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38. } d d y t } 5 } d d y x }} d d x t } 5 (3 x 2 2 4) } d d x t } Since } d d x t }52 2 cm/sec, we have } d d y t } 5 8 2 6 x 2 cm/sec. (a) } d d y t } 5 8 2 6( 2 3) 2 52 46 cm/sec (b) } d d y t } 5 8 2 6(1) 2 5 2 cm/sec (c) } d d y t } 5 8 2 6(4) 2 88 cm/sec 39. (a) The point being plotted would correspond to a point on the edge of the wheel as the wheel turns. (b) One possible answer is u 5 16 p t , where t is in seconds. (An arbitrary constant may be added to this expression, and we have assumed counterclockwise motion.) (c) In general, assuming counterclockwise motion: } d d x t 2 sin u } d d u t 2(sin u )(16 p ) 32 p sin u } d d y t } 5 2 cos u } d d u t } 5 2(cos u )(16 p ) 5 32 p cos u At u 5 } p 4 } : } d d x t 32 p sin } p 4 16 p ( ˇ 2 w ) < 2 71.086 ft/sec } d d y t } 5 32 p cos } p 4 } 5 16 p ( ˇ 2 w ) < 71.086 ft/sec At u 5 } p 2 } : } d d x t 32 p sin } p 2 32 p < 2 100.531 ft/sec } d d y t } 5 32 p cos } p 2 } 5 0 ft/sec At u 5 p : } d d x t 32 p sin p 5 0 ft/sec } d d y t } 5 32 p cos p 32 p < 2 100.531 ft/sec 40. (a) One possible answer: x 5 30 cos u , y 5 40 1 30 sin u (b) Since the ferris wheel makes one revolution every 10 sec, we may let u 5 0.2 p t and we may write x 5 30 cos 0.2 p t , y 5 40 1 30 sin 0.2 p t . (This assumes that the ferris wheel revolves counterclockwise.) In general: } d d x t 30(sin 0.2 p t )(0.2 p ) 6 p sin 0.2 p t } d d y t } 5 30(cos 0.2 p t )(0.2 p ) 5 6 p cos 0.2 p t At t 5 5: } d d x t 6 p sin p 5 0 ft/sec } d d y t } 5 6 p cos p 5 6 p ( 2 1) < 2 18.850 ft/sec At t 5 8: } d d x t 6 p sin 1.6 p < 17.927 ft/sec } d d y t } 5 6 p cos 1.6 p < 5.825 ft/sec 41. (a) } d d y t } 5 } d d t } ( uv ) 5 u } d d v t } 1 v } d d u t } 5 u (0.05 v ) 1 v (0.04 u ) 5 0.09 uv 5 0.09 y Since } d d y t } 5 0.09 y , the rate of growth of total production is 9% per year. (b) } d d y t } 5 } d d t } ( uv ) 5 u } d d v t } 1 v } d d u t } 5 u (0.03 v ) 1 v ( 2 0.02 u ) 5 0.01 uv 5 0.01 y The total production is increasing at the rate of 1% per year. Chapter 4 Review (pp. 242–245) 1. y 5 x ˇ 2 w 2 w x w y 95 x 1 } 2 ˇ 2 w 1 2 w x w } 2 ( 2 1) 1 ( ˇ 2 w 2 w x w )(1) 5 } 2 x 2 1 ˇ 2 2 w ( 2 w 2 2 x w x ) } 5 } 2 ˇ 4 2 2 w 2 w 3 x x w } The first derivative has a zero at } 4 3 } .

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