thomasET_226348_ism51

thomasET_226348_ism51 - 656 48(a Chapter 10 Conic Sections...

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656 Chapter 10 Conic Sections and Polar Coordinates 48. (a) (b) (c) (d) (e) 49. (a) r 4 cos cos ; r 1 cos r 1 0 r 4r 4 (r 2) 0 # # # œ Ê œ œ Ê œ Ê œ Ê œ ) ) ) r r 4 4 Š r 2; therefore cos 1 (2 ) is a point of intersection Ê œ œ œ Ê œ Ê ß ) ) 1 1 2 4 (b) r 0 0 4 cos cos 0 , or is on the graph; r 0 0 1 cos œ Ê œ Ê œ Ê œ Ê œ Ê œ # # # # # ) ) ) ) 1 1 1 1 3 3 ˆ ˆ cos 1 0 (0 0) is on the graph. Since ( 0) for polar coordinates, the graphs Ê œ Ê œ Ê ß œ ) ) ˆ 1 # intersect at the origin. 50. (a) Let r f( ) be symmetric about the x-axis and the y-axis. Then (r ) on the graph (r ) is on the œ ß Ê ß ) ) ) graph because of symmetry about the x-axis. Then ( r ( )) ( r ) is on the graph because of ß œ ß ) ) symmetry about the y-axis. Therefore r f( ) is symmetric about the origin. œ ) (b) Let r f( ) be symmetric about the x-axis and the origin. Then (r ) on the graph (r ) is on the œ ß Ê ß ) ) ) graph because of symmetry about the x-axis. Then ( r ) is on the graph because of symmetry about ß ) the origin. Therefore r f( ) is symmetric about the y-axis. œ ) (c) Let r f( ) be symmetric about the y-axis and the origin. Then (r ) on the graph ( r ) is on the œ ß Ê ß ) ) ) graph because of symmetry about the y-axis. Then ( ( r) ) (r ) is on the graph because of ß œ ß ) ) symmetry about the origin. Therefore r f( ) is symmetric about the x-axis. œ ) 51. The maximum width of the petal of the rose which lies along the x-axis is twice the largest y value of the curve on the interval 0 . So we wish to maximize 2y 2r sin 2 cos 2 sin on 0 . Let Ÿ Ÿ œ œ Ÿ Ÿ ) ) ) ) ) 1 1 4 4 f( ) 2 cos 2 sin 2 1 2 sin (sin ) 2 sin 4 sin f ( ) 2 cos 12 sin cos . Then ) ) ) ) ) ) ) ) ) ) ) œ œ œ Ê œ a b # $ w # f ( ) 0 2 cos 12 sin cos 0 (cos ) 1 6 sin 0 cos 0 or 1 6 sin 0 or w # # # # ) ) ) ) ) ) ) ) ) œ Ê œ Ê œ Ê œ œ Ê œ a b 1 sin . Since we want 0 , we choose sin f( ) 2 sin 4 sin ) ) ) ) ) ) œ Ÿ Ÿ œ Ê œ " " $ 1 6 6 4 È È 1 Š 2 4 . We can see from the graph of r cos 2 that a maximum does occur in the œ œ œ Š " " È È È 6 6 6 2 6 9 ) interval 0 . Therefore the maximum width occurs at sin , and the maximum width Ÿ Ÿ œ ) ) 1 4 6 " " Š È is . 2 6 9 È 52. We wish to maximize y r sin 2(1 cos )(sin ) 2 sin 2 sin cos . Then œ œ œ ) ) ) ) ) ) 2 cos 2(sin )( sin ) 2 cos cos 2 cos 2 sin 2 cos 2 cos 4 cos 2; thus dy d ) œ œ œ ) ) ) ) ) ) ) ) ) ) # # # 0 4 cos 2 cos 2 = 0 2 cos cos 1 0 (2 cos 1)(cos 1) 0 cos dy d ) œ Ê Ê œ Ê œ Ê œ # # " # ) ) ) ) ) ) ) or cos 1 , , . From the graph, we can see that the maximum occurs in the first quadrant so ) ) 1 œ Ê œ 1 1 3 3 5 we choose . Then y 2 sin 2 sin cos . The x-coordinate of this point is x r cos ) œ œ œ œ 1 1 1 1 1 3 3 3 3 3 3 3 È # 2 1 cos cos . Thus the maximum height is h occurring at x . œ œ œ œ ˆ ‰ ˆ 1 1 3 3 3 3 3 3 # # # È
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Section 10.7 Area and Lengths in Polar Coordinates 657 10.7 AREA AND LENGTHS IN POLAR COORDINATES 1. A (4 2 cos ) d 16 16 cos 4 cos d 8 8 cos 2 d œ œ œ ' ' ' 0 0 0 2 2 2 " " # # # # # ) ) ) ) ) ) ) a b ˆ 1 cos 2 ) (9 8 cos cos 2 ) d 9 8 sin sin 2 18 œ œ œ ' 0 2 ) ) ) ) ) ) 1 " # !
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