Pre-Calc Homework Solutions 107

Pre-Calc Homework Solutions 107 - Section 3.7 37(a y4 d 4(y...

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37. (a) y 4 5 y 2 2 x 2 } d d x } ( y 4 ) 5 } d d x } ( y 2 ) 2 } d d x } x 2 4 y 3 } d d y x } 5 2 y } d d y x } 2 2 x (4 y 3 2 2 y ) } d d y x }52 2 x } d d y x } 5 } 4 y 2 3 2 2 x 2 y } 5 } y 2 x 2 y 3 } At 1 } ˇ 4 3 w } , } ˇ 2 3 w } 2 : Slope 5 5 ? 5 } 2 2 1 3 1 At 1 } ˇ 4 3 w } , } 1 2 } 2 : Slope 55 ? } 4 4 } 5 } ˇ 1 3 w } 5 ˇ 3 w (b) [ 2 1.8, 1.8] by [ 2 1.2, 1.2] Parameter interval: 2 1 # t # 1 38. (a) y 2 (2 2 x ) 5 x 3 } d d x } [ y 2 (2 2 x )] 5 } d d x } ( x 3 ) ( y 2 )( 2 1) 1 (2 2 x )(2 y ) } d d y x } 5 3 x 2 2 y (2 2 x ) } d d y x } 5 3 x 2 1 y 2 } d d y x } 5 } 2 3 y x ( 2 2 1 2 y x 2 ) } Slope at (1, 1): } 3 2 ( ( 1 1 ) ) 2 (2 1 2 (1 1 ) ) 2 }5} 4 2 } 5 2 Tangent: y 5 2( x 2 1) 1 1 or y 5 2 x 2 1 Normal: y 52} 1 2 } ( x 2 1) 1 1 or y 1 2 } x 1 } 3 2 } (b) One way is to graph the equations y 56 ! } 2 § x 2 § 3 § x } § . 39. (a) ( 2 1) 3 (1) 2 5 cos ( p ) is true since both sides equal 2 1. (b) x 3 y 2 5 cos( p y ) } d d x } ( x 3 y 2 ) 5 } d d x } cos ( p y ) ( x 3 )(2 y ) } d d y x } 1 ( y 2 )(3 x 2 ) 5 ( 2 sin p y )( p ) } d d y x } (2 x 3 y 1 p sin p y ) } d d y x } 52 3 x 2 y 2 } d d y x } 2 x 3 y 1 3 x p 2 y 2 sin p y } Slope at ( 2 1, 1): 25 } 2 2 3 2 } 5 } 3 2 } The slope of the tangent line is } 3 2 } . 40. (a) When x 5 2, we have y 3 2 2 y 52 1, or y 3 2 2 y 1 1 5 0. Clearly, y 5 1 is one solution, and we may factor y 3 2 2 y + 1 as ( y 2 1)( y 2 1 y 2 1). The solutions of y 2 1 y 2 1 = 0 are y } 2 1 6 2 ˇ 5 w } . Hence, there are three possible y -values: 1, } 2 1 2 2 ˇ 5 w } , and } 2 1 1 2 ˇ 5 w } . (b) y 3 2 xy 1 } d d x } ( y 3 ) 2 } d d x } ( xy ) 5 } d d x } ( 2 1) 3 y 2 y 92 xy ( y )(1) 5 0 (3 y 2 2 x ) y 95 y y } 3 y 2 y 2 x } y 05 } d d x } } 3 y 2 y 2 x } 5 5 Since we are working with numerical information, there is no need to write a general expression for y 0 in terms of x and y. To evaluate f 9 (2), evaluate the expression for y
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