# Direction of arrows in a phase portrait.

If I have a system like:

$\frac{dx}{dt} = x, \frac{dy}{dt} = -y+x^2$

and I am asked to draw a phase portrait, once I have found the type of portrait (saddle point, node, spiral, etc.) from the eigenvalues and have found the $\infty$-isocline and $0$-isocline, how do I determine the direction of the arrows on the portrait?

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What's wrong with plugging and chugging? Find a point on the relevant curve, and see in which direction the vector field points. Or, solve the equation $\mathrm{d}x/\mathrm{d}t = 0$, which shows that the vector field has a 0 x-component when and only when $x = 0$. In particular, you see that in the first and fourth quadrants the vector field has positive $x$ component and in the second and third quadrants the vector field has negative $x$ component. Similarly solving $\mathrm{d}y/\mathrm{d}t = 0$ gives you a curve that divides the plane into two regions... – Willie Wong Apr 27 '12 at 10:32
Basically (I think) each vector points in the direction of increasing t. – Adam Rubinson Apr 27 '12 at 10:37
Is the second instance supposed to be $\frac{dy}{dt}$? – anon Apr 27 '12 at 11:26

I assume that the second equation should start with $\frac{dy}{dt}\equiv \dot y$.
An easy way to figure out the directions is to put arrows on $0$-isoclines, which are vertical for $\dot x=0$ and horizontal for $\dot y=0$. To determine on which end to put an arrow you simply check the sign. For example, if $\dot y>0$ then your arrow on the $x$ isocline is $\uparrow$. Otherwise it is $\downarrow$. The directions on other parts of the portrait follow by continuity.