20E - HW 5 Solutions

20E - HW 5 Solutions - Math 20E Homework #6 10 February...

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Unformatted text preview: 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 s x- y 2 2 + x + y 2 2 = r ( x 2 + 2 xy + y 2 ) + ( x 2- 2 xy + y 2 ) 2 = r 2 x 2 + 2 y 2 2 = q x 2 + y 2 , T preserves magnitude. Let be the angle between ( x , y ) and T ( x , y ) . Then cos = ( x , y ) x- y 2 , x + y 2 k ( x , y ) k k x- y 2 , x + y 2 k = 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 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 = ( u- x )( u + x ) + 4 ( x- u ) = ( 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 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 = 4 a + 12 b = 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....
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This note was uploaded on 05/22/2010 for the course MATH 20E taught by Professor Enright during the Fall '07 term at UCSD.

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20E - HW 5 Solutions - Math 20E Homework #6 10 February...

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