Quick clarification on the definition of vector field I am having a class on differential geometry and another on ODEs
In the ODE class, if we were given something of the type 
$$\dot x = x^2$$
The professor refers to $x^2$ as the vector field.
In the differential geometry class, we refer to 
$$X = x^2 \frac{\partial}{\partial x}$$ as the vector field
Can someone clarify the relationship between the two equations above?
 A: The integral curve to a given vector field is the one for which the derivatives of the component functions match the components of the given vector field.
$$ X(\Phi(t)) = \frac{d\phi}{dt}$$
but, to make sense of this, suppose $X = a\partial_x +b\partial_y$ then
$$ X(\Phi(t)) = a(\Phi(t))\partial_x +b(\Phi(t))\partial_y$$
and the tangent vector to the curve must be expressed in terms of the derivation-basis for vector fields; for $\phi = (\phi_1, \phi_2)$
$$ \frac{d\phi}{dt} = \frac{d\phi_1}{dt}\partial_x+\frac{d\phi_2}{dt}\partial_y$$
so, putting these together, $ X(\Phi(t)) = \frac{d\phi}{dt}$ yields:
$$ a(\Phi(t))\partial_x +b(\Phi(t))\partial_y = \frac{d\phi_1}{dt}\partial_x+\frac{d\phi_2}{dt}\partial_y $$
Now, if we set $\phi_1=x$ and $\phi_2=y$ then we get:
$$ a(x,y) = \frac{dx}{dt}, \qquad \& \qquad b(x,y) = \frac{dy}{dt}$$
For example, if $X = -y\partial_x+x\partial_y$ then we face
$$ -y = \frac{dx}{dt} \qquad \& \qquad x = \frac{dy}{dt} $$
Solutions are circles, $x= R\cos(t), y = R\sin t$.
