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Calculus

Calculus - Chapter 14 Vector Calculus Chapter 14 Section...

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Chapter 14 Vector Calculus 436 Chapter 14 Section 14.1 5. –2 2 –2 2 7. –2 2 –2 2 9. –2 2 –2 2 11. 13. 15. 1 22 (, ) : , xy = + F Graph D, because all vectors point away from the origin and have the same length. 2 : , = F Graph B, because all vectors point away from the origin and the lengths are proportional to the distances from the origin. 3 : , y ex = F Graph A, because the vectors point upward when 0 x > and downward when 0. x < 4 : , y ey = F Graph C, because the vectors depend only on y. 17. 2 2 2,2 fxy x y f x x f y y x y =+ = = ∇= –2 2 2

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Chapter 14 Vector Calculus 437 19 . 22 1 / 2 1 / 2 1 / 2 (, ) ( ) 1 () ( 2 ) 2 1 ( 2 ) 2 , fxy x y x y f x xy x x f x y x =+ = + = + = + ∇= + –2 2 2 21. , y y y yy x e f e x f xe y e x e −− = = =− 2 2 23. 222 1 / 2 (, , ) fxyz x y z xyz + + 1 / 2 1 / 2 1 ( 2 ) 2 1 ( 2 ) 2 f x x x f y y y + = ++ + = 1 / 2 1 ( 2 ) 2 f z z z + = ,, 25. 2 2 2 2 = 2 , , xy y z f xy x f xz y f y z x yx zy = ∇=+
Chapter 14 Vector Calculus 438 27. If ( , ) , , then and . ff fxy yx y x xy ∂∂ ∇= = = (, ) () yd x x y gy f xgy x y == + =+ = Therefore, ( ) 0 and ( ) . gy gy c The vector field is conservative: ( , ) . x y c 29. If ( , ) , , then and . f y x y x f x y = =− ,s o 2. f xgy x gy x y + ′′ = + Since this is not possible, the vector field is not conservative. 31. 22 If ( , ) 2 , , then 2 a n d . 1 ( 2 ) 2 x x yy x xx y x x x x xy gy = = −+ 2 2 23 , so ( ) 1 3 f xg y y g y y y y c + = + The vector field is conservative; 3 11 . x ++ 33 . If ( , ) sin , sin , then sin and sin . s i n c o s sin ( ) sin yx x y y y f x y y xy dx xy g y f y g yxx y y + = Therefore, ( ) 0 and ( ) . The vector field is conservative; c o s . x + 35. If ( , , ) 4 , 3 , , then 4, 3, and . fxyz x z y zy x xz yz f z + 2 2 (, ,) (4 ) 2 (,) 00 3 3 ( 3 ) 2 3 (, ,) 2 2 , so ( ) 0 and ( ) . f xyz x zd x x x z gyz fg gyz y zd y y y z hz x x z y y f xyh z yx z hz hz c =−= + =−+ = + =+=+ + + + + =− + + = The vector field is conservative: 3 . 2 x x z y y z c + + + 37. If 2 3 1 , 2 , 4 t h e n f x y z y z xyz z 2 3 32 1, 2 , and 4 . (,,) ( 1 ) , so 0 and ( , ) ( ). 2( ) 4 , so ( ) 4 2 . f y z xyz z z d x gyz xyz xyz g y x x hz f xy z h z z z z x = = + = + = Since this is not possible, the vector field is not conservative. 39. cos 2 1 cos 2 1 sin 2 dy x dx dy x dx c = = ∫∫ 41. 2 2 2 3 2 dy x dx y dy dx ydy x dx c = = =

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Chapter 14 Vector Calculus 439 43. y y y dy xe dx y dy ye x dx ye dy x dx = = = ∫∫ Integrate by parts: uy du dy = = y y dv e dy ve = =− () yy y ye dy ye e dy ye e c −− + 2 2 1 2 1 (1 ) 2 y ye e c x c ye xc + =+ += + 45. 2 2 2 2 2 22 2 1 1 1 1 1 ln( 1) 2 ln( 2 2 1 dy y dx y yd y dx y y dy dx y yx c c + + = = + = + +=+ += + Since 2 c e is an arbitrary positive constant, we may write 1 ( 0 ) . x yc e c > 47. 00 0 (,,) xy z f xyz f udu gudu hudu c ++ 49. Using the result of exercise 19, , . r r ∇= = + r 51. 32 2 3 / 2 1 / 2 1 / 2 (, ) ( ) 3 ( 2 ) 3 2 3 ( 2 ) 3 2 fxy r x y f x x x f y y y == + = + = + 2 2 2 () 3 , 3 3, 3 fr x x y y x y xy x y r + + = r 53.
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Calculus - Chapter 14 Vector Calculus Chapter 14 Section...

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