Question:
Find the equation of the tangent to the curve $x=\sin 3 t, y=\cos 2 t$ at $t=\frac{\pi}{4}$.
Solution:
finding the slope of the tangent by differentiating $x$ and $y$ with respect to $t$
$\frac{\mathrm{dx}}{\mathrm{dt}}=3 \cos 3 \mathrm{t}$
$\frac{\mathrm{dy}}{\mathrm{dt}}=-2 \sin 2 \mathrm{t}$
Dividing the above equations to obtain the slope of the given tangent
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2 \sin 2 \mathrm{t}}{3 \cos 3 \mathrm{t}}$
$\mathrm{m}$ (tangent) at $\frac{\pi}{4}$ is $\frac{2 \sqrt{2}}{3}$
equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$
therefore, equation of tangent is
$y-0=\frac{2 \sqrt{2}}{3}\left(x-\frac{1}{\sqrt{2}}\right)$