Question

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

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

Answer 1

The Correct Option is D

$\frac{dx}{dt} = 3\sec^{2} t$
$ \frac{dy}{dt} = 3 \sec t \tan t $
$ \frac{dy}{dx} = \frac{\tan t}{\sec t} = \sin t $
$ \frac{d^{2}y}{dx^{2}} =\cos t \frac{dt}{dx} $
$= \frac{\cos t}{3 \sec^{2} t} = \frac{\cos^{3}t}{3} = \frac{1}{3.2\sqrt{2}} = \frac{1}{6\sqrt{2}} $