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Unformatted text preview: Mathematics 250 Fall 2007 Selected Solutions, Assignment 4 MVM 4.5.1 : a) The equation xy = 1 can be solved explicitly to find the expression y = 1 x , on the domain R { } of nonzero real numbers. b) The equation 2sin( xy ) = 1 can be converted to sin( xy ) = 1 2 , which means that xy = π 6 + 2 mπ or xy = 5 π 6 + 2 mπ . This set comprises the graphs of the infinite number of functions y = k x , where k is one of the values just specified. Any given point x o y o satisfying 2sin( xy ) = 1 will lie on exactly one of these graphs, so locally near that point, the set of solutions of the equation can be represented as the graph of the function y = x o y o x . MVM 4.5.2 : Decide whether each of the following is a smooth curve (= 1dimensional manifold). If not, what are the trouble points? (a) y 2 x 3 + x = 0. quad (d) x 2 + y 2 + z 2 1 = 0 = x 2 x + y 2 = 0. (b) y 2 x 3 x 2 = 0. (e) x 2 + y 2 + z 2 1 = 0 = z 2 xy. (c) z xy = 0 = y x 2 . Response: We can use the Implicit Function Theorem to investigate this. The ImpFT says that, at a given point p o , the level set of a function F : R m → R ‘ us a smooth manifold of dimension m ‘ providing that DF ( p o ) is onto, that is, of rank ‘ . Let us see what this tells us about these examples. (a). Here we are dealing with the 0 level set of the function F ( x y ) = y 2 x 3 + x . We compute DF = [ 3 x 2 + 1 2 y ] . The condition DF = 0 gives the equations 3 x 2 + 1 = 0 2 y . This has solutions ± √ 3 3 0 . But the points of the curve with ycoordinate equal to zero have x values satisfying x 3 + x = 0, or x = 0 , ± 1. Hence DF ( p ) = neq 0 for all points on the curve, so this curve is smooth. (b) The analysis of this curve is similar to case (a). We take F (( x y ) = y 2 x 3 x 2 . Then we compute DF = [ 3 x 2 + 2 x 2 y ] . Again for DF to vanish, we must have y = 0, and 3 x 2 + 2 x = 0, or x = 0, or x = 2 3 . The points of the curve when y = 0 are the solutions of x 3 + x 2 = 0, or x = 0 and x = 1. This time, we see that the origin is a point on the curve at which DF vanishes. Thus, this may be a singular point of the curve. In fact, if we rewrite the equation defining the curve as y 2 = x 3 + x 2 = x 2 ( x + 1), we see that, for x near zero but not equal to zero, the function x 3 + x 2 is positive, so there are two values of y , namely y = ± √ x 3 + x 2 , that give points on the curve. However, at x = 0, the positive and negative branches of this curve come together, and the curve has a crossing, or double point, at the origin. So it is not a manifold there. (c) We can regard this curve as the level set of the function F ( x y z ) = z xy y x 2 . We compute DF = y x 1 2 x 1 0 ....
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This note was uploaded on 09/24/2009 for the course MATH 250 taught by Professor Rogerhowe during the Fall '06 term at Yale.
 Fall '06
 RogerHowe
 Real Numbers

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