Question

Find $\frac{dy}{dx}$, when

$x=a\left(\cos \mathrm{\theta }+\mathrm{\theta }\sin \mathrm{\theta }\right)\mathrm{and}y=a\left(\sin \mathrm{\theta }-\mathrm{\theta }\cos \mathrm{\theta }\right)$

Answer 1

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