Question

If $x=3\tan t$ and $y=3\sec t$, then the value of $\frac{d^{2}y}{d{x}^2}\text{at}t=\frac{\pi }{4}$, is :

• A. $\frac{1}{6}$
• B. $\frac{1}{6\sqrt{2}}$
• C. $\frac{1}{3\sqrt{2}}$
• D. $\frac{3}{2\sqrt{2}}$

Answer 1

The correct option is B $\frac{1}{6\sqrt{2}}$

$x=3\tan t,y=3\sec t$
$\frac{dx}{dt}=3{\sec }^{2}t,\frac{dy}{dt}=3\sec t\tan t$

$\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{3\sec t\tan t}{3{\sec }^{2}t}$

$\Rightarrow \frac{dy}{dx}=\sin t$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\cos t\cdot \frac{dt}{dx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\cos t\cdot \frac{1}{3{\sec }^{2}t}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{{\cos }^{3}t}{3}$

${\frac{d^{2}y}{d{x}^2}|}_{t=\pi /4}=\frac{1}{6\sqrt{2}}$