finf98s - S in the Form oF a graph z-1 = g ( x, y ) =- 4-x...

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Solutions to MATH 20E FINAL EXAM Fall 98, Lindblad, Problems 1-5 only. 1. (a) f =6 xye x 2 y - 2 i +3 x 2 e x 2 y - 2 j so it increases most in the direction 12 i +3 j . b) z - z 0 = f x ( x 0 ,y 0 )( x - x 0 )+ f y ( x 0 ,y 0 )( y - y 0 ) so z - 3 = 12( x - 1) + 3( y - 2). 2. a) ∇× F = 0. b ) ± C F · d R = ± C xdx + ydy x 2 + y 2 = ± 2 π 0 r cos tr ( - sin t )+ r sin tr cos t r 2 dt =0 . c) F = φ , where φ = ln ² x 2 + y 2 . d) The fow lines are lines From the origin: dx x/ ( x 2 + y 2 ) = dy y/ ( x 2 + y 2 ) so dx x = dy y and hence ln | x | + C = ln | y | which is equivalent to | y | = e C | x | . 3. a) ∇× F = 0. b ) ± C F · d R = ± C ydx - xdy x 2 + y 2 = ± 2 π 0 r sin tr ( - sin t ) - r cos tr cos t r 2 dt = - ± 2 π 0 dt = - 2 π c) F is not conservative since the integral around the closed curve in (b) is ± = 0. d) The fow lines are circles centered at the origin: dx y/ ( x 2 + y 2 ) = dy - x/ ( x 2 + y 2 ) so - xdx = ydy and hence y 2 = - x 2 + C which is equivalent to x 2 + y 2 = C . 4. a) F (0 , 2) = (2 , 0), F (0 , 4) = (0 , 4), F ( π/ 2 , 2) = (0 , 2) and F ( π/ 2 , 4) = (0 , 4). b) ( x, y ) ( u, v ) =2 v . c ) Area = ±± S dxdy = ±± R ( x, y ) ( u, v ) dudv = ± π/ 2 0 ± 4 2 2 vdudv = ... = π 2 ³ 4 2 - 2 2 ´ 5. a) 0 z 1 implies that 3 x 2 + y 2 4 so we can write
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Unformatted text preview: S in the Form oF a graph z-1 = g ( x, y ) =- 4-x 2-y 2 where ( x, y ) D = { ( x, y ); 3 x 2 + y 2 4 } . dS = dxdy/ | n k | = 1 + g 2 x + g 2 y dxdy = ... = 2 dxdy/ 4-x 2-y 2 so Area( S ) = S dS = D dxdy 4-x 2-y 2 = 2 3 2 rdrd 4-r 2 =-2 4-r 2 2 3 = 2 B x 2 dxdydz = 2 2 r 2 sin 2 cos 2 r 2 sin dddr = 2 2 -cos 3 3 cos 2 r 4 ddr = 2 3 r 5 5 2 2 cos 2 d = 2 3 2 5 5 NOTE: THERE ARE NO SOLUTIONS TO PROBLEMS 6-9!...
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This note was uploaded on 03/16/2010 for the course MATH 223B taught by Professor Staff during the Spring '09 term at University of Arizona- Tucson.

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