Question

Let $x\left(t\right)=2\sqrt{2}\cos t\sqrt{\sin 2t}$ and $y\left(t\right)=2\sqrt{2}\sin t\sqrt{\sin 2t},t\in (0,\frac{\pi }{2}).$ Then $\frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}$ at $t=\frac{\pi }{4}$ is equal to

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

Answer 1

The correct option is B $\frac{-2}{3}$

$x=2\sqrt{2}\cos t\sqrt{\sin 2t},y=2\sqrt{2}\sin t\sqrt{\sin 2t}$
$\therefore \frac{dx}{dt}=\frac{2\sqrt{2}\cos 3t}{\sqrt{\sin 2t}},\frac{dy}{dt}=\frac{2\sqrt{2}\sin 3t}{\sqrt{\sin 2t}}$
$\therefore \frac{dy}{dx}=\tan 3t,(\text{at}t=\frac{\pi }{4},\frac{dy}{dx}=-1)$
and $\frac{d^{2}y}{d{x}^2}=3{\sec }^{2}3t.\frac{dt}{dx}=\frac{3{\sec }^{2}3t.\sqrt{\sin 2t}}{2\sqrt{2}\cos 3t}$
$(\text{At}t=\frac{\pi }{4},\frac{d^{2}y}{d{x}^2}=-3)$
$\therefore \frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}=\frac{2}{-3}=\frac{-2}{3}$