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Unformatted text preview: . 2 , , 3 dt dy dx dt dz = = = Hence . 2 / 5 5 1 1 = = dt t d P P r F (f) This straight line path can be parametrized by . 1 , , 2 , = = = t z t y t x Thus . 2 , , dt dy dt dx dz = = = Hence . 3 / 4 ) 4 2 ( 1 2 1== dt t t d P P r F One cannot deduce these answers since F is not a conservative vector field. Let . 2 2 += C y x ydx xdy I Along C , ; 2 2 2 a y x = + C can be parametrized by . 2 , sin , cos = = t t a y t a x Hence . cos , sin ; 1 2 t a dy t a dx ydx xdy a I C === Thus...
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This note was uploaded on 04/01/2010 for the course MATH 317 MATH 317 taught by Professor Bluman during the Spring '10 term at The University of British Columbia.
 Spring '10
 BLUMAN

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