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Unformatted text preview: Math 2220 Problem Set 9 Solutions Spring 2010 Section 3.3 : 1. For the vector field F ( x, y ) = ( − x, y ) do the following things: (a) Sketch the vector field. Solution. Along the xaxis we have F ( x, 0) = ( − x, 0), pointing toward the origin. Along the yaxis we have F (0 , y ) = (0 , y ), pointing away from the origin. This gives a start in sketching the vector field, as shown in the figure. Looking on the lines y = x and y = − x we get F ( x, x ) = ( − x, x ) and F ( x, − x ) = ( − x, x ), which gives us a bit more, enough to predict what the rest will look like. (b) Determine the flow lines by solving the differential equations dx dt = − x and dy dt = y . Your answer should be functions x ( t ) and y ( t ) each of which depends on an arbitrary constant. Solution. By inspection, dx dt = − x has solutions x = Ae t for A an arbitrary constant. More systematically, we can rewrite the equation dx dt = − x as dx x = − dt . Integrating both sides of this equation gives ln x = − t + C , which implies x = e C e t = Ae t . In a similar way the equation dy dt = y gives y = Be t for B an arbitrary constant. (c) Check that the flow lines are hyperbolas by eliminating the variable t from your formulas in part (b). Solution. We have xy = Ae t Be t = AB . This can be written as xy = C for C = AB . These curves are the hyperbolas y = C/x . (d) Find the values of the arbitrary constants that give the flow line c ( t ) such that c (0) = (2 , 1). Solution. For t = 0 the equation ( x, y ) = ( Ae t , Be t ) = (2 , 1) gives ( A, B ) = (2 , 1) so A = 2 and B = 1 and we have c ( t ) = (2 e t , e t ). 2. Sketch the vector field F ( x, y ) = ( x, x 2 ), determine the flow lines, and show the flow lines are parabolas....
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This note was uploaded on 04/21/2010 for the course MATH 2220 taught by Professor Parkinson during the Spring '08 term at Cornell.
 Spring '08
 PARKINSON
 Math, Multivariable Calculus

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