Tangled with differentials

$\begin{cases} x = t \cos t \\ y = t + \sin t \end{cases}$

$\implies \begin{cases} \dfrac {dx}{dt} = \cos t - t\sin t = \dfrac xt - t(y-t) \\ \dfrac {dy}{dt} = 1 + \cos t = 1 + \dfrac xt \end{cases}$

$\begin{aligned} \frac {dx}{dy} & = \frac {dx}{dt} \times \frac {dt}{dy} = \frac {x-t^2(y-t)}{t+x} \\ \frac {d^2x}{dy^2} & = \frac {\left(\frac {dx}{dy} -2t \frac {dt}{dy}(y-t)-t^2\left(1-\frac {dt}{dy}\right)\right)(t+x) -(x-t^2(y-t))\left(\frac {dt}{dy}+\frac {dx}{dy}\right)}{(t+x)^2} & \small \color{#3D99F6} \text{See note.} \\ \frac {d^2x}{dy^2} \bigg|_{t = \frac \pi 2} & = \frac {\left(-\frac \pi 2 -2 \cdot \frac \pi 2 \left(\frac \pi 2 + 1-\frac \pi 2 \right)-\frac {\pi^2} 4 \left(1-1\right)\right) \left(\frac \pi 2+0\right) - \left(0-\frac {\pi^2}4 \left(\frac \pi 2+1-\frac \pi 2\right)\right)\left(1-\frac \pi 2\right)}{\left(\frac \pi 2+0\right)^2} \\ & = \frac {- \frac {3\pi^2}4 + \frac {\pi^2}4 - \frac {\pi^3}8}{\frac {\pi^2}4} \\ & = - \frac {\pi + 4}2 \end{aligned}$

$\implies a+b+c = 1+4+2 = \boxed{7}$

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Note: When $t = \dfrac \pi 2$ , $x = 0$ , $y= \dfrac \pi 2 + 1$ , $\dfrac {dx}{dt} = - \dfrac \pi 2$ , $\dfrac {dt}{dy} = 1$ and $\dfrac {dx}{dy} = - \dfrac \pi 2$ .

$x=t\cos t,~~~~~~~~y=t+\sin t.\\ \implies~~\dfrac{dx}{dt}=\cos t-t\sin t~~~~~~~~\dfrac{dy}{dt}=1+\cos t.\\ \therefore~~\dfrac{dx}{dy}=\dfrac{\cos t - t\sin t}{1+\cos t}=1~-~\dfrac{1 + t\sin t}{1+\cos t}\\ So~~\dfrac{d^2x}{d^2y}= \dfrac{ \dfrac{ d \frac{dx } {dy}}{dt} }{ \dfrac{dy}{dt}} = \dfrac { \dfrac{d \Big (1~-~ \frac{1 + t\sin t } {1+\cos t} \Big ) } {dt} } { 1+\cos t } \\ = \dfrac{ - \dfrac { ( \sin t+t \cos t )( 1+\cos t) +(1 + t \sin t)(\sin t) } {(1+\cos t )^2} } { 1+\cos t} \\ So~~\dfrac{d^2x}{d^2y}_{t=\frac12\pi} = - \dfrac {(1+\frac 12 \pi*0)(1+0) +(1+\frac 12 \pi*1)(1) } {(1+0 )^3} \\ =(-1)^1*(2+\frac12\pi)=(-1)^1*\dfrac{4+\pi} 2=(-1)^a*\dfrac{b+\pi} c. \\ a+b+c=1+4+2=\Large~~~\color{#D61F06}{7}.$