FinalExamPartB - + Therefore on the circle described above...

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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 = = = constant. (b) Use the result found in (a) to find the temperature in the shaded region shown for steady heat conduction (Laplaces 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 = + = + = = - =-+...
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This note was uploaded on 12/15/2011 for the course EGM 4313 taught by Professor Mei during the Fall '08 term at University of Florida.

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FinalExamPartB - + Therefore on the circle described above...

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