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Last edited by hbghlyj 2025-4-25 09:36回复 20# lemondian
用这个可以得到“K神”的切线结论，还是觉得麻烦了此，是不是还有更简单的方法呢？
［问题］：求曲线 $\rho=\rho(\theta)$ 在对应于 $\theta=\theta_0$ 的点处的切线方程。
［推导］：将曲线 $\rho=\rho(\theta)$ 的方程改写为直角坐标系下的参数方程形式
$$
\left\{\begin{array}{l}
x=\rho(\theta) \cos \theta \\
y=\rho(\theta) \sin \theta
\end{array}\right.
$$
从而得到极坐标系下曲线 $\rho=\rho(\theta)$ 上对应于 $\theta=\theta_0$ 的点的直角坐标为
$$
\left\{\begin{array}{l}
x_0=\rho\left(\theta_0\right) \cos \theta_0 \\
y_0=\rho\left(\theta_0\right) \sin \theta_0
\end{array}\right.
$$
根据参数方程所确定的函数的求导方法有
$$
\frac{d y}{d x}=\frac{y^{\prime}(\theta)}{x^{\prime}(\theta)}=\frac{\rho^{\prime}(\theta) \sin \theta+\rho(\theta) \cos \theta}{\rho^{\prime}(\theta) \cos \theta-\rho(\theta) \sin \theta}
$$
所以，所求的切线在直角坐标系下的方程为
$$
y-\rho\left(\theta_0\right) \sin \theta_0=\frac{\rho^{\prime}\left(\theta_0\right) \sin \theta_0+\rho\left(\theta_0\right) \cos \theta_0}{\rho^{\prime}\left(\theta_0\right) \cos \theta_0-\rho\left(\theta_0\right) \sin \theta_0}\left[x-\rho\left(\theta_0\right) \cos \theta_0\right]
$$于是，所求的切线在极坐标系下的方程为
$$
\rho \sin \theta-\rho\left(\theta_0\right) \sin \theta_0=\frac{\rho^{\prime}\left(\theta_0\right) \sin \theta_0+\rho\left(\theta_0\right) \cos \theta_0}{\rho^{\prime}\left(\theta_0\right) \cos \theta_0-\rho\left(\theta_0\right) \sin \theta_0}\left[\rho \cos \theta-\rho\left(\theta_0\right) \cos \theta_0\right]
$$
下面对其变形整理
$$
\begin{gathered}
\rho \cos \theta \frac{\rho^{\prime}\left(\theta_0\right) \sin \theta_0+\rho\left(\theta_0\right) \cos \theta_0}{\rho^{\prime}\left(\theta_0\right) \cos \theta_0-\rho\left(\theta_0\right) \sin \theta_0}-\rho \sin \theta=\rho\left(\theta_0\right) \cos \theta_0 \frac{\rho^{\prime}\left(\theta_0\right) \sin \theta_0+\rho\left(\theta_0\right) \cos \theta_0}{\rho^{\prime}\left(\theta_0\right) \cos \theta_0-\rho\left(\theta_0\right) \sin \theta_0}-\rho\left(\theta_0\right) \sin \theta_0 \\
\rho\left[\rho^{\prime}\left(\theta_0\right) \cos \theta \sin \theta_0+\rho\left(\theta_0\right) \cos \theta \cos \theta_0-\rho^{\prime}\left(\theta_0\right) \sin \theta \cos \theta_0+\rho\left(\theta_0\right) \sin \theta \sin \theta_0\right] \\
=\rho\left(\theta_0\right)\left[\rho^{\prime}\left(\theta_0\right) \cos \theta_0 \sin \theta_0+\rho\left(\theta_0\right) \cos ^2 \theta_0-\rho^{\prime}\left(\theta_0\right) \cos \theta_0 \sin \theta_0+\rho\left(\theta_0\right) \sin ^2 \theta_0\right] \\
\rho\left[\rho\left(\theta_0\right) \cos \left(\theta-\theta_0\right)-\rho^{\prime}\left(\theta_0\right) \sin \left(\theta-\theta_0\right)\right]=\left[\rho\left(\theta_0\right)\right]^2 \\
\rho=\frac{\left[\rho\left(\theta_0\right)\right]^2}{\rho\left(\theta_0\right) \cos \left(\theta-\theta_0\right)-\rho^{\prime}\left(\theta_0\right) \sin \left(\theta-\theta_0\right)}
\end{gathered}
$$
所以，所求的切线在极坐标系下的方程为
$$
\rho=\frac{\left[\rho\left(\theta_0\right)\right]^2}{\rho\left(\theta_0\right) \cos \left(\theta-\theta_0\right)-\rho^{\prime}\left(\theta_0\right) \sin \left(\theta-\theta_0\right)}
$$ |
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Author: lemondian
isee
Posted 2018-4-27 18:39
