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Engineering Calculus Notes 261

Engineering Calculus Notes 261 - y 2 r x 2 5 xy y 2 so ∂f...

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3.3. DERIVATIVES 249 Approximation and Estimation Just as for functions of one variable, the linearization of a function can be used to get “quick and dirty” estimates of the value of a function when the input is close to one where the exact value is known. For example, consider the function f ( x,y ) = radicalbig x 2 + 5 xy + y 2 ; you can check that f (3 , 1) = 5; what is f (2 . 9 , 1 . 2)? We calculate the partial derivatives at (3 , 1): ∂f ∂x ( x,y ) = 2 x + 5 y 2 radicalbig x 2 + 5 xy + y 2 ∂f ∂y ( x,y ) =
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Unformatted text preview: y 2 r x 2 + 5 xy + y 2 so ∂f ∂x (3 , 1) = 11 10 = 1 . 1 ∂f ∂y (3 , 1) = 17 10 = 1 . 7; since (2 . 9 , 1 . 2) = (3 , 1) + ( − . 1 , . 2) we use △ x = − . 1 , △ y = 0 . 2 to calculate the linearization T (3 , 1) f (2 . 9 , 1 . 2) = f (3 , 1) + ∂f ∂x (3 , 1) △ x + ∂f ∂y (3 , 1) △ y = 5 + (1 . 1)( − . 1) + (1 . 7)(0 . 2) = 5 − . 11 + 0 . 34 = 5 . 23 ....
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