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Engineering Calculus Notes 366

# Engineering Calculus Notes 366 - 354 CHAPTER 3 REAL-VALUED...

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Unformatted text preview: 354 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION → Evaluating these at − = 1, 1 , 2 yields a 2 1 f 1, , 2 2 1 2 3 = (1)2 1 fx 1, , 2 2 1 2 3 = 2(1) 1 fy 1, , 2 2 = 3(1)2 1 fz 1, , 2 2 = (1)2 1 2 1 fxx 1, , 2 2 =2 1 fxy 1, , 2 2 = 6(1)2 1 fxz 1, , 2 2 1 fyy 1, , 2 2 1 fyz 1, , 2 2 1 fzz 1, , 2 2 = 2(1) = 6(1)2 = 3(1)2 (2) = 1 4 (2) = 1 2 2 (2) = 3 2 = 1 2 1 8 = 1 2 3 1 2 3 (2) 2 1 2 1 2 1 2 1 2 (2) = 3 3 = 1 4 (2) = 6 2 = 3 4 = 0. The degree two Taylor polynomial is 1 1 3 1 → T2 f (− ) △x, △y, △z = + a △x + △y + △z 4 2 2 8 1 2△x2 + 6△y 2 + 0△z 2 + (6△x△y + 2△x△z + 3△y △z ) + 2 = 0.25 + 0.5△x + 1.5△y + 0.125△z + 0.25△x2 + 0.25△x△z + 3△x△y + 3△y 2 + 0.75△y △z. → Let us compare the value f (− ) = f (1, 0.5, 2.0) = .25 with f (1.1, 0.4, 1.8): a • The exact value is f (1.1, 0.4, 1.8) = (1.1)2 (0.4)3 (1.8) = 0.139392. ...
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