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ansch11review

# ansch11review - Answers to Even Recommended problems from...

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Answers to Even Recommended problems from Ch. 11 Review 2. The domain of 4 x 2 y 2 is x 2 y 2 4, while the domain of 1 x 2 is 1 x 1, so the domain of\ f ( x , y ) is { x , y | x 2 y 2 4, 1 x 1} 8. 32. (a) We have A = 1 2 bh where b is the base and h is the height of the triangle. dA = A b db A h dh = 1 2 hdb 1 2 b dh t hus the maximum error in the calculated area is dA = 6 0.002  5 2 0.002 = 0.0017m 2 (b) Let z be the hypotenuse, then z = b 2 h 2 , dz = b b 2 h 2 db h b 2 h 2 dh . Thus the maximum error in the calculated hypotenuse length is about dz = 5 13 0.002  12 13 0.002 = 0.17 65 0.0026 m 38: A = 1 2 x y sin , dx dt = 3, dy dt =− 2, d dt = 0.05 , and dA dt = 1 2 [ y sin  dx dt  x sin  dy dt  xy cos  d dt ] . So, when x = 40, y = 50 and = 6 , dA dt = 1 2 [ 25  3  20 − 2  1000 3  0.05 ] = 35 50 3 2 60.8in 2 / s.

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