JEE PYQ: Differentiation Question 17
Question 17 - 2019 (09 Jan Shift 2)
If $x = 3\tan t$ and $y = 3\sec t$, then the value of $\frac{d^2y}{dx^2}$ at $t = \frac{\pi}{4}$ is:
(1) $\frac{1}{3\sqrt{2}}$ (2) $\frac{1}{6\sqrt{2}}$ (3) $\frac{3}{2\sqrt{2}}$ (4) $\frac{1}{6}$
Show Answer
Answer: (2) $\frac{1}{6\sqrt{2}}$
Solution
$\frac{dy}{dx} = \sin t$. $\frac{d^2y}{dx^2} = \frac{\cos^3 t}{3}$. At $t = \frac{\pi}{4}$: $\frac{1}{6\sqrt{2}}$.
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