Find the equation of the tangent


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


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

$y-0=\frac{2 \sqrt{2}}{3}\left(x-\frac{1}{\sqrt{2}}\right)$

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