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Unformatted text preview: 7 Integration 7 ea e Up to now we have been concerned with extracting information about how a function changes from the function itself. Given knowledge about an objects position, for example, we want to know the objects speed. Given information about the height of a curve we want to know its slope. We now consider problems that are, whether obviously or not, the reverse of such problems. EXAMPLE 7.1 An object moves in a straight line so that its speed at time t is given by v ( t ) = 3 t in, say, cm/sec. If the object is at position 10 on the straight line when t = 0, where is the object at any time t ? There are two reasonable ways to approach this problem. If s ( t ) is the position of the object at time t , we know that s ( t ) = v ( t ). Because of our knowledge of derivatives, we know therefore that s ( t ) = 3 t 2 / 2+ k , and because s (0) = 10 we easily discover that k = 10, so s ( t ) = 3 t 2 / 2 +10. For example, at t = 1 the object is at position 3 / 2+ 10 = 11 . 5. This is certainly the easiest way to deal with this problem. Not all similar problems are so easy, as we will see; the second approach to the problem is more difficult but also more general. We start by considering how we might approximate a solution. We know that at t = 0 the object is at position 10. How might we approximate its position at, say, t = 1? We know that the speed of the object at time t = 0 is 0; if its speed were constant then in the first second the object would not move and its position would still be 10 when t = 1. In fact, the object will not be too far from 10 at t = 1, but certainly we can do better. Lets look at the times 0 . 1, 0 . 2, 0 . 3, . . . , 1 . 0, and try approximating the location of the object 133 134 Chapter 7 Integration at each, by supposing that during each tenth of a second the object is going at a constant speed. Since the object initially has speed 0, we again suppose it maintains this speed, but only for a tenth of second; during that time the object would not move. During the tenth of a second from t = 0 . 1 to t = 0 . 2, we suppose that the object is traveling at 0 . 3 cm/sec, namely, its actual speed at t = 0 . 1. In this case the object would travel (0 . 3)(0 . 1) = 0 . 03 centimeters: 0 . 3 cm/sec times 0 . 1 seconds. Similarly, between t = 0 . 2 and t = 0 . 3 the object would travel (0 . 6)(0 . 1) = 0 . 06 centimeters. Continuing, we get as an approximation that the object travels (0 . 0)(0 . 1) + (0 . 3)(0 . 1) + (0 . 6)(0 . 1) + + (2 . 7)(0 . 1) = 1 . 35 centimeters, ending up at position 11.35. This is a better approximation than 10, certainly, but is still just an approximation. (We know in fact that the object ends up at position 11 . 5, because weve already done the problem using the first approach.) Presumably, we will get a better approximation if we divide the time into one hundred intervals of a hundredth of a second each, and repeat the process: (0 . 0)(0 . 01) + (0 . 03)(0 . 01) + (0...
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This note was uploaded on 12/01/2011 for the course MATH 305 taught by Professor Guichard during the Fall '11 term at Whitman.
- Fall '11