notes001 (82)

notes001 (82) - @f=@y , @ 2 f=@x@y , and @ 2 f=@y@x . Do...

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October 4, 2010 NAME____________________________ 1) By considering di/erent paths of approach, show that lim ( x;y ) ! (0 ; 0) x + y x y does not exist. Solution: First we try approaching (0 ; 0) along the y axis (where x = 0 ). This gives us lim ( x;y ) ! (0 ; 0) with x =0 x + y x y = lim y ! 0 0 + y 0 y = 1 . Now let us try approaching (0 ; 0) along the x axis (where y = 0 ). This gives us lim ( x;y ) ! (0 ; 0) with y =0 x + y x y = lim x ! 0 x + 0 x & 0 = 1 . Since two di/erent results were obtained for di/erent paths of approach, we conclude that lim ( x;y ) ! (0 ; 0) x + y x y does not exist. 2) For the function f ( x; y ) = sin (2 x 3 y ) , ±nd @f=@x ,
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Unformatted text preview: @f=@y , @ 2 f=@x@y , and @ 2 f=@y@x . Do not just write down an-swers. You must include intermediate steps in your calculations. Solution: @f @x = cos (2 x & 3 y ) (2) = 2 cos (2 x & 3 y ) . @f @y = cos (2 x & 3 y ) ( & 3) = & 3 cos (2 x & 3 y ) . 1 @ 2 f @x@y = @ @x ( & 3 cos (2 x & 3 y )) = & 3 ( & sin (2 x & 3 y )) (2) = 6 sin (2 x & 3 y ) . @ 2 f @y@x = @ @y (2 cos (2 x & 3 y )) = 2 ( & sin (2 x & 3 y )) ( & 3) = 6 sin (2 x & 3 y ) . 2...
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This note was uploaded on 02/05/2012 for the course MATH 2203 taught by Professor Ellermeyer during the Fall '10 term at FIU.

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notes001 (82) - @f=@y , @ 2 f=@x@y , and @ 2 f=@y@x . Do...

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