Question

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

• A. 2
• B. $-\frac{1}{2}$
• C. $-\frac{3}{4}$
• D. $-\frac{3}{2}$

Answer 1

Answer: (B)

$-\frac{1}{2}$

$x=2\sin t+\sin 2t$

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

$y=2\cos t-\cos 2t$

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

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

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

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

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

$=\frac{2\sin \left(\frac{2t-t}{2}\right)\cos \left(\frac{2t+t}{2}\right)}{2\cos \left(\frac{2t-t}{2}\right)\cos \left(\frac{2t+t}{2}\right)}$

$=\frac{2\sin \left(\frac{t}{2}\right)\cos \left(\frac{3t}{2}\right)}{2\cos \left(\frac{t}{2}\right)\cos \left(\frac{3t}{2}\right)}$

$=\frac{\sin \left(\frac{t}{2}\right)}{\cos \left(\frac{t}{2}\right)}$

$\therefore \frac{dy}{dx}=\tan \left(\frac{t}{2}\right)$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{2}{\sec }^2\left(\frac{t}{2}\right)\frac{dt}{dx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{2}{\sec }^2\left(\frac{t}{2}\right)\frac{1}{2\cos t+2\cos 2t}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{4}{\sec }^2\left(\frac{t}{2}\right)\frac{1}{\cos t+\cos 2t}$

$\Rightarrow {\left[\frac{d^{2}y}{d{x}^2}\right]}_{t=\frac{\pi }{2}}=\frac{1}{4}{\sec }^2\left(\frac{\frac{\pi }{2}}{2}\right)\frac{1}{\cos \frac{\pi }{2}+\cos 2\times \frac{\pi }{2}}$

$=\frac{1}{4}{\sec }^2\frac{\pi }{4}\frac{1}{0+\cos \pi }$

$=\frac{1}{4}\times 2\times -1$

$=-\frac{1}{2}$