求解参数方程任意一点处的曲率

一、题目

已知 $y=y(x)$ 由参数方程 $\left\{\begin{array}{l}x=\frac{1}{2} \ln \left(1+t^{2}\right)
\\ y=\arctan t\end{array}\right.$ 确定，则 $\frac{\mathrm{d} y}{\mathrm{~d} x}=?$, $\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=?$, $y=y(x)$ 在任意点处的曲率 $K=?$

难度评级：

二、解析

1

$$
\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{\mathrm{d} t}{\mathrm{d} x} \Rightarrow
$$

$$
\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{1}{1+t^{2}}
$$

$$
\frac{\mathrm{d} x}{\mathrm{d} t}=\frac{1}{2} \cdot \frac{2 t}{1+t^{2}}=\frac{t}{1+t^{2}}
$$

$$
\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{1+t^{2}} \cdot \frac{1+t^{2}}{t}=\frac{1}{t}
$$

2

$$
\frac{d^{2} y}{\mathrm{d} x^{2}}=\frac{d}{\mathrm{d} x}\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right)=\frac{d}{\mathrm{d} t}\left(\frac{\mathrm{d} y}{\mathrm{d} x}\right) \cdot \frac{\mathrm{d} t}{\mathrm{d} x}=
$$

$$
\frac{d}{\mathrm{d} t}\left(\frac{1}{t}\right) \cdot \frac{\mathrm{d} t}{\mathrm{d} x}=\frac{-1}{t^{2}} \cdot \frac{1+t^{2}}{t}=\frac{-1-t^{2}}{t^{3}}
$$

3

$$
K=\frac{\left|y^{\prime \prime}\right|}{\left(1+y^{\prime 2}\right)^{\frac{3}{2}}}=\left|\frac{-1-t^{2}}{t^{3}}\right| \times \frac{1}{\left(1+\frac{1}{t^{2}}\right)^{\frac{3}{2}}} =
$$

$$
\frac{1+t^{2}}{\left|t^{3}\right|} \times \frac{1}{\left(\frac{t^{2}+1}{\textcolor{red}{t^{2}}}\right)^{\textcolor{red}{\frac{3}{2}}}}=\frac{1+t^{2}}{\left|t^{3}\right|} \times \frac{\textcolor{red}{|t|^{3}}}{\left(t^{2}+1\right)^{\frac{3}{2}}}=
$$

Tips:

平方运算自带绝对值效果，所以：$(t^{2})^{\frac{3}{2}} = |t|^{3}$

$$
\frac{1+t^{2}}{\left(1+t^{2}\right)^{\frac{3}{2}}}=\left(1+t^{2}\right)^{-\frac{1}{2}}=\frac{1}{\sqrt{t^{2}+1}} \Rightarrow
$$

$$
k=\frac{1}{\sqrt{t^{2}+1}}.
$$

拓展资料

1. 《什么是曲率？》
2. 《曲率半径的公式》

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