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FinalExam compiled - EGM 4313 Spring 2009 Final Exam NAME...

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EGM 4313 NAME:_____________________________ Spring 2009 Final Exam Open book/Closed notes/No calculators or computers 2 hour time limit (1) Solve the following heat equation problem: ( 29 ( 29 ( 29 ( 29 2 2 5 , 0, 10, , 20 , ,0 10 5cos 2 u u u x t u t u x x t x x π π π = = = = + + ÷ . First find the steady-state solution that meets the boundary conditions: ( 29 ( 29 ( 29 ( 29 ( 29 0 10 10 20 10 10 ss ss ss ss u x Ax B du A dx u B u x x π π π π = + = = = = = + ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 , , 5 , 0, 0, , 0, ,0 5cos 2 ss u x t u x x t x t t x t x x φ φ φ φ φ π φ = + = = = = ÷ The problem is greatly simplified if you notice that the initial condition on φ meets the boundary conditions and is in the form that one gets from separation of variables hence you should look for a solution of the form: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 25 25 4 4 25 4 5 5 125 5 , 5cos 0 1, 5cos cos 2 2 4 2 25 5 , 5cos 4 2 5 , 10 5cos 2 t t t x x x x t T t T T t T t T t x T t e x t e T t x u x t x e φ φ π - - - = = = - ÷ ÷ ÷ = - = = ÷ = + + ÷

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EGM 4313 NAME:_____________________________ Spring 2009 Final Exam (2) Consider the complex potential ( 29 ( 29 ( ) ln 1 ln 1 F z z z = - - + . Find the velocity potential and the streamfunction as functions of x and y . (Hint: First write the velocity potential and streamfunction in terms of 1 1 2 , , , r r θ and 2 θ as shown in the figure. Then write 1 1 2 , , , r r θ and 2 θ in terms of x and y .) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 1 2 2 2 2 2 1 2 1 2 2 2 2 1 1 2 2 2 2 1 ln ln ln ln ln ln 1 tan ln 1 tan 1 1 ln 1 ln 1 tan tan 1 1 , ln 1 ln 1 , tan tan 1 i w re r i F z r i r i y y x y i x y i x x y y x y x y i x x x y x y x y y x y x θ θ θ θ φ ψ - - - - - - = = + = + - - = - + + - + + - ÷ ÷ - + = - + - + + + - ÷ ÷ ÷ - + = - + - + + = - ÷ - 1 1 y x ÷ +
EGM 4313 NAME:_____________________________ Spring 2009 Final Exam (3) Find the rank of the matrix 1 2 2 3 1 1 3 2 1 2 3 4 2 4 5 7 . Put in row reduced echelon form: 1 2 2 3 1 2 2 3 1 2 2 3 1 1 3 2 0 1 1 1 0 1 1 1 1 2 3 4 0 0 1 1 0 0 1 1 2 4 5 7 0 0 1 1 0 0 0 0 - - - - Hence the rank is 3.

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EGM 4313 NAME:_____________________________ Spring 2009 Final Exam (4) The inverse of a Laplace transform is given by ( 29 1 lim 2 c iT st T c iT e f s ds i π + →∞ - where c is chosen so that all the singular points of f(s) lie to the left of the line s = c ( c is a real number). If ( 29 as e f s s - = show that ( 29 0, 0 1, t a f t t a < < = by appropriate integrals in the complex plane as shown. (1 point extra credit: Explain why the paths must be closed in the manner shown in each case.) ( 29 t a s as st e e e s s - - = has a simple pole at s =0 with reside 1 (a) t<a case: Here the integrand is analytic everywhere inside and on the contour hence the contour integral is 0. Therefore the inverse Laplace transform is 0 if the integrand vanishes on the arc as R → ∞ .
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FinalExam compiled - EGM 4313 Spring 2009 Final Exam NAME...

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