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

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SOLUTIONS FOR ASSIGNMENT 5 Problem Set 4, A2: Determine de values of the constants α and β for which the function u ( x, t ) = e α t sin β x satisfies the 1D heat equation ∂ u ∂ t = 2 u ∂ x 2 . (1) Given u ( x, t ) = e α t sin β x , we have: ∂ u ∂ t = α e α t sin β x ∂ u ∂ x = β e α t cos β x 2 u ∂ x 2 = β 2 e α t sin β x. If we replace the above partial derivative in the heat equation (1), we get: α e α t sin β x = β 2 e α t sin β x α = β 2 . Thus we need α and β such that α + β 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 , 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 = 2cos 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 = 2cos 2 y (sin 2 x + cos 2 y ) sin 2 2 y (sin 2 x + cos 2 y ) 2 . 1
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Thus: f (0 , 0) = f x (0 , 0) = f y (0 , 0) = 0 , f xx (0 , 0) = 2 , f xy (0 , 0) = 0 , f yy (0 , 0) = 2 , and the second degree Taylor polynomial (2) becomes: P 2 , (0 , 0) ( x, y ) = x 2 y 2 .
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