2017年考研数二第10题解析

题目

设函数 $y=y(x)$ 由参数方程 $\left\{\begin{matrix}
x = t + e^{t},\\
y = \sin t
\end{matrix}\right.$ 确定,则 $\frac{d^{2}y}{dx^{2}}|_{t=0}$ = $?$

解析

本题就是考察参数方程求导。

由于:

$$
\frac{d^{2}y}{dx^{2}}=
$$

$$
\frac{d}{dx}(\frac{dy}{dx}).
$$

$$
\frac{dy}{dx} =
$$

$$
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.
$$

$$
\frac{dy}{dt} =
$$

$$
\cos t.
$$

$$
\frac{dx}{dt} =
$$

$$
1 + e^{t}.
$$

于是:

$$
\frac{dy}{dx} =
$$

$$
\frac{\cos t}{1+e^{t}} = y^{‘}.
$$

则:

$$
\frac{d^{2}y}{dx^{2}}=
$$

$$
\frac{d y^{‘}}{dx} =
$$

$$
\frac{\frac{dy^{‘}}{dt}}{\frac{dx}{dt}}.
$$

又:

$$
\frac{dy^{‘}}{dt} =
$$

$$
(\frac{\cos t}{1+e^{t}})^{‘} =
$$

$$
\frac{- \sin t (1+e^{t}) – \cos t (e^{t})}{(1+e^{t})^{2}}.
$$

于是:

$$
\frac{d^{2}y}{dx^{2}}=
$$

$$
\frac{- \sin t (1+e^{t}) – \cos t (e^{t})}{(1+e^{t})^{2}} \cdot \frac{1}{1+e^{t}} =
$$

$$
\frac{- \sin t (1+e^{t}) – \cos t (e^{t})}{(1+e^{t})^{3}}.
$$

当 $t=0$ 时:

$$
\frac{d^{2}y}{dx^{2}}=
$$

$$
\frac{0-1 \cdot 1}{(1+1)^{3}} = – \frac{1}{8}.
$$

综上可知,正确答案为 $- \frac{1}{8}$.

EOF