basic settings
The point is to calculate $A_i(t)$ from the other data, in fact, we care only about $\frac{d}{dt}\big|_0A_i(t)$:
$$ \frac{d}{dt}\big|_0A_i=-\cos\theta_i\frac{d}{dt}\big|0a_i-\cos(\pi-\theta{i+1})\frac{d}{dt}\big|0b{i+1}. $$
Facts from hyperbolic geometry:
$$ \cos\gamma=\frac{\cosh A\cosh B-\cosh C}{\sinh A\sinh B}. $$
$$ \frac{\sinh A}{\sin \alpha}=\frac{\sinh B}{\sin \beta}=\frac{\sinh C}{\sin \gamma}. $$
$$ \cosh l=\cosh d\cosh b_{i+1}-\cos(\pi-\theta_{i+1})\sinh d\sinh b_{i+1}. $$
$$ \cos(\theta_i-\alpha)=\frac{\cosh l\cosh a_i-\cosh A_i}{\sinh l\sinh a_i}. $$
$$ \sin\alpha=\sinh b_{i+1}\frac{\sin(\pi-\theta_{i+1})}{\sinh l}. $$
$$ \cos\alpha=\frac{\cosh l\cosh d-\cosh b_{i+1}}{\sinh l\sinh d}. $$
Thus we solve
$$ \cosh A_i=\cosh a_i\cosh l-\sinh a_i\left(\cos\theta_i\frac{\cosh l\cosh d-\cosh b_{i+1}}{\sinh d}+\sin\theta_i\sin\theta_i\sin(\pi-\theta_{i+1})\sinh b_{i+1}\right). $$
Evaluating at $t=0$, we have $(\cosh l)(0)=\cosh d$, $(\cosh a_i)(0)=(\cosh b_{i+1})(0)=1$, and
$(\cosh l)'(0)=-\cos(\pi-\theta_{i+1})\sinh d\cdot b_{i+1}'(0)$, $(\sinh a_i)'(0)=a_i'(0)$. So
$$ A_i'(0)=\frac{(\cosh A_i)'(0)}{\sinh d}=-\cos(\pi-\theta_{i+1})\cdot b_{i+1}'(0)-\cos\theta_i\cdot a_i'(0). $$