Problem Statement
If $x = a(\theta + \sin\theta)$ and $y = a(1 - \cos\theta)$, find $\frac{d^2y}{dx^2}$ at $\theta = \frac{\pi}{2}$.
Verified Solution & Marking Scheme
Find dx/dθ and dy/dθ
$$\frac{dx}{d\theta} = a(1 + \cos\theta)$$
$$\frac{dy}{d\theta} = a(0 - (-\sin\theta)) = a\sin\theta$$
Compute dy/dx using chain rule
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin\theta}{a(1 + \cos\theta)} = \frac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \tan\left(\frac{\theta}{2}\right)$$
Differentiate dy/dx with respect to x
$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[ \tan\left(\frac{\theta}{2}\right) \right] = \frac{d}{d\theta}\left[ \tan\left(\frac{\theta}{2}\right) \right] \cdot \frac{d\theta}{dx}$$
$$= \sec^2\left(\frac{\theta}{2}\right) \cdot \frac{1}{2} \cdot \frac{1}{a(1 + \cos\theta)} = \frac{1}{2a} \sec^2\left(\frac{\theta}{2}\right) \cdot \frac{1}{2\cos^2(\theta/2)} = \frac{1}{4a}\sec^4\left(\frac{\theta}{2}\right)$$
Evaluate at θ = π/2
At $\theta = \frac{\pi}{2}$, $\frac{\theta}{2} = \frac{\pi}{4}$:
$$\sec\left(\frac{\pi}{4}\right) = \sqrt{2} \implies \sec^4\left(\frac{\pi}{4}\right) = (\sqrt{2})^4 = 4$$
$$\left. \frac{d^2y}{dx^2} \right|_{\theta = \pi/2} = \frac{1}{4a} \times 4 = \frac{1}{a}$$