Question

If $x=2\cos t-\cos 2t,y=2\sin t-\sin 2t$, then the value of ${\frac{d^{2}y}{d{x}^2}|}_{t=\frac{\pi }{2}}$ is

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. $\frac{-5}{2}$
• D. $\frac{-3}{2}$

Answer 1

Answer: (D)

$\frac{-3}{2}$

$\frac{dx}{dt}=-2\sin t+2\sin 2t$
$\frac{dy}{dt}=2\cos t-2\cos 2t$

$\therefore \frac{dy}{dx}=\frac{2\cos t-2\cos 2t}{-2\sin t+2\sin 2t}$
$=\frac{\cos t-\cos 2t}{\sin 2t-\sin t}$
$=\frac{2\sin \frac{3t}{2}\cdot \sin \frac{t}{2}}{2\cos \frac{3t}{2}\cdot \sin \frac{t}{2}}$
$=\tan \frac{3t}{2}$
$\frac{d^{2}y}{d{x}^2}={\sec }^2\frac{3t}{2}\cdot \frac{3}{2}\cdot \frac{dt}{dx}$
$=\frac{3}{2}{\sec }^2\frac{3t}{2}\frac{1}{(2\sin 2t-2\sin t)}$

$\Rightarrow {\frac{d^{2}y}{d{x}^2}|}_{t=\pi /2}=\frac{-3}{2}$