同济高数 第2章 第2-4-8题
📝 题目
8．写出下列曲线在所给参数值相应的点处的切线方程和法线方程：
（1）$\left\{\begin{array}{l}x=\sin t, \\ y=\cos 2 t,\end{array}\right.$ 在 $\displaystyle t=\frac{\pi}{4}$ 处；
（2）$\displaystyle \left\{\begin{array}{l}x=\frac{3 a t}{1+t^{2}}, \\ y=\frac{3 a t^{2}}{1+t^{2}},\end{array}\right.$ 在 $t=2$ 处．
💡 答案解析
[AI解答]
[AI解答]
(1) 曲线参数方程为
$$
\begin{cases}
x = \sin t, \\
y = \cos 2t,
\end{cases}
\quad t = \frac{\pi}{4}.
$$
先求对应点坐标：
$$
x_0 = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad
y_0 = \cos\frac{2\pi}{4} = \cos\frac{\pi}{2} = 0.
$$
求导数：
$$
\frac{dx}{dt} = \cos t, \quad \frac{dy}{dt} = -2\sin 2t.
$$
在 $t=\frac{\pi}{4}$ 处：
$$
\frac{dx}{dt}\Big|_{t=\frac{\pi}{4}} = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2},
\quad
\frac{dy}{dt}\Big|_{t=\frac{\pi}{4}} = -2\sin\frac{\pi}{2} = -2.
$$
切线斜率
$$
k = \frac{dy}{dx} = \frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}} = \frac{-2}{\frac{\sqrt{2}}{2}} = -2\sqrt{2}.
$$
切线方程：
$$
y - 0 = -2\sqrt{2}\left(x - \frac{\sqrt{2}}{2}\right)
\quad\Rightarrow\quad
y = -2\sqrt{2}\,x + 2.
$$
法线斜率 $k_{\perp} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$，法线方程：
$$
y = \frac{\sqrt{2}}{4}\left(x - \frac{\sqrt{2}}{2}\right)
\quad\Rightarrow\quad
y = \frac{\sqrt{2}}{4}x - \frac{1}{4}.
$$
(2) 曲线参数方程为
$$
\begin{cases}
x = \dfrac{3at}{1+t^{2}}, \$$6pt]
y = \dfrac{3at^{2}}{1+t^{2}},
\end{cases}
\quad t=2.
$$
先求对应点：
$$
x_0 = \frac{3a\cdot 2}{1+4} = \frac{6a}{5},\quad
y_0 = \frac{3a\cdot 4}{5} = \frac{12a}{5}.
$$
求导：
$$
\frac{dx}{dt} = 3a\cdot\frac{(1+t^2) - t(2t)}{(1+t^2)^2}
= 3a\cdot\frac{1 - t^2}{(1+t^2)^2},
$$
$$
\frac{dy}{dt} = 3a\cdot\frac{2t(1+t^2) - t^2(2t)}{(1+t^2)^2}
= 3a\cdot\frac{2t}{(1+t^2)^2}.
$$
在 $t=2$ 处：
$$
\frac{dx}{dt}\Big|_{t=2} = 3a\cdot\frac{1-4}{(1+4)^2} = 3a\cdot\frac{-3}{25} = -\frac{9a}{25},
$$
$$
\frac{dy}{dt}\Big|_{t=2} = 3a\cdot\frac{4}{(5)^2} = \frac{12a}{25}.
$$
切线斜率
$$
k = \frac{dy}{dx} = \frac{\frac{12a}{25}}{-\frac{9a}{25}} = -\frac{4}{3}.
$$
切线方程：
$$
y - \frac{12a}{5} = -\frac{4}{3}\left(x - \frac{6a}{5}\right).
$$
化简：
$$
y = -\frac{4}{3}x + \frac{8a}{5} + \frac{12a}{5}
= -\frac{4}{3}x + 4a.
$$
法线斜率 $k_{\perp} = \frac{3}{4}$，法线方程：
$$
y - \frac{12a}{5} = \frac{3}{4}\left(x - \frac{6a}{5}\right)
\quad\Rightarrow\quad
y = \frac{3}{4}x - \frac{9a}{10} + \frac{12a}{5}
= \frac{3}{4}x + \frac{3a}{2}.
$$
难度：★★☆☆☆