How Fast is the Lighthouse Beam Moving?Date: 10/13/2002 at 20:29:57 From: Leigh Subject: Rates and derivatives A lighthouse is located on a small island 3 km away from the nearest point P on a straight shoreline, and its light makes four revolutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km away from P? Date: 10/15/2002 at 13:45:29 From: Doctor Roy Subject: Re: Rates and derivatives Hi, Thanks for writing to Dr. Math. Let's draw a quick picture to make the situation clearer: 3km L ---------------P \ | \ | \ | \ | 1km x \ | \ | \ | A Please note that the picture is not to scale. Point A is where the beam of light hits the shore when it is 1 km away from P, and x represents the length of the beam of light from the lighthouse. We can find the value of x by using the Pythagorean theorem: x^2 = 1^2 + 3^2 x^2 = 1 + 9 x^2 = 10 x = sqrt(10) So, x is sqrt(10) km long, or roughly 3.16 km long. Another fact we need to know is the rotational speed of the lighthouse. The lighthouse beam makes 4 full revolutions in a minute. So, we can express this as: w = 4 rev/min Do you remember the formula for the circumference of a circle? It is given by: C = 2*pi*r The circumference is the same as the distance traveled around the circle, or 1 revolution, so 1 rev = 2*pi*r So the next step is to find the speed along the circumference of a circle of an object moving at 4 rev/min. The radius will be sqrt(10). It is sqrt(10), because that is the "current" distance from the lighthouse to the point it lights on the shoreline. Remember that this distance will change as the light moves. So, the light "at this current moment" travels: 4 rev/min = 4*1 rev/min = 4 * (2*pi*r km) / min = 4 * (2 * pi * sqrt(10) km) / min = 8 pi sqrt(10) km/min Note again that the distance of A from P makes a difference in the speed. The lighthouse turns at the same rate, but the distance from the lighthouse to the shoreline changes. This means the light will appear to move faster the farther from P we get. Note also that we're not done yet! We have the speed of the light along the circumference of the circle. But we _want_ to find the speed along the shorline. 3km L ---------------P \ | \ | \ | \ | 1km x \ | \ | \ | / A/ /| / | / / <---- The speed in _this_ direction is 8 pi sqrt(10) km Trigonometry works with speeds as well as with distances, 3km L ---------------P \ | \ | \ | \ | 1km x \ | \ | \ | / A/ /| v / a| ? / | /______| so once you've figured out the measure of the angle marked 'a', you can find the speed along the shore using the equation ? cos(a) = - v Does this help? - Doctor Roy, The Math Forum http://mathforum.org/dr.math/ Date: 10/15/2002 at 22:14:33 From: leigh Subject: Thank you (Rates and derivatives) Thank you so much! I understand it now. Leigh |
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