= Solution
The <divergence> is $\partial_x(kty)+\partial_y0+\partial_z0=0$, so this is an <incompressible flow>. With the planar convention $(u,v)=(\psi_y,-\psi_x)$, one <stream function> is
$$
\boxed{\psi(x,y,t)=\frac12kty^2}.
$$
At a fixed nonzero $kt$, the <streamlines> have $y=\mathrm{constant}$ and $z=\mathrm{constant}$, since the instantaneous <velocity> is purely axial. If $kt=0$, every point is instantaneously stationary and there is no distinguished <streamline> direction.
Follow a marked particle initially at $(0,y_0,0)$. Its trajectory satisfies $\dot y=\dot z=0$ and $\dot x=kty_0$, so
$$
(x,y,z)=\left(\frac12kt^2y_0,y_0,0\right),\qquad0\leq y_0\leq a.
$$
These points form a straight segment on $x=(kt^2/2)y$, $z=0$, not an instantaneous horizontal <streamline>. Its length is
$$
\boxed{\ell(t)=a\sqrt{1+\frac14k^2t^4}}.
$$
The directed angle of its vector from the origin to the marked endpoint is
$$
\boxed{\theta(t)=\operatorname{atan2}(1,kt^2/2)}.
$$
For $k>0$ and $t>0$, this is $\arctan[2/(kt^2)]$; it equals $\pi/2$ at $t=0$. If an unoriented acute angle to the axis is intended, use $\arctan[2/(|k|t^2)]$, with $\pi/2$ when $kt=0$. This distinction handles a negative value of the given constant as well. If $a=0$, the marked set is a single point: the length formula gives zero and no line direction is defined.
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