Pathline and streamlines for velocity field

Velocity field: $u = x(1+0.5t)$, $v = y$ (A = 0.5 s⁻¹, B = C = 1 s⁻¹).

(i) Pathline of the particle at (1,1) at t = 0

$$\frac{dy}{dt} = y\Rightarrow y = e^{t}$$ $$\frac{dx}{dt} = x(1+0.5t)\Rightarrow \ln x = t + 0.25t^2\Rightarrow x = e^{t+0.25t^2}$$

Eliminating $t = \ln y$:

$$\ln x = \ln y + 0.25(\ln y)^2$$

(ii) Streamlines through (1,1,0)

$$\frac{dy}{dx} = \frac{v}{u} = \frac{y}{x(1+0.5t)}\Rightarrow y = x^{1/(1+0.5t)}$$

 y
 4 |    /  <- t=0 (y = x)
   |   / .-' t=1
 2 |  / .'  t=2
   | /.'
 1 |*-----------------------> x
   1    2    3    4

The pathline starts along the $t=0$ streamline (slope 1 at $t=0$) but bends below it: as time passes the $x$-velocity grows, so the particle crosses successive streamlines that become flatter. The pathline and streamlines differ because the flow is unsteady.

Answer: pathline $\ln x = \ln y + 0.25(\ln y)^2$ (parametric $x = e^{t+0.25t^2}$, $y = e^t$); streamlines $y = x$, $y = x^{2/3}$, $y = x^{1/2}$ at $t = 0, 1, 2$ s.