Engineering Calculus Notes 366

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

Ask a homework question - tutors are online