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

Here, $\cfrac{{dx}}{{dt}} = a( - \sin t + t\cos t + \sin t)$
therefore, $\cfrac{{dx}}{{dt}} = at\cos t$

$\Rightarrow$ $\cfrac{{dy}}{{dt}} = a[\cos t - \{ t( - \sin t) + \cos t(1)\} ]$

$\Rightarrow$ $\cfrac{{dy}}{{dt}} = a(\cos t + t\sin t - \cos t) = at\sin t$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{{dy}}{{dt}} \times \cfrac{{dt}}{{dx}} = at\sin t \times \cfrac{1}{{at\cos t}} = \tan t$

$\Rightarrow$ $\cfrac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}t \cdot \cfrac{{dt}}{{dx}}$

$\Rightarrow$ $\cfrac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}t \cdot \cfrac{1}{{at\cos t}} = \cfrac{{{{\sec }^2}t}}{{at\cos t}} = \cfrac{{{{\sec }^3}t}}{{at}},0 < t < \cfrac{\pi }{2}$