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Unformatted text preview: The University of Sydney Math3061 Geometry and Topology Web page: www.maths.usyd.edu.au/u/UG/SM/MATH3061/ 2009 Tutorial 1 1. (Revision) ( i ) Given that A = (1 , 1), B = (2 , 3) and C = (4 , 5), find AB , BC and CA . Check your answers by calculating the sum of these three vectors. ( ii ) Find an equation for the line through the point (1 , 1) in the direction given by the vector bracketleftbigg 2 3 bracketrightbigg . Solution. ( i ) AB = bracketleftbigg 2 1 3 1 bracketrightbigg = bracketleftbigg 1 2 bracketrightbigg , BC = bracketleftbigg 4 2 5 3 bracketrightbigg = bracketleftbigg 2 2 bracketrightbigg , CA = bracketleftbigg 1 4 1 5 bracketrightbigg = bracketleftbigg 3 4 bracketrightbigg . This can be checked by finding the sum of AB + BC + CA. We have bracketleftbigg 1 2 bracketrightbigg + bracketleftbigg 2 2 bracketrightbigg + bracketleftbigg 3 4 bracketrightbigg = bracketleftbigg bracketrightbigg . ( ii ) The line ax + by + c = 0 has direction vector bracketleftbigg b a bracketrightbigg . So the line through (1 , 1) in the direction bracketleftbigg 2 3 bracketrightbigg has an equation of the form 3 x 2 y + c = 0 , where 3 2 + c = 0, that is, the equation of the line is 3 x 2 y 1 = 0. 2. Determine (in each case) if the given formula defines as a transformation of the plane, and if it does, find a formula for the inverse transformation 1 . ( i ) ( x, y ) = ( y, x + 2), ( ii ) ( x, y ) = (ln x, y ), ( iii ) ( x, y ) = ( x 2 y, y ), ( iv ) ( x, y ) = ( x, y + x 2 ). Solution. ( i ) Let ( u, v ) be any point in E . Then ( x, y ) = ( u, v ) ( x, y ) = ( v 2 , u ) . That is, given any point ( u, v ) E , there is a unique point ( x, y ), namely ( x, y ) = ( v 2 , u ), such that ( x, y ) = ( u, v ).This shows that is bijective. So is a transformation and 1 ( u, v ) = ( v 2 , u ), or in more usual notation, 1 ( x, y ) = ( y 2 , x ). ( ii ) ( x, y ) is not defined for all ( x, y ) E . Hence is not a transformation of the plane. 2 ( iii ) Let ( u, v ) be any point in E . Then ( x, y ) = ( u, v ) ( x 2 y, y ) = ( u, v ) ( x, y ) = ( u 2 v, v ). This shows that is bijective. So is a transformation and 1 ( x, y ) = ( x 2 y, y ). (In this case, = 1 .) ( iv ) Let ( u, v ) be any point in E . Then ( x, y ) = ( u, v ) ( x, y + x 2 ) = ( u, v ) ( x, y ) = ( u, v u 2 ). This shows that is bijective. So is a transformation and 1 ( x, y ) = ( x, y x 2 )....
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