Streamlines, streaklines, and pathlines I still have difficulty distinguishing the streamlines, streaklines and pathlines of a vector fields. Can you help me to understand this concepts by explaining to me the difference between them?

By example, consider the following vector fields
$$
\vec{V}(x,y,z) := \omega x \: \vec{\textbf{i}} + \omega y \: \vec{\textbf{j}} + \left( -\omega x + \alpha t \right) \: \vec{\textbf{k}}
$$
where $\alpha$ is a constant and the parameter $\omega \ne 0$.
(a) Calculate the pathline of the particule at $(x_0, y_0, z_0)$ when $t=0$. 
(b) Calculate the streamline passing by the point $(x_0, y_0, z_0)$.
(c) Calculate the streaklines of passing by the point $(x_0, y_0, z_0)$.
 A: Since the vector field is time dependent, the pathline, streamline and streaklines are different concepts, see Wikipedia. Note that these notions are frame dependent.
The pathline is the actual path in space followed by a particle which is moved by the flow of the vector field. The trajectory of the particle as a function of time, $(x_p(t),y_p(t),z_p(t))$, therefore obeys the dynamical system
\begin{align}
 \frac{\text{d} x_p}{\text{d} t} &= \omega\,x_p, \\
 \frac{\text{d} y_p}{\text{d} t} &= \omega\,y_p, \\
 \frac{\text{d} z_p}{\text{d} t} &= -\omega\,x_p +\alpha t,
\end{align}
with initial condition $(x_p(0),y_p(0),z_p(0)) = (x_0,y_0,z_0)$. Solving this system should be no problem.
A streamline is a curve which exists in its entirety for every time $t$, but changes with time. To calculate a streamline, we first fix time $t = t_*$. A streamline is a contour of the fixed velocity field. Such a contour line is everywhere tangent to this fixed vector field. To find a parametric representation for such a streamline, denoted by $(x_s(\sigma),y_s(\sigma),z_s(\sigma))$, the fact that the streamline is everywhere tangent to the fixed vector field means that
\begin{align}
 \frac{\text{d} x_s}{\text{d} \sigma} &= \omega\,x_s, \\
 \frac{\text{d} y_s}{\text{d} \sigma} &= \omega\,y_s, \\
 \frac{\text{d} z_s}{\text{d} \sigma} &= -\omega\,x_s +\alpha t_*.
\end{align}
We can fix our parametrisation by choosing $(x_s(0),y_s(0),z_s(0)) = (x_0,y_0,z_0)$. You can solve this dynamical system, where the parameter $\sigma$ plays the role of the 'time variable', and obtain the requested stream line.
The streakline is the most difficult one to calculate. The line itself changes in time, and is determined by the entire history of both the path and the time dependent vector field. You can view it as a collection of end points of path lines of particles released in the flow between time $0$ and time $t$. The first particle to be released has traced out our previously calculated path line, and the last particle to be released is still at the initial point $(x_0,y_0,z_0)$. Therefore, the end points of the path line and streakline coincide.
Suppose that at some earlier time $\tau$ (so $0<\tau<t$), a particle was released at $(x_0,y_0,z_0)$. This particle follows the flow of the vector field until time $t$. Let's parametrise the path of this particle with the parameter $\theta$, such that the curve $(x_\tau(\theta),y_\tau(\theta),z_\tau(\theta))$ is always tangent to the vector field. That means that this curve obeys the dynamical system
\begin{align}
 \frac{\text{d} x_\tau}{\text{d} \theta} &= \omega\,x_\tau, \\
 \frac{\text{d} y_\tau}{\text{d} \theta} &= \omega\,y_\tau, \\
 \frac{\text{d} z_\tau}{\text{d} \theta} &= -\omega\,x_\tau +\alpha (\tau + \theta).
\end{align}
Here, we fixed the parametrisation such that at $\theta = 0$, the particle starts at $(x_0,y_0,z_0)$; at that moment (so at time $\tau$), the vector field is given by $(\omega x_0,\omega y_0, -\omega x_0 + \alpha \tau)$. We've chosen $\theta$ such that it increases the same way the global time increases. You could view $\theta$ as 'the particle's own time'. Anyway, remember that we're only interested at the end point of the path of this particle which was released at time $\tau$. That is, we're interested in $(x_\tau(t),y_\tau(t),z_\tau(t))$ (note that you can find the parametrised path $(x_\tau(\theta),y_\tau(\theta),z_\tau(\theta))$ by solving the associated dynamical system). The collection of all these end points for all intermediate times $\tau$ (for which $0<\tau<t$) is the streakline we're after. So, the streakline at time $t$, parametrised by $\tau$, is given by $(x_\tau(t),y_\tau(t),z_\tau(t))$.
