role stringclasses 2
values | content stringlengths 0 2.1k | session_id int64 10 21.7k | sequence_id int64 0 2.38k | annotations listlengths 0 8 |
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volunteer | because the definition of a tangent line means that the, the line can only intersect with a function at one point | 16,993 | 248 | [] |
volunteer | and we define that point as X1. | 16,993 | 249 | [] |
volunteer | X1Y1 | 16,993 | 250 | [] |
student | ok ok | 16,993 | 251 | [] |
volunteer | Any more questions before I continue? | 16,993 | 252 | [] |
volunteer | That's a very good question. | 16,993 | 253 | [] |
student | nope | 16,993 | 254 | [] |
volunteer | Sorry, my handwriting can get a little bad | 16,993 | 255 | [] |
volunteer | Please let me know if I need to, um, kind of enlarge something. | 16,993 | 256 | [] |
volunteer | or rewrite it | 16,993 | 257 | [] |
volunteer | Um | 16,993 | 258 | [] |
volunteer | OK. So back to our um solution for Y2. | 16,993 | 259 | [] |
volunteer | OK. | 16,993 | 260 | [] |
volunteer | So if we remember back to our slope equation | 16,993 | 261 | [] |
volunteer | right? I didn't finish writing it, but you know, uh, if you look at the denominator | 16,993 | 262 | [] |
volunteer | this X2 minus X1. Doesn't that just look like our delta X? | 16,993 | 263 | [] |
student | yes | 16,993 | 264 | [] |
volunteer | Yeah. So we factor this back in here. | 16,993 | 265 | [] |
volunteer | right? So we find that M is equal to | 16,993 | 266 | [] |
volunteer | um we'll say Y to uh | 16,993 | 267 | [] |
volunteer | yeah, I will say F(X2) minus | 16,993 | 268 | [] |
volunteer | F of X1 over delta X. | 16,993 | 269 | [] |
volunteer | times delta X + F(X1) | 16,993 | 270 | [] |
volunteer | Um, what does that give us | 16,993 | 271 | [] |
volunteer | I guess it's just give us X of X1 is equal to X1, that doesn't help us. | 16,993 | 272 | [] |
volunteer | But I guess we can leave this | 16,993 | 273 | [] |
volunteer | leave that as is. | 16,993 | 274 | [] |
volunteer | Um, cause I, I fly plug it back in. This is just going to give me that F of X1 is equal to Y1, which we already known, doesn't really help us. | 16,993 | 275 | [] |
volunteer | but we could rewrite this in terms of | 16,993 | 276 | [] |
volunteer | uh instead of this being Y2, we can rewrite this as M delta X + F of X1. | 16,993 | 277 | [] |
volunteer | So that | 16,993 | 278 | [] |
volunteer | if we | 16,993 | 279 | [] |
volunteer | sorry, go ahead | 16,993 | 280 | [] |
volunteer | Did you have a question | 16,993 | 281 | [] |
volunteer | OK. So, if we leave it | 16,993 | 282 | [] |
volunteer | of this form | 16,993 | 283 | [] |
volunteer | um what Y2 is and we factor this back into our error equation. | 16,993 | 284 | [] |
volunteer | right? We find that | 16,993 | 285 | [] |
volunteer | uh | 16,993 | 286 | [] |
volunteer | M delta X. | 16,993 | 287 | [] |
volunteer | plus F of X1. | 16,993 | 288 | [] |
volunteer | minus F of X2, right? And then we write it all out, we find that | 16,993 | 289 | [] |
volunteer | um | 16,993 | 290 | [] |
volunteer | the error function is just the, whatever our step side is, tell them to slope. | 16,993 | 291 | [] |
volunteer | minus, we could say Delta F. | 16,993 | 292 | [] |
volunteer | right, with the actual function is actually, let me, let me just try to solve for. | 16,993 | 293 | [] |
volunteer | no F of X1 | 16,993 | 294 | [] |
volunteer | Yeah, we said | 16,993 | 295 | [] |
volunteer | Yeah, that's fine. You can just say F of X1. | 16,993 | 296 | [] |
volunteer | plus F of X1. | 16,993 | 297 | [] |
volunteer | which is just equal to | 16,993 | 298 | [] |
volunteer | M delta X is equal to minus. | 16,993 | 299 | [] |
volunteer | RDF | 16,993 | 300 | [] |
volunteer | All right, which is basically just the difference in a function is equal to whatever our change in why is, whatever our slope is. | 16,993 | 301 | [] |
volunteer | Um. | 16,993 | 302 | [] |
volunteer | and | 16,993 | 303 | [] |
volunteer | if we can get | 16,993 | 304 | [] |
volunteer | this is where we start taking, yeah, sorry. | 16,993 | 305 | [] |
volunteer | Let me enlarge this. | 16,993 | 306 | [] |
volunteer | Give me one second I think. I did this, I did this exact derivation | 16,993 | 307 | [] |
volunteer | a couple of weeks ago. I'm think I'm trying to make sure I remember everything correctly. | 16,993 | 308 | [] |
volunteer | Uh, don't forget anything. | 16,993 | 309 | [] |
student | ok take your time] | 16,993 | 310 | [] |
volunteer | I found a tangent line um found the air functioning, found the error value. | 16,993 | 311 | [] |
volunteer | I think we just want the error value in terms of | 16,993 | 312 | [] |
volunteer | why one and S's. | 16,993 | 313 | [] |
volunteer | Yeah. | 16,993 | 314 | [] |
volunteer | Yeah, sorry. I feel like this delta, this, this error function is wrong. | 16,993 | 315 | [] |
volunteer | Somehow it's close to being around. I feel like I'm missing something. | 16,993 | 316 | [] |
volunteer | Cause once you consider that er cause like the air function | 16,993 | 317 | [] |
volunteer | if you take a limit | 16,993 | 318 | [] |
volunteer | as delta X | 16,993 | 319 | [] |
volunteer | goes to 0, right? That we make our step size infinitesimally small. | 16,993 | 320 | [] |
volunteer | right? We make our X2 | 16,993 | 321 | [] |
volunteer | kind of really close to X1, right? You'll notice that our air function | 16,993 | 322 | [] |
volunteer | should equal 0 | 16,993 | 323 | [] |
volunteer | right? At that, at that point, you would say that Y2 is equal to f of X2. | 16,993 | 324 | [] |
volunteer | right? If your, if your air function is absolutely zero. And this is where we can um | 16,993 | 325 | [] |
volunteer | say that our approach using the tangent line. | 16,993 | 326 | [] |
volunteer | is an effective approach to evaluating um | 16,993 | 327 | [] |
volunteer | the next values in the function, right? That it can be used effectively and if you had um | 16,993 | 328 | [] |
volunteer | if you had the | 16,993 | 329 | [] |
volunteer | tangent, if you had the slope | 16,993 | 330 | [] |
volunteer | at any point in X. So you had M | 16,993 | 331 | [] |
volunteer | for every X | 16,993 | 332 | [] |
volunteer | in the domain, right? All real numbers that the function lies. | 16,993 | 333 | [] |
volunteer | then you should be able to approximate | 16,993 | 334 | [] |
volunteer | all the values of F of X. | 16,993 | 335 | [] |
volunteer | Um, there is | 16,993 | 336 | [] |
volunteer | Yeah | 16,993 | 337 | [] |
volunteer | All right, I'm gonna cheat | 16,993 | 338 | [] |
volunteer | Actually, I can't, I don't know if I can find that. | 16,993 | 339 | [] |
volunteer | Um. | 16,993 | 340 | [] |
volunteer | anyway, and if you had, and if you did the sum, | 16,993 | 341 | [] |
volunteer | or if you | 16,993 | 342 | [] |
volunteer | if you did the set of all slopes for every X. | 16,993 | 343 | [] |
volunteer | you would have the function F of X, which would give you all the information of the change in behavior. | 16,993 | 344 | [] |
volunteer | of F | 16,993 | 345 | [] |
volunteer | right? So what a derivative actually is | 16,993 | 346 | [] |
volunteer | is just, it tells you the behavior of the function F. | 16,993 | 347 | [] |
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