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# MIDTER2 - MATH 2401 Fall 2009 Practice Exam 2 Solutions...

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MATH 2401, Fall 2009 Practice Exam 2, Solutions Problem 1 . Calculations. (a) Find the directional derivative of f ( x, y, z ) = xy + yz + zx at P (1 , - 1 , 1) in the direction of i + 2 j + k Solution: f = ( y + z ) i + ( x + z ) j + ( y + x ) k , f (1 , - 1 , 1) = 2 j . u = 6 6 ( i + 2 j + k ), so f u (1 , - 1 , 1) = f (1 , - 1 , 1) u = 2 3 6 . (b) Find the rate of change of f ( x, y ) = xe y + ye - x along the curve r ( t ) = ( lnt ) i + t ( lnt ) j . Solution: f = ( e y - ye - x ) i + ( xe y + e - x ) j , f ( r ( t )) = ( t t - lnt ) i + ( t t lnt + 1 t ) j , df dt = f ( r ( t )) r ( t ) = t t ( 1 t + lnt + ( lnt ) 2 ) + 1 t . (c) Find ∂u ∂s for u = x 2 - xy , x = scost , y = tsins .

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Solution: ∂u ∂s = ∂u ∂x ∂x ∂s + ∂u ∂y ∂y ∂s = (2 x - y )( cost )+( - x )( tcoss ) = 2 scos 2 t - t sins cost - st coss cost. (d) Find dy dx if xcos ( xy ) + ycos ( x ) = 2. Solution: Set u = xcos ( xy ) + ycos ( x ) - 2, ∂u ∂x = cos ( xy ) - xysin ( xy ) - ysin ( x ) . ∂u ∂y = - x 2 sin ( xy ) + cos ( x ) . dy dx = - ∂u/∂x ∂u/∂y = cos ( xy ) - xysin ( xy ) - ysin ( x ) x 2 sin ( xy ) - cos ( x ) . (e) Is F ( x, y ) = ( x + siny ) i + ( xcosy - 2 y ) j a gradient of a function f ( x, y )? If yes, find the general form of f ( x, y ).
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