we obtain g all second 0 2128 10 21 81

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Unformatted text preview: ng: = 2x + k1 y + ⇠y = k 8x + 8, (c) Let ⇠ = y + y x, 2x == +1 8x,+ k1 ⇠x = 2, 8x = 1, 2⌘x = k2 ⌘y = 1 and 2 + ⌘ y = k 2 then y y= all second partial 2derivatives are zero. Noting thatk2 = x, e = 0, f = y+ x = k y + 8x = d (c) Put the equation in its 1canonical form. = 2, ⇠y = 1, ⌘x = 8, ⌘y = 1 and c) Let ⇠ = y + 2x, ⌘ = y + 18x, then ⇠x 1 sin(y ), g = 0, and x = 6 (⌘ ⇠ ), y = 3 (4⇠ ⌘ ), we obtain g = 0 and ˆ x⌘ =, = (c) all secondy partial ⌘ = y + 8x,are zero. = 2, ⇠y = 1, d x== 8,, e y= 01fand Let ⇠ = + 2x, derivatives then ⇠x Noting that ⌘ 1 1 B partial = 6 ( 6 sin(y ), g = = and xderivatives), y = 3 (4⇠ Noting 2(16)(1)(1)=x0 e 3= 0, f = ⇠ are zero. ⌘ ), we obtain g all second 0, 2(1)(2)(8) ⌘ 10 ((2)(1) + (8)(1)) + that d =ˆ = , and 1 x = 1 (⌘ ⇠ )1 y =11 (4⇠ ⌘ ), we obtain g = 0 and sin(y ), g == and(⌘ ⇠ )(2) =((2)(1) + (8)(1)) + 2(16)(1)(1) = 36 ˆ ⌘ D 0, 2(1)(2)(8) 10 , ⇠ 6 3 B 6 3 3 1 4 4 B = 1 (⌘ ⇠ )(2) = 1 ⇠ 1 ⌘ D = 2(1)(2)(8) 10 ((...
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This document was uploaded on 02/09/2014.

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