通过函数乘法求导运算法则，经计算可得结果： y = y ( x )    ,    x = x ( t ) x ′ ( t ) = d x d t   ,   y ′ ( t ) = d y d t   ,   x ′ ′ ( t ) = d 2 x d t 2   ,   y ′ ′ ( t ) = d 2 y d t 2 d 2 y d x 2 = [    ] ⋅ y ′ ′ ( t ) − [    ] ⋅ x ′ ′ ( t ) d 2 y d x 2 = [ x ′ ( t ) x ′ ( t ) 3 ] ⋅ y ′ ′ ( t ) − [ y ′ ( t ) x ′ ( t ) 3 ] ⋅ x ′ ′ ( t ) d 2 y d x 2 = [ x ′ ( t ) x ′ ( t ) 3 ] ⋅ y ′ ′ ( t ) − [ y ′ ( t ) x ′ ( t ) 3 ] ⋅ x ′ ′ ( t ) = ∣ x ′ ( t ) x ′ ′ ( t ) y ′ ( t ) y ′ ′ ( t ) ∣ x ′ ( t ) 3 \begin{aligned} & y=y(x) \ \ , \ \ x=x(t)\\ \\ & x'(t)=\frac{dx}{dt} \ , \ y'(t)=\frac{dy}{dt} \ , \ x''(t)=\frac{d^2x}{dt^2} \ , \ y''(t)=\frac{d^2y}{dt^2}\\ \\ & \frac{d^2y}{dx^2}=[\ \ ]\cdot y''(t)-[\ \ ]\cdot x''(t)\\ \\ & \frac{d^2y}{dx^2}=[\frac{x'(t)}{x'(t)^3}]\cdot y''(t)-[\frac{y'(t)}{x'(t)^3}]\cdot x''(t)\\ & \frac{d^2y}{dx^2}=[\frac{x'(t)}{x'(t)^3}]\cdot y''(t)-[\frac{y'(t)}{x'(t)^3}]\cdot x''(t)=\frac{\left|\begin{array}{ll} x'(t) & x''(t) \\ y'(t) & y''(t) \end{array}\right|}{x'(t)^3} \end{aligned} ​y=y(x)  ,  x=x(t)x′(t)=dtdx​ , y′(t)=dtdy​ , x′′(t)=dt2d2x​ , y′′(t)=dt2d2y​dx2d2y​=[  ]⋅y′′(t)−[  ]⋅x′′(t)dx2d2y​=[x′(t)3x′(t)​]⋅y′′(t)−[x′(t)3y′(t)​]⋅x′′(t)dx2d2y​=[x′(t)3x′(t)​]⋅y′′(t)−[x′(t)3y′(t)​]⋅x′′(t)=x′(t)3∣ ∣​x′(t)y′(t)​x′′(t)y′′(t)​∣ ∣​​​

曲率公式为： k = ∣ y ′ ′ ( 1 + y ′ 2 ) 3 2 ∣ \displaystyle{ k=\left|\frac{y^{\prime \prime}}{\left(1+y^{\prime 2}\right)^{\frac{3}{2}}}\right| }% k=∣ ∣​(1+y′2)23​y′′​∣ ∣​ ，这意味着，我们求解曲率的核心诉求转变为求解一阶导和二阶导的值。

由于在复合函数求二阶导的过程中，我们计算二阶导是将 x x x， y y y 分别看作 t t t 的函数，而这也恰恰符合参数方程的形式，于是，对于参数方程的二阶导数，我们依然有与【复合函数的二阶导数】相同的结论。

然后代入曲率公式，经计算化简可以得到： y = y ( x )    ,    x = x ( t ) x ′ ( t ) = d x d t   ,   y ′ ( t ) = d y d t   ,   x ′ ′ ( t ) = d 2 x d t 2   ,   y ′ ′ ( t ) = d 2 y d t 2 d 2 y d x 2 = [ x ′ ( t ) x ′ ( t ) 3 ] ⋅ y ′ ′ ( t ) − [ y ′ ( t ) x ′ ( t ) 3 ] ⋅ x ′ ′ ( t ) = ∣ x ′ ( t ) x ′ ′ ( t ) y ′ ( t ) y ′ ′ ( t ) ∣ x ′ ( t ) 3 k = ∣ x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) ∣ ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3 2 = ∣ ∣ x ′ ( t ) x ′ ′ ( t ) y ′ ( t ) y ′ ′ ( t ) ∣ ∣ ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3 \begin{aligned} & y=y(x) \ \ , \ \ x=x(t)\\ \\ & x'(t)=\frac{dx}{dt} \ , \ y'(t)=\frac{dy}{dt} \ , \ x''(t)=\frac{d^2x}{dt^2} \ , \ y''(t)=\frac{d^2y}{dt^2}\\ & \frac{d^2y}{dx^2}=[\frac{x'(t)}{x'(t)^3}]\cdot y''(t)-[\frac{y'(t)}{x'(t)^3}]\cdot x''(t)=\frac{\left|\begin{array}{ll} x'(t) & x''(t) \\ y'(t) & y''(t) \end{array}\right|}{x'(t)^3} \\ \\ & k = \frac{|x'(t)y''(t)-x''(t)y'(t)|}{(x'(t)^2+y'(t)^2)^{\frac32}} =\frac{\left|\left|\begin{array}{ll} x'(t) & x''(t) \\ y'(t) & y''(t) \end{array}\right|\right|}{(\sqrt{x'(t)^2+y'(t)^2})^3} \\ \end{aligned} ​y=y(x)  ,  x=x(t)x′(t)=dtdx​ , y′(t)=dtdy​ , x′′(t)=dt2d2x​ , y′′(t)=dt2d2y​dx2d2y​=[x′(t)3x′(t)​]⋅y′′(t)−[x′(t)3y′(t)​]⋅x′′(t)=x′(t)3∣ ∣​x′(t)y′(t)​x′′(t)y′′(t)​∣ ∣​​k=(x′(t)2+y′(t)2)23​∣x′(t)y′′(t)−x′′(t)y′(t)∣​=(x′(t)2+y′(t)2 ​)3∣ ∣​∣ ∣​x′(t)y′(t)​x′′(t)y′′(t)​∣ ∣​∣ ∣​​​ 在代入方程时，我们可以通过表格法参数方程作为辅助求解曲率（代入数值的手段），

遵循【交叉相乘再相减，平方求和开根号】的原则，

以题目作为示例： 曲线 { x = t 2 + 2 t y = 3 ln ⁡ t  上对应于  t = 1  的点处的曲率是 曲线 \left\{\begin{array}{l}x=t^{2}+2 t \\ y=3 \ln t\end{array}\right.\ 上对应于\ t=1\ 的点处的曲率是 曲线{x=t2+2ty=3lnt​ 上对应于 t=1 的点处的曲率是 解答过程： k = ∣ ∣ x ′ ( t ) x ′ ′ ( t ) y ′ ( t ) y ′ ′ ( t ) ∣ ∣ ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3 = ∣ ∣            ∣ ∣ ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3    ( 草稿纸写这个即可 ) k =\frac{\left|\left|\begin{array}{ll} x'(t) & x''(t) \\ y'(t) & y''(t) \end{array}\right|\right|}{(\sqrt{x'(t)^2+y'(t)^2})^3}= \frac{\left|\left|\begin{array}{ll} & \\ & \ \ \ \ \ \ \ \ \ \ \end{array}\right|\right|}{(\sqrt{x'(t)^2+y'(t)^2})^3}\ \ (草稿纸写这个即可) k=(x′(t)2+y′(t)2 ​)3∣ ∣​∣ ∣​x′(t)y′(t)​x′′(t)y′′(t)​∣ ∣​∣ ∣​​=(x′(t)2+y′(t)2 ​)3∣ ∣​∣ ∣​​          ​∣ ∣​∣ ∣​​  (草稿纸写这个即可)

∣ 4     2 3 − 3 ∣ = − 18 →   18 4 2 + 3 2 = 5   →   5 3 = 125 k =    18    125 \begin{aligned} & \left|\begin{array}{ll} 4 & \ \ \ 2 \\ 3 & -3 \end{array}\right| = -18 \rightarrow \ 18\\ \\ & \sqrt{4^2+3^2}=5\ \rightarrow\ 5^3=125\\ \\ & k=\frac{\ \ 18}{\ \ 125} \end{aligned} ​∣ ∣​43​   2−3​∣ ∣​=−18→ 1842+32 ​=5 → 53=125k=  125  18​​

解答完毕。

特别鸣谢：公式推导与表格法灵感来源于 C C L CCL CCL .