20E - HW 2 Solutions

20E - HW 2 Solutions - Math 20E Homework HW 2 Fall 2009#2...

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Math 20E Homework #2 13 January 2005 Section 2.3, Problem 6: Using the respective functions from exercise 1, compute the tangent plane to the graphs at the indicated points: (a) (0,0) (b) (0,1) (c) (0, π ) (d) (0,1) Solution. (a) f = xy , so the partials are y and x , respectively. At ( 0, 0 ) , these (as well as f itself) evaluate to 0, so the tangent plane is the plane z = 0. (b) f = e xy , so the partials are ye xy and xe xy . At ( 0, 1 ) these evaluate to 1 and 0, and f ( 0, 1 ) = 1, so the tangent plane is z = 1 + 1 ( x - 0 ) + 0 ( y - 1 ) = x + 1. (c) f = x cos x cos y , so the partials are ( cos x - x sin x ) cos y and - x cos x sin y . At ( 0, π ) these evaluate to - 1 and 0, and f ( 0, π ) = 0, so the tangent plane is z = 0 - 1 ( x - 0 ) + 0 ( y - π ) = - x . (d) f = ( x 2 + y 2 ) log ( x 2 + y 2 ) . The partials are 2 x ( 1 + log ( x 2 + y 2 )) and 2 y ( 1 + log ( x 2 + y 2 )) , respectively. At ( 0, 1 ) these evaluate to 0 and 2, and f ( 0, 1 ) = 0. So we get a tangent plane of z = 0 + 0 ( x - 0 ) + 2 ( y - 1 ) = 2 y - 2. Section 2.3, Problem 8: Compute the matrix of partial derivatives of (a) f ( x , y ) = ( e x , sin xy ) (b) f ( x , y , z ) = ( x - y , y - z ) (c) f ( x , y ) = ( x + y , x - y , xy ) (d) f ( x , y , z ) = ( x + z , y - 5 z , x - y ) Solution. (a) ± e x 0 y cos xy x cos xy ² (b) ± 1 - 1 0 0 1 1 ² (c) 1 1 1 - 1 y x (d) 1 0 1 0 1 - 5 1 - 1 0 Section 2.3, Problem 12: Use the linear approximation formula to approximate a suitable function f ( x , y ) and thereby estimate the following: (a) ( 0.99 e 0.02 ) 8 (b) ( 0.99 ) 3 + ( 2.01 ) 3 - 6 ( 0.99 )( 2.01 ) (c) p ( 4.01 ) 2 + ( 3.98 ) 2 + ( 2.02 2 ) Solution. (a) Using the approximation for
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20E - HW 2 Solutions - Math 20E Homework HW 2 Fall 2009#2...

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