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October 25, 2014

October 25, 2014

Posted by **Jon** on Sunday, November 11, 2007 at 2:32pm.

I need to show work, so formatting answers in this manner would be most appreciated. Thanks in advance! :) :)

- Calc -
**drwls**, Sunday, November 11, 2007 at 3:09pm3 rev/min = 6 pi radians/min

Multiply that by the 1 km radial distance to get the speed the beam apparears to move at that distance:

6 pi x 1 = 18.85 km/min

Multiply that by 60 min/hr to get the spped in km per hour

- Calc -
**Reiny**, Sunday, November 11, 2007 at 4:57pmI beg to differ with the usually correct drwls.

This is a related rate problem dealing with angular velocity, so trig will be needed.

Let the point on the shore be x km down the shoreline.

so we have a right angled triangle, let the angle at the lighthouse be ß

we are given dß/dt = 6pi/min

The tanß = x/1

and sec^2 ß(dß/dt) = dx/dt

when x = 1/2 and using Pythagoras I found sec^2 ß = 1.25

so 1.25(6pi) = dx/dt

so dx/dt = 23.5629 km/min

= 1413.72 km/h

- Calc -
**drwls**, Sunday, November 11, 2007 at 6:30pmReiny is correct; I did not fully read the problem and assumed the shore was perpendicular to the beam

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