sln10 - EAS4510 Homework #10 Solution #1) Given: f ( x, y )...

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EAS4510 Homework #10 Solution #1) Given: 32 2 2 (, ) 5 8 2 5 0 31 0 8 0 22 0 fxy x x x y y f xx x f y y =+ ++−+ = =++ = =− = The only point that satisfies all three of these equations is: (, ) (2 , 1 ) xy The point (-1,1) also resembles a double point graphically but it does not satisfy all requirements. #2) ( ) cos f = (i) Taylor series approximation about 0 x : 00 () ( ) ( ) f xf x f x x x = +− Î 0 ( ) cos( ) sin( )( ) cos( ) sin( ) f xxx x x x δ (ii) Substite 0 x : 0 ( ) cos( ) cos( )cos( ) sin( )sin( ) f δδ = += Notice that these two results are very similar. Recall for small angles: cos( ) 1 sin( ) x x x Î 0 ( ) cos( ) cos( ) sin( ) f x = So we see that the two term Taylor approximation about 0 x is the same as a approximating ( ) cos f = as a small perturbation from the point 0 x . #3) a) 2 12 33 2 11 2 2 44 2 2 3( ) ) x x x y Ux x x x n x Uy y n y U yyy U yy µ ρρ µµ ρµ ρ µρ + + + ∂∂ ∂∂∂ 1 1 y y = 2 2 y y = 1 x x x + = 2 x x x ∂− = Î 1 2 1 2 5 5 2 2 1 2 2 21 2 5 5 1 2 ) ) 3 ( ) ( ) 3 ( ) ( ) x y U x x x x x x y y U x x x x y x ⎛⎞ = + + ⎜⎟ ⎝⎠ = + + Î
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sln10 - EAS4510 Homework #10 Solution #1) Given: f ( x, y )...

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