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15.5.Ex50-54

# 15.5.Ex50-54 - yx we conclude that such a function f does...

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694 C H A P T E R 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 50. Find a function f ( x , y , z ) such that f = z , 2 y , x . SOLUTION f ( x , y , z ) = xz + y 2 is a good choice. 51. Find a function f ( x , y ) such that f = y , x . SOLUTION We must find a function f ( x , y ) such that f = f x , f y = y , x That is, f x = y , f y = x We integrate the first equation with respect to x . Since y is treated as a constant, the constant of integration is a function of y . We get f ( x , y ) = y dx = yx + g ( y ) (1) We differentiate f with respect to y and substitute in the second equation. This gives f y = y ( yx + g ( y )) = x + g ( y ) Hence, x + g ( y ) = x g ( y ) = 0 g ( y ) = C Substituting in (1) gives f ( x , y ) = yx + C One of the solutions is f ( x , y ) = yx (obtained for C = 0). 52. Show that there does not exist a function f ( x , y ) such that f = y 2 , x . Hint: Use Clairaut’s Theorem f xy = f yx . SOLUTION Suppose that for some differentiable function f ( x , y ) , f = f x , f y = y 2 , x That is, f x = y 2 and f y = x . Therefore, f xy = y f x = y y 2 = 2 y and f yx = x f y = x x = 1 Since f xy and f yx are both continuous, they must be equal by Clairaut’s Theorem. Since
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Unformatted text preview: yx we conclude that such a function f does not exist. 53. Let 1 f = f ( a + h , b + k ) − f ( a , b ) be the change in f at P = ( a , b ) . Set 1 v = h h , k i . Show that the linear approximation can be written 1 f ≈ ∇ f P · 1 v 6 SOLUTION The linear approximation is 1 f ≈ f x ( a , b ) h + f y ( a , b ) k =-f x ( a , b ), f y ( a , b ) ® · h h , k i = ∇ f P · 1 v 54. Use Eq. (6) to estimate 1 f = f ( 3 . 53 , 8 . 98 ) − f ( 3 . 5 , 9 ) assuming that ∇ f ( 3 . 5 , 9 ) = h 2 , − 1 i . SOLUTION By Eq. (6), 1 f ≈ ∇ f P · 1 v The vector 1 v is the following vector: 1 v = h 3 . 53 − 3 . 5 , 8 . 98 − 9 i = h . 03 , − . 02 i Hence, 1 f ≈ ∇ f ( 3 , 5 , 9 ) · 1 v = h 2 , − 1 i · h . 03 , − . 02 i = . 08...
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