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Unformatted text preview: Math 237 Assignment 9 Solutions 1. Consider the following maps T : R 2 → R 2 . Find the image under T of the rectangle D = { ( x,y ) ∈ R 2  ≤ x ≤ 2 , 2 ≤ y ≤ 2 } . a) 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) . For x = 0 , 2 ≤ y ≤ 2 we have ( u,v ) = T (0 ,y ) = (0 , 0) . So, the image of the line is the point (0 , 0) . For the line x = 2 , 2 ≤ y ≤ 2 we have ( u,v ) = T (2 ,y ) = (2 cos( πy/ 3) , 2 sin( πy/ 3)) , 2 ≤ y ≤ 2 . We recognize this as a parametric equation for an arc of a circle of radius 2 from angle 2 3 π to 2 3 π . Thus, the image is the arc of the circle u 2 + v 2 = 4 over these angles. For the line y = 2 , ≤ x ≤ 2 we have ( u,v ) = F ( x, 2) = ( 1 2 x, √ 3 2 x ) . Hence, we have v = √ 3 2 x = √ 3 u . Moreover, since ≤ x ≤ 2 , then 1 ≤  1 2 x ≤ so 1 ≤ u ≤ . Hence, the image is the line v = √ 3 u , for 1 ≤ u ≤ . For the line y = 2 , ≤ x ≤ 2 we have ( u,v ) = F ( x, 2) = ( 1 2 x, √ 3 2 x ) . Hence, we have v = √ 3 2 x = √ 3 u . Moreover, since ≤ x ≤ 2 , then 1 ≤  1 2 x ≤ so 1 ≤ u ≤ . Hence, the image is the line v = √ 3 u , for 1 ≤ u ≤ . b) T ( x,y ) = ( x 2 y 2 ,xy ) . Solution: We have u = x 2 y 2 and v = xy . For x = 0 , 2 ≤ y ≤ 2 we have ( u,v ) = T (0 ,y ) = ( y 2 , 0) . Thus, we have v = 0 and since 2 ≤ y ≤ 2 and u = y 2 , we get that u runs from 4 to and then from to 4 . For x = 2 , 2 ≤ y ≤ 2 we have ( u,v ) = T (2 ,y ) = (4 y 2 , 2 y ) . Thus, we have v 2 = y and so u = 4 y 2 = 4 v 2 4 . Since 2 ≤ y ≤ 2 , we get 2 ≤ v 2 ≤ 2 ⇒  4 ≤ v ≤ 4 . 2 For y = 2 , ≤ x ≤ 2 we have ( u,v ) = T ( x, 2) = ( x 2 4 , 2 x ) . Thus, we have v 2 = x and so u = x 2 4 = v 2 4 4 . Since ≤ x ≤ 2 , we get ≤ v 2 ≤ 2 ⇒ ≤ v ≤ 4 . For y = 2 , ≤ x ≤ 2 we have ( u,v ) = T ( x, 2) = ( x 2 4 , 2 x ) . Thus, we have v 2 = x and so u = x 2 4 = v 2 4 4 . Since ≤ x ≤ 2 , we get ≤  v 2 ≤ 2 ⇒  4 ≤ v ≤ ....
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 Spring '08
 CELMIN
 Math, Linear Algebra, Algebra, Derivative, Sin, Cos, dr dθ, dxy

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