Question:

<b>Parametric Second Derivative [JEE Main 2019]</b><br>$\frac{dx}{dt} = 3\sec^2 t$, $\frac{dy}{dt} = 3\sec t\tan t$<br>$\frac{dy}{dx} = \frac{3\sec t\tan t}{3\sec^2 t} = \frac{\sin t}{1} = \sin t$<br>Wait: $\frac{dy}{dx} = \frac{\tan t}{\sec t} = \sin t$.<br>$\frac{d^2y}{dx^2} = \frac{d(\sin t)/dt}{dx/dt} = \frac{\cos t}{3\sec^2 t} = \frac{\cos^3 t}{3}$<br>At $t=\pi/4$: $\cos(\pi/4) = 1/\sqrt{2}$<br>$\frac{d^2y}{dx^2} = \frac{(1/\sqrt{2})^3}{3} = \frac{1}{3\cdot 2\sqrt{2}} = \frac{1}{6\sqrt{2}}$<br><b>Answer: $\dfrac{1}{6\sqrt{2}}$</b><br><b>Key concept:</b> Parametric second derivative formula; $dy/dx = \sin t$ simplifies everything.<br><b>Trap:</b> Computing $d^2y/dt^2 \div d^2x/dt^2$ instead of using the correct formula.