How is this extrapolation approximation done?? I'm reading a paper with a given approximation that I can't figure out. Consider a particle whose position and velocity at some time $i$ is defined by
$
\textbf{r}_i = \begin{bmatrix} x_i \\ y_i \\ z_i \end{bmatrix} \>\>,
        \>\>\>\>\>\>\> \textbf{v}_i = \dot{\textbf{r}}_i\>\>.
$
Now, we define the particle's position at some later time by linear extrapolation:
$
\begin{align}
||\textbf{r}|| &= ||\textbf{r}_i + \textbf{v}_idt||\\
    &= \left[(x_i + \dot{x}_idt)^2 + (y_i + \dot{y}_idt)^2 + (z_i + \dot{z}_idt)^2\right]^{1/2}\\
    &\approx ||\textbf{r}_i|| + \dfrac{dt(x_i\dot{x}_i + y_i\dot{y}_i + z_i\dot{z}_i)}{||\textbf{r}_i||}
\end{align}
$
The first two lines are fine. In the third line, I do not understand how this approximation comes about. I know that it has something to do with dropping any dt^2 terms. Expanding the root (second line above) gives 
$
\begin{align}
||\textbf{r}|| &= ||\textbf{r}_i + \textbf{v}_idt||\\
    &= \left[(x_i + \dot{x}_idt)^2 + (y_i + \dot{y}_idt)^2 + (z_i + \dot{z}_idt)^2\right]^{1/2}\\
&= \sqrt{x_i^2 + y_i^2 + z_i^2 + x_i\dot{x}_idt + y_i\dot{y}_idt + zi\dot{z}_idt + \dot{x}_i^2dt^2 + \dot{y}_i^2dt^2 + \dot{z}_i^2dt^2}\\
&= \sqrt{||\textbf{r}_i||^2 + dt(\textbf{r}_i \cdot \textbf{v}_i) + dt^2|\textbf{v}_i|^2}
\end{align}
$
Dropping the final term involving $dt^2$ gets me closer, but I don't see how to get rid of the root or where the $\dfrac{1}{
||\textbf{r}_i||}$ should come from in the second term in the approximated equation. 

Note that:
$$||\vec r||^2=||\vec r_i+\vec v_idt||^2=(\vec r_i+\vec v_idt)\cdot(\vec r_i+\vec v_idt)\approx ||\vec r_i||^2+2\vec r_i\vec v_idt=||\vec r_i||^2\left(1+\frac{2\vec r_i\vec v_idt}{||\vec r_i||^2}\right)$$
thus

$$||\vec r||=||\vec r_i||\left(1+\frac{2\vec r_i\vec v_idt}{||\vec r_i||^2}\right)^\frac12\approx||\vec r_i||\left(1+\frac12\frac{2\vec r_i\vec v_idt}{||\vec r_i||^2}\right)=||\vec r_i||+\frac{\vec r_i\vec v_idt}{||\vec r_i||}$$

where we have used the binomial series
$$(1+x)^n=1+nx+o(x) \quad x\to0$$
