A lighthouse is 1km off shore. A marathoner is running along the shore. The search light on the lighthouse is rotating at a rate of 1/6 revolutionper minute when the runner is 3km from the point on...

A lighthouse is 1km off shore. A marathoner is running along the shore. The search light on the lighthouse is rotating at a rate of 1/6 revolution

per minute when the runner is 3km from the point on the shore that is closest to the lighthouse. The runner is illuminated by the beam. How fast is the beam of light moving along the shore at this moment?

Asked on by islnds

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nathanshields | High School Teacher | (Level 1) Associate Educator

Posted on

Get that diagram going!  Another triangle - lighthouse (point A) to shore (point B) is 1 km, then right angle, then shore to runner (point C) is 3 km. This time the runner's speed doesn't matter - we're interested in the speed of the light along the shore.  So forget about the runner - pretend he's a rock called "Point C".

It also doesn't matter which direction the light is moving - let's just says it's getting farther away from point B.  Call the point where the light hits the shore D (so we can move it around), and let's refer to the distance between B and D as length l (that's a lower case L).

Whew, glad that's over.  Now we're interested in how fast length l is changing.

What else is changing?  We don't care about the runner.  The angle BAD (let's call this angle L) is changing, since the light in the lighthouse is spinning around.  How fast is it changing?  1/6 rev/sec.  = 1/6 (360) deg/sec = 1/6 (2pi) rad/sec.

We always use radians, so let's say dL/dt = pi/3

Now we need to relate angle L and distance l.  If you recall your trig, tan L = l/1.  So, L = arctan l.

Drop a derivative on both sides:

`d/dtL=d/dt arctan l`

`(dL)/dt=1/(1+l^2)*(dl)/dt`

Now remember, we're interested in l right when it's 3km from point B, so let l = 3 and let dL/dt = pi/3

`pi/3=1/(1+3^2)*(dl)/dt`

Now you just solve for dl/dt... I hope you get 10pi/3... but what are the units?

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