Problem Set 9 Solution

Therefore 2 x21 2 12 2 5 2y x 5 v2 x 2y xe dx

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Unformatted text preview: ght lines. So D is the parallelogram with vertices T (0, 0) = (0, 0) T (1, 3) = (11, 2) T (−1, 2) = (4, 3) T (0, 5) = (15, 5). 9. Since T is supposed to take (0, 5) to (4, 1), it must take (0, 1) to (4/5, 1/5). Since T is supposed to take (−1, 3) to (3, 2) and (1, 2) to (1, −1) it should take (5, 0) = 3(1, 2) − 2(−1, 3), to 3(3, 2) − 2(1, −1) = (7, 8). Therefore T (1, 0) = (7/5, 8/5). Therefore � �� � 7/5 4/5 u T (u, v ) = . 8/5 1/5 v 10. We have x = u and y = (v + u)/2. The Jacobian is � � �1 ∂ (x, y ) 0� 1 � �= . (u, v ) = � 1/2 1/2� 2 ∂ (u, v ) 5 This is nowhere zero. As t...
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