text: FL2009 Prelim1 solutions
  
  # FL2009 Prelim1 solutions - Math 3210 first prelim Wednesday...
  
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  Unformatted text preview: Math 3210 first prelim Wednesday 30 September 2009 10:10-11:00 pm Solutions 1. (a) d α = dx 1 dx 2 + dx 3 dx 4 . (b) α d α = x 1 dx 2 dx 3 dx 4 + x 3 dx 1 dx 2 dx 4 + dx 1 dx 2 dx 5 + dx 3 dx 4 dx 5 , ( d α ) 2 = 2 dx 1 dx 2 dx 3 dx 4 , α ( d α ) 2 = 2 dx 1 dx 2 dx 3 dx 4 dx 5 . (c) * α =- x 1 dx 1 dx 3 dx 4 dx 5- x 3 dx 1 dx 2 dx 3 dx 5 + dx 1 dx 2 dx 3 dx 4 α ( * α ) = ( x 2 1 + x 2 3 + 1 ) dx 1 dx 2 dx 3 dx 4 dx 5 . 2. (a) d dx Z x ln ( x + y ) dy = ln ( x + x ) + Z x ∂ ∂ x ln ( x + y ) dy = ln 2 x + Z x 1 x + y dy = ln 2 x + ln ( x + y ) fl fl x y = = ln 2 x + ln 2 x- ln x = 2 ln 2 x- ln x = 2 ( ln 2 + ln x )- ln x = 2 ln 2 + ln x . (b) c ( t ) = p c ( t ) · c ( t ) is constant, so c ( t ) · c ( t ) = c 1 ( t ) 2 + ··· + c n ( t ) 2 is constant, so d dt ( c ( t ) · c ( t )) = 2 c 1 ( t ) c 1 ( t ) + ··· + 2 c n ( t ) c n ( t ) = 0, i.e. c ( t ) · c ( t ) = 0. 3. (a) * ( dx 1 dx 3 dx 4 ) = dx 2 , * ( dx 1 dx 3 dx 4 dx 5 ) =- dx 2 ....
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  FL2009 Prelim1 solutions - Math 3210 first prelim Wednesday...
  
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