发布网友 发布时间:2022-08-24 19:26
共1个回答
热心网友 时间:2024-03-05 03:31
ρ=e^θ ( e^(π/2) , π/2 )
dρ/dθ = e^θ
dρ/dθ|(θ =π/2) = e^(π/2)
x=ρ.cosθ
dx/dθ = -ρ.sinθ + cosθ . (dρ/dθ)
dx/dθ|(θ =π/2) = -e^(π/2)
y=ρ.sinθ
dy/dθ = ρ.cosθ + sinθ . dρ/dθ
dy/dθ |(θ =π/2) = e^(π/2)
dy/dx |θ =π/2
=(dy/dθ |θ =π/2 ) /(dx/dθ|θ =π/2)
= e^(π/2)/[-e^(π/2) ]
=-1