A revolving beacon is situated 925 m from a straight shore.

[peblez]

A revolving beacon is situated 925 m from a straight shore. it turns at 2 rev/min. how fast does thte beam sweep along the shore at its nearest point? how fast does it sweep along the shore at a point 1275 m from the nearest point?

I really have no clue on how to tackle this problem. Can someone show me how to do this one? Thank you.

 

[soroban]

Re: Beacon question

Hello, peblez!

Did you make a sketch?

A revolving beacon is situated 925 m from a straight shore.
It turns at 2 rev/min.
(a) How fast does the beam sweep along the shore at its nearest point?
(b) How fast does it sweep along the shore at a point 1275 m from the nearest point?

          A
          *
          |   *
          |  θ    *
      925 |           *
          |               *
          |                   *
          * - - - - - - - - - - - *
          B           x           C

The beacon is at A. .The nearest point on shore is B. .\(\displaystyle AB\,=\,925\) m.

The beam shines to point C. .Let \(\displaystyle x\,=\,BC.\)

Let \(\displaystyle \theta\,=\,\angle A.\)

We are told that: \(\displaystyle \:\frac{d\theta}{dt}\:=\:2\text{ rev/min} \:=\:4\pi\text{ radians/min}\)

From right triangle \(\displaystyle ABC\), we have: \(\displaystyle \:\tan\theta \:=\:\frac{x}{925}\;\;\Rightarrow\;\;x\:=\:925\cdot\tan\theta\)

Differentiate with respect to time: \(\displaystyle \:\frac{dx}{dt}\:=\:925\cdot\sec^2\theta\cdot\frac{d\theta}{dt}\)


(a) When the beam is at \(\displaystyle B:\:\theta\,=\,0\)

Hence, we have: \(\displaystyle \:\frac{dx}{dt}\:=\:925(\sec^20)(4\pi) \:=\:3700\pi\) m/min.


(b) When \(\displaystyle x\,=\,1275:\;\tan\theta\,=\,\frac{1275}{925}\:=\:\frac{51}{37}\)

Using the identity: \(\displaystyle \,\sec^2\theta\:=\:\tan^2\theta\,+\,1\),
. . we have: \(\displaystyle \sec^2\theta\:=\:\left(\frac{51}{37}\right)^2\,+\,1\;\;\Rightarrow\;\;\sec^2\theta\,=\,\frac{3970}{1369}\)

Then: \(\displaystyle \:\frac{dx}{dt}\:=\:425\left(\frac{3910}{1369\right)(4\pi) \:=\:\frac{14,689,000\pi}{1369}\) m/min.