Question

Find the equations of the tangent and normal to the curve $x=a{\sin }^3\theta$ and $y=a{\cos }^3\theta$ at $\theta =\frac{\pi }{4}.$

Answer 1

Let $x=asi{n}^3\theta ,y=aco{s}^3\theta$

$\frac{dx}{d\theta }=3asi{n}^2\theta cos\theta$

$\frac{dy}{d\theta }=-3aco{s}^2\theta sin\theta$

$\frac{dy}{dx}=\frac{-3aco{s}^2\theta sin\theta }{3asi{n}^2\theta cos\theta }=-cot\theta$

${\begin{array}{c}\frac{dy}{dx}\end{array}|}_{\theta =\frac{\pi }{4}}=-1$

Equation of tangent at $\theta =\frac{\pi }{4}$

$y-aco{s}^3\frac{\pi }{4}=-1(x-asi{n}^3\frac{\pi }{4})$

$y+x=\frac{2a}{2\sqrt{2}}\Rightarrow y+x=\frac{a}{\sqrt{2}}$

Equation of normal at $\theta =\frac{\pi }{4}$

$y-\frac{a}{2\sqrt{2}}=1(x-\frac{a}{2\sqrt{2}})$

Therefore, $y-x=0.$