灯塔光束沿海岸移动有多快

设海岸线上距离最近点的距离为 $x$，设光束与垂线夹角为 $\theta$，则

$$\tan\theta=\frac{x}{L}\Rightarrow x=L\tan\theta.$$

灯光角速度为 $\omega=2\pi$（弧度/分钟）。对时间求导：

$$\frac{dx}{dt}=L\sec^2\theta\cdot\frac{d\theta}{dt}=\omega L\sec^2\theta =\omega L\left(1+\tan^2\theta\right)=\omega L\left(1+\left(\frac{x}{L}\right)^2\right).$$

因此沿海岸线速度随 $x$ 增大而增大，并在 $\theta\to \pi/2$（光束趋于平行海岸线）时趋于无穷大。

当 $x=3L$ 时，$\sec^2\theta=1+(3)^2=10$，所以

$$\frac{dx}{dt}=\omega L\cdot 10=2\pi\cdot 3\cdot 10=\boxed{60\pi\ \text{英里/分钟}}.$$

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英文解析

Let $x$ be the distance from the closest point on the coastline, and let $\theta$ be the angle between the light beam and the vertical. Then

$$\tan\theta=\frac{x}{L}\Rightarrow x=L\tan\theta.$$

The angular velocity of the light is $\omega=2\pi$ (radians/minute). Differentiating with respect to time:

$$\frac{dx}{dt}=L\sec^2\theta\cdot\frac{d\theta}{dt}=\omega L\sec^2\theta =\omega L\left(1+\tan^2\theta\right)=\omega L\left(1+\left(\frac{x}{L}\right)^2\right).$$

Thus, the speed along the coastline increases as $x$ increases and approaches infinity as $\theta\to \pi/2$ (the light beam becomes parallel to the coastline).

When $x=3L$, $\sec^2\theta=1+(3)^2=10$, so

$$\frac{dx}{dt}=\omega L\cdot 10=2\pi\cdot 3\cdot 10=\boxed{60\pi\ \text{miles/minute}}.$$