If \( x=a(\theta+\sin\theta)\ ,y=a(1-\cos\theta) \) , then \( \displaystyle\frac{d^2y}{dx^2}= \)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{4a}sec^4\displaystyle\frac{\theta}{2} \)
Here x and y are given in terms of the parameter \( \theta \).
\( \displaystyle \frac{dx}{d\theta}=a(1+\cos\theta)=2a\cos^2\frac{\theta}{2} \)
\( \displaystyle \frac{dy}{d\theta}=a\sin\theta=2a\sin\frac{\theta}{2}\cos\frac{\theta}{2} \)
\( \displaystyle \frac{dy}{dx}=\frac{2a\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2a\cos^2\frac{\theta}{2}} \)
\( \displaystyle =\tan\frac{\theta}{2} \)
Differentiating again with respect to x, with \( \frac{d\theta}{dx}=\frac{1}{2a\cos^2\frac{\theta}{2}} \):
\( \displaystyle \frac{d^2y}{dx^2}=\frac{1}{2}\sec^2\frac{\theta}{2}\cdot\frac{d\theta}{dx} \)
\( \displaystyle =\frac{1}{2}\sec^2\frac{\theta}{2}\cdot\frac{1}{2a}\sec^2\frac{\theta}{2} \)
\( \displaystyle =\frac{1}{4a}\sec^4\frac{\theta}{2} \)
Why the other options are incorrect
Option (2) \( \tan\displaystyle\frac{\theta}{2} \) — Incorrect. This is only \( \frac{dy}{dx} \); the question asks for the second derivative.
Option (3) \( -\cot\theta \) — Incorrect. The second derivative is \( \frac{1}{4a}\sec^4\frac{\theta}{2} \), not \( -\cot\theta \).
Option (4) \( \tan\theta \) — Incorrect. The second derivative is \( \frac{1}{4a}\sec^4\frac{\theta}{2} \), not \( \tan\theta \).