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74. (a) r ( t ) 5 (155 cos 18 8 2 11.7) t i 1 (4 1 155 sin 18 8 t 2 16 t 2 ) j x ( t ) 5 (155 cos 18 8 2 11.7) t y ( t ) 5 4 1 155 sin 18 8 t 2 16 t 2 (b) y max 5 } (155 2 s ( i 3 n 2) 18 8 ) 2 } 1 4 < 39.847 feet, reached at t max 5 } 155 3 si 2 n 18 8 } < 1.497 sec (c) y ( t ) 5 0 when t < 3.075 sec (found using the quadratic formula), and then x < (155 cos 18 8 2 11.7)(3.075) < 417.307 ft. (d) Solve y ( t ) 5 25 using the quadratic formula: t 5 < 0.534 and 2.460 seconds. At those times, x 5 (155 cos 18 8 2 11.7) t equals < 72.406 and < 333.867 feet from home plate. (e) Yes, the batter has hit a home run. When the ball is 380 feet from home plate (at t < 2.800 seconds), it is approximately 12.673 feet off the ground and therefore clears the fence by at least two feet. 75. (a) r ( t ) 5 3 (155 cos 18 8 2 11.7) } 0. 1 09 } (1 2 e 2 0.09 t ) 4 i 1 3 4 1 1 } 155 0 s .0 in 9 18 8 } 2 (1 2 e 2 0.09 t ) 1 } 0. 3 0 2 9 2 } (1 2 0.09 t 2 e 2 0.09 t ) 4 j x ( t ) 5 (155 cos 18 8 2 11.7) } 0. 1 09 } (1 2 e 2 0.09 t ) y ( t ) 5 4 1 1 } 155 0 s .0 in 9 18 8 } 2 (1 2 e 2 0.09 t ) 1 } 0. 3 0 2 9 2 } (1 2 0.09 t 2 e 2 0.09 t ) (b) Plot y ( t ) and use the maximum function to find y < 36.921 feet at t < 1.404 seconds. (c) Plot y ( t ) and find that y ( t ) 5 0 at t < 2.959, then plug this into the expression for x ( t ) to find x (2.959) < 352.520 ft. (d) Plot y ( t ) and find that y ( t ) 5 30 at t < 0.753 and 2.068 seconds. At those times, x < 98.799 and 256.138 feet (from home plate). (e) No, the batter has not hit a home run. If the drag coefficient k is less than < 0.011, the hit will be a home run. (This result can be found by trying different k -values until the parametrically plotted curve has y \$ 10 for x 5 380.) 76. (a) E BD 5 E AD 2 E AB (b) E AP 5 E AB 1 } 1 2 } E BD = } 1 2 } E AB 1 } 1 2 } E AD (c) E AC 5 E AB 1 E AD , so by part (b), E AP 5 } 1 2 } E AC . 77. The widths between the successive turns are constant and are given by 2 p a . Cumulative Review Exercises (pp. 573–576) 1. Since the function has no discontinuity at x 5 1, the limit is 5 0. 2. By l’Hôpital’s Rule, lim x 0 } sin 4 x 3 x } 5 lim x 0 } 3 co 4 s 3 x } 5 } 3 4 } . 3. By l’Hôpital’s Rule, lim x 0 5 lim x 0 5 2 1. 4. By l’Hôpital’s Rule, lim x } x x 1 2 e e x x } 5 lim x } 1 1 1 2 e e x x } 5 lim x } 2 e e x x } 5 2 1. 5. By l’Hôpital’s Rule, lim t 0 } t (1 t 2 2 s c i o n s t t ) } 5 lim t 0 5 lim t 0 } t cos t si 1 n t 2 sin t } 5 lim t 0 } 2 t sin c t o 1 s t 3 cos t } 5 3 6. By l’Hôpital’s Rule, lim x 0 1 } ln ( e ln x 2 x 1) } 5 lim x 0 1 5 lim x 0 1 } e x x 2 e x 1 } 5 lim x 0 1 } xe x e 1 x e x } 5 1 7. Use f ( x ) 5 ( e x 1 x ) 1/ x . Then ln f ( x ) 5 } ln ( e x x 1 x ) } , and lim x 0 } ln ( e x x 1 x ) } 5 lim x 0 } ( e x 1 1) 1 /( e x 1 x ) } 5 2. So lim x 0 ( e x 1 x ) 1/ x 5 lim x 0 e ln f ( x ) 5 e 2 . 8. lim x 0 1 } 3 x x 1 1 } 2 } si 1 n x } 2 5 lim x 0 5 lim x 0 5 lim x 0 5 3 2 (3 x 1 1) sin x 1 6 cos x }}} 2 x sin x 1 2 cos x (3 x 1 1) cos x 1 3 sin x 2 1 }}} x cos x 1 sin x (3 x 1 1) sin x 2 x }} x sin x } ( e x e 2 x 1) } } } 1 x } t sin t 1 (1 2 cos t ) }}} 1 2 cos t 2 } ( x 1 1 1) 2 } }} 1 } x 1 1 1 } 2 1 }} x 2(1) 2 2 1 2 1 }} 1 2 1 1 2 12 2 155 sin 18 8 6 ˇ 1 w 5 w 5 w 2 w si w n w 2 w 1 w 8 w 8 w 2 w 4 w (1 w 6 w )( w 2 w 1 w ) w }}}}} 2 32 Cumulative Review 441

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9. (a) 2(1) 2 1 2 5 1 (b) 2 2 1 5 1 (c) 1 [from (a) and (b)] (d) Yes, since lim x 1 f ( x ) 5 f (1) 5 1 (e) No.
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