For the flow $\vec{u}=xt\vec{i}-y\vec{j}$ I have been trying to find and sketch the streamlines at $t=0$ and $t=\frac{1}{2}$, and also the pathline for the particle released at the point $(1,1)$ at $t=0$
My attempt at the solution for when $t=0$, $\vec{u}=-y\vec{j}$ hence $\vec{u}||\vec{j}$ is tangent to the lines $||$ to the y-axis. so the diagram would just be vertical lines, with the direction going downwards on each of the lines, since the $j$ is negative? however would the direction when $y$ is negative be the opposite way (since the $j$ component would then be positive)?
For when $t=\frac{1}{2}$ I derived that the equation of the streamline would be just $y=Ax^2$ where $A$ is a constant. (is this correct?) and then I drew the streamline in three parts, I considered when $A>0$ and when $A=0$ and finally when $A<0$, when $A<0$ the direction would be going downwards, when $A>0$ the direction on the graph would be going upwards, and when $A=0$ it would just be a line on the x axis and the direction would be going in the way of the positive $x$ axis.
Now for finding the pathline for the particle released at the point $(1,1)$ at $t=0$ I did the following.
Let $x_{0}=1,y_{0}=1$ at $t=0$ then $\frac{d{x}}{d{t}}=xt$, so $x=e^{\frac{1}{2}t^2+c}$ $\implies x=Ae^{\frac{1}{2}t^2}$ ,Now when $x_{0}=1$, $t=0$ so $A=1$ and so $x=e^{\frac{1}{2}t^2}$ and $\frac{dy}{dt}=-y$ so $y=\frac{1}{Be^t}$ $B$ is constant, now when $y_{0}=1$, $t=0$ so, $B=1$, and thus $y=\frac{1}{e^t}$
Now eliminating $t$ to obtain the pathline for the particle which passes through $(1,1)$ at $t=0$ $x=e^{\frac{1}{2}t^2}$ $\implies t=\sqrt{tln(x)}\implies t=ln(x)$ and so we get that $y=\frac{1}{x}$, I was just wondering if this is the correct equation I have calculated for the pathline for the particle being released at the point $(1,1)$ at $t=0$. And also what direction would it be travelling in, I assume there are two cases one when $x<0$ and one when $x>0$ although in the examples i have they haven't indicated a direction for the pathlines.