Unformatted text preview: . 2 , , 3 dt dy dx dt dz = = = Hence . 2 / 5 5 1 1 = = â‹… âˆ« âˆ« dt t d P P r F (f) This straight line path can be parametrized by . 1 , , 2 , â‰¤ â‰¤ = = = t z t y t x Thus . 2 , , dt dy dt dx dz = = = Hence . 3 / 4 ) 4 2 ( 1 2 1== â‹… âˆ« âˆ« dt t t d P P r F One cannot deduce these answers since F is not a conservative vector field. Let . 2 2 âˆ« += C y x ydx xdy I Along C , ; 2 2 2 a y x = + C can be parametrized by . 2 , sin , cos Ï€ â‰¤ â‰¤ = = t t a y t a x Hence . cos , sin ; 1 2 t a dy t a dx ydx xdy a I C === âˆ« Thus...
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 Spring '10
 BLUMAN
 Trigraph, straight line path

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