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# ps8sol - 18.02 Fall 2008 – Problem Set 8 Part B Solutions...

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Unformatted text preview: 18.02 Fall 2008 – Problem Set 8, Part B Solutions 1. a) For θ ( x, y ) = tan − 1 ( y/x ): ∂θ ∂x = 1 1 + ( y/x ) 2 − y x 2 = − y x 2 + y 2 ; ∂θ ∂y = 1 1 + ( y/x ) 2 1 x = x x 2 + y 2 ⇒ ∇ θ = vector F. b) Because θ ( x, y ) = tan − 1 ( y/x ) is well-defined in the right half-plane ( x > 0) and vector F = ∇ θ , the fundamental theorem implies integraltext C vector F · dvector r = θ ( x 2 , y 2 ) − θ ( x 1 , y 1 ) = θ 2 − θ 1 . c) integraldisplay C 1 vector F · dvector r = integraldisplay C 1 − y dx + x dy x 2 + y 2 = integraldisplay π ( − sin θ )( − sin θ ) + cos θ cos θ cos 2 θ + sin 2 θ dθ = integraldisplay π dθ = π . Similarly, integraldisplay C 2 vector F · dvector r = integraldisplay − π dθ = − integraldisplay − π dθ = − π . (Or geometrically: length( C 1 ) = length( C 2 ) = π , vector F · ˆ T = 1 on C 1 ; vector F · ˆ T = − 1 on C 2 ) d) curl vector F = ∂ ∂x parenleftbigg x x 2 + y 2 parenrightbigg − ∂ ∂y parenleftbigg − y x 2 + y 2 parenrightbigg = ( x 2 + y 2 ) − 2 x 2 ( x 2 + y 2 ) 2 + ( x 2 + y 2 ) − 2 y 2 ( x 2 + y 2 ) 2 = 0....
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## This note was uploaded on 04/28/2009 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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ps8sol - 18.02 Fall 2008 – Problem Set 8 Part B Solutions...

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