If \(x = 2(\theta - \sin\theta)\) and \(y = 2(1 - \cos\theta)\), \(0 \lt \theta \lt \pi\), then the value of \(\displaystyle\frac{d^{2}y}{dx^{2}}\) at \(\theta = \displaystyle\frac{\pi}{2}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-\displaystyle\frac{1}{2}\)
Differentiating each with respect to \(\theta\):
\( \displaystyle \frac{dx}{d\theta} = 2(1 - \cos\theta), \quad \frac{dy}{d\theta} = 2\sin\theta \)
Using the half-angle formulas \(1 - \cos\theta = 2\sin^{2}\frac{\theta}{2}\) and \(\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\):
\( \displaystyle \frac{dy}{dx} = \frac{2\sin\theta}{2(1 - \cos\theta)} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^{2}\frac{\theta}{2}} = \cot\frac{\theta}{2} \)
For the second derivative, differentiate with respect to \(\theta\) and divide by \(\frac{dx}{d\theta}\):
\( \displaystyle \frac{d}{d\theta}\cot\frac{\theta}{2} = -\frac{1}{2}\mathrm{cosec}^{2}\frac{\theta}{2}, \quad \frac{dx}{d\theta} = 2(1 - \cos\theta) = 4\sin^{2}\frac{\theta}{2} \)
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{-\frac{1}{2}\mathrm{cosec}^{2}\frac{\theta}{2}}{4\sin^{2}\frac{\theta}{2}} = -\frac{1}{8}\mathrm{cosec}^{4}\frac{\theta}{2} \)
At \(\theta = \frac{\pi}{2}\), \(\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}\), so \(\mathrm{cosec}^{4}\frac{\pi}{4} = (\sqrt{2})^{4} = 4\):
\( \displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{1}{8} \times 4 = -\frac{1}{2} \)