If \( x=a(1-\cos\theta),y=a(\theta+\sin\theta) \) then \( \displaystyle\frac{d^2y}{dx^2} \) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -\displaystyle\frac{1}{4a}\mathrm{cosec}^3\displaystyle\frac{\theta}{2}\sec\displaystyle\frac{\theta}{2} \)
Since x and y are given in terms of \( \theta \), we use parametric differentiation:
\( \displaystyle \frac{dx}{d\theta}=a\sin\theta \)
\( \displaystyle \frac{dy}{d\theta}=a(1+\cos\theta) \)
With the half-angle formulas \( 1+\cos\theta=2\cos^2\frac{\theta}{2} \) and \( \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \):
\( \displaystyle \frac{dy}{dx}=\frac{1+\cos\theta}{\sin\theta} \)
\( \displaystyle =\frac{2\cos^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \)
\( \displaystyle =\cot\frac{\theta}{2} \)
Differentiating again with respect to x, the derivative with respect to \( \theta \) is multiplied by \( \frac{d\theta}{dx} \):
\( \displaystyle \frac{d^2y}{dx^2}=\frac{d}{d\theta}\left(\cot\frac{\theta}{2}\right)\cdot\frac{d\theta}{dx} \)
\( \displaystyle =-\frac{1}{2}\mathrm{cosec}^2\frac{\theta}{2}\cdot\frac{1}{a\sin\theta} \)
\( \displaystyle =-\frac{1}{2}\mathrm{cosec}^2\frac{\theta}{2}\cdot\frac{1}{2a\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \)
\( \displaystyle =-\frac{1}{4a}\mathrm{cosec}^3\frac{\theta}{2}\sec\frac{\theta}{2} \)
Why the other options are incorrect
Option (1) \( -\displaystyle\frac{1}{2}\mathrm{cosec}^2\displaystyle\frac{\theta}{2} \) — Incorrect. This is only \( \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \); the factor \( \frac{d\theta}{dx}=\frac{1}{a\sin\theta} \) is missing.
Option (3) \( \displaystyle\frac{1}{4a}\sec^3\displaystyle\frac{\theta}{2}\mathrm{cosec}\displaystyle\frac{\theta}{2} \) — Incorrect. The answer is negative and has \( \mathrm{cosec}^3\frac{\theta}{2}\sec\frac{\theta}{2} \), not \( \sec^3\frac{\theta}{2}\mathrm{cosec}\frac{\theta}{2} \).
Option (4) \( \displaystyle\frac{1}{2}\sec^2\displaystyle\frac{\theta}{2} \) — Incorrect. It has no factor \( \frac{1}{a} \), but x and y are both proportional to a, so \( \frac{d^2y}{dx^2} \) must contain \( \frac{1}{a} \); its sign is also wrong.
Tip: In parametric second derivatives, \( \frac{d^2y}{dx^2}\ne\frac{d^2y/d\theta^2}{d^2x/d\theta^2} \): differentiate \( \frac{dy}{dx} \) with respect to \( \theta \) and then divide by \( \frac{dx}{d\theta} \).