Question

Show that the displacements of a fluid particle occupying the position $(x_0, y_0)$ at $t = t_0 = 0$ in the Lagrangian system are then $x = \frac{5}{6}x_0 - \frac{1}{6}y_0 - \frac{1}{6}z_0 + \frac{1}{6}(x_0 + y_0 + z_0 + \frac{1}{12})e^{6t} - \frac{1}{12}t + \frac{1}{4}t^2 - \frac{1}{72}$ $y = -\frac{1}{3}x_0 + \frac{2}{3}y_0 - \frac{1}{3}z_0 + \frac{1}{3}(x_0 + y_0 + z_0 + \frac{1}{12})e^{6t} - \frac{1}{6}t - \frac{1}{36}$ $z = -\frac{1}{2}x_0 - \frac{1}{2}y_0 + \frac{1}{2}z_0 + \frac{1}{2}(x_0 + y_0 + z_0 + \frac{1}{12})e^{6t} - \frac{1}{4}t - \frac{1}{4}t^2 - \frac{1}{24}$

          Show that the displacements of a fluid particle occupying the position $(x_0, y_0)$ at $t = t_0 = 0$ in the Lagrangian system are then
$x = \frac{5}{6}x_0 - \frac{1}{6}y_0 - \frac{1}{6}z_0 + \frac{1}{6}(x_0 + y_0 + z_0 + \frac{1}{12})e^{6t} - \frac{1}{12}t + \frac{1}{4}t^2 - \frac{1}{72}$
$y = -\frac{1}{3}x_0 + \frac{2}{3}y_0 - \frac{1}{3}z_0 + \frac{1}{3}(x_0 + y_0 + z_0 + \frac{1}{12})e^{6t} - \frac{1}{6}t - \frac{1}{36}$
$z = -\frac{1}{2}x_0 - \frac{1}{2}y_0 + \frac{1}{2}z_0 + \frac{1}{2}(x_0 + y_0 + z_0 + \frac{1}{12})e^{6t} - \frac{1}{4}t - \frac{1}{4}t^2 - \frac{1}{24}$

Show that the displacements of a fluid particle occupying the position (x0, y0) at t = t0 = 0 in the Lagrangian system are then
x = (5)/(6)x0 - (1)/(6)y0 - (1)/(6)z0 + (1)/(6)(x0 + y0 + z0 + (1)/(12))e^6t - (1)/(12)t + (1)/(4)t^2 - (1)/(72)
y = -(1)/(3)x0 + (2)/(3)y0 - (1)/(3)z0 + (1)/(3)(x0 + y0 + z0 + (1)/(12))e^6t - (1)/(6)t - (1)/(36)
z = -(1)/(2)x0 - (1)/(2)y0 + (1)/(2)z0 + (1)/(2)(x0 + y0 + z0 + (1)/(12))e^6t - (1)/(4)t - (1)/(4)t^2 - (1)/(24)

Added by Kara M.

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