solution-A6

solution-A6 - SOLUTIONS OF ASSIGNMENT #6 1. (a) For F (x,...

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SOLUTIONS OF ASSIGNMENT #6 1. (a) For F ( x,y,z ) = x 2 + y 2 + 4 z 2 + z 4 - 64, we have z x = - F x F z = x 4 z + 2 z 3 , z y = - F y F z = y 4 z + 2 z 3 At P = (4 , 4 , 2), z x = z y = 1 / 4. Now, z xx = ∂x ( z x ) = ∂x ( - x 4 z + 2 z 3 ) = - 4 z + z 3 - x (4 + 3 z 2 ) z x (4 z + z 3 ) 2 = - 1 / 8 . (b) The normal is proportional to < - 1 / 4 , - 1 / 4 , - 1 > , the equation of the tangent plane is ( x - 4) + ( y - 4) + 4( z - 2) = 0 . (c) z = < - 1 / 4 , - 1 / 4 > , the direction of the maximum increase is < 1 , 1 > . 2. z x = ye - x 2 +4 y 2 2 (1 - x 2 ) ,z y = xe - x 2 +4 y 2 2 (1 - 4 y 2 ). Solution for z x = 0 is either y = 0 or x = 1 , - 1. y = 0 and z y = 0 will force x = 0. If x = 1 or - 1, z y = 0 will give y = 1 / 2 or - 1 / 2. We have five critical points (0 , 0) , (1 , 1 / 2) , (1 , - 1 / 2) , ( - 1 , 1 / 2) , ( - 1 , - 1 / 2). We now calculate 2 z : z xx = xye - x 2 +4 y 2 2 ( - 2 - (1 - x 2 )), z yy = xye - x 2 +4 y 2 2 ( - 8 - 4(1 - 4 y 2 )), z xy = e - x 2 +4 y 2 2 (1 - x 2 )(1 - 4 y
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This note was uploaded on 05/01/2008 for the course MATH 222 taught by Professor Karlpeterrussell during the Winter '08 term at McGill.

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solution-A6 - SOLUTIONS OF ASSIGNMENT #6 1. (a) For F (x,...

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