Find the slopes of the tangent and the normal to the curve `x=acos^3theta, y=a sin^3 theta` at `theta=pi//4`

Explanation

Given, `x=acos^3theta, y=a sin^3 theta`


`(dx)/(d theta)=-3acos^2 theta sin theta`


`(dy)/(d theta)=-3asin^2 theta cos theta`


Now, `(dy)/(dx)=(3asin^2 theta cos theta)/(-3acos^2 theta sin theta)=-tan theta`


Slope of normal at any point `=-(1)/((dy)/(dx))=-(1)/(-tan theta)=cot theta`


Slope of normal at `theta=pi//4`


` = cot pi//4=1`