Limit Continuity And Differentiability Ques 64

Answer:

Correct Answer: 64.(b)

Solution: (b) We have, $x=3 \tan t$ and $y=3 \sec t$

Clearly, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{d}{d t}(3 \sec t)}{\frac{d}{d t}(3 \tan t)}$

$ \begin{aligned} & =\frac{3 \sec t \tan t}{3 \sec ^2 t}=\frac{\tan t}{\sec t}=\sin t \\ & \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d t}\left(\frac{d y}{d x}\right) \cdot \frac{d t}{d x} \\ & =\frac{\frac{d}{d t}\left(\frac{d y}{d x}\right)}{\frac{d x}{d t}}=\frac{\frac{d}{d t}(\sin t)}{\frac{d}{d t}(3 \tan t)}=\frac{\cos t}{3 \sec ^2 t}=\frac{\cos ^3 t}{3} \end{aligned} $

Now, $\frac{d^2 y}{d x^2}\left(\right.$ at $\left.t=\frac{\pi}{4}\right)=\frac{\cos ^3 \frac{\pi}{4}}{3}=\frac{1}{3(2 \sqrt{2})}=\frac{1}{6 \sqrt{2}}$