47 change of z with respect to y. OK, any questions about that? 124 00:09:47 --> 00:09:57 No? What, yes? 125 00:09:57 --> 00:09:59 Yes, OK, let me explain that again. 126 00:09:59 --> 00:10:03 So we found an expression for dz in terms of dx and dy. 127 00:10:03 --> 00:10:07 That means that this thing, the differential, 128 00:10:07 --> 00:10:11 is the total differential of z viewed as a function of x and y. 129 00:10:11 --> 00:10:16 OK, and so the coefficients of dx and dy are the partials. 130 00:10:16 --> 00:10:20 Or, another way to think about it, if you want to know partial 131 00:10:20 --> 00:10:22 z partial x, it means you set y to be constant. 132 00:10:22 --> 00:10:27 Setting y to be constant means that you will put zero in the 133 00:10:27 --> 00:10:30 place of dy. So, you will be left with dz 134 00:10:30 --> 00:10:35 equals minus four sixths dx. And, that will give you the 135 00:10:35 --> 00:10:41 rate of change of z with respect to x when you keep y constant, 136 00:10:41 --> 00:10:46 OK? So, there are various ways to 137 00:10:46 --> 00:10:53 think about this, but hopefully it makes sense. 138 00:10:53 --> 00:11:03 OK, so how do we think about this in general? 139 00:11:03 --> 00:11:15 Well, if we know that g of x, y, z equals a constant, 140 00:11:15 --> 00:11:27 then dg, which is gxdx gydy gzdz should be set equal to 141 00:11:27 --> 00:11:32 zero. OK, and now we can solve for 142 00:11:32 --> 00:11:37 whichever variable we want to express in terms of the others. 143 00:11:37 --> 00:11:47 So, for example, if we care about z as a 144 00:11:47 --> 00:12:02 function of x and y -- -- we'll get that dz is negative gx over 145 00:12:02 --> 00:12:17 gz dx minus gy over gz dy. And, so if we want partial z 146 00:12:17 --> 00:12:23 over partial x, well, so one way is just to say 147 00:12:23 --> 00:12:26 that's going to be the coefficient of dx in here, 148 00:12:26 --> 00:12:29 or just to write down the other way. 149 00:12:29 --> 00:12:34 We are setting y equals constant. 150 00:12:34 --> 00:12:39 So, that means we set dy equal to zero. 151 00:12:39 --> 00:12:48 And then, we get dz equals negative gx over gz dx. 152 00:12:48 --> 00:12:56 So, that means partial z over partial x is minus gx over gz. 153 00:12:56 --> 00:12:59 And, see, that's a very counterintuitive 154 00:12:59 --> 00:13:02 formula because you have this minus sign that you somehow 155 00:13:02 --> 00:13:06 probably couldn't have seen come if you hadn't actually derived 156 00:13:06 --> 00:13:11 things this way. I mean, it's pretty surprising 157 00:13:11 --> 00:13:17 to see that minus sign come out of nowhere the first time you 158 00:13:17 --> 00:13:22 see it. OK, so now we know how to find 159 00:13:22 --> 00:13:26 the rate of change of constrained variables with 160 00:13:26 --> 00:13:31 respect to each other. You can apply the same to find, 161 00:13:31 --> 00:13:35 if you want partial x, partial y, or any of them, 162 00:13:35 --> 00:13:41 you can do it. Any questions so far? 163 00:13:41 --> 00:13:48 No? OK, so, before we proceed 164 00:13:48 --> 00:13:55 further, I should probably expose some problem with the 165 00:13:55 --> 00:14:03 notations that we have so far. So, let me try to get you a bit 166 00:14:03 --> 00:14:07 confused, OK? So, let's take a very simple 167 00:14:07 --> 00:14:10 example. Let's say I have a function, 168 00:14:10 --> 00:14:15 f of x, y equals x y. OK, so far it doesn't sound 169 00:14:15 --> 00:14:20 very confusing. And then, I can write partial f 170 00:14:20 --> 00:14:24 over partial x. And, I think you all know how 171 00:14:24 --> 00:14:28 to compute it. It's going to be just one. 172 00:14:28 --> 00:14:34 OK, so far we are pretty happy. Now let's do a change of 173 00:14:34 --> 00:14:44 variables. Let's set x=u and y=u v. 174 00:14:44 --> 00:14:46 It's not very complicated change of variables. 175 00:14:46 --> 00:14:54 But let's do it. Then, f in terms of u and v, 176 00:14:54 --> 00:15:02 well, so f, remember f was x y becomes u plus u plus v. 177 00:15:02 --> 00:15:13 That's twice u plus v. What's partial f over partial u? 178 00:15:13 --> 00:15:18 It's two. So, x and u are the same thing. 179 00:15:18 --> 00:15:21 Partial f over partial x, and partial f over partial u, 180 00:15:21 --> 00:15:24 well, unless you believe that one equals two, 181 00:15:24 --> 00:15:26 they are really not the same thing, OK? 182 00:15:26 --> 00:15:36 So, that's an interesting, slightly strange phenomenon. 183 00:15:36 --> 00:15:46 x equals u, but partial f partial x is not the same as 184 00:15:46 --> 00:15:52 partial f partial u. So, how do we get rid of this 185 00:15:52 --> 00:15:55 contradiction? Well, we have to think a bit 186 00:15:55 --> 00:15:59 more about what these notations mean, OK? 187 00:15:59 --> 00:16:03 So, when we write partial f over partial x, 188 00:16:03 --> 00:16:08 it means that we are varying x, keeping y constant. 189 00:16:08 --> 00:16:11 When we write partial f over partial u, it means we are 190 00:16:11 --> 00:16:15 varying u, keeping v constant. So, varying u or varying x is 191 00:16:15 --> 00:16:17 the same thing. But, keeping v constant, 192 00:16:17 --> 00:16:20 or keeping y constant are not the same thing. 193 00:16:20 --> 00:16:23 If I keep y constant, then when I change x, 194 00:16:23 --> 00:16:27 so when I change u, then v will also have to change 195 00:16:27 --> 00:16:29 so that their sum stays the same. 196 00:16:29 --> 00:16:32 Or, if you prefer the other way around, when I do this one I 197 00:16:32 --> 00:16:35 keep v constant. If I keep v constant and I 198 00:16:35 --> 00:16:39 change u, then y will change. It won't be constant. 199 00:16:39 --> 00:16:43 So, that means, well, life looked quite nice 200 00:16:43 --> 00:16:49 and easy with these notations. But, what's dangerous about 201 00:16:49 --> 00:16:55 them is they are not making explicit what it is exactly that 202 00:16:55 --> 00:17:01 we are keeping constant. OK, so just to write things, 203 00:17:01 --> 00:17:08 so here we change u and x that are the same thing. 204 00:17:08 --> 00:17:14 But we keep y constant, while here we change u, 205 00:17:14 --> 00:17:19 which is still the same thing as x. 206 00:17:19 --> 00:17:26 But, what we keep constant is v, or in terms of x and y, 207 00:17:26 --> 00:17:33 that's y minus x constant. And, that's why they are not 208 00:17:33 --> 00:17:36 the same. So, whenever there's any risk 209 00:17:36 --> 00:17:39 of confusion, OK, so not in the cases that we 210 00:17:39 --> 00:17:42 had before because what we've done until now, 211 00:17:42 --> 00:17:46 we didn't really have a problem. But, in a situation like this, 212 00:17:46 --> 00:17:50 to clarify things, we'll actually say explicitly 213 00:17:50 --> 00:17:53 what it is that we want to keep constant. 214 00:17:53 --> 00:18:04 215 00:18:04 --> 00:18:07 OK, so what's going to be our new notation? 216 00:18:07 --> 00:18:14 Well, so it's not particularly pleasant because it uses, 217 00:18:14 --> 00:18:16 now, a subscript not to indicate what you are 218 00:18:16 --> 00:18:18 differentiating, but rather what you were 219 00:18:18 --> 00:18:22 holding constant. So, that's quite a conflict of 220 00:18:22 --> 00:18:25 notation with what we had before. 221 00:18:25 --> 00:18:32 I think I can safely blame it on physicists or chemists. 222 00:18:32 --> 00:18:43 OK, so this one means we keep y constant, and partial f over 223 00:18:43 --> 00:18:51 partial u with v held constant, similarly. 224 00:18:51 --> 00:18:54 OK, so now what happens is we no longer have any 225 00:18:54 --> 00:18:59 contradiction. We have partial f over partial 226 00:18:59 --> 00:19:06 x with y constant is different from partial f over partial x 227 00:19:06 --> 00:19:12 with v constant, which is the same as partial f 228 00:19:12 --> 00:19:18 over partial u with v constant. OK, so this guy is one. 229 00:19:18 --> 00:19:28 And these guys are two. So, now we can safely use the 230 00:19:28 --> 00:19:33 fact that x equals u if we are keeping track of what is 231 00:19:33 --> 00:19:36 actually held constant, OK? 232 00:19:36 --> 00:19:39 So now, that's going to be particularly important when we 233 00:19:39 --> 00:19:41 have variables that are related because, 234 00:19:41 --> 00:19:45 let's say now that I have a function that depends on x, 235 00:19:45 --> 00:19:48 y, and z. But, x, y, and z are related. 236 00:19:48 --> 00:19:54 Then, it means that I look at, say, x and y as my independent 237 00:19:54 --> 00:19:59 variables, and z as a function of x and y. 238 00:19:59 --> 00:20:01 Then, it means that when I do partials, say, 239 00:20:01 --> 00:20:04 with respect to x, I will hold y constant. 240 00:20:04 --> 00:20:08 But, I will let z vary as a function of x and y. 241 00:20:08 --> 00:20:10 Or, I could do it the other way around. 242 00:20:10 --> 00:20:12 I could vary x, keep z constant, 243 00:20:12 --> 00:20:15 and let y be a function of x and z. 244 00:20:15 --> 00:20:24 And so, I will need to use this kind of notation to indicate 245 00:20:24 --> 00:20:34 which one I mean. OK, any questions? 246 00:20:34 --> 00:20:39 No? All right, so let's try to do 247 00:20:39 --> 00:20:42 an example where we have a function that depends on 248 00:20:42 --> 00:20:46 variables that are related. OK, so I don't want to do one 249 00:20:46 --> 00:20:50 with PV=nRT because probably, I mean, if you've seen it, 250 00:20:50 --> 00:20:53 then you've seen too much of it. 251 00:20:53 --> 00:20:58 And, if you haven't seen it, then maybe it's not the best 252 00:20:58 --> 00:21:02 example. So, let's do a geometric 253 00:21:02 --> 00:21:08 example. So, let's look at the area of 254 00:21:08 --> 00:21:14 the triangle. So, let's say I have a 255 00:21:14 --> 00:21:21 triangle, and my variables will be the sides a and b. 256 00:21:21 --> 00:21:26 And the angle here, theta. OK, so what's the area of this 257 00:21:26 --> 00:21:29 triangle? Well, its base times height 258 00:21:29 --> 00:21:34 over two. So, it's one half of the base 259 00:21:34 --> 00:21:39 is a, and the height is b sine theta. 260 00:21:39 --> 00:21:45 OK, so that's a function of a, b, and theta. 261 00:21:45 --> 00:21:47 Now, let's say, actually, there is a relation 262 00:21:47 --> 00:21:49 between a, b, and theta that I didn't tell 263 00:21:49 --> 00:21:52 you about, namely, actually, 264 00:21:52 --> 00:21:58 I want to assume that it's a right triangle, 265 00:21:58 --> 00:22:05 OK? So, let's now assume it's a 266 00:22:05 --> 00:22:16 right triangle with, let's say, the hypotenuse is b. 267 00:22:16 --> 00:22:19 So, we have the right angle here, actually. 268 00:22:19 --> 00:22:23 So, a is here. b is here. Theta is here. 269 00:22:23 --> 00:22:28 So, saying it's a right triangle is the same thing as 270 00:22:28 --> 00:22:31 saying that b equals sine theta, OK? 271 00:22:31 --> 00:22:37 So that's our constraint. That's the relation between a, 272 00:22:37 --> 00:22:46 b, and theta. And, this is a function of a, 273 00:22:46 --> 00:22:53 b, and theta. And, let's say that we want to 274 00:22:53 --> 00:22:57 understand how the area depends on theta. 275 00:22:57 --> 00:23:00 OK, what's the rate of change of the area of this triangle 276 00:23:00 --> 00:23:06 with respect to theta? So, I claim there's various 277 00:23:06 --> 00:23:09 answers. I can think of at least three 278 00:23:09 --> 00:23:10 possible answers. 279 00:23:10 --> 00:23:44 280 00:23:44 --> 00:23:52 So, what can we possibly mean by the rate of change of A with 281 00:23:52 --> 00:23:57 respect to theta? So, these are all things that 282 00:23:57 --> 00:23:59 we might want to call partial A partial theta. 283 00:23:59 --> 00:24:03 But of course, we'll have to actually use 284 00:24:03 --> 00:24:06 different notations to distinguish them. 285 00:24:06 --> 00:24:11 So, the first way that we actually already know about is 286 00:24:11 --> 00:24:17 if we just forget about the fact that the variables are related, 287 00:24:17 --> 00:24:20 OK? So, if we just think of little 288 00:24:20 --> 00:24:23 a, b, and theta as independent 289 00:24:23 --> 00:24:25 variables, and we just change theta, 290 00:24:25 --> 00:24:48 keeping a and b constant -- So, that's exactly what we meant by 291 00:24:48 --> 00:24:51 partial A, partial theta, right? 292 00:24:51 --> 00:24:59 I'm not putting any constraints. So, just to use some new 293 00:24:59 --> 00:25:03 notation, that would be the rate of change of A with respect to 294 00:25:03 --> 00:25:07 theta, keeping a and b fixed at the same time. 295 00:25:07 --> 00:25:11 Of course, if we are keeping a and b fixed, and we are changing 296 00:25:11 --> 00:25:14 theta, it means we completely ignore this property of being a 297 00:25:14 --> 00:25:16 right triangle. So, in fact, 298 00:25:16 --> 00:25:20 it corresponds to changing the area by changing the angle, 299 00:25:20 --> 00:25:23 keeping these lengths fixed. And, of course, 300 00:25:23 --> 00:25:27 we lose the right angle. When we rotate this side here, 301 00:25:27 --> 00:25:32 but the angle doesn't stay at a right angle. 302 00:25:32 --> 00:25:35 And that one, we know how to compute, 303 00:25:35 --> 00:25:40 right, because it's the one we've been computing all along. 304 00:25:40 --> 00:25:44 So, that means we keep a and b fixed. 305 00:25:44 --> 00:25:51 And then, so let's see, what's the derivatives of A 306 00:25:51 --> 00:26:02 with respect to theta? It's one half ab cosine theta. 307 00:26:02 --> 00:26:11 OK, now that one we know. Any questions? 308 00:26:11 --> 00:26:14 No? OK, the two other guys will be 309 00:26:14 --> 00:26:18 more interesting. So far, I'm not really doing 310 00:26:18 --> 00:26:23 anything with my constraint. Let's say that actually I do 311 00:26:23 --> 00:26:27 want to keep the right angle. Then, when I change theta, 312 00:26:27 --> 00:26:31 there's two options. One is I keep a constant, 313 00:26:31 --> 00:26:35 and then of course b will have to change because if this width 314 00:26:35 --> 00:26:38 stays the same, then when I change theta, 315 00:26:38 --> 00:26:41 the height increases, and then this side length 316 00:26:41 --> 00:26:45 increases. The other option is to change 317 00:26:45 --> 00:26:47 the angle, keeping b constant. So, actually, 318 00:26:47 --> 00:26:49 this side stays the same length. 319 00:26:49 --> 00:26:53 But then, a has to become a bit shorter. 320 00:26:53 --> 00:26:56 And, of course, the area will change in 321 00:26:56 --> 00:26:59 different ways depending on what I do. 322 00:26:59 --> 00:27:05 So, that's why I said we have three different answers. 323 00:27:05 --> 00:27:10 So, the next one is keep, I forgot which one I said 324 00:27:10 --> 00:27:17 first. Let's say keep a constant. 325 00:27:17 --> 00:27:26 And, that means that b will change. 326 00:27:26 --> 00:27:30 b is going to be some function of a and theta. 327 00:27:30 --> 00:27:34 Well, in fact here, we know what the function is 328 00:27:34 --> 00:27:37 because we can solve the constraint, namely, 329 00:27:37 --> 00:27:45 b is a over cosine theta. But we don't actually need to 330 00:27:45 --> 00:27:55 know that so that the triangle, so that the right angle, 331 00:27:55 --> 00:28:05 so that we keep a right angle. And, so the name we will have 332 00:28:05 --> 00:28:11 for this is partial a over partial theta with a held 333 00:28:11 --> 00:28:14 constant, OK? And, the fact that I'm not 334 00:28:14 --> 00:28:17 putting b in my subscript there means that actually b will be a 335 00:28:17 --> 00:28:20 dependent variable. It changes in whatever way it 336 00:28:20 --> 00:28:26 has to change so that when theta changes, a stays the same while 337 00:28:26 --> 00:28:29 b changes so that we keep a right triangle. 338 00:28:29 --> 00:28:38 339 00:28:38 --> 00:28:46 And, the third guy is the one where we actually keep b 340 00:28:46 --> 00:28:51 constant, and now a, 341 00:28:51 --> 00:28:54 we think a as a function of b and theta, 342 00:28:54 --> 00:28:58 and it changes so that we keep the right angle. 343 00:28:58 --> 00:29:01 So actually as a function of b and theta, it's given over 344 00:29:01 --> 00:29:06 there. A equals b cosine theta. 345 00:29:06 --> 00:29:13 And so, this guy is called partial a over partial theta 346 00:29:13 --> 00:29:19 with b held constant. OK, so we've just defined them. 347 00:29:19 --> 00:29:21 We don't know yet how to compute these things. 348 00:29:21 --> 00:29:22 That's what we're going to do now. 349 00:29:22 --> 00:29:25 That is the definition, and what these things mean. 350 00:29:25 --> 00:29:33 Is that clear to everyone? Yes, OK. 351 00:29:33 --> 00:29:41 Yes? OK, so the second answer, 352 00:29:41 --> 00:29:46 again, so one way to ask ourselves, 353 00:29:46 --> 00:29:48 how does the area depend on theta, 354 00:29:48 --> 00:29:53 is to say, well, actually look at the area of 355 00:29:53 --> 00:29:59 the right triangle as a function of a and theta only by solving 356 00:29:59 --> 00:30:03 for b. And then, we'll change theta, 357 00:30:03 --> 00:30:06 keep a constant, and ask, how does the area 358 00:30:06 --> 00:30:08 change? So, when we do that, 359 00:30:08 --> 00:30:11 when we change theta and keep a the same, 360 00:30:11 --> 00:30:14 then b has to change so that it stays a right triangle, 361 00:30:14 --> 00:30:18 right, so that this relation still holds. 362 00:30:18 --> 00:30:22 That requires us to change b. So, when we write partial a 363 00:30:22 --> 00:30:26 over partial theta with a constant, it means that, 364 00:30:26 --> 00:30:30 actually, b will be the dependent variable. 365 00:30:30 --> 00:30:35 It depends on a and theta. And so, the area depends on 366 00:30:35 --> 00:30:40 theta, not only because theta is in the formula, 367 00:30:40 --> 00:30:46 but also because b changes, and b is in the formula. 368 00:30:46 --> 00:30:53 Yes? No, no, we don't keep theta 369 00:30:53 --> 00:30:54 constant. We vary theta, right? 370 00:30:54 --> 00:30:58 The goal is to see how things change when I change theta by a 371 00:30:58 --> 00:31:01 little bit. OK, so if I change theta a 372 00:31:01 --> 00:31:04 little bit in this one, if I change theta a little bit 373 00:31:04 --> 00:31:07 and I keep a the same, then b has to change also in 374 00:31:07 --> 00:31:09 some way. There's a right triangle. 375 00:31:09 --> 00:31:16 And then, because theta and b change, that causes the area to 376 00:31:16 --> 00:31:18 change. OK, so maybe I should 377 00:31:18 --> 00:31:23 re-explain that again. So, theta changes. 378 00:31:23 --> 00:31:30 A is constant. But, we have the constraint, 379 00:31:30 --> 00:31:37 a equals be plus sine theta. That means that b changes. 380 00:31:37 --> 00:31:43 And then, the question is, how does A change? 381 00:31:43 --> 00:31:46 Well, it will change in part because theta changes, 382 00:31:46 --> 00:31:50 and in part because b changes. But, we want to know how it 383 00:31:50 --> 00:31:54 depends on theta in this situation. 384 00:31:54 --> 00:32:04 Yes? Ah, that's a very good question. 385 00:32:04 --> 00:32:08 So, what about, I don't keep a and b constant? 386 00:32:08 --> 00:32:10 Well, then there's too many choices. 387 00:32:10 --> 00:32:13 So I have to decide actually how I'm going to change things. 388 00:32:13 --> 00:32:17 See, if I just say I have this relation, that means I have two 389 00:32:17 --> 00:32:20 independent variables left, whichever two of the three I 390 00:32:20 --> 00:32:23 want. But, I still have to specify 391 00:32:23 --> 00:32:27 two of them to say exactly which triangle I mean. 392 00:32:27 --> 00:32:31 So, I cannot ask myself just how will it change if I change 393 00:32:31 --> 00:32:34 theta and do random things with a and b? 394 00:32:34 --> 00:32:36 It depends what I do with a and b. 395 00:32:36 --> 00:32:40 Of course, I could choose to change them simultaneously, 396 00:32:40 --> 00:32:45 but then I have to specify how exactly I'm going to do that. 397 00:32:45 --> 00:32:49 Ah, yes, if you wanted to, indeed, we could also change 398 00:32:49 --> 00:32:53 things in such a way that the third side remains constant. 399 00:32:53 --> 00:32:55 And that would be, yet, a different way to attack 400 00:32:55 --> 00:32:57 the problem. I mean, we don't have good 401 00:32:57 --> 00:33:00 notation for this, here, because we didn't give it 402 00:33:00 --> 00:33:01 a name. But, yeah, I mean, we could. 403 00:33:01 --> 00:33:07 We could call this guy c, and then we'd have a different 404 00:33:07 --> 00:33:11 formula, and so on. So, I mean, I'm not looking at 405 00:33:11 --> 00:33:17 it for simplicity. But, you could have many more. 406 00:33:17 --> 00:33:19 I mean, in general, you will want, 407 00:33:19 --> 00:33:22 once you have a set of nice, natural variables, 408 00:33:22 --> 00:33:25 you will want to look mostly at situations where one of the 409 00:33:25 --> 00:33:29 variables changes. Some of them are held fixed, 410 00:33:29 --> 00:33:33 and then some dependent variable does whatever it must 411 00:33:33 --> 00:33:36 so that the constraint keeps holding. 412 00:33:36 --> 00:33:39 OK, so let's try to compute one of them. 413 00:33:39 --> 00:33:44 Let's say I decide that we will compute this one. 414 00:33:44 --> 00:33:46 OK, let's see how we can compute partial a, 415 00:33:46 --> 00:33:49 partial theta with a held fixed. 416 00:33:49 --> 00:34:21 417 00:34:21 --> 00:34:27 [APPLAUSE] OK, so let's try to compute 418 00:34:27 --> 00:34:34 partial A, partial theta with a held constant. 419 00:34:34 --> 00:34:40 So, let's see three different ways of doing that. 420 00:34:40 --> 00:34:45 So, let me start with method zero. 421 00:34:45 --> 00:34:50 OK, it's not a real method. That's why I'm not getting a 422 00:34:50 --> 00:34:54 positive number. So, that one is just, 423 00:34:54 --> 00:34:58 we solve for b, and we remove b from the 424 00:34:58 --> 00:35:01 formulas. OK, so here it works well 425 00:35:01 --> 00:35:04 because we know how to solve for b. 426 00:35:04 --> 00:35:07 But I'm not considering this to be a real method because in 427 00:35:07 --> 00:35:08 general we don't know how to do that. 428 00:35:08 --> 00:35:12 I mean, in the beginning I had this relation that was an 429 00:35:12 --> 00:35:16 equation of degree three. You don't really want to solve 430 00:35:16 --> 00:35:19 your equation for the dependent variable usually. 431 00:35:19 --> 00:35:33 Here, we can. So, solve for b and substitute. 432 00:35:33 --> 00:35:38 So, how do we do that? Well, the constraint is a=b 433 00:35:38 --> 00:35:45 cosine theta. That means b is a over cosine 434 00:35:45 --> 00:35:48 theta. Some of you know that as a 435 00:35:48 --> 00:35:56 secan theta. That's the same. 436 00:35:56 --> 00:36:04 And now, if we express the area in terms of a and theta only, 437 00:36:04 --> 00:36:13 A is one half of ab cosine, sorry, ab sine theta is now one 438 00:36:13 --> 00:36:20 half of a^2 sine theta over cosine theta. 439 00:36:20 --> 00:36:29 Or, if you prefer, one half of a^2 tangent theta. 440 00:36:29 --> 00:36:32 Well, now that it's only a function of a and theta, 441 00:36:32 --> 00:36:35 I know what it means to take the partial derivative with 442 00:36:35 --> 00:36:38 respect to theta, keeping a constant. 443 00:36:38 --> 00:36:51 I know how to do it. So, partial A over partial 444 00:36:51 --> 00:36:55 theta, a held constant, 445 00:36:55 --> 00:36:59 well, if a is a constant, then I get this one half a^2 446 00:36:59 --> 00:37:03 coming out times, what's the derivative of 447 00:37:03 --> 00:37:09 tangent? Secan squared, very good. 448 00:37:09 --> 00:37:12 If you're European and you've never heard of secan, 449 00:37:12 --> 00:37:15 that's one over cosine. And, if you know the derivative 450 00:37:15 --> 00:37:18 as one plus tangent squared, that's the same thing. 451 00:37:18 --> 00:37:24 And, it's also correct. OK, so, that's one way of doing 452 00:37:24 --> 00:37:26 it. But, as I've already said, 453 00:37:26 --> 00:37:30 it doesn't get us very far if we don't know how to solve for 454 00:37:30 --> 00:37:33 b. We really used the fact that we 455 00:37:33 --> 00:37:36 could solve for b and get rid of it. 456 00:37:36 --> 00:37:45 So, there's two systematic methods, and let's say the basic 457 00:37:45 --> 00:37:53 rule is that you should give both of them a chance. 458 00:37:53 --> 00:37:56 You should see which one you prefer, and you should be able 459 00:37:56 --> 00:37:59 to use one or the other on the exam. 460 00:37:59 --> 00:38:04 OK, most likely you'll actually have a choice between one or the 461 00:38:04 --> 00:38:06 other. It will be up to you to decide 462 00:38:06 --> 00:38:10 which one you want to use. But, you cannot use solving in 463 00:38:10 --> 00:38:14 substitution. That's not fair. 464 00:38:14 --> 00:38:25 OK, so the first one is to use differentials. 465 00:38:25 --> 00:38:29 By the way, in the notes they are called also method one and 466 00:38:29 --> 00:38:32 method two. I'm not promising that I have 467 00:38:32 --> 00:38:32 the same one, am I? 468 00:38:32 --> 00:38:35 I mean, I might have one and two switched. 469 00:38:35 --> 00:38:39 It doesn't really matter. So, how do we do things using 470 00:38:39 --> 00:38:43 differentials? Well, first, 471 00:38:43 --> 00:38:52 we know that we want to keep a fixed, and that means that we'll 472 00:38:52 --> 00:38:56 set da equal to zero, OK? 473 00:38:56 --> 00:39:00 The second thing that we want to do is we want to look at the 474 00:39:00 --> 00:39:04 constraint. The constraint is a equals b 475 00:39:04 --> 00:39:08 cosine theta. And, we want to differentiate 476 00:39:08 --> 00:39:10 that. Well, differentiate the 477 00:39:10 --> 00:39:15 left-hand side. You get da. 478 00:39:15 --> 00:39:18 And, differentiate the right-hand side as a function of 479 00:39:18 --> 00:39:20 b and theta. You should get, 480 00:39:20 --> 00:39:23 well, how many db's? Well, that's the rate of change 481 00:39:23 --> 00:39:28 with respect to b. That's cosine theta db minus b 482 00:39:28 --> 00:39:35 sine theta d theta. That's a product rule applied 483 00:39:35 --> 00:39:47 to b times cosine theta. So -- Well, now, 484 00:39:47 --> 00:39:51 if we have a constraint that's relating da, db, 485 00:39:51 --> 00:39:54 and d theta, OK, so that's actually what we 486 00:39:54 --> 00:39:56 did, right, that's the same sort of thing 487 00:39:56 --> 00:39:59 as what we did at the beginning when we related dx, 488 00:39:59 --> 00:40:02 dy, and dz. That's really the same thing, 489 00:40:02 --> 00:40:05 except now are variables are a, b, and theta. 490 00:40:05 --> 00:40:07 Now, we know that also we are keeping a fixed. 491 00:40:07 --> 00:40:10 So actually, we set this equal to zero. 492 00:40:10 --> 00:40:18 So, we have zero equals da equals cosine theta db minus b 493 00:40:18 --> 00:40:23 sine theta d theta. That means that actually we 494 00:40:23 --> 00:40:33 know how to solve for db. OK, so cosine theta db equals b 495 00:40:33 --> 00:40:45 sine theta d theta or db is b tangent theta d theta. 496 00:40:45 --> 00:40:47 OK, so in fact, what we found, 497 00:40:47 --> 00:40:50 if you want, is the rate of change of b with 498 00:40:50 --> 00:40:53 respect to theta. Why do we care? 499 00:40:53 --> 00:40:59 Well, we care because let's look, now, at dA, 500 00:40:59 --> 00:41:03 the function that we want to look at. 501 00:41:03 --> 00:41:12 OK, so the function is A equals one half ab sine theta. 502 00:41:12 --> 00:41:15 Well, then, dA, so we had to use the product 503 00:41:15 --> 00:41:18 rule carefully, or we use the partials. 504 00:41:18 --> 00:41:21 So, the coefficient of d little a will be partial with respect 505 00:41:21 --> 00:41:26 to little a. That's one half b sine theta da 506 00:41:26 --> 00:41:36 plus coefficient of db will be one half a sine theta db plus 507 00:41:36 --> 00:41:45 coefficient of d theta will be one half ab cosine theta d 508 00:41:45 --> 00:41:48 theta. But now, what do I do with that? 509 00:41:48 --> 00:41:52 Well, first I said a is constant. 510 00:41:52 --> 00:41:56 So, da is zero. Second, well, 511 00:41:56 --> 00:41:59 actually we don't like b at all, right? 512 00:41:59 --> 00:42:03 We want to view a as a function of theta. 513 00:42:03 --> 00:42:13 So, well, maybe we actually want to use this formula for db 514 00:42:13 --> 00:42:18 that we found in here. OK, and then we'll be left only 515 00:42:18 --> 00:42:20 with d thetas, which is what we want. 516 00:42:20 --> 00:42:56 517 00:42:56 --> 00:43:06 So, if we plug this one into that one, we get da equals one 518 00:43:06 --> 00:43:16 half a sine theta times b tangent theta d theta plus one 519 00:43:16 --> 00:43:26 half ab cosine theta d theta. And, if we collect these things 520 00:43:26 --> 00:43:35 together, we get one half of ab times sine theta times tangent 521 00:43:35 --> 00:43:41 theta plus cosine theta d theta. And, if you know your trig, 522 00:43:41 --> 00:43:44 but you'll see that this is sine squared over cosine plus 523 00:43:44 --> 00:43:49 cosine squared over cosine. That's the same as secan theta. 524 00:43:49 --> 00:43:54 So, now you have expressed da as something times d theta. 525 00:43:54 --> 00:43:59 Well, that coefficient is the rate of change of A with respect 526 00:43:59 --> 00:44:04 to theta with the understanding that we are keeping a fixed, 527 00:44:04 --> 00:44:10 and letting b vary as a dependent variable. 528 00:44:10 --> 00:44:11 Not enough space: sorry. 529 00:44:11 --> 00:44:26 530 00:44:26 --> 00:44:29 OK, in case it's clearer for you, let's think about it 531 00:44:29 --> 00:44:32 backwards. So, we wanted to find how A 532 00:44:32 --> 00:44:35 changes. To find how A changes, 533 00:44:35 --> 00:44:38 we write da. But now, this tells us how A 534 00:44:38 --> 00:44:41 depends on little a, little b, and theta. 535 00:44:41 --> 00:44:45 Well, we know actually we want to keep little a constant. 536 00:44:45 --> 00:44:49 So, we set this to be zero. Theta, well, 537 00:44:49 --> 00:44:52 we are very happy because we want to express things in terms 538 00:44:52 --> 00:44:55 of theta. Db we want to get rid of. 539 00:44:55 --> 00:45:00 How do we get rid of db? Well, we do that by figuring 540 00:45:00 --> 00:45:05 out how b depends on theta when a is fixed. 541 00:45:05 --> 00:45:08 And, we do that by differentiating the constraint 542 00:45:08 --> 00:45:12 equation, and setting da equal to zero. 543 00:45:12 --> 00:45:31 OK, so -- I guess to summarize the method, we wrote dA in terms 544 00:45:31 --> 00:45:41 of da, db, d theta. Then, we say that a is constant 545 00:45:41 --> 00:45:50 means we set da equals zero. And, the third thing is that 546 00:45:50 --> 00:45:57 because, well, we differentiate the 547 00:45:57 --> 00:46:06 constraint. And, we can solve for db in 548 00:46:06 --> 00:46:19 terms of d theta. And then, we plug into dA, 549 00:46:19 --> 00:46:32 and we get the answer. OK, oops. 550 00:46:32 --> 00:46:38 So, here's another method to do the same thing differently is to 551 00:46:38 --> 00:46:43 use the chain rule. So, we can use the chain rule 552 00:46:43 --> 00:46:45 with dependent variables, OK? 553 00:46:45 --> 00:46:48 So, what does the chain rule tell us? 554 00:46:48 --> 00:46:54 The chain rule tells us, so we will want to 555 00:46:54 --> 00:47:02 differentiate -- -- the formula for a with respect to theta 556 00:47:02 --> 00:47:06 holding a constant. So, I claim, 557 00:47:06 --> 00:47:10 well, what does the chain rule tell us? 558 00:47:10 --> 00:47:14 It tells us that, well, when we change things, 559 00:47:14 --> 00:47:19 a changes because of the changes in the variables. 560 00:47:19 --> 00:47:24 So, part of it is that A depends on theta and theta 561 00:47:24 --> 00:47:28 changes. How fast does theta change? 562 00:47:28 --> 00:47:31 Well, you could call that the rate of change of theta with 563 00:47:31 --> 00:47:33 respect to theta with a constant. 564 00:47:33 --> 00:47:35 But of course, how fast does theta depend to 565 00:47:35 --> 00:47:38 itself? The answer is one. 566 00:47:38 --> 00:47:44 So, that's pretty easy. Plus, then we have the partial 567 00:47:44 --> 00:47:49 derivative, formal partial derivative, of A with respect to 568 00:47:49 --> 00:47:55 little a times the rate of change of a in our situation. 569 00:47:55 --> 00:47:58 Well, how does little a change if a is constant? 570 00:47:58 --> 00:48:08 Well, it doesn't change. And then, there is Ab, 571 00:48:08 --> 00:48:14 the formal partial derivative times, sorry, 572 00:48:14 --> 00:48:20 the rate of change of b. OK, and how do we find this one? 573 00:48:20 --> 00:48:27 Well, here we have to use the constraint. 574 00:48:27 --> 00:48:30 OK, and we can find this one from the constraint as we've 575 00:48:30 --> 00:48:32 seen at the beginning either by differentiating the constraint, 576 00:48:32 --> 00:48:36 or by using the chain rule on the constraint. 577 00:48:36 --> 00:48:39 So, of course the calculations are exactly the same. 578 00:48:39 --> 00:48:44 See, this is the same formula as the one over there, 579 00:48:44 --> 00:48:48 just dividing everything by partial theta and with 580 00:48:48 --> 00:48:54 subscripts little a. But, if it's easier to think 581 00:48:54 --> 00:48:59 about it this way, then that's also valid. 582 00:48:59 --> 00:49:03 OK, so tomorrow we are going to review for the test, 583 00:49:03 --> 00:49:06 so I'm going to tell you a bit more about this also as we go 584 00:49:06 --> 00:49:09 over one practice problem on that. 585 00:49:09 --> 00:49:10