# If x=$acos^3\theta$ and y=$asin^3\theta,$then find the value of $\Large \frac{d^2y}{dx^2}$ at $\theta=\Large \frac{\pi}{6}$.

(Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

Given $x=a \cos^3\theta$ and $y = a \sin^3 \theta$
$\textbf{Step 1: Calculate} \;$ $\mathbf{\large \frac{dy}{dx}}$:
$\Rightarrow$ $x=a \cos^3\theta \rightarrow \large \frac{dx}{d\theta}\normalsize $$=3a\;\cos^2\theta(-sin\theta) \Rightarrow Similarly y=asin^3\theta \rightarrow \large \frac{dy}{d\theta}\normalsize$$=3a\;\sin^2\theta(cos\theta)$
$\Rightarrow \large \frac{dy}{dx}=\frac{dy}{d\theta}.\frac{d\theta}{dx}=\frac{3a sin^2\theta.cos\theta}{-3a cos^2\theta sin\theta}$
$\Rightarrow \large \frac{dy}{dx}$$=-\tan\theta \textbf{Step 2: Calculate} \; \mathbf{\large \frac{d^2y}{dx^2}}: \large \frac{d^2y}{dx^2}=\frac{d}{d\theta}\big(\frac{dy}{dx}\big)\frac{d\theta}{dx} \Rightarrow \large \frac{d^2y}{dx^2}=\frac{d}{d\theta}$$(-\tan\theta)\times -3a \sin\theta \cos^2\theta$
$\Rightarrow \large \frac{d^2y}{dx^2}=\large \frac{sec^2\theta}{3asin\theta cos^2\theta}$
Given that $\theta=\frac{\pi}{6}$,
$\Rightarrow \large \frac{d^2y}{dx^2}=\frac{sec^2\frac{\pi}{6}}{3asin\frac{\pi}{6} cos^2\frac{\pi}{6}}=\frac{4/3}{3a(1/2)(3/4)}$
$\Rightarrow \large \frac{d^2y}{dx^2}=\frac{32}{27a}$
edited Mar 22, 2013