Pre-Calc Homework Solutions 125

Pre-Calc Homework Solutions 125 - Chapter 3 Review d g(x dx...

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(f) } d d x } g ( x 1 f ( x )) 5 g 9 ( x 1 f ( x )) } d d x } ( x 1 f ( x )) 5 g 9 ( x 1 f ( x ))(1 1 f 9 ( x )) At x 5 0, the derivative is g 9 (0 1 f (0))[1 1 f 9 (0)] 5 g 9 (0 2 1)[1 1 ( 2 2)] 5 (1)( 2 1) 5 2 1 68. } d d w s } 5 } d d w r } } d d r s } 5 } d d r } [sin ( ˇ r w 2 2)] } d d s } 3 8 sin 1 s 1 } p 6 } 24 5 3 cos ( ˇ r w 2 2) } 2 ˇ 1 r w } 43 8 cos 1 s 1 } p 6 } 24 At s 5 0, we have r 5 8 sin 1 0 1 } p 6 } 2 5 4 and so } d d w s } 5 3 cos ( ˇ 4 w 2 2) } 2 ˇ 1 4 w } 43 8 cos 1 0 1 } p 6 } 24 5 1 } co 4 s 0 } 21 8 cos } p 6 } 2 5 1 } 1 4 } 21 } 8 ˇ 2 3 w } 2 5 ˇ 3 w 69. Solving u 2 t 1 u 5 1 for t , we have t 5 } 1 u 2 2 u } 5 u 2 2 2 u 2 1 , and we may write: } d d u r } 5 } d d r t } } d d u t } } d d u } ( u 2 1 7) 1/3 5 } d d r t } } d d u } ( u 2 2 2 u 2 1 ) } 1 3 } ( u 2 1 7) 2 2/3 (2 u ) 5 1 } d d r t } 2 ( 2 2 u 2 3 1 u 2 2 ) } d d r t } 5 5 At t 5 0, we may solve u 2 t 1 u 5 1 to obtain u 5 1, and so } d d r t } 5 5 } 2(8 2 ) 2 3 2/3 } 5 2 } 1 6 } . 70. (a) One possible answer: x ( t ) 5 10 cos 1 t 1 } p 4 } 2 , y ( t ) 5 1 (b) s (0) 5 10 cos } p 4 } 5 5 ˇ 2 w (c) Farthest left: When cos 1 t 1 } p 4 } 2 5 2 1, we have s ( t ) 5 2 10.
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