S05mt2ansl - MATH 226 CALCULUS III Prof Baxendale Corrected...

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MATH 226 CALCULUS III Prof Baxendale 3/28/05 Corrected Answers for Sample Second Midterm The answer to question 7 has been changed. BEWARE: The following are just the numerical answers. Since the justification is missing, they would not receive full credit on the real exam. 1: f xx = 4 + 6 xy + 2 y 4 , f xy = 3 x 2 + 8 xy 3 , f yy = 12 x 2 y 2 ; saddle point. 2: ± 0 . 19. 3: ∂z ∂s = 11 e 2 + 7 and ∂z ∂t = 5 e 2 + 1 4: h 2 xyz 3 + z, x 2 z 3 , 3 x 2 yz
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Unformatted text preview: 2 + x i ; 129 / √ 6; 50 x + 8 y + 37 z = 148. 5: Maximum value is 21, attained at (1,6,-1); minimum value is -21, attained at (-1,-6,1). 6: 639 / 20. 7. The region is given by 0 ≤ θ ≤ π/ 2 , 2cos θ ≤ r ≤ 2. Then Z Z D y dA = Z π/ 2 Z 2 2cos θ ( r sin θ ) rdrdθ = 8 3 Z π/ 2 sin θ (1-cos 3 θ ) dθ = 2 ....
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