Problem Statement
If $x = a(\cos\theta + \theta \sin\theta)$ and $y = a(\sin\theta - \theta \cos\theta)$, find $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{4}$.
Verified Solution & Marking Scheme
Differentiate x and y with Respect to θ
$\frac{dx}{d\theta} = a(-\sin\theta + \sin\theta + \theta \cos\theta) = a\theta \cos\theta$
$\frac{dy}{d\theta} = a(\cos\theta - (\cos\theta - \theta \sin\theta)) = a\theta \sin\theta$
Compute First Derivative dy/dx
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\theta \sin\theta}{a\theta \cos\theta} = \tan\theta$
Compute Second Derivative via Chain Rule
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan\theta) = \frac{d}{d\theta}(\tan\theta) \cdot \frac{d\theta}{dx} = \sec^2\theta \cdot \frac{1}{a\theta \cos\theta} = \frac{\sec^3\theta}{a\theta}$
At $\theta = \frac{\pi}{4}$:
$\sec\left(\frac{\pi}{4}\right) = \sqrt{2} \implies \sec^3\left(\frac{\pi}{4}\right) = (\sqrt{2})^3 = 2\sqrt{2}$
$\frac{d^2y}{dx^2} = \frac{2\sqrt{2}}{a(\pi/4)} = \frac{8\sqrt{2}}{\pi a}$