Question

If $x=a(\theta -\sin \theta ),y=a(1-\cos \theta )$, then $\frac{d^{2}y}{d{x}^2}$ at $\theta =\frac{\pi }{3}$ is

• A. $-\frac{2}{a}$
• B. $\frac{2}{a}$
• C. $-\frac{4}{a}$
• D. $\frac{4}{a}$

Answer 1

Answer: (C)

$-\frac{4}{a}$

$x=a(\theta -\sin \theta )\Rightarrow \frac{dx}{d\theta }=a(1-\cos \theta )$
$y=a(1-\cos \theta )\Rightarrow \frac{dy}{d\theta }=a(1+\sin \theta )$
$\therefore \frac{dy}{dx}=\frac{1+\sin \theta }{1-\cos \theta }$
$\Rightarrow \frac{d^{2}y}{{dx}^2}=\frac{(1-\cos \theta )\left(\cos \theta \right)-(1+\sin \theta )\left(\sin \theta \right)}{(1-\cos \theta )}\times \left(\frac{dt}{dx}\right)$
$=\frac{\cos \theta -\sin \theta -1}{{(1-\cos \theta )}^2}\times \frac{1}{a(1-\cos \theta )}$
At $\theta =\frac{\pi }{3}\frac{d^{2}y}{{dx}^2}=-\frac{4}{a}$