If $x=3 \tan t$ and $y=3 \sec t$, then the value of $\frac{d^{2} y}{d x^{2}}$ at
$\mathrm{t}=\frac{\pi}{4}$, is:
Correct Option: 2,
$\because \quad x=3 \tan t \Rightarrow \frac{d x}{d t}=3 \sec ^{2} t$
and $y=3 \sec t \Rightarrow \frac{d y}{d t}=3 \sec t \cdot \tan t$
$\because \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t} \quad \therefore \quad \frac{d y}{d x}=\frac{\tan t}{\sec t}=\sin t$
$\therefore \quad \frac{d^{2} y}{d x^{2}}=\frac{d}{d t}(\sin t) \cdot \frac{d t}{d x}$
$=\cos t \cdot \frac{1}{3 \sec ^{2} t}$
$\therefore \quad \frac{d^{2} y}{d x^{2}}\left(\right.$ at $\left.t=\frac{\pi}{4}\right)=\frac{1}{3} \cdot\left(\frac{1}{\sqrt{2}}\right)^{3}$
$=\frac{1}{6 \sqrt{2}}$
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