同济高数 第2章 第2-4-*10题
📝 题目
＊10．求下列参数方程所确定的函数的三阶导数 $\displaystyle \frac{\mathrm{d}^{3} y}{\mathrm{~d} x^{3}}$ ：
（1）$\left\{\begin{array}{l}x=1-t^{2}, \\ y=t-t^{3} ;\end{array}\right.$
（2）$\left\{\begin{array}{l}x=\ln \left(1+t^{2}\right), \\ y=t-\arctan t .\end{array}\right.$
💡 答案解析
[AI解答]
[AI解答]
（1） 已知参数方程
$$
\begin{cases}
x = 1 - t^2, \\
y = t - t^3.
\end{cases}
$$
先求一阶导数：
$$
\frac{dy}{dx} = \frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}} = \frac{1 - 3t^2}{-2t} = \frac{3t^2 - 1}{2t}.
$$
再求二阶导数：
$$
\frac{d^2y}{dx^2} = \frac{\displaystyle\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\displaystyle\frac{dx}{dt}}
= \frac{\displaystyle\frac{d}{dt}\left( \frac{3t^2 - 1}{2t} \right)}{-2t}.
$$
先计算分子：
$$
\frac{d}{dt}\left( \frac{3t^2 - 1}{2t} \right)
= \frac{(6t)(2t) - (3t^2 - 1)(2)}{4t^2}
= \frac{12t^2 - 6t^2 + 2}{4t^2}
= \frac{6t^2 + 2}{4t^2}
= \frac{3t^2 + 1}{2t^2}.
$$
因此
$$
\frac{d^2y}{dx^2} = \frac{\displaystyle\frac{3t^2 + 1}{2t^2}}{-2t} = -\frac{3t^2 + 1}{4t^3}.
$$
最后求三阶导数：
$$
\frac{d^3y}{dx^3} = \frac{\displaystyle\frac{d}{dt}\left( \frac{d^2y}{dx^2} \right)}{\displaystyle\frac{dx}{dt}}
= \frac{\displaystyle\frac{d}{dt}\left( -\frac{3t^2 + 1}{4t^3} \right)}{-2t}.
$$
先计算分子导数：
$$
\frac{d}{dt}\left( -\frac{3t^2 + 1}{4t^3} \right)
= -\frac{1}{4} \cdot \frac{(6t)(t^3) - (3t^2+1)(3t^2)}{t^6}
= -\frac{1}{4} \cdot \frac{6t^4 - 9t^4 - 3t^2}{t^6}
= -\frac{1}{4} \cdot \frac{-3t^4 - 3t^2}{t^6}
= -\frac{1}{4} \cdot \frac{-3t^2(t^2+1)}{t^6}
= \frac{3(t^2+1)}{4t^4}.
$$
因此
$$
\frac{d^3y}{dx^3} = \frac{\displaystyle\frac{3(t^2+1)}{4t^4}}{-2t} = -\frac{3(t^2+1)}{8t^5}.
$$
所以（1）的结果为：
$$
\boxed{-\frac{3(t^2+1)}{8t^5}}.
$$
---
（2） 已知参数方程
$$
\begin{cases}
x = \ln(1+t^2), \\
y = t - \arctan t.
\end{cases}
$$
先求一阶导数：
$$
\frac{dx}{dt} = \frac{2t}{1+t^2}, \quad \frac{dy}{dt} = 1 - \frac{1}{1+t^2} = \frac{t^2}{1+t^2}.
$$
因此
$$
\frac{dy}{dx} = \frac{\displaystyle\frac{t^2}{1+t^2}}{\displaystyle\frac{2t}{1+t^2}} = \frac{t}{2}.
$$
再求二阶导数：
$$
\frac{d^2y}{dx^2} = \frac{\displaystyle\frac{d}{dt}\left( \frac{t}{2} \right)}{\displaystyle\frac{dx}{dt}}
= \frac{\displaystyle\frac{1}{2}}{\displaystyle\frac{2t}{1+t^2}}
= \frac{1}{2} \cdot \frac{1+t^2}{2t} = \frac{1+t^2}{4t}.
$$
最后求三阶导数：
$$
\frac{d^3y}{dx^3} = \frac{\displaystyle\frac{d}{dt}\left( \frac{1+t^2}{4t} \right)}{\displaystyle\frac{dx}{dt}}.
$$
先计算分子：
$$
\frac{d}{dt}\left( \frac{1+t^2}{4t} \right)
= \frac{1}{4} \cdot \frac{(2t)(t) - (1+t^2)(1)}{t^2}
= \frac{1}{4} \cdot \frac{2t^2 - 1 - t^2}{t^2}
= \frac{1}{4} \cdot \frac{t^2 - 1}{t^2}.
$$
因此
$$
\frac{d^3y}{dx^3} = \frac{\displaystyle\frac{t^2 - 1}{4t^2}}{\displaystyle\frac{2t}{1+t^2}}
= \frac{t^2 - 1}{4t^2} \cdot \frac{1+t^2}{2t}
= \frac{(t^2 - 1)(1+t^2)}{8t^3}
= \frac{t^4 - 1}{8t^3}.
$$
所以（2）的结果为：
$$
\boxed{\frac{t^4 - 1}{8t^3}}.
$$
难度：★★☆☆☆