r(t) = ((4cost)^3, (4sint)^3) for -pi<=pi
i tried just finding T(t) by using r'(t)/|r'(t)| but i keep getting a giant mess which is no good given i have to find the normal vector and curvature later. Please help.
i am having one idea i.e first obtain locus of above coordinate system
x^2/3 +y^2/3=16 now obtain dy/dx for this curve that comes out to be -tan(t)
now obtain equation of tangent to this curve
but what i understand that you want to apply concept of gradient in above problem but that always gives equation of normal