Week 10 - Solution

# Week 10 - Solution - Tutorial Note Week 10 Solution Curl...

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Tutorial Note - Week 10 - Solution Curl and Divergent 1. (a) i. curl F ( x, y, z ) = v v v v v v v v i j k ∂x ∂y ∂z x y z v v v v v v v v = b ( z ) ( y ) , ( x ) ( z ) , ( y ) ( x ) B = [0 , 0 , 0] = 0 ii. div F ( x, y, z ) = b , , B [ x, y, z ] = ( x ) + ( y ) + ( z ) = 3 (b) i. curl F ( x, y, z ) = ± 0 2 yz, 0 2 xz, 0 x 2 ² = ± 2 2 xz, x 2 ² ii. div F ( x, y, z ) = 2 xy + z 2 + x 2 (c) i. curl F ( x, y, z ) = ± x 2 z 0 , y 2 + 2 xyz, 0 2 yz ² = ± x 2 z, y 2 + 2 2 yz ² ii. div F ( x, y, z ) = x 2 y (d) i. curl F ( x, y, z ) = [0 0 , 0 0 , sin x 0] = [0 , 0 , sin x ] ii. div F ( x, y, z ) = cos x + 2 z 2. (a) curl F ( x, y, z ) = [0 0 , 0 0 , 1 1] = 0 So, F is conservative and the potential function f ( x, y, z ), where f = F , exists. F = [ y, x, 1] = [ f x ( x, y, z ) , f y ( x, y, z ) , f z ( x, y, z )] f ( x, y, z ) = i f x ( x, y, z ) dx = i y dx = xy + g ( y, z ) f y ( x, y, z ) = f ( x, y, z ) = [ xy + g ( y, z )] = x + g y ( y, z ) Comparing the component in F and f y ( x, y, z ), x = x + g y ( y, z ) = g y ( y, z ) = 0 and g ( y, z ) = h ( z ) . 1

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So, f ( x, y, z ) = xy + h ( z ) and f z ( x, y, z ) = h ( z ) . Hence, h ( z ) = 1 and h ( z ) = z + C . Therefore, f ( x, y, z ) = xy + z + C . (b) curl F ( x, y, z ) = [0 0 , 0 1 , 0 0] = [0 , 1 , 0] n = 0 So, F is not conservative and the potential function f ( x, y, z ) does not exist. (c) curl F ( x, y, z ) = [0 0 , 0 0 , cos x + sin y ] = [0 , 0 , cos x + sin y ] n = 0 So, F is not conservative and the potential function f ( x, y, z ) does not exist. (d) curl F ( x, y, z ) = [ x x, y y, z z ] = 0 So, F is conservative and the potential function f ( x, y, z ) exists. f x ( x, y, z ) = yz = f ( x, y, z ) = i yz dx = xyz + g ( y, z ) f y ( x, y, z ) = xz + g y ( y, z ) = xz + g y ( y, z ) = y 2 + xz = g y ( y, z ) = y 2 = g ( y, z ) = 1 3 y 3 + h ( z ) = f ( x, y, z ) = xyz + 1 3 y 3 + h ( z ) f z ( x, y, z ) = xy + h ( z ) = h ( z ) = 0 = h ( z ) = C Therefore, f ( x, y, z ) = xyz + 1 3 y 3 + C . Parameter Equations of Straight Lines 1. (a) x ( t ) = (6 2) t + 2 = 2 + 4 t y ( t ) = (0 1) t + 1 = 1 t z ( t ) = (3 8) t + 8 = 8 5 t , where 0 t 1 (b) x ( t ) = 1 + 5 t , y ( t ) = 3 t , z ( t ) = 5 2 t
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