Question

If $x=a(2\theta -\sin 2\theta )$ and $y=a(1-\cos 2\theta )$, find $\frac{dy}{dx}$ when $\theta =\frac{\pi }{3}$.

Answer 1

$x=a(2\theta -\sin 2\theta )$

$\frac{dx}{d\theta }=a(2-\cos 2\theta \cdot 2)=2a(1-\cos 2\theta )$

$y=a(1-\cos 2\theta )$

$\frac{dy}{d\theta }=a(\sin 2\theta \cdot 2)=2a\cdot \sin 2\theta$

Now, $\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{2a.\sin 2\theta }{2a(1-\cos 2\theta )}=\frac{\sin 2\theta }{1-\cos 2\theta }$

$\frac{dy}{dx}{]}_{\theta =\frac{\pi }{3}}=\frac{\sin \frac{2\pi }{3}}{1-\cos \frac{2\pi }{3}}=\frac{\frac{\sqrt{3}}{2}}{1+\frac{1}{2}}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$