math237_a5 - SOLUTIONS FOR ASSIGNMENT 5 Problem Set 4, A2:...

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Unformatted text preview: SOLUTIONS FOR ASSIGNMENT 5 Problem Set 4, A2: Determine de values of the constants and for which the function u ( x, t ) = e t sin x satisfies the 1D heat equation u t = 2 u x 2 . (1) Given u ( x, t ) = e t sin x , we have: u t = e t sin x u x = e t cos x 2 u x 2 = 2 e t sin x. If we replace the above partial derivative in the heat equation (1), we get: e t sin x = 2 e t sin x = 2 . Thus we need and such that + 2 = 0 in order for the given function to be a solution to the 1D heat equation. Problem Set 4, A7: a). Use the second degree Taylor polynomial to derive the approximation ln(sin 2 x + cos 2 y ) x 2 y 2 , for ( x, y ) sufficiently close to (0 , 0) . b). Sketch the surface z = ln(sin 2 x + cos 2 y ) near the point (0 , , 0) . The second degree Taylor polynomial at the point (0 , 0) is given by: P 2 , (0 , 0) ( x, y ) = f (0 , 0) + f x (0 , 0) x + f y (0 , 0) y + 1 2 [ f xx (0 , 0) x 2 + 2 f xy xy + f yy y 2 ] . (2) Since f ( x, y ) = ln(sin 2 x + cos 2 y ) , we have: f x = sin 2 x sin 2 x + cos 2 y , f y = sin 2 y sin 2 x + cos 2 y , and f xx = 2 cos 2 x (sin 2 x + cos 2 y ) sin 2 2 x (sin 2 x + cos 2 y ) 2 , f xy = sin 2 x cos 2 x (sin 2 x + cos 2 y ) 2 , f yy = 2 cos 2 y (sin 2 x + cos 2 y ) sin 2 2 y (sin 2 x + cos 2 y ) 2 ....
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math237_a5 - SOLUTIONS FOR ASSIGNMENT 5 Problem Set 4, A2:...

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