15.5%20Ex%2050%20-%2056

15.5%20Ex%2050%20-%2056 - 694 C H A P T E R 15 D I F F E R...

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694 CHAPTER 15 DIFFERENTIATION IN SEVERAL VARIABLES (ET CHAPTER 14) 50. Find a function f ( x , y , z ) such that f = h z , 2 y , x i . SOLUTION f ( x , y , z ) = xz + y 2 is a good choice. 51. Find a function f ( x , y ) such that f = h y , x i . We must ±nd a function f ( x , y ) such that f = ¿ f x , f y À = h y , x i That is, f x = y , f y = x We integrate the ±rst equation with respect to x .Since y is treated as a constant, the constant of integration is a function of y .Weget f ( x , y ) = Z ydx = yx + g ( y ) (1) We differentiate f with respect to y and substitute in the second equation. This gives f y = y ( + g ( y )) = x + g ± ( y ) Hence, x + g ± ( y ) = x g ± ( y ) = 0 g ( y ) = C Substituting in (1) gives f ( x , y ) = + C One of the solutions is f ( x , y ) = (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 . Suppose that for some differentiable function f ( x , y ) , f = - f x , f y ® = D y 2 , x E That is, f x = y 2 and f y = x . Therefore, f = y f x = y y 2 = 2 y and f = x f y = x x = 1 Since f and f are both continuous, they must be equal by Clairaut’s Theorem. Since
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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15.5%20Ex%2050%20-%2056 - 694 C H A P T E R 15 D I F F E R...

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