Consider
$\begin{cases} x'=v\cos\theta\\ y'=v\sin\theta\\ \theta'=v\frac{\tan\psi}{L}\\ v'=a-\gamma v \end{cases}$
where $\gamma=1/6,\, L=11.5,\, a=100,\, \psi=1$.
Let $\vec{y}(t)=(x(t),y(t),\theta(t),v(t))$ and $\vec{p}(t)=(\psi,L,a,\gamma)$. So system can be restated as $\vec{y}'=\vec{f}(t,\vec{y},\vec{p}),\, \vec{y}(0)=\vec{c}$.
Since we are only perturbing one coefficient, $\psi$, we are only interested in first column of perturbation matrix $P=\frac{\partial\vec{y}}{\partial\vec{p}}$, that is, vector $P=(\frac{\partial x}{\partial\psi},\frac{\partial y}{\partial\psi},\frac{\partial\theta}{\partial\psi},\frac{\partial v}{\partial\psi})$.
Derive system of ODE's for perturbation vector P:
$\begin{equation*}\begin{cases} P_1'=-vP_3\sin\theta+P_4\cos\theta \\ P_2'=vP_3\cos\theta+P_4\sin\theta \\ P_3'=[P_4\tan\psi+\frac{v}{\cos^2\psi}]/L \\ P_4'=-\gamma P_4 \end{cases}\end{equation*}$
This is how far I've gotten, I'm a bit rusty but if you can provide idea for how to get over the hump I'll much appreciate it.
I start with $v'=a-\gamma v$ which I solve to get $v(t)=\frac{a}{\gamma}+Ce^{-\gamma t}$ From this you can obtain $\frac{\partial v}{\partial\psi}=-\gamma e^{-\gamma t}$ so we get $P_4'=-\gamma P_4$ where $P_4=e^{-\gamma t}$ if we take $C=1$ for simplicity.
From here, I try to solve for $\theta$ but run into a wall, $\theta'=v\frac{\tan\psi}{L}=\frac{a}{\gamma}\frac{\tan\psi}{L}+\frac{\tan\psi}{L}e^{-\gamma t}\implies \theta(t)=\frac{a}{\gamma}\frac{\tan\psi}{L}t-\frac{\tan\psi}{\gamma L}e^{-\gamma t}$ but $\frac{\partial\theta}{\partial\psi}=[\frac{a}{\gamma}\frac{1}{\cos^2\psi}t-\frac{1}{\gamma\cos^2\psi}e^{-\gamma t}]/L$ which is not of the form $P_3'=[P_4\tan\psi+\frac{v}{\cos^2\psi}]/L$
Update: I was able to deduce
$\begin{equation*} \begin{cases} \vec{x}'=v\cos\theta\\ \vec{y}'=v\sin\theta\\\theta'=v\frac{\tan\theta}{L}\\ v'=a-\gamma v \end{cases} \end{equation*}$\implies $\begin{equation*} \begin{cases} \frac{d}{d\psi}\vec{x}'=-v\frac{\partial\theta}{\partial\psi}\sin\theta+\frac{\partial v}{\partial\psi}\cos\theta\\ \frac{d}{d\psi}\vec{y}'=v\frac{\partial\theta}{\partial\psi}\cos\theta+\frac{\partial v}{\partial\psi}\sin\theta\\\frac{d}{d\psi}\theta'=\frac{[\frac{\partial v}{\partial\psi}\tan\psi+\frac{v}{\cos^2\psi}]}{L}\\ \frac{d}{d\psi}v'=-\gamma\frac{\partial v}{\partial\psi} \end{cases} \end{equation*}$\implies $P_4=\frac{dv}{d\psi}$ and $P_3=\frac{d\theta}{d\psi}$