Question

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

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

Answer 1

$\mathrm{We}\mathrm{have},x=a{e}^{\theta }\left(\sin \theta -\cos \theta \right)\mathrm{and}y=a{e}^{\theta }\left(\sin \theta +\cos \theta \right)$

$$\begin{gathered} \Rightarrow \frac{dx}{d\theta }=a\left[e^{\theta }\frac{d}{d\theta }\left(\sin \theta -\cos \theta \right)+\left(\sin \theta -\cos \theta \right)\frac{d}{d\theta }\left(e^{\theta }\right)\right]\mathrm{and}\frac{dy}{d\theta }=a\left[e^{\theta }\frac{d}{d\theta }\left(\sin \theta +\cos \theta \right)+\left(\sin \theta +\cos \theta \right)\frac{d}{d\theta }\left(e^{\theta }\right)\right] \\ \Rightarrow \frac{dx}{d\theta }=a\left[e^{\theta }\left(\cos \theta +\sin \theta \right)+\left(\sin \theta -\cos \theta \right)e^{\theta }\right]\mathrm{and}\frac{dy}{d\theta }=a\left[e^{\theta }\left(\cos \theta -\sin \theta \right)+\left(\sin \theta +\cos \theta \right)e^{\theta }\right] \\ \Rightarrow \frac{dx}{d\theta }=a\left[2e^{\theta }\sin \theta \right]\mathrm{and}\frac{dy}{d\theta }=a\left[2e^{\theta }\cos \theta \right] \end{gathered}$$

$\therefore \frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}=\frac{a\left(2e^{\theta }\cos \theta \right)}{a\left(2e^{\theta }\sin \theta \right)}=\cot \theta$