01test2

# 5 let f xyz x 2 xy y 2 z 3 4 a find a point x y z

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5. Let f ( x,y,z ) = x 2 + xy + y 2 + z 3 - 4. a ) Find a point ( x 0 ,y 0 ,z 0 ) satisfying f ( x 0 ,y 0 ,z 0 ) = 0 . Answer: There are many solutions: The point ( x 0 ,y 0 ,z 0 ) = (1 , 1 , 1) seemed to be the consensus choice, but (0 , 2 , 0), (2 , 0 , 0), (5 , 1 , - 3), (1 , 5 , - 3), ( - 2 , 1 , 1), (1 , - 2 , 1), and (2 , 2 , 2) are other integral solutions. Of course, you are not restricted to integers, but they often make calculation easier. b ) Can x be expressed as a function g ( y,z ) in some neighborhood of ( x 0 ,y 0 ,z 0 )? Answer: We use the point ( x 0 ,y 0 ,z 0 ) = (1 , 1 , 1). Here ∂f/∂x = 2 x 0 + y 0 = 3 6 = 0. The Implicit Function Theorem now implies that we can write x = g ( y,z ) in some neighborhood of (1 , 1 , 1). Note: In this case, we can use the quadratic formula to ﬁnd an expression for g , g ( y,z ) = - y + p 16 - 3 y 2 - 4 z 3 2 . Of course, this is only valid for 16 - 3 y 2 - 4 z 3 0. Notice that the positive square root is chosen so that g ( y 0 ,z 0 ) = g (1 , 1) = 1 = x 0 . Had we used ( - 2 , 1 , 1) as our reference point, we would use the negative square root. c ) Compute dg ( y 0 ,z 0 ). Answer: By the Implicit Function Theorem, dg ( y 0 ,z 0 ) = - ± ∂f ∂x ² - 1 ³ ∂f/∂y ∂f/∂z ´ = - 1 2 x 0 + y 0 ³ x 0 + 2 y 0 3 z 0 ´ = - 1 3 ³ 3 3 ´ = ³ - 1 - 1 ´ . Notice that this is an easier computation than using the formula for g given in part ( b ).
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