ch06_06 - 546 CHAPTER 6 Integration Techniques 6-38 0 x a...

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546 CHAPTER 6 . . Integration Techniques 6-38 for 0 x a . Thus, the area of the ellipse is 4 b a 0 1 x 2 a 2 dx . Try this integral on your CAS. The (im- plicit) assumption we usually make is that a > 0, but your CAS should not make this assumption for you. Does your CAS give you π ab or π b | a | ? EXPLORATORY EXERCISES 1. This exercise explores two aspects of a very famous prob- lem (solved in the 1600s; when you finish the problem, think about solving this problem before calculators, computers or even much of calculus was invented). The idea is to imagine a bead sliding down a thin wire that extends in some shape from the point (0, 0) to the point ( π, 2). Assume that grav- ity pulls the bead down but that there is no friction or other force acting on the bead. This situation is easiest to analyze using parametric equations where we have functions x ( u ) and y ( u ) giving the horizontal and vertical position of the bead in terms of the variable u . Examples of paths the bead might follow are x ( u ) = π u y ( u ) = − 2 u and x ( u ) = π u y ( u ) = 2( u 1) 2 2 and x ( u ) = π u sin π u y ( u ) = cos π u 1 . In each case, the bead starts at (0, 0) for u = 0 and finishes at ( π, 2) for u = 1. You should graph each path on your graphing calculator. The first path is a line, the second is a parabola and the third is a special curve called a brachistochrone. For a given path, the time it takes the bead to travel the path is given by T = 1 g 1 0 [ x ( u )] 2 + [ y ( u )] 2 2 y ( u ) du , where g is the gravitational constant. Compute this quantity for the line and the parabola. Explain why the parabola would be a faster path for the bead to slide down, even though the line is shorter in distance. (Think of which would be a faster hill to ski down.) It can be shown that the brachistochrone is the fastest path possible. Try to get your CAS to compute the optimal time. Comparing the graphs of the parabola and brachistochrone, what important advantage does the brachis- tochrone have at the start of the path? 2. It turns out that the brachistochrone in exploratory exercise 1 has an amazing property, along with providing the fastest time (which is essentially what the term brachistochrone means). The path is shown in the figure. y x 1 2 3 2 1
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6-39 SECTION 6.6 . . Improper Integrals 547 There is something fundamentally wrong with this “calculation.” Note that f ( x ) = 1 / x 2 is not continuous over the interval of integration. (See Figure 6.2.) Since the Fundamental Theorem assumes a continuous integrand, our use of the theorem is invalid and our answer is incorrect . Further, note that an answer of 3 2 is especially suspicious given that the integrand 1 x 2 is always positive. y x 2 4 2 4 2 4 FIGURE 6.2 y = 1 x 2 Recall that in Chapter 4, we defined the definite integral by b a f ( x ) dx = lim n →∞ n i = 1 f ( c i ) x , where c i was taken to be any point in the subinterval [ x i 1 , x i ], for i = 1 , 2 ,..., n and where the limit had to be the same for any choice of these c i ’s. So, if f ( x ) → ∞ [or f ( x ) → −∞ ] at some point in [ a , b ], then the limit defining b a f ( x ) dx is meaningless.
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