2017年考研数二第10题解析

一、题目

二、解析

本题就是考察参数方程求导。

由于：

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

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

$$
\frac{\mathrm{d} y}{\mathrm{d} t} = \cos t
$$

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

于是：

$$
\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\cos t}{1 + \mathrm{e}^{t}} = y^{\prime}
$$

则：

$$
\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} = \frac{\mathrm{d} y^{\prime}}{\mathrm{d} x} = \frac{\frac{\mathrm{d} y^{\prime}}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}}
$$

又：

$$
\frac{\mathrm{d} y^{\prime}}{\mathrm{d} t} = (\frac{\cos t}{1 + \mathrm{e}^{t}})^{\prime} = \frac{- \sin t (1 + \mathrm{e}^{t}) – \cos t (\mathrm{e}^{t})}{(1 + \mathrm{e}^{t})^{2}}
$$

于是：

$$
\begin{aligned}
\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} \\ \\
& = \frac{- \sin t (1 + \mathrm{e}^{t}) – \cos t (\mathrm{e}^{t})}{(1 + \mathrm{e}^{t})^{2}} \cdot \frac{1}{1 + \mathrm{e}^{t}} \\ \\
& = \frac{- \sin t (1 + \mathrm{e}^{t}) – \cos t (\mathrm{e}^{t})}{(1 + \mathrm{e}^{t})^{3}}
\end{aligned}
$$

当 $t=0$ 时：

$$
\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}} = \frac{0-1 \cdot 1}{(1+1)^{3}} = – \frac{1}{8}
$$

综上可知，正确答案为 $- \frac{1}{8}$.

EOF

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