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20E - HW 5 Solutions

20E - HW 5 Solutions - Math 20E HW 5 Homework#6 x y x y 2 2...

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Math 20E Homework #6 10 February 2005 Section 6.1, Problem 2: Define T ( x , y ) = x - y 2 , x + y 2 . Show that T rotates the unit square [ 0, 1 ] × [ 0, 1 ] . Solution. First of all, since x - y 2 2 + x + y 2 2 = ( x 2 + 2 xy + y 2 ) + ( x 2 - 2 xy + y 2 ) 2 = 2 x 2 + 2 y 2 2 = x 2 + y 2 , T preserves magnitude. Let θ be the angle between ( x , y ) and T ( x , y ) . Then cos θ = ( x , y ) · x - y 2 , x + y 2 ( x , y ) · x - y 2 , x + y 2 = x 2 - xy 2 + xy + y 2 2 x 2 + y 2 = 1 2 · x 2 + y 2 x 2 + y 2 = 1 2 . Since T sends every point ( x , y ) to one of equal magnitude at an angle of θ counterclock- wise, T is a rotation of the entire square. Section 6.1, Problem 3: Let D * = [ 0, 1 ] × [ 0, 1 ] and define T on D * by T ( u , v ) = ( - u 2 + 4 u , v ) . Find the image D . Is T one-to-one? Solution. Since both - u 2 + 4 u and v are increasing functions on [ 0, 1 ] × [ 0, 1 ] , the end- points of the range are determined by the endpoints of D * . So D = [ 0, 3 ] × [ 0, 1 ] . Next, we show that T is one-to-one. To show this, suppose T ( x , y ) = T ( u , v ) . Then ( - x 2 + 4 x , y ) = ( - u 2 + 4 u , v ) , which implies that y = v and - x 2 + 4 x = - u 2 + 4 u . Rearranging terms, we have - x 2 + 4 x = - u 2 + 4 u u 2 - x 2 + 4 x - 4 u = 0 ( u - x )( u + x ) + 4 ( x - u ) = 0 ( u - x )( u + x - 4 ) = 0, so either u = x or u + x = 4. Since u and x are in D * , they are both between 0 and 1. So we cannot have u + x = 4, which implies that u = x . So we can conclude that T is one-to-one. Section 6.1, Problem 4: Let D * be the parallelogram bounded by the lines y = 3 x - 4, y = 3 x , y = 1 2 x , and y = 1 2 ( x + 4 ) . Let D = [ 0, 1 ] × [ 0, 1 ] . Find a T such that D is the image of D * under T . Solution. We need the corners of D * to be sent to the corners of D . Set T = a b c d . The corners of D * are ( 0, 0 ) , 8 5 , 4 5 , 12 5 , 16 5 , and 4 5 , 12 5 . 1

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Math 20E Homework #6 10 February 2005 Setting T ( 8/5, 4/5 ) = ( 1, 0 ) , T ( 4/5, 12/5 ) = ( 0, 1 ) gives 8 a + 4 b = 5 8 c + 4 d = 0 4 a + 12 b = 0 4 c + 12 d = 5 Solving yields a = 3/4, b = - 1/4, c = - 1/4, and d = 1/2, so T = 3/4 - 1/4 - 1/4 1/2 is a valid transformation. Section 6.1, Problem 10: Suppose T : R 2 R 2 is linear and is given by T ( x ) = A x , where A is a 2 × 2 matrix. Show that if det A = 0, then T takes parallelograms onto parallelograms. Solution. Say we have a parallelogram P . It can be described as the set of points of the form p + λ v + μ w , where v and w are not scalar multiples of each other and λ , μ [ 0, 1 ] .
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20E - HW 5 Solutions - Math 20E HW 5 Homework#6 x y x y 2 2...

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