Find the equation of tangent and normal to the curve $x = s i n 3 t$ , $y = c o s 2 t$ at $t = \frac{π}{4}$ .

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Solution

$\begin{matrix} \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 2 sinh 2 t}{3 cos 3 t} t = \frac{π}{4} = \frac{- 2}{3} \times \frac{1}{- 1 / \sqrt{3}} = \frac{+ 2 \sqrt{2}}{3} \end{matrix}$
Point $x = \frac{1}{\sqrt{2}}, y = 0$
Tangent
$\begin{matrix} y - y_{1} = \frac{d y}{d x} (x - x_{1}) = y - 0 = \frac{2 \sqrt{2}}{3} (x - \frac{1}{\sqrt{2}}) \end{matrix}$
Normal
$\begin{matrix} y - y_{1} = - \frac{d y}{d x} (x - x_{1}) = y - 0 = \frac{- 3}{2 \sqrt{2}} (x - \frac{1}{\sqrt{2}}) \end{matrix}$

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t = \frac{π}{4}

x = \frac{{sin}^{3} t}{\sqrt{(cos 2 t)}}, y = \frac{{cos}^{3} t}{\sqrt{(cos 2 t)}},

\frac{d y}{d x}

t = \frac{π}{6} .

Find the equation of the tangent and normal to the given curve at the given points

$x = c o s t, y = s i n t = \frac{π}{4}$

x = {sin}^{3} θ

y = a c o s^{3} θ

θ = \frac{π}{4}

x = 1 - cos θ

y = 1 - sin θ

θ = \frac{π}{4}

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