Chpt 14 Review Solns-sm.dvi

Chpt 14 Review Solns-sm.dvi - 802 C H A P T E R 15...

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Unformatted text preview: 802 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 1 + ln x n = + E n We subtract the last equation from the other equations to obtain ln x i- ln x n = ( E i- E n ) , i = 1 , . . . , n- 1 or ln x i x n = ( E i- E n ) x i = x n e ( E i- E n ) , i = 1 , . . . , n- 1 (1) Substituting x 1 , . . . , x n- 1 in the equation of the constraint x 1 + + x n- 1 + x n = N and solving for x n , we get x n e ( E 1- E n ) + x n e ( E 2- E n ) + + x n e ( E n- 1- E n ) + x n = N We multiply the equation by e E n to obtain x n e E 1 + x n e E 2 + + x n e E n- 1 + x n e E n = N e E n x n e E 1 + e E 2 + + e E n = N e E n x n = N e E n e E 1 + e E 2 + + e E n We next use (1) to determine x i for i = 1 , . . . , n- 1: x i = N e E n e E 1 + e E 2 + + e E n e ( E i- E n ) = N e E i e E 1 + e E 2 + + e E n Letting A = e E 1 + e E 2 ++ e En N , we obtain x i = A- 1 e e i , i = 1 , . . . , n- 1 The value of is determined by the second constraint h ( x 1 , . . . , x n ) = 0, although it would be very difficult to calculate. Chapter Review Exercises 803 CHAPTER REVIEW EXERCISES 1. Given f ( x , y ) = x 2- y 2 x + 3 , (a) Sketch the domain of f . (b) Calculate f ( 3 , 1 ) and f (- 5 ,- 3 ) . (c) Find a point satisfying f ( x , y ) = 1. SOLUTION (a) f is defined where x 2- y 2 0 and x + 3 = 0. We solve these two inequalities: x 2- y 2 x 2 y 2 | x | | y | x + 3 = x = - 3 Therefore, the domain of f is the following set: D = { ( x , y ) : | x | | y | , x = - 3 } x- 3 y (b) To find f ( 3 , 1 ) we substitute x = 3, y = 1 in f ( x , y ) . We get f ( 3 , 1 ) = 3 2- 1 2 3 + 3 = 8 6 = 2 3 Similarly, setting x = - 5, y = - 3, we get f (- 5 ,- 3 ) = (- 5 ) 2- (- 3 ) 2- 5 + 3 = 16- 2 = - 2 . (c) We must find a point ( x , y ) such that f ( x , y ) = x 2- y 2 x + 3 = 1 We choose, for instance, y = 1, substitute and solve for x . This gives x 2- 1 2 x + 3 = 1 x 2- 1 = x + 3 x 2- 1 = ( x + 3 ) 2 = x 2 + 6 x + 9 6 x = - 10 x = - 5 3 Thus, the point - 5 3 , 1 satisfies f - 5 3 , 1 = 1. 2. Find the domain and range of: (a) f ( x , y , z ) = x- y + y- z (b) f ( x , y ) = ln ( 4 x 2- y ) SOLUTION (a) f ( x , y , z ) is defined where the differences under the root signs are nonnegative. That is, x- y 0 and y- z 0. We solve the inequalities x- y y x y- z y z z y x 804 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) The domain of f is the following set: D = { ( x , y , z ) | z y x } The range is the set of all nonnegative numbers....
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Chpt 14 Review Solns-sm.dvi - 802 C H A P T E R 15...

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