高等数学笔记：复合函数的二阶导数与参数方程求解曲率

繁星数学随想录·技巧卷

复合函数的二阶导数与参数方程求解曲率

复合函数的二阶导数

通过函数乘法求导运算法则，经计算可得结果：
$$\begin{array}{rl}& y=y(x),x=x(t)\\ & \\ & x^{\prime }(t)=\frac{dx}{dt},y^{\prime }(t)=\frac{dy}{dt},x^{\prime \prime }(t)=\frac{d^{2}x}{d{t}^2},y^{\prime \prime }(t)=\frac{d^{2}y}{d{t}^2}\\ & \\ & \frac{d^{2}y}{d{x}^2}=[]\cdot y^{\prime \prime }(t)-[]\cdot x^{\prime \prime }(t)\\ & \\ & \frac{d^{2}y}{d{x}^2}=[\frac{x^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot y^{\prime \prime }(t)-[\frac{y^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot x^{\prime \prime }(t)\\ & \frac{d^{2}y}{d{x}^2}=[\frac{x^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot y^{\prime \prime }(t)-[\frac{y^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot x^{\prime \prime }(t)=\frac{\left|\begin{array}{ll}{x}^{\prime }(t)& x^{\prime \prime }(t)\\ y^{\prime }(t)& y^{\prime \prime }(t)\end{array}\right|}{x^{\prime }(t{)}^3}\end{array}$$

参数方程求解曲率

曲率公式为： $k=\left|\frac{y^{\prime \prime }}{{\left(1+y^{\prime 2}\right)}^{\frac{3}{2}}}\right|$ ，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 $x$，$y$ 分别看作 $t$ 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：
$$\begin{array}{rl}& y=y(x),x=x(t)\\ & \\ & x^{\prime }(t)=\frac{dx}{dt},y^{\prime }(t)=\frac{dy}{dt},x^{\prime \prime }(t)=\frac{d^{2}x}{d{t}^2},y^{\prime \prime }(t)=\frac{d^{2}y}{d{t}^2}\\ & \frac{d^{2}y}{d{x}^2}=[\frac{x^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot y^{\prime \prime }(t)-[\frac{y^{\prime }(t)}{x^{\prime }(t{)}^3}]\cdot x^{\prime \prime }(t)=\frac{\left|\begin{array}{ll}{x}^{\prime }(t)& x^{\prime \prime }(t)\\ y^{\prime }(t)& y^{\prime \prime }(t)\end{array}\right|}{x^{\prime }(t{)}^3}\\ & \\ & k=\frac{|x^{\prime }(t)y^{\prime \prime }(t)-x^{\prime \prime }(t)y^{\prime }(t)|}{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2{)}^{\frac{3}{2}}}=\frac{\left|\left|\begin{array}{ll}{x}^{\prime }(t)& x^{\prime \prime }(t)\\ y^{\prime }(t)& y^{\prime \prime }(t)\end{array}\right|\right|}{(\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}{)}^3}\end{array}$$
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，

	${}^{\prime }$	${}^{\prime \prime }$
$x$	$x^{\prime }(t)=x^{\prime }(t_0)$	$x^{\prime \prime }(t)=x^{\prime \prime }(t_0)$
$y$	$y^{\prime }(t)=y^{\prime }(t_0)$	$y^{\prime \prime }(t)=y^{\prime \prime }(t_0)$

以题目作为示例：
$$曲线\{\begin{array}{l}x=t^2+2t\\ y=3\ln t\end{array}上对应于t=1的点处的曲率是$$
解答过程：
$$k=\frac{\left|\left|\begin{array}{ll}{x}^{\prime }(t)& x^{\prime \prime }(t)\\ y^{\prime }(t)& y^{\prime \prime }(t)\end{array}\right|\right|}{(\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}{)}^3}=\frac{\left|\left|\begin{array}{ll}& \\ & \end{array}\right|\right|}{(\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}{)}^3}(草稿纸写这个即可)$$

t=1	${}^{\prime }$	${}^{\prime \prime }$
$x$	$2t+2=4$	$2$
$y$	$3/t=3$	$-3/t^2=-3$

$$\begin{array}{rl}& \left|\begin{array}{ll}4& 2\\ 3& -3\end{array}\right|=-18\to 18\\ & \\ & \sqrt{4^2+3^2}=5\to 5^3=125\\ & \\ & k=\frac{18}{125}\end{array}$$

解答完毕。

特别鸣谢：公式推导与表格法灵感来源于 $CCL$ .