Assuming right ascension is $\alpha$, declination is $\delta$, radius $r=|\vec{r}_{ECI}|$.
$\vec{r}_{ECI}= \begin{bmatrix} r COS(\delta)COS(\alpha) \\ rCOS(\delta)SIN(\alpha) \\ rSIN(\delta) \end{bmatrix}$
The velocity vector can be derived in a similar way, which is eq. 4-2.
Note that right ascension / declination are in sky-fixed coordinates (rotate over Earth); while longitude / latitude are in Earth-fixed coordinates (rotate with Earth).
The next questions are to find:
$\dot{r} = \frac{d|\vec{r}|}{dt} = \frac{d}{dt} (\vec{r} \cdot \vec{r})^{1/2}=\frac{1}{2}(\vec{r} \cdot \vec{r})^{-1/2}\frac{d}{dt}(\vec{r} \cdot \vec{r})=\frac{1}{2}(\vec{r} \cdot \vec{r})^{-1/2} *2\vec{r}\cdot \vec{v}=\frac{\vec{r}\cdot \vec{v}}{r}$
$\dot{\alpha}$ and $\dot{\delta}$ can be derived in the similar approach. The results are listed in Algorithm 25 (Geocentric Radec).