Question

Find the slope of the normal to the curve x = acos3 θ, y = asin3 θ at θ=π/4.

Answer 1

It is given that x = acos3 θ and y = asin3 θ.

$\frac{dx}{dθ}$=3a cos2θ(-sinθ)=-3a cos2θ sinθ

$\frac{dy}{dθ}$=3a sin2θ(cosθ)

$\frac{dy}{dx}$=$\frac{(\frac{dy}{dθ}) }{ (\frac{dx}{dθ}) }$=$\frac{3asin^2θ\,cosθ}{-3acos^2θ\,sinθ}$= $\frac{-sinθ}{cosθ}$=-tanθ

Therefore, the slope of the tangent at $θ=\frac{π}{4}$ is given by,

$(\frac{dy}{dx}) \bigg]_{θ=\frac{π}{4}}$=$-tanθ\bigg]_{θ=\frac{π}{4}}$=-tan$\frac{ π}{4}$=-1

Hence, the slope of the normal at $θ=\frac{π}{4}$ is given by,

$\frac{1}{\text{slop of the tangent at} \,θ=\frac{π}{4}}= (\frac{-1}{-1}) =1$