Question

If $x=a{\cos }^3\theta ,y=a{\sin }^3\theta$, then find $\frac{d^{2}y}{d{x}^2}$ at $\theta =\frac{\pi }{4}$.

Answer 1

Given

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

Diff. w.r.t $x$ on both

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

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

$\because \frac{dx}{dy}=\frac{dy}{d\theta }/\frac{dx}{d\theta }$

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

$\therefore \frac{dy}{dx}=-\tan \theta$

Again Diff. w.r.t $x$ on both

$\frac{d}{d\theta }\left(\frac{dy}{dx}\right)=-{\sec }^2\theta$

$\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)\frac{dx}{d\theta }=se{c}^2\theta$

$\Rightarrow \left(\frac{d^{2}y}{d{x}^2}\right)(-3a{\cos }^2\theta )\sin =\theta -{\sec }^2\theta$

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

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-{\sec }^2\theta }{-3a{\cos }^2\theta \sin \theta }$

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

$\therefore {\left(\frac{d^{2}y}{d{x}^2}\right)}_{\theta =\pi /4}=\frac{(\sqrt{2}{)}^2}{3a{\left(\frac{1}{\sqrt{2}}\right)}^2\left(\frac{1}{\sqrt{2}}\right)}$

$\therefore {\left(\frac{d^{2}y}{d{x}^2}\right)}_{\theta =\pi /4}=\frac{2}{3a\times \frac{1}{2}\times \frac{1}{\sqrt{2}}}=\frac{4\sqrt{2}}{3a}$