18-01F07-L10

18-01F07-L10 - MIT OpenCourseWare http:/ocw.mit.edu 18.01...

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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus, Fall 2007 Please use the following citation format: David Jerison, 18.01 Single Variable Calculus, Fall 2007 . (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms
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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus, Fall 2007 Transcript – Lecture 10 The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROF. JERISON: So, we're ready to begin Lecture 10, and what I'm going to begin with is by finishing up some things from last time. We'll talk about approximations, and I want to fill in a number of comments and get you a little bit more oriented in the point of view that I'm trying to express about approximations. So, first of all, I want to remind you of the actual applied example that I wrote down last time. So that was this business here. There was something from special relativity. And the approximation that we used was the linear approximation, with a - 1/2 power that comes out to be t( 1 1/2 v^2 / C^2). I want to reiterate why this is a useful way of thinking of things. And why this is that this comes up in real life. Why this is maybe more important than everything that I've taught you about technically so far. So, first of all, what this is telling us is the change in t / t, if you do the arithmetic here and subtract t that's using the change in t is t' - t here. If you work that out, this is approximately the same as 1/2 (v^2 / C^2). So what is this saying? This is saying that if you have this satellite, which is going at speed v, and little c is the speed of light, then the change in the watch down here on earth, relative to the time on the satellite, is going to be proportional to this ratio here. So, physically, this makes sense. This is time divided by time. And this is velocity squared divided by velocity squared. So, in each case, the units divide out. So this is a dimensionless quantity. And this is a dimensionless quantity. And the only point here that we're trying to make is just this notion of proportionality. So I want to write this down. Just, in summary. So the error fraction, if you like, which is sort of the number of significant digits that we have in our measurement, is proportional, in this case, to this quantity. It happens to be proportional to this quantity here. And the factor is, happens to be, 1/2. So these proportionality factors are what we're looking for. Their rates of change. Their rates of change of something with respect to something else. Now, on your homework, you have something rather similar to this. So in Problem,
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This note was uploaded on 01/18/2012 for the course MATH 18.01 taught by Professor Brubaker during the Fall '08 term at MIT.

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18-01F07-L10 - MIT OpenCourseWare http:/ocw.mit.edu 18.01...

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