121616949-math.153 - k = 10 so s t = 3 t 2 2 10 For example...

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7 Integration Up to now we have been concerned with extracting information about how a function changes from the function itself. Given knowledge about an object’s position, for example, we want to know the object’s 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
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Unformatted text preview: 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. Let’s 139...
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