Question

Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at $t=\frac{\mathrm{\pi }}{4}$.

Answer 1

$$\begin{gathered} x=\sin 3t\mathrm{and}y=\cos 2t \\ \frac{dx}{dt}=3\cos 3t\mathrm{and}\frac{dy}{dt}=-2\sin 2t \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-2\sin 2t}{3\cos 3t} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{t=\frac{\mathrm{\pi }}{4}}\text{=-}\frac{-2\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{3\cos \left({\displaystyle \frac{3\mathrm{\pi }}{4}}\right)}\text{=}\frac{-2}{{\displaystyle \frac{-3}{\sqrt{2}}}}\text{=}\frac{\text{2}\sqrt{2}}{3} \\ {x}_1=\sin \left(3\times \frac{\mathrm{\pi }}{4}\right)=\frac{1}{\sqrt{2}}\mathrm{and}{y}_1=\cos \left(2\times \frac{\mathrm{\pi }}{4}\right)=0 \\ \mathrm{So},\left(x_1,y_1\right)=\left(\frac{1}{\sqrt{2}},0\right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-0=\frac{\text{2}\sqrt{2}}{3}\left(x-\frac{1}{\sqrt{2}}\right) \\ \Rightarrow 3y=2\sqrt{2}x-2 \\ \Rightarrow 2\sqrt{2}x-3y-2=0 \end{gathered}$$