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Numerade Educator



Problem 44 Hard Difficulty

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 from $ P? $


The light is moving at the speed of $\frac{80 \pi}{3}$ kilometres/minute


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Video Transcript

All right. We've got a lighthouse that is located on a small island three kilometers from the nearest point P on a straight shoreline and it's light. Makes four revolution for a minute were asked to calculate how fast the beam of light moving along the shoreline. How fast is the beam of light moving along the shoreline when it is one kilometers from P? That's the first. Let's go ahead and draw out a figure. Mhm. You've got our cue here that are p here with that point A here. We know it's three kilometers way and we've got angle data. All right, now have exhale. Okay, Now we know that Q is a point on the shoreline. We know X is the distance from Point P and A is the lighthouse. So and we're told that it makes four revolutions per minute so we can calculate for a change and data over tea as being four times two pi. So we'll get eight. Let's write about. We got four revolutions multiplied by, um, two pi and then we can get the units and radiance per minute. So we'll be at eight. I radiance per All right now we calculate for X. We know our X would be the same thing as tangent data opposite over adjacent, which is X over three. So if we calculate for DX DT, we could do X d t, for example. He did the what I mean to say is if we took the derivative of this equation here, we could get the change of X over time. And if we multiply both sides by three, we have a three on this side. We can cancel it out from here, and then we take the derivative of tan data. We know that it's the same thing seeking to swear data. And then we have a D data over DT. All right now we could start to plug in our values. We know that our seeking data can be written. Seeking square data can be written as 8 10 squared data. If we use identities plus one and we have a deep data we've already calculated for debate over DT, which is eight pi radiance. Okay, so now if we can substitute Pan data for X over three and we know X is gonna be one in this situation because we're told that is moving along the shoreline when it when it is one kilometer from P. So we could substitute one for X. And we could substitute this whole thing for X or three. So we have 1/3 squared. That's one times three. I'm a pie radiance equal to the x T. T. Go ahead and remove the units for now, because you have to also account for the fact that this is this right here would be a kilometer. We have DX DT actually get 1/9 plus one, which is 10/9. And then if you multiplied by three, you'll have 10/3 and then 10/3 times ate pie. It's just 80 over three pie. And then for our units, the radiance would cancel out. You have the kilometer over a minutes that is equal to your DX over DT. All right, so that'll be the change in X over time. All right, well, I hope that clarifies the question there, and we could say the light is moving at the speed of 8. 80. Pie over three kilometers permit. All right. Thank you for watching