a7_soln

# a7_soln - Math 237 Assignment 7 Due: Friday, July 18th 1....

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Unformatted text preview: Math 237 Assignment 7 Due: Friday, July 18th 1. Consider the following maps T : R 2 → R 2 . Find the image under T of the square D = { ( x, y ) ∈ R 2 | 1 ≤ x ≤ 2 , 2 ≤ y ≤ 3 } . a) T ( x, y ) = (2 x + 3 y, x − y ). Solution: We have u = 2 x + 3 y and v = x − y , so u + 3 v = 2 x + 3 y + 3 x − 3 y = 5 x ⇒ x = u + 3 v 5 , u − 2 v = 2 x + 3 y − (2 x − 2 y ) = 5 y ⇒ y = u − 2 v 5 . For the line x = 1, 2 ≤ y ≤ 3 we get 1 = u +3 v 5 ⇒ v = − 1 3 u + 5 3 and 2 ≤ u − 2 v 5 ≤ 3 10 ≤ u − 2 parenleftbigg − 1 3 u + 5 3 parenrightbigg ≤ 15 40 3 ≤ 5 3 u ≤ 55 3 8 ≤ u ≤ 11 For the line x = 2, 2 ≤ y ≤ 3 we get 2 = u +3 v 5 ⇒ v = − 1 3 u + 10 3 and 2 ≤ u − 2 v 5 ≤ 3 10 ≤ u − 2 parenleftbigg − 1 3 u + 10 3 parenrightbigg ≤ 15 50 3 ≤ 5 3 u ≤ 65 3 10 ≤ u ≤ 13 For the line y = 2, 1 ≤ x ≤ 2 we get 2 = u- 2 v 5 ⇒ v = 1 2 u − 5 and 1 ≤ u + 3 v 5 ≤ 2 5 ≤ u + 3 parenleftbigg 1 2 u − 5 parenrightbigg ≤ 10 20 ≤ 5 2 u ≤ 25 8 ≤ u ≤ 10 For the line y = 3, 1 ≤ x ≤ 2 we get 3 = u- 2 v 5 ⇒ v = 1 2 u − 15 2 and 2 1 ≤ u + 3 v 5 ≤ 2 5 ≤ u + 3 parenleftbigg 1 2 u − 15 2 parenrightbigg ≤ 10 55 2 ≤ 5 2 u ≤ 65 2 11 ≤ u ≤ 13 b) T ( x, y ) = ( x cos( πy/ 3) , x sin( πy/ 3)). Solution: We have u = x cos πy/ 3 and v = x sin πy/ 3. Thus, we observe that u 2 + v 2 = x 2 For the line x = 1, 2 ≤ y ≤ 3 we have u 2 + v 2 = 1 and when y = 2 we get u = cos 2 π 3 = − 1 2 and v = sin 2 π 3 = √ 3 2 and when y = 3 we get u = − 1 and v = 0. Hence − 1 ≤ u ≤ − 1 2 and ≤ v ≤ √ 3 2 ....
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## This note was uploaded on 04/18/2010 for the course MATH 237 taught by Professor Wolczuk during the Spring '08 term at Waterloo.

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a7_soln - Math 237 Assignment 7 Due: Friday, July 18th 1....

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