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Lighthouse problem A lighthouse stands 500 m Off a straigh shore. and the focused beam Of its light revolves (at constant rale) four limes each minule. As shown in ...

Question

Lighthouse problem A lighthouse stands 500 m Off a straigh shore. and the focused beam Of its light revolves (at constant rale) four limes each minule. As shown in the figure . P is the point on shore closest t0 the lighthouse and Q is point On the shore 200 m from P What is the speed of the beam along lhe shore when it strikes the point Q?

lighthouse problem A lighthouse stands 500 m Off a straigh shore. and the focused beam Of its light revolves (at constant rale) four limes each minule. As shown in the figure . P is the point on shore closest t0 the lighthouse and Q is point On the shore 200 m from P What is the speed of the beam along lhe shore when it strikes the point Q?



Answers

A lighthouse stands 500 m off of a straigh shore, the focused beam of its light revolving four times each minute, As shown in the figure, $P$ is the point on shore closest to the lighthouse and $Q$ is a point on the shore $200 \mathrm{m}$ from $P .$ What is the speed of the beam along the shore when it strikes the point $Q ?$ Describe how the speed of the beam along the shore varies with the distance between $P$ and $Q .$ Neglect the height of the lighthouse. (IMAGE CANT COPY)

Given the situation here, where we have a lighthouse that's 500 m away from the shore and we have the distance between Q and P, for instance, is 200 m. We basically want to find out how fast the the beam is moving along the shore. And so we want to set up the proper variable here because this is changing. We could go ahead and call this why and so what we're really interested in is we want to find out what do y DT is? How fast is that light moving up the shore? And that's what we're going to be solving for here. Um And so in order to do that, we need a couple of things. We need to first come up with a relationship between these variables, and so we'll need data here, and then we're going to end up using tangent of data is equal to the opposite. Which is why, because it changes, which is why it's not just 200. Divided by the 500 is constant, so opposite divided by adjacent, which is where we came up with that. Of course, that's our right angle there. Okay, so from here. We're gonna take the derivative with respect to T. But in order to do that, we're going to need to know what one of the rates is. And we do because we're giving the information up top here. Another way to re translate this is that this tells us how fast of data is changing over time. And any time you have revolutions or four times, you know, per unit time, it's gonna be four times two pi since two pi and radiance is one full revolution. So four times to pipe gives us a pie. Radiance per minute is our detail. Okay, so now that we know this, we could go ahead and continue on with, um differentiating on the left side down here. So we're going to come back and substitute this in in just a minute. So this will give us seeking squared of data that is the derivative of tangent multiplied by. We just differentiated data with respect to time. So it has a deep data. DT is equal to leave the one over 500. It's just a constant. And then that multiplied by d Y d t. And this is what we want to solve for here b Y d t. So let's go ahead and circle that. Let's also multiply the 500 over to the other side, the one over 500. So then it becomes just the reciprocal 500 over one recalculated purposes. Let's also rearrange so you can squared. That's the same thing as one over cosine squared. Yeah, And then we know deep Data DT in this given instance is gonna be a pie so we could go ahead and replace that with a pie. So up top. We've got 500 times a pie, all divided by CO since word of data. But we need to know what data is, and that's where we're going to set up tangent of data and then given the specific situation that we have, where point q is 200. So why is 200 divided by the adjacent just 500? All those zeros then canceled. Just cancel those. And then if we want to isolate data, we could take inverse tangent of both sides and so theta what equal inverse tangent of to threats to fifth. And then from here. Just, um, when you substitute this end or data here. Just make sure that you're in radiant mode as well, since we're talking about radiance. So we're gonna have 500 times ate pie, all divided by co sign of inverse tangent. 2/5. And then all of that cosine term squared. And so that will end up giving us our final solution. And I'm just gonna go back to it here. So when I did that, I got 4000 600 40 Hi. And that would be in the units of meters, her minutes or as an approximation, we could rewrite that as 14,000 five, 76 0.9, which would actually just rammed it up to 14,577 basically there. And that would have the same units of meters per minute. So that is how fast the light is shining up the shoreline for that part For part A Here, basically, for the first question, they're asking the speed of the beam when it strikes. Que Okay. So from there, then we want to find out how the speed of the beam along the shore varies with the distance between P and Q. So what would happen if Q. Was closer, right? If there was down here, for instance Well, then the angle Fada would be smaller. If that's true, then let's see. Um because we're dividing by that. What if data is very, very small? So, for instance, what if we have 500 times eight Hi and then divided by co sign of I'm just gonna enter in a really small value like 0.0 one. Um and then that squared. It would actually give us a slower rate of change here, and so this would end up giving us something like 12,556. So it's gonna be slower at first. And then the bigger that data gets, the faster it's gonna be. So it's going to be faster at Cube. And it would be at any point between Q and P a swell, and you could just test that again with some different test values and then enter those in and find out how that then effects d y d t. Given that whole equation that we have over here. So hopefully that helped to summarize the last one. Monta. It's bigger. In other words, then we could say do Y T is also greater. And so these have some type of proportional relationship where when one gets bigger data, de y DT also gets greater.

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

So if we sketched a diagram of our situation, we have a blue shoreline, a red light house that is three kilometers away from the shore. And we have a point down here that is X equals one kilometer from this point p and so we can create a high pod news and a right triangle right here where this distances D and so x squared it was three squared equals he squared. So we have one squared plus three squared equals C squared is going to give us d equals the square root of one plus nine or D equals route 10. Now, thinking about this angle theta, we're told that the lighthouse makes four revolutions permanent and we know that there's two pi radiance in each revolution. So de Sade anti tea is going to equal eight High radiance permanent. Now, if we look at our triangle, we have that Qianjin data is equal to x over three. If you take the derivative we have seeking square data times di stato DT is equal to d x t t over three and so seeking square data is going to be one over to send square data. So we have Route 10 over three squared times de Seda DT, which is a pie. And if we multiply both sides by three, we get that DX DT is equal to three times Route 10/3 squared times a pie, which is going to simplify to 10 over three times a pie or DX DT equals 80 pie over three kilometers permanent.

Thanks to chicken deserve it. Over the following we get by prime. Well, um, seeking or hyperbolic speaking in verse. That's your evidence is equal to negative one over or ex terms adults or into intact. But you did a negative X and then we have one minus or inside term squared. So what does that give us? Forget? Yeah, To the X negative. Each attacks over the square roots of one minus you to the negative two acts.


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