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

Given $x=a(2\theta -\sin 2\theta )$ and $y=a(1-\cos 2\theta )$

Differentiating $x$ and $y$ w.r.t $\theta$ we get

$\frac{dx}{d\theta }=2a-2a\cos 2\theta$ and $\frac{dy}{d\theta }=0+2a\sin 2\theta$

Using chain rule, we get

$\frac{dy}{dx}=\frac{dy}{d\theta }\cdot \frac{d\theta }{dx}$

$=2a\sin 2\theta \times \frac{1}{2a-2a\cos 2\theta }$

$=\frac{\sin 2\theta }{1-\cos 2\theta }$

$\because \cos 2\theta =1-2{\sin }^2\theta =2{\cos }^2\theta -1$

$\Rightarrow 1-\cos 2\theta =2{\sin }^2\theta$

Also, $\sin 2\theta =2\sin \theta \cos \theta$

$\Rightarrow \frac{dy}{dx}=\frac{2\sin \theta \cos \theta }{2{\sin }^2\theta }=\cot \theta$

$\Rightarrow \frac{dy}{dx}{|}_{\theta =\frac{\pi }{3}}=\cot \frac{\pi }{3}=\frac{1}{\sqrt{3}}$