prac exam 2A solutions

Therefore fy 2x3 3xy 2 g y comparing this with q

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Unformatted text preview: x + g (y ). Therefore, fy = 2x3 + 3xy 2 + g ′ (y ). Comparing this with Q, we get 2x3 + 3xy 2 + g ′ (y ) = 2x3 + 3xy 2 + 2 so g ′ (y ) = 2 and g = 2y + c. So f = 2x3 y + xy 3 + x + 2y (+constant). F · dr = f (−eπ , 0) − f (1, 0) = −eπ − 1. c) C starts at (1, 0) and ends at (−eπ , 0), so C ux uy v x vy 5. a) hence dx dy = 2x/y −x2 /y 2 y x = 1 du dv. 3|u| 4 b) 5 dx dy = R 2 1 1 dudv = 3u 4 1 2 ln 5 dv = ln 5. 3 3 2 −My dA. (Green’s theorem) M dx = 6. a) = 3x2 /y . Therefore, dudv = |3x2 /y | dxdy = 3|u| dxdy and R C 1 b) We want...
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