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

$\frac{4\sqrt{2}b}{3a^2}$

The correct answer is Option (1) → $\frac{4\sqrt{2}b}{3a^2}$ ##

Given, $x = a \sin^3 \theta$ and $y = b \cos^3 \theta$

$\frac{dx}{d\theta} = 3a \sin^2 \theta \cos \theta$

$\frac{dy}{d\theta} = -3b \cos^2 \theta \sin \theta$

$∴\frac{dy}{dx} = \frac{-3b \cos^2 \theta \sin \theta}{3a \sin^2 \theta \cos \theta} = -\frac{b}{a} \cot \theta$

And, $\frac{d^2y}{dx^2} = \frac{d}{d\theta} \left( -\frac{b}{a} \cot \theta \right) \frac{d\theta}{dx}$

$= -\frac{b}{a} (-\text{cosec}^2 \theta) \cdot \frac{1}{3a \sin^2 \theta \cos \theta}$

$= \frac{b}{3a^2 \sin^4 \theta \cos \theta}$

Now, $\left. \frac{d^2y}{dx^2} \right|_{\theta = \frac{\pi}{4}} = \frac{b}{3a^2 \left( \frac{1}{\sqrt{2}} \right)^4 \left( \frac{1}{\sqrt{2}} \right)}$

$= \frac{b}{3a^2 \cdot \frac{1}{4} \cdot \frac{1}{\sqrt{2}}} = \frac{4b\sqrt{2}}{3a^2}$