Business Calc Homework w answers_Part_36

Business Calc Homework w answers_Part_36 - 176 Section 4.6...

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4. Use implicit differentiation. } d d x } ( x sin y ) 5 } d d x } (1 2 xy ) ( x )(cos y ) } d d y x } 1 (sin y )(1) 5 2 x } d d y x } 2 y (1) ( x 1 x cos y ) } d d y x } 5 2 y 2 sin y } d d y x } 5 } x 2 1 y 2 x c s o in s y y } } d d y x } 5 2 } x y 1 1 x s c in os y y } 5. Use implicit differentiation. } d d x } x 2 5 } d d x } tan y 2 x 5 sec 2 y } d d y x } } d d y x } 5 } se 2 c x 2 y } } d d y x } 5 2 x cos 2 y 6. Use implicit differentiation. } d d x } ln ( x 1 y ) 5 } d d x } (2 x ) } x 1 1 y } 1 1 1 } d d y x } 2 5 2 1 1 } d d y x } 5 2( x 1 y ) } d d y x } 5 2 x 1 2 y 2 1 7. Using A ( 2 2, 1) we create the parametric equations x 5 2 2 1 at and y 5 1 1 bt , which determine a line passing through A at t 5 0. We determine a and b so that the line passes through B (4, 2 3) at t 5 1. Since 4 5 2 2 1 a , we have a 5 6, and since 2 3 5 1 1 b , we have b 5 2 4. Thus, one parametrization for the line segment is x 5 2 2 1 6 t , y 5 1 2 4 t , 0 # t # 1. (Other answers are possible.) 8. Using A (0, 2 4), we create the parametric equations x 5 0 1 at and y 5 2 4 1 bt , which determine a line passing through A at t 5 0. We now determine a and b so that the line passes through B (5, 0) at t 5 1. Since 5 5 0 1 a , we have a 5 5, and since 0 5 2 4 1 b , we have b 5 4. Thus, one parametrization for the line segment is x 5 5 t , y 5 2 4 1 4 t , 0 # t # 1. (Other answers are possible.) 9. One possible answer: } p 2 } # t # } 3 2 p } 10. One possible answer: } 3 2 p } # t # 2 p Section 4.6 Exercises 1. Since } d d A t } 5 } d d A r } } d d r t } , we have } d d A t } 5 2 p r } d d r t } . 2. Since } d d S t } 5 } d d S r } } d d r t } , we have } d d S t } 5 8 p r } d d r t } . 3. (a) Since } d d V t } 5 } d d V h } } d d h t } , we have } d d V t } 5 p r 2 } d d h t } . (b) Since } d d V t } 5 } d d V r } } d d r t } , we have } d d V t } 5 2 p rh } d d r t } . (c) } d d V t } 5 } d d t } p r 2 h 5 p } d d t } ( r 2 h ) } d d V t } 5 p 1 r 2 } d d h t } 1 h (2 r ) } d d r t } 2 } d d V t } 5 p r 2 } d d h t } 1 2 p rh } d d r t } 4. (a) } d d P t } 5 } d d t } ( RI 2 ) } d d P t } 5 R } d d t } I 2 1 I 2 } d d R t } } d d P t } 5 R 1 2 I } d d I t } 2 1 I 2 } d d R t } } d d P t } 5 2 RI } d d I t } 1 I 2 } d d R t } (b) If P is constant, we have } d d P t } 5 0, which means 2 RI } d d I t } 1 I 2 } d d R t } 5 0, or } d d R t } 5 2 } 2 I R } } d d I t } 5 2 } 2 I P 3 } } d d I t } . 5. } d d s t } 5 } d d t } ˇ x 2 w 1 w y w 2 w 1 w z w 2 w } d d s t } 5 } 2 ˇ x 2 w 1 w 1 y w 2 w 1 w z w 2 w } } d d t } ( x 2 1 y 2 1 z 2 ) } d d s t } 5 } 2 ˇ x 2 w 1 w 1 y w 2 w 1 w z w 2 w } 1 2 x } d d x t } 1 2 y } d d y t } 1 2 z } d d z t } 2 } d d s t } 5 6. } d d A t } 5 } d d t } 1 } 1 2 } ab sin u 2 } d d A t } 5 } 1 2 } 1 } d d a t } ? b ? sin u 1 a ? } d d b t } ? sin u 1 ab ? } d d t } sin u 2 } d d A t } 5 } 1 2 } 1 b sin u } d d a t } 1 a sin u } d d b t } 1 ab cos u } d d u t } 2 7. (a) Since V is increasing at the rate of 1 volt/sec, } d d V t } 5 1 volt/sec. (b) Since I is decreasing at the rate of } 1 3 } amp/sec, } d d I t } 5 2 } 1 3 } amp/sec. (c) Differentiating both sides of V 5 IR , we have } d d V t } 5 I } d d R t } 1 R } d d I t } .
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