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Unformatted text preview: 1 Mathematics 1c: Solutions, Homework Set 6 Due: Monday, May 17 at 10am. 1. (10 Points) Section 6.1, Exercise 6 Let D * be the parallelogram with vertices ( 1 , 3) , (0 , 0) , (2 , 1) and (1 , 2) and D be the rectangle D = [0 , 1] × [0 , 1] . Find a transformation T such that D is the image set of D * under T . Solution. We are required to find a linear mapping T with T ( D * ) = D . To do this, we seek a linear mapping T ( u,v ) = ( x,y ) of the form x = au + bv and y = cu + dv. We require vertices to be mapped to vertices in the same clockwise order and observe that we already have T (0 , 0) = (0 , 0). Thus, we suppose T (1 , 2) = (1 , 1) ,T ( 1 , 3) = (1 , 0) and T (2 , 1) = (0 , 1). This gives us three sets of equations 1 = a + 2 b and 1 = c + 2 d 1 = a + 3 b and 0 = c + 3 d 0 = 2 a b and 1 = 2 c d. From the last line, b = 2 a and so from the first equation, we find a = 1 / 5 ,b = 2 / 5 and similarly from the second line, c = 3 d and so from one of the other two equations for c,d , we get c = 3 / 5 ,d = 1 / 5. Therefore, we conclude that T is given by T ( u,v ) = ( u + 2 v, 3 u + v ) / 5. 2. (10 Points) Section 6.2, Exercise 6 Define T ( u,v ) = ( u 2 v 2 , 2 uv ) . Let D * be the set of ( u,v ) with u 2 + v 2 ≤ 1 ,u ≥ ,v ≥ . Find T ( D * ) = D and evaluate ZZ D dxdy. Solution. One trick to finding D is to use the fact that the boundary of D * gets mapped into the boundary of D (assuming that T is one to one). Thus, let us first show that T is one to one. There are two ways to do this, one using polar coordinates, or the other by algebraic brute force. Taking the brute force route, assume that u 2 v 2 = x, 2 uv = y . We must show that there is a unique ( u,v ) ∈ D * solving this equation. Squaring and adding we get that ( u 2 + v 2 ) 2 = ( u 2 v 2 ) 2 + 4 u 2 v 2 = x 2 + y 2 ....
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This note was uploaded on 07/19/2010 for the course MA 1C taught by Professor Ramakrishnan during the Spring '08 term at Caltech.
 Spring '08
 Ramakrishnan
 Math

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