What's the covariant derivative of a normalized, timelike Killing vector? I'm reading The large scale structure of spacetime and in page 72 the author says:

A static metric admits a timelike killing vector $K$. We define the timelike unit vector $V$ as $V=K/f$, where $f^2=-g(K,K)$.

Then, $\nabla_{b}V^a = - V^c V_b \nabla_c V^a$
But the math doesn't check out, and I'm not sure how he got that result.
 A: The previous answer to this question is actually incorrect, the last contraction with $V_b$ doesn't make sense mathematically. The formulation of the problem misses one crucial detail present in Ellis Hawking p. 72, the killing vector field $K$ is orthogonal to a family of spacelike surfaces. This introduces an additional condition coming from Frobenius' theorem
$$
K_{[a;b} K_{c]} = 0.
$$
This identity gives the missing piece $K_{[a;b]}$, which is not constrained by the killing equation
$$
K_{a;b} + K_{b;a} = 0.
$$
Using both the Frobenius' theorem and the Killing equation we indeed can reproduce the result from Ellis Hawking. Indeed, using the Killing property we get
$$
K_{a;b} K^b = - K_{b;a} K^b = - \frac{1}{2} (K_b K^b)_{,a} = \frac{1}{2} 
(f^2)_{,a} = f f_{,a}.
$$
From that we get another identity
$$
K^a K^b K_{a;b} = f f_{,a} K^{a} = K_{(a;b)} K^a K^b = 0,
$$
because, again, $K$ is a Killing vector. From that we see that
$$
f_{,a} K^a = f_{,a} V^a = 0.
$$
We are ready to establish the identity for $\dot{V}_a$
$$
\dot{V}_a = V_{a;b} V^b = (f^{-1} K_{a})_{;b} V^b = f^{-2} K_{a;b} K^b - f^{-2} K_a f_{,b} V^b  = f^{-1} f_{,a}.
$$
Finally we take the Frobenius' theorem, using the Killing equation we can rewrite it in the following way
$$
K_{a;b} K_c + K_{b;c} K_a + K_{c;a} K_b = 0.
$$
We contract this identity with $K^c$ to get
$$
K_{a;b} f^2 = K_{b;c} K^c K_a + K^c K_{c;a} K_b = f f_{,b} K_a - f f_{,a} K_b = f^3 (\dot{V}_b V_a - \dot{V}_a V_b). 
$$
Finally,
$$
V_{a;b} = (f^{-1} K_a)_{;b} = f^{-1} K_{a;b} - f^{-2} f_{,b} K_a = \dot{V}_b V_a - \dot{V}_a V_b - \dot{V}_b V_a = -\dot{V}_a V_b.
$$
I spent quite a lot of time trying to figure out the missing piece,  Frobenius' theorem, it is surprisingly easy to miss. I hope I'll save someone else that time.
A: On p. 72 of the Hawking and Ellis book, a timelike Killing vector $\pmb{K}$ of magnitude $K^aK_a = -f^2$, and a unit vector $\pmb{V} = f^{-1}\pmb{K}$ along $\pmb{K}$, with $V^aV_a = -1$, are introduced. Using component notation throughout, we make note of the fact that since $\pmb{K}$ is a Killing vector, its components satisfy the relation
\begin{align}
K_{b{;c}} + K_{c{;b}} = 0 \quad \Longrightarrow \quad K_{c{;b}} = -K_{b{;c}}\,. \tag{1ed}
\end{align}
The covariant derivative of the relation $V^aV_a = -1$ yields
\begin{align}
V^a V_{a;c} + V_aV^a_{\,\,\,;c} = V^a V_{a;c} +  V^a V_{a;c} = 2V^a V_{a;c} = 0\,, 
\end{align}
where the first equality is obtained by raising and lowering indices in the second term (noting that the covariant derivative of the metric is zero). It follows that
\begin{align}
V^a V_{a\,;c} =  0\,. \tag{2ed}
\end{align}
This equation implies another useful relation after substituting for the components of the unit vector in terms of the Killing vector:
\begin{align}
V^a V_{a\,;c} = f^{-1}K^a(f^{-1} K_ {a})_{\,;c} = - f^{-1}K^a(f^{-1} K_ {c;a} - f^{-2} f_{,c} K_a)= 0\,, 
\end{align}
hence, interchanging indices on the covariant derivative using Killing's equation (1ed), and recalling that $K^a K_a = -f^2$, we obtain
\begin{align}
K^a K_ {c_;a} = f f_{,c} \,.  \tag{3ed}
\end{align}
With these preliminaries out of the way, we find for the covariant derivative of $V^a$:
\begin{align}
V^a_{\,\,\,;b} = (f^{-1}K^a)_{;b} &= (f^{-1}g^{ac}K_c)_{;b}\,,\\ 
&= -f^{-2}f_{,b}g^{ac}K_c + f^{-1}g^{ac}K_{c{;b}}\,,\\
&= -f^{-1}f_{,b}V^a - f^{-1}g^{ac}K_{b{;c}}\,,\\
&= -f^{-1}f_{,b}V^a - f^{-1}g^{ac}(fV_b)_{;c}\,,\\
&= -f^{-1}f_{,b}V^a - f^{-1} g^{ac}\left[f_{,c}V_b + fV_{b\,;c}\right]\,,\\
&= -f^{-1}f_{,b}g^{ac}V_c -g^{ac}f^{-1}f_{,c}V_b - g^{ac}V_{b\,;c}\,,
\end{align}
where Killing's equation (1ed) was used to get the third equality.  According to Hawking and Ellis, $\dot{V}^a$ is defined by the contraction of the last equation with $V^b$:
\begin{align}
\dot{V}^a \equiv V^a_{\,\,\,;b}V^b &= -f^{-1}f_{,b}g^{ac}V_cV^b - g^{ac}f^{-1}f_{,c}V_bV^b + g^{ac}V_{b;c}V^b\,.
\end{align}
The third term vanishes according to equation (2ed), and we replace $V_bV^b = -1$ in the second term, reducing the last equation to
\begin{align}
\dot{V}^a = V^a_{\,\,\,;b}V^b = -f^{-1}f_{,b}V^bV^a + f^{-1}f_{,b}g^{ab}\,.\tag{4ed}
\end{align}
Contracting this result with $V_a$ yields $\dot{V}^a V_a =  V_aV^a_{\,\,\,;b}V^b = 0$, according to equation (2ed), hence
\begin{align}
0 & = \dot{V}^a V_a, \\ 
&= - f^{-1}f_{,b}V^bV^aV_a + g^{ab}f^{-1}f_{,b}V_a\,,\\
&= f^{-1}f_{,b}V^b + f^{-1}f_{,b}V^b \,,\\
&= 2f^{-1}f_{,b}V^b \,,
\end{align}
from which it follows that
\begin{align}
f_{,b}V^b = 0 \,. \tag{5ed}
\end{align}
The first term on the right-hand side of equation (4ed) then also vanishes, yielding
\begin{align}
\dot{V}^a = V^a_{\,\,\,;b}V^b = f^{-1}f_{,b}g^{ab}  \,, \tag{6ed}
\end{align}
which is one of the equations for $\dot{V}^a$ given by Hawking and Ellis.
As pointed out by SDNick in his answer, the next step I took of contracting this equation with $V_b$ to get the second form of $\dot{V}^a$ is utter nonsense, since $b$ is not a free index. He answers your question correctly and succinctly. I am here nearly two years later, noticing his answer and seeing my mistake. In correcting it I am adding nothing that hasn't appeared in his answer, although my presentation makes use of some results that were correct in my original answer.
Overlooked in my former discussion was the fact that Hawking and Ellis were considering a static metric, in which case the time-like Killing vector is also hypersurface orthogonal. Such a vector satisfies the additional condition
\begin{align}
K_{[a; b} K_{c]} = 0 \equiv K_{a;b} K_c + K_{b;c} K_a + K_{c;a} K_b - K_{b;a} K_c - K_{c;b} K_a - K_{a;c} K_b.
\end{align}
Using the Killing equation (1ed), we can replace the sum of negative terms by the sum of positive terms to obtain twice the sum of positive terms, hence
\begin{align}
K_{a;b} K_c + K_{b;c} K_a + K_{c;a} K_b = 0.
\end{align}
Contracting with $K^c$, and using equation (3ed):
\begin{align}
-f^2 K_{a;b} + K^c K_{b;c} K_a + K^c K_{c;a} K_b  = -f^2 K_{a;b} + f f_{,b} K_a - f f_{,a} K_b = 0,
\end{align}
from which
\begin{align}
K_{a;b}  = f^{-1} f_{,b} K_a - f^{-1} f_{,a} K_b\,.  \tag{7ed}
\end{align}
Raising the index on $K_{a;b}$, and contracting the resulting equation with $g^{ad}$ then gives
\begin{align}
K^d_{\,\,\,;b}  = f^{-1} f_{,b} K^d - f^{-1} f_{,a} g^{ad} K_b,
\end{align}
Replacing the Killing vector components in terms of the unit vector components, we obtain
\begin{align}
(f V^d)_{\,;b} \equiv f V^d_{\,\,\,;b} +  f_{,b} V^d  = f_{,b} V^d - f_{,a} g^{ad} V_b.
\end{align}
Two terms cancel in the last equation.  Then, multiplying the result thru by $f^{-1}$ and using equation (6ed), we finally obtain the Hawking and Ellis equation for the covariant derivative $V^d_{\,\,\,;b}$:
\begin{align}
V^d_{\,\,\,;b} = - f^{-1}f_{,a} g^{ad} V_b = -\dot{V}^d V_b. \tag{8ed}
\end{align}
