Explanation

(a)
Let $y = a\sin^{3}{t\ and\ x = a\cos^{3}{t,}then}$
On differentiating w.r.t. $t$, we get
$\frac{dy}{dt} = 3a\sin^{2}{t\cos t}$
and $\frac{dx}{dt} = 3a\cos^{2}{t( - \sin{t)}}$
$\therefore\frac{dy}{dx} = \frac{\left( \frac{dy}{dt} \right)}{\left( \frac{dx}{dt} \right)} = \frac{3a\sin^{2}{t\cos t}}{3a\cos^{2}{t( - \sin{t)}}} = - \tan t$
Again differentiating w.r.t. $x$, we get
$\frac{d^{2}y}{dx^{2}} = - \sec^{2}t\frac{dt}{dx} = \frac{- \sec^{2}t}{3a\cos^{2}{t( - \sin t})}$
$= \frac{1}{3a}\left( \frac{\sec^{4}t}{\sin t} \right)$
$\therefore\left( \frac{d^{2}y}{dx^{2}} \right)_{t = \frac{\pi}{4}} = \frac{1}{3a}.\frac{4}{\frac{1}{\sqrt{2}}} = \frac{4\sqrt{2}}{3a}$