Light Beam Speed on Shoreline from P - 16km

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SUMMARY

The discussion focuses on calculating the speed of a lighthouse light beam moving along a shoreline, specifically when it shines on a point 3 km from point P, which is 16 km offshore. The light beam completes 5 revolutions per minute, translating to an angular velocity of $\dfrac{\pi}{6}$ rad/sec. By applying trigonometric relationships in the right triangle formed by the light beam, the shoreline, and the perpendicular distance from the lighthouse, the rate of change of the beam's position along the shoreline, $\dfrac{dx}{dt}$, can be determined. The calculation yields a specific speed at the given distance.

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A lighthouse is located on a small island 16 km off-shore from the nearest point P on a straight shoreline. Its light makes 5 revolutions per minute. How fast is the light beam moving along the shoreline when it is shining on a point 3 km along the shoreline from P?
 
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$\dfrac{d\theta}{dt}$ = 5 rpm = $\dfrac{10\pi}{60 \, sec} = \dfrac{\pi}{6}$ rad/sec

consider the right triangle formed by the light beam, the shoreline, and the perpendicular distance from the light house to the shoreline (recommend you make a sketch)

let $\theta$ be the angle between the light beam and the perpendicular distance, and $x$ be the distance from where the light beam intersects the shoreline to where the perpendicular distance segment intersects the shoreline. You are given the fixed perpendicular distance.

Using the aforementioned right triangle, write a trig equation that relates $x$, $\theta$, and the 16 km distance.

Take the time derivative of the equation and determine $\dfrac{dx}{dt}$ when $x=3$ km.

Mind your units.
 
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