Discussion
In a Lagrangian system, the position of a fluid particle in a flow is described as \(x = x_{o}e^{-kt}\) and \(y = y_{o}e^{kt}\) where t is the time while \(x_{o}\), \(y_{o}\), and k are constants. The flow is
Answer: B
The flow is steady and two-dimensional.
Differentiate the particle paths to get the velocity, then express it in terms of position:
\[ \begin{aligned} u &= \frac{dx}{dt} = -kx_{o}e^{-kt} = -kx \\ v &= \frac{dy}{dt} = ky_{o}e^{kt} = ky \end{aligned} \]
The Eulerian velocity field (-kx, ky) does not contain t, so the flow is steady, even though each particle's own velocity changes as it moves. It has two non-zero velocity components that depend on two coordinates, so it is two-dimensional (stagnation-point flow).
Calling it unsteady confuses the Lagrangian time dependence of a particle with the field's, and calling it one-dimensional ignores the second velocity component.
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