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Engineering Calculus Notes 364

# Engineering Calculus Notes 364 - −→ a = π 3 is T 2 f...

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352 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION We consider two examples. The function f ( x,y ) = e 2 x cos y has ∂f ∂x ( x 0 ,y 0 ) = 2 e 2 x cos y ∂f ∂y ( x 0 ,y 0 ) = e 2 x sin y 2 f 2 x ( x 0 ,y 0 ) = 4 e 2 x cos y 2 f ∂x∂y ( x 0 ,y 0 ) = 2 e 2 x sin y 2 f 2 y ( x 0 ,y 0 ) = e 2 x cos y. At −→ a = ( 0 , π 3 ) , these values are f p 0 , π 3 P = e 0 cos π 3 = 1 2 ∂f ∂x p 0 , π 3 P = 2 e 0 cos π 3 = 1 ∂f ∂y p 0 , π 3 P = e 0 sin π 3 = 3 2 2 f 2 x p 0 , π 3 P = 4 e 0 cos π 3 = 2 2 f ∂x∂y p 0 , π 3 P = 2 e 0 sin π 3 = 3 2 f 2 y p 0 , π 3 P = e 0 cos π 3 = 1 2 so the degree two Taylor polynomial at
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Unformatted text preview: −→ a = ( , π 3 ) is T 2 f pp , π 3 PP △ x, △ y = 1 2 + △ x − ± √ 3 2 ² △ y + △ x 2 − 1 4 △ y 2 − √ 3 △ x △ y. Let us compare the value f ( . 1 , π 2 ) with f ( , π 3 ) = 0 . 5:...
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