{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Business Calc Homework w answers_Part_26

Business Calc Homework w answers_Part_26 - 126 Chapter 3...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
74. continued (c) One possible answer: If the current ridership is less than 40, then the proposed plan may be good. If the current ridership is greater than or equal to 40, then the plan is not a good idea. Look at the graph of y 5 r ( x ). [0, 60] by [ 2 50, 200] 75. (a) Since x 5 tan u , we have } d d x t } 5 (sec 2 u ) } d d u t } 5 2 0.6 sec 2 u . At point A , we have u 5 0 and } d d x t } 5 2 0.6 sec 2 0 5 2 0.6 km/sec. (b) 0.6 } r s a e d c } ? } 1 r 2 ev p o r lu ad tion } ? } 6 1 0 m s i e n c } = } 1 p 8 } revolutions per minute or approximately 5.73 revolutions per minute. 76. Let f ( x ) 5 sin ( x 2 sin x ). Then f 9 ( x ) 5 cos ( x 2 sin x ) } d d x } ( x 2 sin x ) 5 cos ( x 2 sin x )(1 2 cos x ). This derivative is zero when cos ( x 2 sin x ) 5 0 (which we need not solve) or when cos x 5 1,which occurs at x 5 2 k p for integers k . For each of these values, f ( x ) 5 f (2 k p ) 5 sin (2 k p 2 sin 2 k p ) 5 sin (2 k p 2 0) 5 0. Thus, f ( x ) 5 f 9 ( x ) 5 0 for x 5 2 k p , which means that the graph has a horizontal tangent at each of these values of x . 77. y 9 ( r ) 5 } d d r } 1 } 2 1 rl } ! } p T § d } § 2 5 1 } 2 1 l } ! } p T § d } § 2 } d d r } 1 } 1 r } 2 5 2 } 2 r 1 2 l } ! } p T § d } § y 9 ( l ) 5 } d d l } 1 } 2 1 rl } ! } p T § d } § 2 5 1 } 2 1 r } ! } p T § d } § 2 } d d l } 1 } 1 l } 2 5 2 } 2 r 1 l 2 } ! } p T § d } § y 9 ( d ) 5 } d d d } 1 } 2 1 rl } ! } p T § d } § 2 5 1 } 2 1 rl } ! } p T } § 2 } d d d } ( d 2 1/2 ) 5 } 2 1 rl } ! } p T } § 1 2 } 1 2 } d 2 3/2 2 5 2 } 4 1 rl } ! } p § T d § 3 } § y 9 ( T ) 5 } d d T } 1 } 2 1 rl } ! } p T § d } § 2 5 1 } 2 1 rl } ! } p 1 § d } § 2 } d d T } ( ˇ T w ) 5 } 2 1 rl } ! } p 1 § d } § 1 } 2 ˇ 1 T w } 2 5 } 4 rl ˇ 1 p w d w T w } Since y 9 ( r ) , 0, y 9 ( l ) , 0, and y 9 ( d ) , 0, increasing r , l , or d would decrease the frequency. Since y 9 ( T ) . 0, increasing T would increase the frequency. 78. (a) P (0) 5 } 1 2 1 00 e 5 } < 1 student (b) lim t P ( t ) 5 lim t } 1 1 20 e 0 5 2 t } 5 } 20 1 0 } 5 200 students (c) P 9 ( t ) 5 } d d t } 200(1 1 e 5 2 t ) 2 1 5 2 200(1 1 e 5 2 t ) 2 2 ( e 5 2 t )( 2 1) 5 } (1 2 1 00 e e 5 5 2 2 t t ) 2 } P 0 ( t ) 5 5 5 Since P 0 5 0 when t 5 5, the critical point of y 5 P 9 ( t ) occurs at t 5 5. To confirm that this corresponds to the maximum value of P 9 ( t ), note that P 0 ( t ) . 0 for t , 5 and P 0 ( t ) , 0 for t . 5. The maximum rate occurs at t 5 5, and this rate is P 9 (5) 5 } (1 2 1 00 e e 0 0 ) 2 } 5 } 2 2 0 2 0 } 5 50 students per day. Note: This problem can also be solved graphically. 79. [ 2 p , p ] by [ 2 4, 4] (a) x k } p 4 } , where k is an odd integer (b) 1 2 } p 2 } , } p 2 } 2 (c) Where it’s not defined, at x 5 k } p 4 } , k an odd integer (d) It has period } p 2 } and continues to repeat the pattern seen in this window. (200 e 5 2 t )( e 5 2 t 2 1) }}} (1 1 e 5 2 t ) 3 (1 1 e 5 2 t )( 2 200 e 5 2 t ) 1 400( e 5 2 t ) 2 }}}} (1 1 e 5 2 t ) 3 (1 1 e 5 2 t ) 2 (200 e 5 2 t )( 2 1) 2 (200 e 5 2 t )(2)(1 1 e 5 2 t )( e 5 2 t )( 2 1) }}}}}}} (1 1 e 5 2 t ) 4 126 Chapter 3 Review
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
80. Use implicit differentiation. x 2 2 y 2 5 1 } d d x } ( x 2 ) 2 } d d x } ( y 2 ) 5 } d d x } (1) 2 x 2 2 yy 9 5 0 y 9 5 } 2 2 x y } 5 } x y } y 0 5 } d d x } } x y } 5 } ( y )(1) 2 y 2 ( x )( y 9 ) } 5 5 } y 2 y 2 3 x 2 } 5 2 } y 1 3 } (since the given equation is x 2 2 y 2 5 1) At (2, ˇ 3 w ), } d dx 2 y 2 } 5 2 } y 1 3 } 5 2 } ( ˇ 1 3 w ) 3 } 5 2 } 3 ˇ 1 3 w } .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}