Question

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

Answer 1

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

$\frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta$

$\frac{dy}{d\theta }=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }$

$=-\tan \theta$

$\frac{d^{2}y}{d{x}^2}=-{\sec }^2\theta -\frac{d\theta }{dx}$

$=-{\sec }^2\theta .\frac{1}{-3a^2{\cos }^2\theta \sin \theta }$

$=\frac{1}{3a^2{\cos }^4\theta \sin \theta }$

$\frac{d^{2}y}{d{x}^2}=\frac{1}{3a^2{\cos }^4\frac{\pi }{6}\sin \frac{\pi }{6}}$

At $\theta =\frac{\pi }{6}$

$=\frac{1}{3a^2\times }{\left(\frac{\sqrt{3}}{2}\right)}^4\times \frac{1}{2}$

$=\frac{1}{3a^2\times \frac{9}{32}}$

$=\frac{32}{27a}$