If $x = a \sec^3 \theta, y = a \tan^3 \theta$, then find $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{4}$.

$\frac{1}{12a}$

The correct answer is Option (1) → $\frac{1}{12a}$ ##

$x = a \sec^3 \theta$

Differentiating with respect to $\theta$:

$\frac{dx}{d\theta} = 3a \sec^2 \theta \sec \theta \tan \theta = 3a \sec^3 \theta \tan \theta$

$y = a \tan^3 \theta$

Differentiating with respect to $\theta$:

$\frac{dy}{d\theta} = 3a \tan^2 \theta \sec^2 \theta$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a \tan^2 \theta \sec^2 \theta}{3a \sec^3 \theta \tan \theta} = \frac{\tan \theta}{\sec \theta} = \sin \theta \quad \dots (i)$

Differentiating equation (i) with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{d\theta}(\sin \theta) \frac{d\theta}{dx}$

$\frac{d^2y}{dx^2} = \frac{\cos \theta}{3a \sec^3 \theta \tan \theta} = \frac{\cos \theta \cdot \cos^3 \theta \cdot \cos \theta}{3a \sin \theta} = \frac{\cos^5 \theta}{3a \sin \theta}$

At $\theta = \frac{\pi}{4}$:

$\left( \frac{d^2y}{dx^2} \right)_{\theta = \pi/4} = \frac{(\frac{1}{\sqrt{2}})^5}{3a (\frac{1}{\sqrt{2}})} = \frac{1}{(\sqrt{2})^4 3a} = \frac{1}{4 \cdot 3a} = \frac{1}{12a}$