一个圆锥 x^2 + y^2 = (z/c)^2，z≧0 用变分法找出在锥面上两点间的最短距离 我的做法: x = ρcosψ，y = ρsinψ，z = cρ，ρ = x^2 + y^2 => dx = (dρ)cosψ-ρsinψ(dψ) dy = (dρ)sinψ+ρcosψ(dψ) dz = c(dρ) => ds = sqrt[(dx)^2+(dy)^2+(dz)^2] = sqrt[(dρ)^2+ρ^2(dψ)^2+c^2(dρ)^2] = sqrt[(1+c^2)(dρ)^2+ρ^2(dψ)^2] = sqrt[(1+c^2)+ρ^2(dψ/dρ)^2](dρ) Define sqrt[(1+c^2)+ρ^2(dψ/dρ)^2] = f，dψ/dρ = ψ' => S = ∫ ds = ∫sqrt[(1+c^2)+ρ^2(dψ/dρ)^2](dρ) C use Euler's equation: δ partial differential operator δf d ┌ δf ┐ δf ── - ──│───│ = 0 ∵ ── = 0 δψ dρ └ δψ'┘ δψ δf ρ^2(ψ') => ─── = constant = a = ─────────────── δψ' [(1+c^2)+ρ^2(ψ')^2]^(1/2) => ψ(ρ) = ......... 解出来的轨迹看起来很复杂 但是在直觉上，最短轨迹在把圆锥摊开之後，轨迹应该为直线 而求得的解看起来却不是这麽一回事 请问我哪里算错了吗? 谢谢 --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 140.112.211.87 ※ 编辑: rachel5566 来自: 140.112.211.87 (01/16 00:15)