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A rotating beacon is located 1 kilometer off a straight shoreline. If the beacon rotates at a rate of 3 revolutions per minute, how fast (in kilometers per hour) does the beam of light appear to be moving to a viewer who is 1/2 kilometer down the shoreline?

Mathematics
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find the radius of the circle by finding the distance between the viewer and the beacon. then find the circumference by using \[2\pi r^2\] this will be the distance the light covers over 1 revolution. then using dimensional analysis: 3 rev/min (7.037 km/rev) (60 min/1 hr) = 1266.7 km/hr
This is similar to the related rates problem you asked earlier. But this one is harder since you have to be able to draw out the picture in order to visualize what’s going on. Since we can’t draw here, I’ll describe for you. The beacon is a point somewhere, call it B. From the beacon , draw a line segment representing 1 km. Let the endpoint of that segment be called S. That point S is the closest point on the Shore to the Beacon. Now draw a line perpendicular to SB through S. This line represents the shoreline. Now somewhere on that line, place a point X on it. This point X represents the location of the person. So now you should have a right triangle SXB. The idea of the solution is this. Imagine a wall directly on the shoreline. Imagine the beacon slowly rotating and watch what the light does as its moving along the wall. At point S, the light seems to be moving more slowly that it is at point X. Although the beacon is rotating at a constant speed, the apparent speed of the beam shining on the wall changes depending on where along the wall you are. If you’re really far from S, the beam moves really fast past you. That’s the intuition. To set up the related rates problem, we need to define our variables. Let Theta be the angle SBX. We know that d Theta / dt is given in the problem as 3 revolutions per minute. Now let the length SX be called x. This represents the distance the person is down the shoreline from the closest point on the shore to the beacon, S. We are also given that SB = 1 km. If you draw all this into your triangle and want a relation between Theta and x, you might see that Tan Theta = x / 1. That’s how Theta and x are related. How are their rates related? Well take the derivatives of both sides with respect to t. (sec Theta)^2 * Theta’ = x’. Now we want values for Theta and Theta’. We are already given Theta’. To find Theta, we know we’re given the value for x was ½ km But tan Theta = x, so Theta = arctan (x). So now we know Theta = arctan(1/2). Finally, plug both Theta and Theta’ in to find x’.

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