Question

If $x=a\sin 2t\left(1+\cos 2t\right)\mathrm{and}y=b\cos 2t\left(1-\cos 2t\right)$, show that at $t=\frac{\mathrm{\pi }}{4},\frac{dy}{dx}=\frac{b}{a}$$t=\frac{\mathrm{\pi }}{4},\frac{dy}{dx}=\frac{b}{a}$

Answer 1

$$\begin{gathered} x=a\sin 2t\left(1+\cos 2t\right)\mathrm{and}y=b\cos 2t\left(1-\cos 2t\right) \\ \Rightarrow \frac{dx}{dt}=2a\cos 2t\left(1+\cos 2t\right)+2a\sin 2t\left(1-\cos 2t\right)\mathrm{and}\frac{dy}{dt}=-2b\sin 2t\left(1-\cos 2t\right)+2b\cos 2t\left(1+\cos 2t\right) \\ \Rightarrow \frac{dx}{dt}=2a\left(\cos 2t+{\cos }^{2}2t+\sin 2t-\sin 2t\cos 2t\right)\mathrm{and}\frac{dy}{dt}=2b\left(-\sin 2t+\sin 2t\cos 2t+\cos 2t+{\cos }^{2}2t\right) \\ \therefore \frac{dy}{dt}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{-2b\left(-\sin 2t+\sin 2t\cos 2t+\cos 2t+{\cos }^{2}2t\right)}{2a\left(\cos 2t+{\cos }^{2}2t+\sin 2t-\sin 2t\cos 2t\right)} \\ \Rightarrow {\left(\frac{dy}{dt}\right)}_{t=\frac{\mathrm{\pi }}{4}}=\frac{-2b\left(-\sin {\displaystyle \frac{\mathrm{\pi }}{2}}+\sin {\displaystyle \frac{\mathrm{\pi }}{2}}\cos {\displaystyle \frac{\mathrm{\pi }}{2}}+\cos {\displaystyle \frac{\mathrm{\pi }}{2}}+{\cos }^2{\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{2a\left(\cos {\displaystyle \frac{\mathrm{\pi }}{2}}+{\cos }^2{\displaystyle \frac{\mathrm{\pi }}{2}}+\sin {\displaystyle \frac{\mathrm{\pi }}{2}}-\sin {\displaystyle \frac{\mathrm{\pi }}{2}}\cos {\displaystyle \frac{\mathrm{\pi }}{2}}\right)}=\frac{b}{a} \end{gathered}$$