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