s5 - University of Toronto Scarborough Department of...

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University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT B42H 2010/2011 Solutions #5 1. (a) ω = F 1 dx + F 2 dy + F 3 dz = ( ye xy + e z + 2 xy ) dx + ( xe xy + x 2 + 2 yz + 1) dy + ( xe z + y 2 + z ) dz Since ∂F 1 ∂y = e xy + xy e xy + 2 x = 2 ∂x , 1 ∂z = e z = 3 and 2 = 2 y = 3 , ω is closed. Since the domain of ω is R 3 , which is simply connected, ω closed = ω exact. There is a function g : R 3 R such that dg = ω . We now Fnd g . F 1 = xy + e z + 2 xy = ∂g , so g ( x, y, z ) = i ( xy + e z + 2 xy ) dx + h ( y, z ) = e xy + xe z + x 2 y + h ( y, z ). F 2 = xe xy + x 2 + 2 + 1 = = xe xy + x 2 + ∂h so = 2 + 1 = h ( y, z ) = i (2 + 1) dy + k ( z ) = y 2 z + y + k ( z ) giving g ( x, y, z ) = e xy + xe z + x 2 y + y 2 z + y + k ( z ). F 3 = xe z + y 2 + z = = xe z + y 2 + dk dz so dk dz = z = k ( z ) = i z dz = 1 2 z 2 + C . Hence g ( x, y, z ) = e xy + xe z + x 2 y + y 2 z + y + 1 2 z 2 + C . (b) ω = F 1 dx + F 2 dy + F 3 dz = x sin y dx + y cos z dy - z cos xdz Since 1 = x cos y while 2 = 0, 1 n = 2 , hence, ω is not closed and, consequently, not exact. There is no g such that dg = ω . (c) ω = F 1 dx + F 2 dy + F 3 dz = ( y 2 + z ln 3) dx +(2 xy +sin z ) dy + ( y cos z +( x +1) ln 3 ) dz Since 1 = 2 y = 2 , 1 = ln 3 = 3 and 2 = cos z = 3 , ω is closed. Since the domain of ω is R 3 , which is simply connected, ω closed = ω exact. There is a function g : R 3 R such that dg = ω . We now Fnd g . F 1 = y 2 + z ln 3 = , so g ( x, y, z ) = i ( y 2 + z ln 3) dx + h ( y, z ) = xy 2 + xz ln 3+ h ( y, z ) F 2 = 2 xy + sin z = = 2 xy + so = sin z = h ( y, z ) = i sinz dy + k ( z ) = y sin z + k ( z ) giving g ( x, y, z ) = xy 2 + xz ln 3 + y sin z + k ( z ).
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MATB42H Solutions # 5 page 2 F 3 = y cos z + ( x + 1) ln 3 = ∂g ∂z = x ln 3 + y cos z + dk dz so dk dz = ln 3 = k ( z ) = i ln 3 dz = z ln 3 + C . Hence g ( x, y, z ) = xy 2 + xz ln 3 + y sin z + z ln 3 + C , where C is some constant. 2. We note that F is conservative with potential function g ( x, y, z ) = xy + z . Hence the work done is independant of the path chosen. Work done is W = i 2 π 0 F · d s = g ( B ) - g ( A ) = g (0 , 0 , 2 π ) - g (
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s5 - University of Toronto Scarborough Department of...

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