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...

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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?