Question

极坐标曲线ρ=θ,在θ=π对应点处法线方程的直角坐标形式

Answer 1

这是阿基米德螺线,
√(x^2+y^2)=arctan(y/x),
隐函数求导,
(1/2)(x^2+y^2)^(-1/2)*2x+(1/2)(x^2+y^2)^(-1/2)*2y*dy/dx=(1/x)(dy/dx)/[1+(y/x)^2]+y/(-x^2)/[1+(y/x)^2]
x/√(x^2+y^2)+y*(dy/dx)/√(x^2+y^2)=x*dy/dx/(x^2+y^2)-y/(x^2+y^2)
dy/dx[x/(x^2+y^2)-y/√(x^2+y^2)]=x/√(x^2+y^2)+y/+y(x^2+y^2)
dy/dx[x-y√(x^2+y^2)]=x√(x^2+y^2)+y,
dy/dx=[x√(x^2+y^2)+y]/[x-y√(x^2+y^2)],
当θ=π时,y=0,√(x^2+0)=π,
∴x=π,
dy/dx|x=π,y=0|=π,
在(π,0)处切线斜率为π,则法线斜率为-1/π,
∴法线方程为:y/(x-π)=-1/π,
∴法线方程为:y=-x/π+1.