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,

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