Direction of a Point in a Vector Field

Question

Consider the system of first order ODEs $$x'=x(y-1) \ \ y'=y(2-x^2-y). \tag{1}$$

For the equilibrium points $(\pm 1,1)$, these give stable spirals. I am trying to determine the direction of rotation of this spiral. I considered a point, $$(x,y)=(2,2),$$ and by substituting this into $(1)$, I found that $$x'>0, \ \ y'<0.$$ By constructing an arrow diagram as shown below, I believed that the point $(x,y)=(2,2)$ rotates anticlockwise (and hence the spiral rotates anticlockwise). But the solution in my books states this is in fact a clockwise rotation.

Answer 1

Since $x' \gt 0$ and $y' \lt 0$, the arrow on your diagram for the sum of the $2$ vectors should be in the opposite direction, i.e., to the lower right, and going clockwise, instead, just like your book says.

Answer 2

No, the vector field at $(2,2)$ should be $(2,-8)$, it is in the opposite direction with respect to your drawing. A better view is given in this picture:

$\hskip 1 in$

Note that the spirals around $(1,1)$ are clockwise, and, by symmetry, the spirals around $(-1,1)$ are anti-clockwise