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Calculus: One and Several Variables

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 788 SECTION 16.1 CHAPTER 16 SECTION 16.1 1. f = (6 x y ) i + (1 x ) j 2. f = (2 Ax + By ) i + ( Bx + 2 Cy ) j 3. f = e xy [( xy + 1) i + x 2 j ] 4. f = 1 ( x 2 + y 2 ) 2 [( y 2 x 2 + 2 xy ) i + ( y 2 x 2 2 xy ) j ] 5. f = 2 y 2 sin( x 2 + 1) + 4 x 2 y 2 cos( x 2 + 1) i + 4 xy sin( x 2 + 1) j 6. f = 2 x x 2 + y 2 i + 2 y x 2 + y 2 j 7. f = ( e x y + e y x ) i + ( e x y e y x ) j = ( e x y + e y x )( i j ) 8. f = AD BC ( Cx + Dy ) 2 [ y i x j ] 9. f = ( z 2 + 2 xy ) i + ( x 2 + 2 yz ) j + ( y 2 + 2 zx ) k 10. f = x x 2 + y 2 + z 2 i + y x 2 + y 2 + z 2 j + z x 2 + y 2 + z 2 k 11. f = e z (2 xy i + x 2 j x 2 y k ) 12. f = xyz x + y + z + yz ln( x + y + z ) i + xyz x + y + z + xz ln( x + y + z ) j + xyz x + y + z + xy ln( x + y + z ) k 13. f = e x +2 y cos ( z 2 + 1 ) i + 2 e x +2 y cos ( z 2 + 1 ) j 2 ze x +2 y sin ( z 2 + 1 ) k 14. f = e yz 2 /x 3 3 yz 2 x 4 i + z 2 x 3 j + 2 yz x 3 k 15. f = 2 y cos(2 xy ) + 2 x i + 2 x cos(2 xy ) j + 1 z k 16. f = 2 xy z 3 z 4 i + x 2 z j x 2 y z 2 + 12 xz 3 k 17. f = (4 x 3 y ) i + (8 y 3 x ) j ; at (2 , 3) , f = i + 18 j 18. f = 1 ( x y ) 2 ( 2 y i + 2 x j ) , f (3 , 1) = 1 2 i + 3 2 j 19. f = 2 x x 2 + y 2 i + 2 y x 2 + y 2 j ; at (2 , 1) , f = 4 5 i + 2 5 j
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 SECTION 16.1 789 20. f = tan 1 ( y/x ) xy x 2 + y 2 i + x 2 x 2 + y 2 j , f (1 , 1) = π 4 1 2 i + 1 2 j 21. f = (sin xy + xy cos xy ) i + x 2 cos xy j ; at (1 , π/ 2) , f = i 22. f = e ( x 2 + y 2 ) [( y 2 x 2 y ) i + ( x 2 xy 2 ) j ] , f (1 , 1) = e 2 ( i j ) 23. f = e x sin( z + 2 y ) i + 2 e x cos( z + 2 y ) j + e x cos( z + 2 y ) k ; at (0 , π/ 4 , π/ 4) , f = 1 2 2( i + 2 j + k ) 24. f = cos πz i cos πz j π ( x y )sin πz k , f 1 , 0 , 1 2 = π k 25. f = i y y 2 + z 2 j z y 2 + z 2 k ; at (2 , 3 , 4) , f = i + 3 5 j 4 5 k 26. f = sin( xyz 2 )( yz 2 i + xz 2 j + 2 xyz k ) , f π, 1 4 , 1 = 2 2 1 4 i + π j π 2 k 27. (a) f (0 , 2) = 4 i (b) f ( 1 4 π, 1 6 π ) = 1 1 + 3 2 2 i + 1 2 1 + 3 2 j (c) f (1 , e ) = (1 2 e ) i 2 j 28. (a) f (1 , 2 , 3) = 1 8 2 i + 1 2 2 j 27 8 2 k (b) f (1 , 2 , 3) = 5 18 i + 1 9 j + 1 18 k (c) f (1 , e 2 , π/ 6) = 3 2 i + π 12 e 2 j + k 29. For the function f ( x, y ) = 3 x 2 xy + y, we have f ( x + h ) f ( x ) = f ( x + h 1 , y + h 2 ) f ( x, y ) = 3( x + h 1 ) 2 ( x + h 1 )( y + h 2 ) + ( y + h 2 ) 3 x 2 xy + y = [(6 x y ) i + (1 x ) j ] · ( h 1 i + h 2 j ) + 3 h 2 1 h 1 h 2 = [(6 x y ) i + (1 x ) j ] · h + 3 h 2 1 h 1 h 2 The remainder g ( h ) = 3 h 2 1 h 1 h 2 = (3 h 1 i h 1 j ) · ( h 1 i + h 2 j ) , and | g ( h ) | h = 3 h 1 i h 1 j · h · cos θ h 3 h 1 i h 1 j Since 3 h 1 i h 1 j 0 as h 0 it follows that f = (6 x y ) i + (1 x ) j 30. f ( x + h ) f ( x ) = [( x + 2 y ) i + (2 x + 2 y ) j ] · [ h 1 i + h 2 j ] + 1 2 h 2 1 + 2 h 1 h 2 + h 2 2 ; g ( h ) = 1 2 h 2 1 + 2 h 1 h 2 + h 2 2 is o ( h ).
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36 790 SECTION 16.1 31. For the function f ( x, y, z ) = x 2 y + y 2 z + z 2 x, we have f ( x + h ) f ( x ) = f ( x + h 1 , y + h 2 , z + h 3 ) f ( x, y, z ) = ( x + h 1 ) 2 ( y + h 2 ) + ( y + h 2 ) 2 ( z + h 3 ) + ( z + h 3 ) 2 ( x + h 1 ) ( x 2 y + y 2 z + z 2 x ) = ( 2 xy + z 2 ) h 1 + ( 2 yz + x 2 ) h 2 + ( 2 xz + y 2 ) h 3 + (2 xh 2 + yh 1 + h 1 h 2 ) h 1 + (2 yh 3 + zh 2 + h 2 h
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