multivariable_07_Integration_2up

multivariable_07_Integration_2up - 7 Integration 7 ea e Up...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 12/01/2011 for the course MATH 305 taught by Professor Guichard during the Fall '11 term at Whitman.

Page1 / 8

multivariable_07_Integration_2up - 7 Integration 7 ea e Up...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online