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Unformatted text preview: Mathematics 218: Analysis in Several Variables Professor Robert Gunning Fall Term 2009 Fourth Problem Set Rik Sengupta October 26, 2009 1 Group 1 1. (a) Sketch the plane curve defined by the equation f ( x ) = x 2 1 3 x 2 2 3 = 0. (b) At which points can this curve be described parametrically by x 2 = g ( x 1 )? Solution. Figure 1: Parametric Description Since f : R 2 R , we know from Corollary 3.3 of the Implicit Function Theorem that we can define g when 2 f ( x 1 ,x 2 ) 6 = 0. Here we obviously have 2 f ( x 1 ,x 2 ) = 6 x 2 , and this is zero exactly when x 2 = 0, i.e. x 1 = 3. Hence it follows readily that we can define g as in the problem iff 1  x 1  3. We cannot define g when x 2 = 0 because any neighborhood of the point ( 3 , 0) has points from both the upper half and the lower half of the curve, and hence, no matter how small, it always contains points of the form ( x 1 ,x 2 ) and ( x 1 , x 2 ), and so g cannot possibly be defined. 2. (a) At which points is the plane curve defined by the equation f ( x ) = x 3 1 + x 3 2 3 x 1 x 2 = 0 a submanifold of R 2 ? (b) Find the tangent vector to this curve at the point ( 3 2 , 3 2 ) and the equation of the tangent line to the curve at this point. Solution. It is immediate that f is continuously differentiable, and S = { x : f ( x ) = 0 } . From Theorem 3.4, we know that a sufficient condition for f ( x ) = 0 to define a submanifold in a neighborhood of a point a is that f ( a ) 6 = 0. Note that f ( x ) = 3 x 2 1 x 2 x 2 2 x 1 which vanishes iff x 2 1 x 2 = x 2 2 x 1 = 0. But this only happens at (0 , 0) and (1 , 1), and (1 , 1) is not even on the submanifold. It follows that S is definitely a manifold away from the origin. At (0 , 0), however, the curve crosses itself. It looks as follows: Figure 2: Crossing Over at the Origin So, since it crosses itself at the origin, there is a singularity here, and the curve cannot preserve the nice structure required for a manifold. Hence, the curve is a manifold exactly everywhere except the origin. The tangent vector at any point is the vector at that point perpendicular to the gradient vector at that point. From our equation, we know the gradient of f at the point ( 3 2 , 3 2 ) is f = ( 9 4 , 9 4 ). This clearly points in the direction of the vector (1 , 1). The tangent line is thus perpendicular 2 to (1 , 1), i.e. it is parallel to (1 , 1). So, the equation of the tangent line is l ( t ) = ( 3 2 , 3 2 ) + t (1 , 1) = ( 3 2 + t, 3 2 t ). 3. Show that the mapping f : R 2 R 2 for which f ( x ) = { x 2 1 x 2 2 , 2 x 1 x 2 } is surjective and has an inverse in an open neighborhood of any point of R 2 other than the origin, but that this mapping does not have an inverse in the full space R 2 ....
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This note was uploaded on 05/03/2010 for the course MAT 218 taught by Professor Robertc.gunning during the Fall '09 term at Princeton.
 Fall '09
 RobertC.Gunning
 Math

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