Question

求曲线; x=acos^3Θ:y=asin^3Θ ,在Θ=π/6时的切线方程,法线方程,

Answer 1

dy/dx=(dy/dΘ)/(dx/dΘ)

=(3asin^2ΘcosΘ)/(-3acos^2ΘsinΘ)

=-tanΘ

dy(π/6)/dx=-根号3/3

x(π/6)=3a根号3/8 y(π/6)=a/8

切线方程：y-a/8=-根号3/3*(x-3a根号3/8)

y=-根号3/3*x+a/2

法线方程：y-a/8=根号3*(x-3a根号3/8)

y=根号3*x-a