Question

if $x=\cos \theta ,y={\sin }^3\theta ,then{\left(\frac{dy}{dx}\right)}^2+y\frac{d^{2}y}{d{x}^2}$ at $\theta =\frac{\pi }{2}$ is

• A. $1$
• B. $2$
• C. $-2$
• D. $-3$

Answer 1

The correct option is D $-3$

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

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

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

$=-3\sin \theta \cos \theta$

$\therefore \frac{dy}{dx}=-3\sin \theta \cos \theta$

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

$=\frac{-3}{2}\times \sin 2\theta$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{-3}{2}\times 2\times \cos 2\theta \times \frac{d\theta }{dx}=-3\cos 2\theta \times \frac{-1}{\sin \theta }$

$\therefore {\left(\frac{dy}{dx}\right)}^2+y\frac{d^{2}y}{d{x}^2}$

$=\frac{9}{4}{\sin }^2\theta +{\sin }^3\theta \times -3\cos 2\theta \times \frac{-1}{\sin \theta }$

at $\theta =\frac{\pi }{2}$

$=\frac{9}{4}{\sin }^2(2\times \frac{\pi }{2})+{\sin }^3\frac{\pi }{2}-3\times \cos 2\times \frac{\pi }{2}\times \frac{-1}{\sin \frac{\pi }{2}}$

$=\frac{9}{4}{\sin }^2\pi +1\times -3\cos \pi \times -1$

$=0+1\times -3\cos \pi \times -1$

$=3\cos \pi =3\times -1=-3$ $\to$ option D