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 ² 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 ³ x ( u ) = π u y ( u ) =− 2 u and ³ x ( u ) = π u y ( u ) = 2( u 1) 2 2 and ³ 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 ( 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. Foragiven 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 ± Suppose that instead of starting the bead at the point (0, 0), you start the bead partway down the path at x = c .How would the time to reach the bottom from x = c compare to the total time from x = 0? Note that the answer is not obvious, since the farther down you start, the less speed the bead will gain. If x = c corresponds to u = a , the time to reach the bottom is given by π g ± 1 a ² 1 cos π u cos a π cos π u .If a = 0 (that is, the bead starts at the top), the time is π/ g (the integral equals 1).
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This note was uploaded on 10/03/2010 for the course CHE 2C CHE 2C taught by Professor Atsumi during the Spring '10 term at UC Davis.

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

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