Question

If $x=(\cos 2\theta -2\cos \theta );y=(\sin \theta -2\sin 2\theta )$, find ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{\theta =\frac{\pi }{2}}$.

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $-\frac{1}{4}$

Answer 1

Answer: (D)

$-\frac{1}{4}$

Here,

$x=\cos 2\theta -2\cos \theta$

$y=\sin \theta -2\sin 2\theta$

Calculate $\frac{d^{2}x}{d{\theta }^2}$.

$\frac{dx}{d\theta }=-2\sin 2\theta +2\sin \theta$

$\frac{d^{2}x}{d{\theta }^2}=-4\cos 2\theta +2\cos \theta$

Calculate $\frac{d^{2}y}{d{\theta }^2}$.

$\frac{dy}{d\theta }=\cos \theta -4\cos 2\theta$

$\frac{d^{2}y}{d{\theta }^2}=-\sin \theta +8\sin 2\theta$

Therefore,

${\left(\frac{d^{2}y}{d{x}^2}\right)}_{\theta =\frac{\pi }{2}}$

$=\frac{-\sin \theta +8\sin 2\theta }{-4\cos 2\theta +2\cos \theta }$

$=\frac{-1+0}{(-4\times -1)+0}$

$=-\frac{1}{4}$

Hence, the required value is $-\frac{1}{4}$.