Exercise · ZW-PHASE-PLANE-010 · ★★★ Advanced
In \(x'=y,\ y'=-x\), in which sense are the circular trajectories traversed?
Choose one option.
Show answer
- B. Clockwise
Explanation
At \((1,0)\) the velocity is \((x',y')=(0,-1)\): it points down, and to the right of the origin that is clockwise. In general \(xy'-yx'=-(x^2+y^2)<0\).
Review the concept: Phase plane
Coordinate plane \((x,y)\) of an autonomous system whose velocities \(f\) and \(g\) do not depend on \(t\); each solution is drawn there as a curve, its trajectory.
\[x'=f(x,y),\quad y'=g(x,y)\;\Longrightarrow\;\dfrac{dy}{dx}=\dfrac{g(x,y)}{f(x,y)}\quad(f\neq0)\]
More practice problems: phase plane
- ZW-PHASE-PLANE-001 For x'=x+y,\ y'=x-y, what is the slope dy/dx of the trajectory through (2,1)?
- ZW-PHASE-PLANE-002 For x'=x-2y,\ y'=3x+y, what is the velocity vector (x',y') at the point (2,1)?
- ZW-PHASE-PLANE-003 What are the equilibria of x'=x(1-y),\ y'=y(x-1)?
- ZW-PHASE-PLANE-004 The trajectories of x'=2x,\ y'=y are the curves:
- ZW-PHASE-PLANE-005 In the system x'=y,\ y'=-x, which expression stays constant along each trajectory?
- ZW-PHASE-PLANE-006 For x'=y,\ y'=x, which point lies on the same trajectory as (1,0)?
- ZW-PHASE-PLANE-007 For x'=-x,\ y'=-2y, what happens as t→∞ to the solution starting at (1,1)?
- ZW-PHASE-PLANE-008 In an autonomous system with smooth f and g, can two different trajectories cross at a point that i…