Math 317 Solutions to Assignment5page1

# Math 317 Solutions to Assignment5page1 - Math 317 Solutions...

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1 Math 317 Solutions to Assignment #5 1. (a) C can be parametrized by . 1 0 , = x x y Along C , . dx dy = Hence . 2 4 )] 2 3 ( ) 2 [ 1 0 1 0 = = - + + = xdx dx x x x x d C r F (b) Here C can be parametrized by . 1 0 , , 2 = = t t y t x Along C , . 2 ) ( , 2 tdt t d dy dt dx = = = Thus . 3 / 7 ] 4 7 2 [ ] 2 ) 2 3 ( ) 2 [ 1 0 3 2 1 0 2 2 = - + = - + + = dt t t t dt t t t t t d C r F 2. (a) The straight line path can be parametrized by . 1 0 , 0 , = = = t z y t x Thus . 0 , = = = dz dy dt dx Hence . 0 ) 0 ( 1 0 1 1 0 1 0 = = = dt dx F d P P P P r F [Alternatively, the answer can be obtained by noting that 0 F along the path. (b) This straight line path can be parametrized by . 2 0 , , 0 , 1 = = = t t y z x Thus . 0 , = = = dz dx dt dy Hence . 2 2 0 2 1 0 1 0 - = - = = tdt dy F d P P P P r F (c) This straight line path can be parametrized by . 3 0 , , 2 , 1 = = = t t z y x Thus . 0 , = = = dy dx dt dz Hence . 2 / 9 3 0 3 1 0 1 0 = = = tdt dz F d P P P P r F (d) This straight line path can be parametrized by . 1 0 , 3 , 2 , = = = t t z t y t x Thus . 2 , , 3 dt dy dt dx dt dz = = = Hence . 6 / 19 )] 3 )( 3 ( ) 2 )( 2 ( 2 [ 1 0 2 1 0 = + - = dt t t t d P P r F (e) This straight line path can be parametrized by . 1 0 , 3 , 2 , 1 = = = t t z t y

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 π ≤ ≤ = = 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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• Spring '10
• BLUMAN
• Trigraph, straight line path

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