Question

If $x=a\left(\mathrm{\theta }-\sin \mathrm{\theta }\right)\mathrm{and},y=a\left(1+\cos \mathrm{\theta }\right),\mathrm{find}\frac{dy}{dx}\mathrm{at}\mathrm{\theta }=\frac{\pi }{3}.$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},x=a\left(\theta -\sin \theta \right)andy=a\left(1+\cos \theta \right) \\ \Rightarrow \frac{dx}{d\theta }=\frac{d}{d\theta }\left[a\left(\theta -\sin \theta \right)\right]\mathrm{and}\frac{dy}{d\theta }=\frac{d}{d\theta }\left[a\left(1+\cos \theta \right)\right] \\ \Rightarrow \frac{dx}{d\theta }=a\left(1-\cos \theta \right)\mathrm{and}\frac{dy}{d\theta }=a\left(-\sin \theta \right) \\ \therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}=\frac{-a\sin \theta }{a\left(1-\cos \theta \right)} \\ \mathrm{Now},{\left[\frac{dy}{dx}\right]}_{\theta =\frac{\mathrm{\pi }}{3}}=-\frac{\sin {\displaystyle \frac{\mathrm{\pi }}{3}}}{1-\cos {\displaystyle \frac{\mathrm{\pi }}{3}}}=-\frac{{\displaystyle \frac{\sqrt{3}}{2}}}{1-{\displaystyle \frac{1}{2}}}=-\sqrt{3} \end{gathered}$$