I'd like to understand the Taylor expansion (second order) of the function $L(q,\dot{q}) = \frac{1}{2} \dot{q} \cdot A(q) \dot{q}-V(q,\dot{q})$, where $A$ is the mass matrix and $V$ is of the form $V(q,\dot{q}) = V_0(q) + a(q)\cdot \dot{q}$. Here $q,\dot{q}$ represent the lagrangian coordinates and velocity.
Let $q_0$ be a configuration of equilibrium, i.e $\frac{\partial V_0}{\partial q}(q_0) = 0$.
It should be true that $$L_0(q,\dot{q}) = \frac{1}{2} \dot{q} \cdot A(q_0) \dot{q} - V_0(q_0) - \frac{1}{2}(q-q_0)\cdot V''(q_0)(q-q_0)- [a(q_0)+\frac{\partial a}{\partial q}(q_0)(q-q_0)]\cdot \dot{q}$$
Is the Taylor expansion until the second order of the function $L$ cited before.
What I don't get are the $\dot{q}$ multiplyng $A(q_0)$, since Taylor in two variables I remember, should be $f(x,y) = f(a,b) +\frac{\partial f}{\partial x}(a,b)(x-a) + \frac{\partial f}{\partial y}(a,b)(y-b)$. Is this really a Taylor exapansion in two variables around $(q_0,0)$ ? Why $\dot{q}$ does even appear ? It seems more like a Taylor expansion around $q_0$ keeping $\dot{q}$ variable, but I don't if this is possible.
Any help would be appreciated.
Take: $$f(x,y) := \frac{1}{2}x^{T}A(y)x - V(x,y) = \frac{1}{2}x^{T}A(y) - V_{0}(y) -a(y)\cdot x$$ Then, taking $(x,y) = (\dot{q},q_{0})$, the first term becomes: $$f(\dot{q},q_{0}) = \frac{1}{2}\dot{q}^{T}A(q_{0})\dot{q} -V_{0}(q_{0})-a(q_{0})\cdot \dot{q}.$$