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finw99s - Solutions to Math 20E Final Winter 99 Lindblad 1...

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Solutions to Math 20E Final Winter 99, Lindblad. 1. Adding the two equations together gives 3 x - y + 2 = 0 so y = 3 x + 2. Inserting this in the first equation gives z = 3 y - 2 x +1 = 3(3 x +2) - 2 x +1 = 7 x +7. Hence the equation of the plane is ( x, y, z ) = ( t, 3 t + 2 , 7 t + 7) = (1 , 3 , 7) t + (0 , 2 , 7). 2. (a). F ( x, y, z ) = φ = y 2 i + 2 xy j . (b). Along the curve we have dx y 2 = dy 2 xy = dz 0 so ydy = 2 xdx and dz = 0. Hence y 2 = 2 x 2 + C so y = ± 2 x 2 + C 1 and z = C 2 Since the curve starts at (1 , - 1 , 1) we must have ( x, y, z ) = ( t, - 2 t 2 - 1 , 1) 3. (a) φ x = ze x - y , φ y = - ze x - y and φ z = e x - y . Integrating these equations give φ = ze x - y + h 1 ( y, z ), φ y = ze x - y + h 2 ( x, z ) and φ = ze x - y + h 3 ( x, y ) which can all be satisfied if we pick h 1 = h 2 = h 3 = 0 in wchich case φ = ze x - y . (b) The end point of the curve is (1 , 0 , 0) and the initial point is (0 , 0 , 1) so C F · R = φ (1 , 0 , 0) - φ (0 , 0 , 1) = - 1. (c) If G denotes the vector field then ∇ × G = 0 and G is continuously di ff eren- tiable in the region obtained by removing the origin from space. Since this region is simply connected it follows that there exists a potential so
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