Question

If $x=\theta +\sin \theta ,y=1-\cos \theta$, then $y^{\prime \prime }$ at $\theta ={30}^{\circ }$ is

• A. $\frac{7+4\sqrt{3}}{4}$
• B. $\frac{7-4\sqrt{3}}{4}$
• C. $7-4\sqrt{3}$
• D. $0$

Answer 1

The correct option is C $7-4\sqrt{3}$

$Given:x=(\theta -\sin \theta )y=(1-\cos \theta )Letusdifferentiatetheseindividualequationwithrespectto\theta \frac{dx}{d\theta }=1-\cos \theta .......\left(1\right)\frac{dy}{d\theta }=-(-\sin \theta )=\sin \theta ........(2)dividingequation1and2\frac{dy}{dx}=\frac{\sin \theta }{1-\cos \theta }=\cot \frac{\theta }{2}.......(3)Differentiatingtheequation3w.r.tx{y}_2=\frac{d}{dx}\left(\cot \frac{\theta }{2}\right)=-\frac{1}{2}{\csc }^2\frac{\theta }{2}\times \frac{d\theta }{dx}=-\frac{1}{2}{\csc }^2\frac{\theta }{2}\times \frac{1}{1-\cos \theta }At\theta =30y_2=-\frac{1}{2}{\csc }^2\frac{30}{2}\times \frac{1}{1-\cos 30}{y}_2=7-4\sqrt{3}HenceOPTION3iscorrectoption$