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Unformatted text preview: EA 4 Workshop questions - week of 11/17/08 1. Consider the system x = x 2 + y 2 , (1) y = x 2 + y 2 . (2) (a) Determine all critical points for system (1). Answer- the only critical point is x = 0 and y = 0 . (b) How does this critical point differ from those discussed in class? Answer- there are no linear terms. (c) Determine an equation for the trajectories of this system. Answer- from (1) we get dy dx = 1 , so that all trajectories are lines with slope 1 in the phase plane. (d) Consider the solution with the initial condition x (0) =- 1 , y (0) =- 2 . Determine the trajectory of the solution and the limits of x ( t ) and y ( t ) as t → t * , where t * is the upper limit of times for which the solution exists (will be finite). Answer- the limit is ∞ for both x and y . The trajectory is simply the line y + 2 x + 1 = 1 . There are no critical points on this trajectory. The only thing that can happen is that x and y increase indefinitely. (e) Consider the solution with the initial condition x (0) = y (0) =- 1 . Determine the limits of x ( t ) and y ( t ) as t → ∞ . Answer- Facilitators this question is designed to test student’s understanding of trajectories and critical points and is the key aspect of this problem. The initial point is on the line y = x . This line has slope 1 and so students might expect this line to be the trajectory of the solution. However, this is not true . The reason is that the trajectory cannot include the critical point. The critical point is a solution (a point not a curve in phase space). Thus, no other trajectory, even if it forms a continuous curve withcurve in phase space)....
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- Spring '08
- Trajectory, Euclidean geometry, single solution, Clockwise