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FinalExamPartB

# FinalExamPartB - Therefore on the circle described above 1...

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EGM 4313 – Summer 2007 NAME:____________________________________ Final Exam – Part B 60 minute time limit. Closed notes – Open book – No calculators 1) Use the Gauss-Jordan method to find the inverse of the matrix 1 4 5 0 1 2 0 0 1 . 1 1 4 5 | 1 0 0 1 4 0 | 1 0 5 0 1 2 | 0 1 0 0 1 0 | 0 1 2 0 0 1 | 0 0 1 0 0 1 | 0 0 1 1 0 0 | 1 4 3 0 1 0 | 0 1 2 0 0 1 | 0 0 1 1 4 5 1 4 3 0 1 2 0 1 2 0 0 1 0 0 1 - - - - - - = -

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EGM 4313 – Summer 2007 NAME:____________________________________ Final Exam – Part B 2) Consider the function ( 29 1 f z z = . (a) Show that the real part of this function is constant on circles of radius | a | centered at the point z=a where a is a real number. The equation of a circle of radius |a| centered at z=a (a real) is ( 29 2 2 2 x a y a - + = . Therefore 2 2 2 2 2 2 2 2 x ax a y a x y ax - + + = + = . 2 2 1 1 1 Re Re Re x iy x z x iy x iy x iy x y - = = = ÷ ÷ ÷ + + - +
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Unformatted text preview: + Therefore on the circle described above 1 1 Re 2 2 x on circle z ax a = = Ã· = constant. (b) Use the result found in (a) to find the temperature in the shaded region shown for steady heat conduction (Laplaceâ€™s Equation). The temperature on the outer part (circle of radius 2) is 100 and the temperature on the inner part (circle of radius 1) is 50. Using superposition of solutions 2 2 x T a b x y = + + To set T to be 50 on the inner circle we can set T=50 at x=2, y=0. Likewise we set T=100 at x=4, y=0. ( 29 2 2 1 1 50 ,100 150, 200 2 4 , 150 200 a b a b a b x T x y x y = + = + â‡’ = = -âˆ´ =-+...
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FinalExamPartB - Therefore on the circle described above 1...

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