Question

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

• A. $ 3/2 $
• B. $ 5/2 $
• C. $ 4/2 $
• D. $ -3/2 $

Answer 1

Answer: (D)

$\frac{d x}{d t}=-2 \sin t+2 \sin 2 t$
$\frac{d y}{d t} ={2} \cos t-2 \cos 2 t$
$\therefore \frac{d y}{d x} =\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t}$
$=\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t}$
$=\frac{2 \sin \frac{3 t}{2} \cdot \sin \frac{t}{2}}{2 \cos \frac{3 t}{2} \cdot \sin \frac{t}{2}}$
$=\tan \frac{3 t}{2}$
$\frac{d^{2} y}{d x^{2}}=\sec ^{2} \frac{3 t}{2} \cdot \frac{3}{2} \cdot \frac{d t}{d x}$
$=\frac{3}{2} \sec ^{2} \frac{3 t}{2} \frac{1}{(2 \sin 2 t-2 \sin t)}$
$\Rightarrow \left.\frac{d^{2} y}{d x^{2}}\right|_{t=\pi / 2}=-3 / 2$