Question

$\mathrm{If}x=\cos t\left(3-2{\cos }^{2}t\right),y=\sin t\left(3-2{\sin }^{2}t\right)\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{dy}{dx}\mathrm{at}t=\frac{\mathrm{\pi }}{4}$

Answer 1

$$\begin{gathered} x=\cos t\left(3-2{\cos }^{2}t\right)\mathrm{and}y=\sin t\left(3-2{\sin }^{2}t\right) \\ \Rightarrow \frac{dx}{dt}=-\sin t\left(3-2{\cos }^{2}t\right)+\cos t\left(4\cos t\sin t\right)\mathrm{and}\frac{dy}{dt}=\cos t\left(3-2{\sin }^{2}t\right)+\sin t\left(-4\sin t\cos t\right) \\ \Rightarrow \frac{dx}{dt}=-3\sin t+6\sin t{\cos }^{2}t\mathrm{and}\frac{dy}{dt}=3\cos t-6{\sin }^{2}t\cos t \\ \Rightarrow \frac{dx}{dt}=-3\sin t\left(1-2{\cos }^{2}t\right)\mathrm{and}\frac{dy}{dt}=3\cos t\left(1-2{\sin }^{2}t\right) \\ \Rightarrow \frac{dx}{dt}=3\sin t\cos 2t\mathrm{and}\frac{dy}{dt}=3\cos t\cos 2t \\ \therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{{\displaystyle 3\cos t\cos 2t}}{{\displaystyle 3\sin t\cos 2t}}=\cot t \\ \mathrm{Now},{\left(\frac{dy}{dx}\right)}_{t=\frac{\mathrm{\pi }}{4}}=\cot \frac{\mathrm{\pi }}{4}=1 \end{gathered}$$