x=acos^3t,y=asin^3t,求dy/dx

显然
dx/dt =d (acos^3t) /dt =3acos²t * (cost)'= -3a*sint *cos²t
而dy/dt =d(asin^3t) /dt =3asin²t * (sint)'= 3a*sin²t *cost
所以
dy/dx
= (dy/dt) / (dx/dt)
= (3a*sin²t *cost) / (-3a*sint *cos²t)
= -tant