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FL2009 Prelim2 solutions

FL2009 Prelim2 solutions - Math 3210 1 second prelim...

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Math 3210 second prelim Wednesday 28 October 2009 10:10-11:00 am Solutions 1. (a) σ - 1 = σ ; (b) στ = ( n - 2, n , n - 1, n - 3, n - 4, . . . , 3, 2, 1 ) ; (c) τσ = ( n , n - 1, n - 2, n - 3, . . . , 6, 5, 4, 2, 1, 3 ) ; (d) στσ - 1 = ( 1, 2, 3, . . . , n - 4, n - 3, n - 1, n , n - 2 ) . 2. (a) φ i ( x 1 , x 2 ) = x n + 1 - i 1 x i - 1 2 , so φ * y i = x n + 1 - i 1 x i - 1 2 and φ * dy i = ( n + 1 - i ) x n - i 1 x i - 1 2 dx 1 + ( i - 1 ) x n + 1 - i 1 x i - 2 2 dx 2 . (b) φ * ( dy i dy j ) = φ * dy i φ * dy j = ( ( n + 1 - i ) x n - i 1 x i - 1 2 dx 1 + ( i - 1 ) x n + 1 - i 1 x i - 2 2 dx 2 ) × ( ( n + 1 - j ) x n - j 1 x j - 1 2 dx 1 + ( j - 1 ) x n + 1 - j 1 x j - 2 2 dx 2 ) = ( n + 1 - i )( j - 1 ) x 2 n + 1 - ( i + j ) 1 x i + j - 3 2 dx 1 dx 2 + ( i - 1 )( n + 1 - j ) x 2 n + 1 - ( i + j ) 1 x i + j - 3 2 dx 2 dx 1 = ( ( n + 1 - i )( j - 1 ) - ( i - 1 )( n + 1 - j ) ) x 2 n + 1 - ( i + j ) 1 x i + j - 3 2 dx 1 dx 2 = ( j - i ) nx 2 n + 1 - ( i + j ) 1 x i + j - 3 2 dx 1 dx 2 . 3. (a) Since cos 2 t = 1 - 2 sin 2 t , the curve is the piece of the parabola x = 1 - 2 y 2 contained in the square - 1 x , y 1. (b) c * dx = ad cos 2 t = - 2 a sin 2 t dt , so c * α = - 2 ab sin t sin 2 t dt = - 4 ab cos t sin 2 t dt , so R c α = - 4 ab R π /2 - π /2 cos t sin 2 t dt = - 4 3 ab sin 3 t fl fl π /2 - π /2 = - 8 3 ab . 4. (a) The lower edge of the square is mapped to the portion 0 x 1 1 of the x 1 -axis; the right edge is mapped to the portion of the parabola x 2 2 = x 3 in the x 1 = 1-plane contained between 0 x 3 1; the upper edge is mapped to the curve ( t 3 , t 2 , t ) ; the left edge is mapped in its entirety to the origin. So the image of c is a surface bounded by three curves. (b) d α = dx 1 dx 2 + dx 2 dx 3 , so c * d α = dt 3 1 d ( t 2 1 t 2 ) + d ( t 2 1 t 2 ) d ( t 1 t 2 2 ) = 3 t 2 1 dt 1 ( t 2 1 dt 2 ) + ( t 2 1 dt 2 + 2 t 1 t 2 dt 1 )( 2 t 1 t 2 dt 2 + t 2 2 dt 1 ) =
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