If \( x=a(\theta+\sin\theta),\ y=a(1-\cos\theta) \) then \( \displaystyle\frac{d^2y}{dx^2} \) is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{\sec^2\left(\frac{\theta}{2}\right)}{2a(1+\cos\theta)} \)
Both x and y are given in terms of the parameter θ, so differentiate each with respect to θ:
\( \displaystyle \frac{dx}{d\theta}=a(1+\cos\theta) \)
\( \displaystyle \frac{dy}{d\theta}=a\sin\theta \)
\( \displaystyle \frac{dy}{dx}=\frac{a\sin\theta}{a(1+\cos\theta)} \)
\( \displaystyle =\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} \)
\( \displaystyle =\tan\frac{\theta}{2} \)
For the second derivative, \( \frac{dy}{dx} \) is differentiated with respect to θ and multiplied by \( \frac{d\theta}{dx} \):
\( \displaystyle \frac{d^2y}{dx^2}=\frac{d}{d\theta}\left(\tan\frac{\theta}{2}\right)\cdot\frac{d\theta}{dx} \)
\( \displaystyle =\frac{1}{2}\sec^2\frac{\theta}{2}\cdot\frac{1}{a(1+\cos\theta)} \)
\( \displaystyle \frac{d^2y}{dx^2}=\frac{\sec^2\left(\frac{\theta}{2}\right)}{2a(1+\cos\theta)} \)
Tip: In parametric form \( \frac{d^2y}{dx^2} \) is not \( \frac{d^2y/d\theta^2}{d^2x/d\theta^2} \); after differentiating \( \frac{dy}{dx} \) with respect to θ, always divide by \( \frac{dx}{d\theta} \).