If x = 2 cos t – cos 2t and y = 2 sin t – sin 2t then the value of $\frac{d^2y}{dx^2}$ at $= \frac{\pi}{2}$ is :
Answer & explanation
Correct answer: option 4
$\frac{d x}{d t}=-2 \sin t+2 \sin 2 t$
$\frac{d y}{d t}=2 \cos t-2 \cos 2 t$
$\frac{d y}{d t}=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t}=\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t}$
$=\frac{2 \sin \frac{3 t}{2} \sin \frac{t}{2}}{2 \cos \frac{3 t}{2} \sin \frac{t}{2}}=\tan \frac{3 t}{2}$
$\frac{d^2 y}{d x^2}=\sec ^2 \frac{3 t}{2} \times \frac{3}{2} \times \frac{d t}{d x}$
$=\frac{3}{2} \sec ^2 \frac{3 t}{2} . \frac{1}{2 \sin 2 t-2 \sin t}$
$\left.\Rightarrow \frac{d^2 y}{d x^2}\right|_{t=\frac{\pi}{2}}=-\frac{3}{2}$
Hence (4) is correct answer.