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灯塔光束沿海岸移动有多快?

how fast is the light traveling along the coast?

专题
Brainteaser / 脑筋急转弯
难度
L4

题目详情

海岸线是一条直线,灯塔距离海岸的最近点为 L=3L=3 英里。灯光每分钟旋转 1 圈。

问:光束在海岸线上扫过的速度有多快?当光斑距离最近点为 3L3L 时,速度是多少?

Suppose there is a straight coastline and a lighthouse that is L=3\mathrm{L} = 3 miles away from the coast. The light revolves at one revolution per minute. How fast is the beam of light traveling along the coastline? When the beam is 3L away from the coastal point closest to the light, how fast is the light traveling along the coast?

解析

设海岸线上距离最近点的距离为 xx,设光束与垂线夹角为 θ\theta,则

tanθ=xLx=Ltanθ.\tan\theta=\frac{x}{L}\Rightarrow x=L\tan\theta.

灯光角速度为 ω=2π\omega=2\pi(弧度/分钟)。对时间求导:

dxdt=Lsec2θdθdt=ωLsec2θ=ωL(1+tan2θ)=ωL(1+(xL)2).\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 xx 增大而增大,并在 θπ/2\theta\to \pi/2(光束趋于平行海岸线)时趋于无穷大。

x=3Lx=3L 时,sec2θ=1+(3)2=10\sec^2\theta=1+(3)^2=10,所以

dxdt=ωL10=2π310=60π 英里/分钟.\frac{dx}{dt}=\omega L\cdot 10=2\pi\cdot 3\cdot 10=\boxed{60\pi\ \text{英里/分钟}}.