Assignment 7 solutions

U in x y now since x r cos y r sin r x2 y

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Unformatted text preview: ∂ − r−2 cos ∂) + 2 (−r cos ∂ − r −1 cos ∂) = (2 − 1 − 1)r −3 cos ∂ + (1 − 1) cos ∂ = 0, r r r as desired. U in (x, y ): Now, since x = r cos ∂, y = r sin ∂ � r = ⎛ x2 + y 2 , ∂ = tan−1 �y� x �u = −r sin ∂ − r−1 sin ∂ �∂ �2u = −r cos ∂ − r −1 cos ∂. � ∂2 . So u = r cos ∂ + r−1 cos ∂ = x + x x(1 + x2 + y 2 ) = . x2 + y 2 x2 + y 2 So �u (x2 + y 2 ) · 1 − x(2x) y 2 − x2 =1+ =1+ 2 . �x (x2 + y 2 )2 (x + y 2 )2 �u −2xy = 0 + x(−1 · (x2 + y 2 )−2 · 2y ) = 2 . �y (x + y 2 )2 So v= � 1+ y 2 − x2 −2xy , 2 + y 2 )2 (x2 + y 2 )2 (x � . v · n = 0: n = (x, y ) is normal to the circle v = (1 + y 2 − x2 , −2xy ) on the circle v · n = x − x3 − xy 2 = x − x(x2 + y 2 ) = 0...
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This note was uploaded on 04/12/2011 for the course MATH 18.085 taught by Professor Staff during the Fall '10 term at MIT.

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