Find the slope of the normal to the curve `x=a\ cos^3theta` , `y=a\ sin^3theta` at `theta=pi/4` .

Explanation

` x = a cos ^(3) theta`
` rArr (dx)/(d theta) = a d/(d theta) cos^(3) theta = a * 3 cos^(2) theta d/(d theta) (cos theta)`
` =- 3a cos^(2) theta sin theta`
and ` y = a sin^(3) theta`
` rArr (dy)/(d theta) = a d/(d theta) sin^(3) theta = a * 3 sin^(2) theta d/(d theta) (sin theta)`
` = 3a sin^(2) theta cos theta`
Now ` (dy)/(dx) = ((dy)//(d theta))/((dx)//(d theta)) = (3a sin^(2) theta cos theta)/(-3a cos^(2) theta sin theta) = - tan theta`
`at theta = pi/4`
Slope of tangent
` m=- tan pi/4 =- 1`
and slope of normal =`-1/m =- 1/(-1) = 1`