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 $-3$

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

So that $\frac{dy}{dx}=-3\sin \theta \cos \theta =-\frac{3}{2}\sin 2\theta .$

Differentiating again, we have

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

Now, ${\left(\frac{dy}{dx}\right)}^2+y\frac{d^{2}y}{{dx}^2}=9{\sin }^2\theta {\cos }^2\theta +{\sin }^3\theta \frac{3\cos 2\theta }{\sin \theta }$

$=9{\sin }^2\theta {\cos }^2\theta +3{\sin }^2\theta \cos 2\theta$

Substitute $\theta =\frac{\pi }{2};$ we get ${\left(\frac{dy}{dx}\right)}^2+y\frac{d^{2}y}{{dx}^2}=-3$