x=acos^3t y=asin^3t 在t=π/6时,求相应点切线方程和法线方程

dx/dt=3a(cost)^2(-sint)=-3asint(cost)^2,dy/dt=3a(sint)^2*(cost),dy/dx=(dy/dt)/(dx/dt)=[3a(sint)^2*(cost)]/(-3asint(cost)^2,]=-sint/cost=-tant,当t=π/6时,dy/dx=-√3/3,当t=π/6时,x=3√3a/8,y=a/8,∴相应...