Question

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

Answer 1

Given values are:
$$\begin{gathered} x=a\left(2\theta -\sin 2\theta \right) \\ \mathrm{and} \\ y=a\left(1-\cos 2\theta \right) \end{gathered}$$

Applying parametric differentiation

$\frac{dx}{d\theta }$= 2a − 2acos2$\theta$

$\frac{dy}{d\theta }$= 0 + 2asin2$\theta$

$\frac{dy}{dx}$= $\frac{dy}{d\theta }\times \frac{d\theta }{dx}=\frac{\sin 2\theta }{1-\cos 2\theta }$

Now putting the value of $\theta$= $\frac{\mathrm{\pi }}{3}$
$$\begin{gathered} {\overline{)\frac{dy}{dx}}}_{\theta =\frac{\mathrm{\pi }}{3}}=\frac{\sin 2\left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)}{1-\cos 2\left(\frac{\mathrm{\pi }}{3}\right)} \\ =\frac{{\displaystyle \frac{\sqrt{3}}{2}}}{1+{\displaystyle \frac{1}{2}}} \\ =\frac{{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{3}{2}}}=\frac{1}{\sqrt{3}} \end{gathered}$$
So, $\frac{dy}{dx}$ is $\frac{1}{\sqrt{3}}$ at $\mathrm{\theta }=\frac{\pi }{3}$.