Question

If $x=3\sin t-\sin 3t,y=3\cos t-\cos 3t\mathrm{find}\frac{dy}{dx}\mathrm{at}t=\frac{\mathrm{\pi }}{3}$

Answer 1

$$\begin{gathered} x=3\sin t-\sin 3t\mathrm{and}y=3\cos t-\cos 3t \\ \Rightarrow \frac{dx}{dt}=3\cos t-3\cos 3t\mathrm{and}\frac{dy}{dt}=-3\sin t+3\sin 3t \\ \therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{{\displaystyle -3\sin t+3\sin 3t}}{{\displaystyle 3\cos t-3\cos 3t}} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{t=\frac{\mathrm{\pi }}{3}}=\frac{-3\sin {\displaystyle \frac{\mathrm{\pi }}{3}}+3\mathrm{sin\pi }}{3\cos {\displaystyle \frac{\mathrm{\pi }}{3}}-3\mathrm{cos\pi }} \\ =\frac{-3\times {\displaystyle \frac{\sqrt{3}}{2}}+0}{3\times {\displaystyle \frac{1}{2}}+3} \\ =\frac{{\displaystyle \frac{-3\sqrt{3}}{2}}}{{\displaystyle \frac{9}{2}}} \\ =-\frac{1}{\sqrt{3}} \end{gathered}$$