If \( x=a(1-\cos\theta) \) and \( y=a(\theta+\sin\theta) \), then find the value of \( \displaystyle\frac{d^2y}{dx^2} \) at \( \theta=\pi/2 \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( -\displaystyle\frac{1}{a} \)
Differentiating both with respect to θ:
\( \displaystyle \frac{dx}{d\theta}=a\sin\theta \)
\( \displaystyle \frac{dy}{d\theta}=a(1+\cos\theta) \)
\( \displaystyle \frac{dy}{dx}=\frac{a(1+\cos\theta)}{a\sin\theta} \)
\( \displaystyle =\frac{2\cos^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} \)
\( \displaystyle =\cot\frac{\theta}{2} \)
The second derivative is taken with respect to x, so multiply by \( \frac{d\theta}{dx}=\frac{1}{a\sin\theta} \):
\( \displaystyle \frac{d^2y}{dx^2}=\frac{d}{d\theta}\left(\cot\frac{\theta}{2}\right)\cdot\frac{d\theta}{dx} \)
\( \displaystyle =-\frac{1}{2}\mathrm{cosec}^2\frac{\theta}{2}\cdot\frac{1}{a\sin\theta} \)
At \( \theta=\frac{\pi}{2} \): \( \mathrm{cosec}^2\frac{\pi}{4}=2 \) and \( \sin\frac{\pi}{2}=1 \).
\( \displaystyle \frac{d^2y}{dx^2}=-\frac{1}{2}\times 2\times\frac{1}{a} \)
\( \displaystyle =-\frac{1}{a} \)
Tip: For a parametric curve, \( \frac{d^2y}{dx^2} \) is the θ-derivative of \( \frac{dy}{dx} \) multiplied by \( \frac{d\theta}{dx} \); it is not \( \frac{d^2y/d\theta^2}{d^2x/d\theta^2} \).