Question

1/求曲线p=cos2θ在θ=π/6所对应点处的切线方程_____________

2、r=e^θ在θ=π/2处的切线方程为____________
求详解！

Answer 1

1、解：θ=π/6时ρ=cos(2*π/6)=1/2，则x=ρcosθ=1/2*cos(π/6)=√3/4，y=ρsinθ=1/2*sin(π/6)=1/4
参数方程为x=ρcosθ=cos2θcosθ,y=cos2θsinθ
故dx/dθ=-2sin2θcosθ-cos2θsinθ
dy/dθ=-2sin2θsinθ+cos2θcosθ
则dy/dx=(dy/dθ)/(dx/dθ)=(-2sin2θsinθ+cos2θcosθ)/(-2sin2θcosθ-cos2θsinθ)
=(-2tan2θtanθ+1)/(-2tan2θ-tanθ)
将θ=π/6代入得切线斜率
dy/dx=(-2√3*√3/3+1)/(-2√3-√3/3)=1/(7√3/3)=√3/7
故切线方程为y-1/4=√3/7*(x-√3/4)
也即y=√3/7*x+5/28
2、解：θ=π/2时r=e^θ=e^(π/2)，则x=rcosθ=e^(π/2)*cos(π/2)=0，y=ρsinθ=e^(π/2)*sin(π/2)=e^(π/2)
参数方程为x=rcosθ=e^θ*cosθ,y=e^θ*sinθ
故dx/dθ=e^θcosθ-e^θsinθ=e^θ(cosθ-sinθ)
dy/dθ=e^θsinθ+e^θcosθ=e^θ(sinθ+cosθ)
则dy/dx=(dy/dθ)/(dx/dθ)=e^θ(sinθ+cosθ)/[e^θ(cosθ-sinθ)]=(sinθ+cosθ)/(cosθ-sinθ)
将θ=π/2代入得切线斜率
dy/dx=(sinπ/2+cosπ/2)/(cosπ/2-sinπ/2)=(1+0)/(0-1)=-1
故切线方程为y-e^(π/2)=-(x-0)
也即y=-x+e^(π/2)
不明白请追问。