Draw streakline and pathline of dye

A pathline is the track of one particle; a streakline is the line joining all the particles that passed through the injection point earlier. Here the flow is unsteady (the velocity changes at $t = 8$ s). The injection point is taken as the origin, and the streak is observed at $t = 16$ s.

Velocity: $0\le t<8$: $(u, v) = (0, 2)$ m/s; $8\le t\le 16$: $(2, -2)$ m/s.

Pathline (particle at the origin at t = 0)

• 0 to 8 s: it moves $\Delta y = 2\times 8 = 16$ m, from $(0, 0)$ to $(0, 16)$.
• 8 to 16 s: it moves $\Delta x = 2\times 8 = 16$ m and $\Delta y = -16$ m, from $(0, 16)$ to $(16, 0)$.

Streakline at t = 16 s

For a particle released at time $\tau$:

• Released in $0\le\tau\le 8$: it moves up for $(8-\tau)$ s, then diagonally for 8 s:

$$x = 16,\qquad y = 2(8-\tau) - 16 = -2\tau$$

So these particles lie on the vertical line $x = 16$, from $y = 0$ ($\tau=0$) to $y = -16$ ($\tau=8$).

• Released in $8\le\tau\le 16$: it moves diagonally for $(16-\tau)$ s:

$$x = 2(16-\tau),\qquad y = -2(16-\tau)$$

So these particles lie on the line $y = -x$, from $(16, -16)$ ($\tau=8$) to the origin ($\tau=16$).

 y (m)
  16 *               pathline: (0,0) -> (0,16) -> (16,0)
  12 |  \
   8 |     \
   4 |        \
   0 O-----------*----> x (m)
  -4    .        :
  -8       .     :   streakline at t = 16 s:
 -12          .  :   (0,0) -> (16,-16) -> (16,0)
 -16             *
     0  4  8  12 16

Answer: the pathline goes from (0,0) to (0,16) and then to (16,0); the streakline at $t = 16$ s goes from (0,0) to (16,-16) (a straight line) and then vertically up the line $x = 16$ to (16,0). The two curves are different because the flow is unsteady. Scale: 1 division = 4 m on both axes.