If $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t),$ find $\frac{d^{2} y}{d x^{2}}.$

(D) Given that,$x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t).$
First,we find $\frac{dx}{dt}$:
$\frac{dx}{dt} = a \left[ \frac{d}{dt}(\cos t) + \frac{d}{dt}(t \sin t) \right] = a [-\sin t + \sin t + t \cos t] = at \cos t.$
Next,we find $\frac{dy}{dt}$:
$\frac{dy}{dt} = a \left[ \frac{d}{dt}(\sin t) - \frac{d}{dt}(t \cos t) \right] = a [\cos t - (\cos t - t \sin t)] = at \sin t.$
Now,calculate $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at \sin t}{at \cos t} = \tan t.$
Finally,find the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan t) = \sec^2 t \cdot \frac{dt}{dx}.$
Since $\frac{dx}{dt} = at \cos t,$ we have $\frac{dt}{dx} = \frac{1}{at \cos t}.$
Therefore,$\frac{d^2y}{dx^2} = \sec^2 t \cdot \frac{1}{at \cos t} = \frac{\sec^3 t}{at}.$