Question

Let $x(t)=2\sqrt{2}\cos t\sqrt{\sin 2t}$ and $y(t)=2\sqrt{2}\sin t\sqrt{\sin 2t},t\in \left(0,\frac{\pi }{2}\right)$. 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. $\frac{-2\sqrt{2}}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{-2}{3}$

Answer 1

Answer: (D)

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