PS7_answers - Math 237 A1. i) Problem Set 7 Answers ii)...

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Unformatted text preview: Math 237 A1. i) Problem Set 7 Answers ii) iii) A2. A3. (4.980, 0.6224). A4. a) A5. 3e −e −e 3e c) 2e2 0 . 4 4 2(e − 1) 2(e + 1) 04 11 u−v 2 1 , 2 (u + v ) . A6. a) F −1 (u, v ) = − ln b) DF = A7. i) ii) −e−x 1 −1/(u − v ) 1(u − v ) , DF −1 = . e−x 1 1/2 1/ 2 y = 2x. = πk or x − y = π 2 A8. i) 5 ∆x∆y . 2 B2. f (x, y ) = x √ +2 2 ∂ (u,v) π π ( , ) = 2. The Jacobian is zero when x + y ∂ (x,y ) 4 4 ∂ (u,v) (0, 1) = −2. The Jacobian is zero when y = 0 or ∂ (x,y ) + πk , k ∈ Z. ii) 1 (x 2 1 √ ∆x∆y . 2 1 B1. u = − 1 (y − 3x), v = 4 (2y − x). 4 + 2 y ), 1 y 2 2 √z +y 9+y B3. u = √ √ 2y 9+y 2 ,v= ,w= √ 2 √ x 2z + √ 2 9+y 2 B4. b) The tangent vectors are (3t2 , 2t − 4t3 ). 1 2 B5. a) ∆u∆v ≈ ∆x∆y . b) The image is u2 + v 2 − 2kuv 2 + k 2 v 4 ≤ 1. The area of image is approximately the same as the area of the disc since ∆u∆v ≈ ∆x∆y . ...
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This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.

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PS7_answers - Math 237 A1. i) Problem Set 7 Answers ii)...

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