定义

K=∣dαdr∣ K = \left|\frac{d\alpha}{dr}\right|

ρ=1/K \rho = 1/K

计算

下面暂时省略绝对值

K=darctan⁡y′dx2+dy2=y′′1+(y′)2dxdx1+(y′)2=y′′[1+(y′)2]3/2 \begin{aligned} K &= \frac{d\arctan y'}{\sqrt{dx^2 + dy^2}} = \frac{\frac{y''}{1+(y')^2}dx}{dx\sqrt{1+(y')^2}} \\ & = \frac{y''}{\left[1+(y')^2\right]^{3/2}} \end{aligned}

如果由参数方程确定

{x=ϕ(t)y=ψ(t) \begin{cases} x = \phi (t) \\ y = \psi(t) \end{cases}

那么有

K=ddx(dydx)[1+(dydx)2]3/2=ddt(ψ′(t)ϕ′(t))1ϕ′(t)[1+(ψ′(t)ϕ′(t))2]3/2=ϕ′(t)ψ′′(t)−ψ′(t)−ϕ′′(t)ϕ′3(t)[1+(ψ′(t)ϕ′(t))2]3/2=ϕ′(t)ψ′′(t)−ψ′(t)−ϕ′′(t)(ϕ′2(t)+ψ′2(t))3/2 \begin{aligned} K &= \frac{ \frac{d}{dx}\left( \frac{dy}{dx} \right) }{\left[1 + (\frac{dy}{dx})^2 \right]^{3/2} } \\ & = \frac{ \frac{d}{dt}\left( \frac{\psi'(t)}{\phi'(t)} \right)\frac{1}{\phi'(t)} }{\left[1 + (\frac{\psi'(t)}{\phi'(t)})^2 \right]^{3/2} } \\ & = \frac{ \frac{\phi'(t)\psi''(t) - \psi'(t) - \phi''(t)}{\phi'^3(t)}}{\left[1 + (\frac{\psi'(t)}{\phi'(t)})^2 \right]^{3/2} } \\ & = \frac{\phi'(t)\psi''(t) - \psi'(t) - \phi''(t)}{\left(\phi'^2(t) + \psi'^2(t)\right)^{3/2} } \end{aligned}