# Easy way to find the streamlines

In a textbook, this problem appears:

Find the streamlines of the vector field $\mathbf{F}=(x^2+y^2)^{-1}(-y\hat{x}+x\hat{y})$.

The system we need to solve, I suppose, is:

$\dfrac{dx}{d\tau}=\dfrac{-y}{x^2+y^2}$

$\dfrac{dy}{d\tau}=\dfrac{x}{x^2+y^2}$

This is a text which just introduced the concept streamlines. It's not about differential equations. But I cannot find a simple way to tell what the streamlines are. The answer is "horizontal circles with the center on the $z$ axis."

I visualized this using Mathematica to confirm the answer, I also solved the system using Mathematica but the answer was very complex. I don't see how I could find that solution by hand.

Is there some trick I can use to solve this easily?

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What is the name / author of the text? By streamlines, do you mean phase portrait? – Amzoti Sep 2 '13 at 21:18
If you're not supposed to do it rigorously, then sketching $\mathbf{F}$ at a few select points can give you an idea. – Javier Sep 2 '13 at 21:21
@Amzoti It's a short text written by my teacher (not in that class anymore though) not published anywhere but at my uni. I'm translating from another language but I don't think it's the same as a phase portrait. It corresponds to "StreamPlot" in Mathematica, the documentation refers to this type of plot as a "stream plot". (Imagine the vector field being that of a velocity, how would a particle in such moving water travel?) – user3680 Sep 2 '13 at 21:25
I want to know if there is an easy way to solve this analytically. The other problems can be solved like that and there is nothing to indicate that we need a different approach here. – user3680 Sep 2 '13 at 21:25
Did you consider polar coordinates? – Amzoti Sep 2 '13 at 21:28

Hint: divide side by side the two equations (for example the second by the first), obtaing$$\frac {dy}{dx}=-\frac x{y}$$The solutions are$$x^2+y^2=c \quad (c>0)$$
A non-calculus answer. The dot product with the radius vector $\mathbf{r}=(x,y)$ is zero, so the streamlines are always perpendicular to the radius, and hence form circles.