# Linear & Angular Velocity Related Rates

1. Dec 24, 2015

### Michele Nunes

1. The problem statement, all variables and given/known data
A rotating beacon is located 1 kilometer off a straight shoreline (see figure). 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?

2. Relevant equations

3. The attempt at a solution
Okay so I used the equation w = 2πf where w is the angular velocity and f is the frequency of rotation. I implicitly differentiated with respect to time, then plugged in 3 rev/min for df/dt, then converted to km/hr, and got 360π km/hr. The answer is 450π km/hr. I think my units are off somewhere or I'm using the wrong equation but I'm not sure.

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2. Dec 24, 2015

### Student100

What was the distance you used in the conversion? Can you type out your work step by step?

3. Dec 24, 2015

### Michele Nunes

d/dt[w] = d/dt[2πf]
dw/dt = 2π(df/dt)
dw/dt = 2π(3 rev/min) = 6π km/min = 360π km/hr

4. Dec 24, 2015

### Student100

What about the distance from the shore to the light house? How are you going from revs to km?

5. Dec 24, 2015

### Michele Nunes

I honestly don't know how to relate the units of revolutions to kilometers, that's what I'm not understanding, like they give me the frequency of revolutions of the beacon but I don't know what to do with that to end up getting kilometers/hour.

6. Dec 24, 2015

### Student100

First use your trig relationship to find the distance from the observer on the shore to the beacon house.

7. Dec 24, 2015

### Michele Nunes

Okay it's sqrt(5)/2 but what do I do with it? I've looked up angular velocity equations and they all have like 1 too many variables that I don't have enough information on.

8. Dec 24, 2015

### LCKurtz

Let the lighthouse be $Q$, let $P$ be the point on the shore nearest the lighthouse, let $x$ be the distance from $P$ to where the spotlight hits the shore, and $\theta$ be the angle of rotation from $QP$. What you need to calculate is $\frac{dx}{dt}$. You don't need the distance from the observer on shore to the beacon house to work this problem.

9. Dec 24, 2015

### Michele Nunes

Okay I got it now, thank you!

10. Dec 24, 2015

### Thewindyfan

If you don't mind, could you tell me how you ended up solving this problem? I was trying it myself but I'm not sure if I have the right equation to differentiate or if I'm using the rate of the rotating beacon light correctly.

11. Dec 24, 2015

### LCKurtz

Show us what you did if you want advice or help.

12. Dec 25, 2015

### Thewindyfan

Okay so based off your guidance in the last post, I basically sorted out what was given and what we want. You mentioned that the hypotenuse or the length of the beam isn't necessary for this problem so I decided to use the equation tan(Θ) = (x)/1
From implicitly differentiating with respect to time, I got the equation sec^2(Θ)*dΘ/dt = dx/dt
Since we know that dr/dt(rate of change of revolution of beacon) = 3 revs/min, I used dimensional analysis to get that dΘ/dt at that moment is 6π rad/min, but then I realized I definitely went wrong somewhere with my related equation since I would still need to know the length of the beam of light and even when I did, I got a pretty small number. I'm stuck on exactly how you have to deal with the dr/dt and what the best equation to relate these rates would be for this specific problem.

13. Dec 25, 2015

### Samy_A

You are almost there.
Hint: $1+\tan^²(\theta)=\sec²(\theta)$

14. Dec 25, 2015

### LCKurtz

That would be $\frac{d\theta}{dt}$, and it is the rate of revolution, not the "rate of change" of it. And don't forget the final units requested in the problem are kilometers/hour for $\frac{dx}{dt}$.

Last edited: Dec 26, 2015
15. Dec 25, 2015

### Thewindyfan

Wow I can't believe I haven't thought of using this identity for these types of related rates problems! Thanks for the tip.

Ah so I just forgot to have $\frac{d\theta}{dt}$ in the proper units before plugging in the numbers into the differentiated equation! Thank you for pointing that out! I haven't seen a related rates problem like this so I guess I was overthinking about it.