Question

If $x=a{\cos }^3\theta ,y=b{\sin }^3\theta$, then the value of $\frac{d^{2}y}{d{x}^2}$ at $\theta =\frac{\pi }{6}$

• A. $\frac{32b}{27a^2}$
• B. $\frac{27b}{32a^2}$
• C. $\frac{32a^2}{27b}$
• D. $\frac{24a^2}{27b}$

Answer 1

Answer: (A)

$\frac{32b}{27a^2}$

Given, $x=a{\cos }^3\theta ,y=b{\sin }^3\theta$

Therefore, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3b{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }$
$\frac{dy}{dx}=\frac{-b}{a}\tan \theta$
Thus $\frac{d^{2}y}{d{x}^2}=\frac{d}{d\theta }\left(\frac{dy}{dx}\right)\frac{d\theta }{dx}$

$=\left(\frac{-b}{a}{\sec }^2\theta \right)\left(\frac{-1}{3a{\cos }^2\theta \sin \theta }\right)$
$=\frac{2b}{3a^{2(\frac{3}{4}\times \frac{3}{4})}}$

$=\frac{32b}{27a^2}$