This is not homework. I have the solutions, I just can't understand why mine are wrong.

The problem states: A revolving beacon $3600$ feet off a straight shore makes $2$ revolutions per minute. How fast does its beam sweep along the shore, (a) at the point on the shore nearest the beacon? (b) at the point on the shore $4800$ feet away from the beacon?

My solutions (second one is wrong): I didn't feel I needed calculus to solve (a). Basically the beacon is travelling 2 times the circumference of a circle of radius 3600 feet at (assumedly) constant speed in 1 minute, so $$C = 2 \pi r = 7200 \pi$$ and therefore $v = 14400 \frac{feet}{minute}$, which is going to be the instantaneous speed at any point on the circumference of that circle, and this agrees with the solution in the book.

For (b), I figured that since the point is $4800$ feet away from the beacon, I could just again take the constant speed of the beacon along the circumference of a circle of radius 4800 this time and have that be the instantaneous speed at any point along the circumference of that circle, and therefore at the point on the shore $4800$ feet away from the beacon.

Which gives me $$C = 2 \pi r = 9600 \pi$$ and $v = 19200 \frac{feet}{minute}$.

However, the book disagrees with that result, and states:

Measuring the distance, $x$, along the shore from the foot of the perpendicular from the beacon to the shore and taking $A$ as the angle between this perpendicular and the beam, we have $x = 3600 \tan{A} $. Hence, $\frac{dx}{dt}=\frac{14400 \pi}{\cos^2{A}}$, making the answer to (b) $25600 \pi \frac{feet}{minute}$.

I don't understand why my answer to (b) is wrong. If constant speed made sense as instantaneous speed at a point given a radius of $3600$, why doesn't it work for a radius of $4800$. I fail to see how the two questions are fundamentally different.


The difference is the angle the shore makes with the light beam at each spot.

Note that the question asks "How fast does its beam sweep along the shore," which implies that the direction is parallel to the shore. Your method calculates the speed along (tangent to) the circle swept out by the light beam.

In your part (a) the shoreline is tangent to the swept-out circle, so your direction is along the shore and your answer agrees with the book's. In part (b) the circle intersects the shoreline at an angle, which your answer did not take into account.

(As pointed out by @Michael in the comments, the following analysis is incorrect. I'll leave it here so the comments make sense.)

It is possible to get from your answer to the book's answer. Your answer is the component of the desired velocity vector that is perpendicular to the ray from the beacon to the intersection point of circle and shore. You can use trigonometry to find the angle of the desired velocity vector to your component and then the magnitude of the desired velocity vector. We do that kind of problem in my 12th-grade Physics class, but I won't bother doing it here. Ask if you want those details, but I think my first three paragraphs answer your stated question.

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  • 2
    $\begingroup$ How can Jeremy use his answer to get the book's answer? If you imagine the velocity vector of the radius-4800 circle, its speed into the beach is $19200\pi$ as Jeremy calculated (of course moving at an angle into the beach), but taking its component in the direction along the shore gives $19200\pi \cos(A) = 14400\pi$, which is less than the book answer. The problem is counter-intuitive since there is a quadratic effect due to the circle radius itself changing. $\endgroup$ – Michael Nov 21 '15 at 21:50
  • $\begingroup$ I guess the moral of the story is to first come up with $x(t)$ as a function of $t$ (the book gives $x(t) = 3600\tan(A(t))$) and then use derivatives. I likely would have gotten this problem wrong myself by being lazy and not deriving $x(t)$ or noticing the quadratic effect. $\endgroup$ – Michael Nov 21 '15 at 21:53
  • $\begingroup$ Rory, thank you for your answer. I see now that I should've imagined the light travelling along the coast, whereas what I was imagining at the time was the light hitting the coast at an angle. However, it's not intuitive to me why there is a difference between those two speeds at that point. Do you have an intuitive way to think about it physically speaking? Or any way, really... $\endgroup$ – jeremy radcliff Nov 21 '15 at 22:21
  • $\begingroup$ Physically, I can see that the speed of the light along the coast slows down as it gets closer to the perpendicular to the coast, whereas the speed of the light along the circumference of the circle of radius d is constant. However, shouldn't they approach the same limit of the ratio distance:time at any point on the coast? $\endgroup$ – jeremy radcliff Nov 21 '15 at 22:26
  • $\begingroup$ @michael your first comment is almost correct, you merely applied the cosine correction in the wrong direction. Where the beam sweeps at an angle to the shore, it sweeps faster. Divide by the cosine rather than multiply (you have the component, you need to calculate the projection) and you get the book's answer... $\endgroup$ – Chris Stratton Nov 21 '15 at 23:23

The position on shore is related to the angle by $x(t) = d \tan \theta(t)$.

enter image description here

Differentiating gives $\dot{x}(t) = d { 1\over \cos^2 \theta(t) } \dot{\theta}(t)$.

In all cases $\dot{\theta}(t) = 4 \pi$ $\text{min}^{-1}$, $d=3,600$ $\text{ft}$.

For Part (a), if $\theta(t) = 0$, then we have $\dot{x}(t) = d \dot{\theta}(t) = 14,400 \pi $ $\text{ft } \text{min}^{-1}$.

For Part (b), we have $\cos \theta(t) = {3600 \over 4800 }= {3 \over 4}$ so $\dot{x}(t) = d ({ 4 \over 3})^2 \dot{\theta}(t) = 25,600 \pi$ $\text{ft } \text{min}^{-1}$.

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  • $\begingroup$ Thanks for your answer. I chose Rory's answer because it really helped my understand why my answer was wrong, but your walkthrough is much more detailed and helpful than the book's solution. Took me a while to walk through it and make sense of everything, but I did and it was very helpful. $\endgroup$ – jeremy radcliff Nov 21 '15 at 22:23
  • $\begingroup$ No explanation needed at all! $\endgroup$ – copper.hat Nov 21 '15 at 22:29
  • $\begingroup$ @copper.hat : Thanks for your suggestion on how to fix my formatting problem. It seems I could neither edit nor delete. I tried taking your suggestion of flagging for moderator help, but it seems I cannot flag my own comment! So, can you flag my comment for removal? Either the one that is causing the problem, or my other one that tries to get a moderator to fix the issue? Thanks again! $\endgroup$ – Michael Nov 22 '15 at 4:26
  • $\begingroup$ @Michael: I flagged your comment when I wrote the above, it seems that a moderator has repaired the MathJax. $\endgroup$ – copper.hat Nov 22 '15 at 6:43

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