Differential equations and direction fields I am told that the equation
$\dfrac{dx}{dt} = f(t, x)$
means that for any point in the tx-plane (for which $f$ is defined) we can evaluate the gradient $\dfrac{dx}{dt}$ and represent this graphically by means of a small arrow representing the vector $\left(1, \dfrac{dx}{dt} \right)$.
However, I do not understand how we get $\left(1, \dfrac{dx}{dt} \right)$ from $\dfrac{dx}{dt} = f(t, x)$. 
Please explain the reasoning behind this and prove if it is true.
Thanks.
 A: A slightly different point of view: When you solve the ode you get a solution in the form $x(t)$ for $t$ in some interval $I$. In the $(t,x)$ plane this solution is represented by the parametrized curve $(t,x(t))$, $t\in I$.
A directional derivative of this curve is:
 $$ \frac{d}{dt} (t,x(t)) = (1,\frac{dx}{dt}) = \left(1,v(t,x(t))\right) $$
So the vector $(1,v(t,x))$ is tangent to every point $(t,x)$ on a solution curve. 
This fact means that drawing the vector field $(t,x)\mapsto (1,v(t,x))$  provides an efficient visual method to 'guess' the look and qualitative behaviour of solutions without knowing them in advance.
A: You are actually considering the direction field on the $xt$-plane give  by the system
\begin{align}
t' =&\ 1\\
x' =&\ f(x, t).
\end{align}
A: Basically what this means is that if you take a solution curve for your differential equation and graph it in the $(t,x)$ plane, the forward tangent to the curve at a point $(t_0, x_0)$  will be in the direction of $(1, f(t_0,x_0))$.
