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A lighthouse is 7 miles from the shore and sweeps at 10 seconds per revolution. At what rate does the beam move along the shore, in mph, when the beam makes a 30 degree angle with the shore?

I've checked my solution a million times and can't find anything wrong with it except the answer: 3,628,800 mph. That can't possibly be right!

Here's my reasoning: 10 seconds per revolution is 36 degrees per second. So $\frac{d\theta}{dt}=36$ degrees/sec. Then we have $$s = 7\tan(\theta)$$ $$\frac{ds}{dt} = 7\sec^2{\theta}\frac{d\theta}{dt} = \text{ 3,628,800 mph}$$ where $s$ is the distance along the shore and $\theta = 60^{\circ}$. Where's my mistake?

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    $\begingroup$ I don't really see how this even requires calculus--I guess I'm wrong (or I don't understand the setup). It would seem to me that the linear velocity would simply be $v = \omega r$ which gives: $v = \frac{2\pi}{10}\cdot 7 = \frac{14\pi}{10} \approx 4.39$ miles per second, which gives $4.39 \frac{3600\text{ seconds}}{1\text{ hour}} \approx 15834$ mph. I could see how, due to the angle, the "actual" angular velocity might be less--but it's just a matter of calculating that... $\endgroup$
    – Jared
    Commented Feb 24, 2016 at 3:14
  • $\begingroup$ Omg no the speed goes to infinity as the beam becomes parallel. $\endgroup$ Commented Feb 24, 2016 at 3:18
  • $\begingroup$ Yes, I know, but 1 revolution per 10 seconds is an angular velocity. So you then calculate the linear velocity given that angular velocity based on the distance (again, doesn't require calculus)--if this is a calculus question then I clearly do not understand the setup because I don't see how calculus comes into play. $\endgroup$
    – Jared
    Commented Feb 24, 2016 at 3:19
  • $\begingroup$ What are your assumptions about how the lighthouse light rotates? Do you have a picture? Based on your solution, you are assuming that the light rotates "vertical" (i.e. goes from horizon to horizon, shining light up in the air) as opposed to rotating "horizontally" (i.e. spinning on a vertical axis) which is what I assume a lighthouse would do. $\endgroup$
    – Jared
    Commented Feb 24, 2016 at 3:28
  • $\begingroup$ This is really a question suitable for a discussion in the physics forum. $\endgroup$
    – jim
    Commented Feb 24, 2016 at 9:33

3 Answers 3

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"Where is my mistake?"

You have to measure $\theta$ in radians, and the time unit is $1$ hour. As the beam makes $360$ full turns per hour one has $$|\dot\theta|=2\pi\cdot 360\ .$$ You therefore obtain $$|\dot s|=7\cdot \sec^2(60^\circ)\cdot 2\pi\cdot 360\doteq 63\,335\qquad({\rm mph})\ .$$

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  • $\begingroup$ Why are you multiplying by $4$? Without the multiply by $4$, you get the same thing I got in my initial comment which is $\frac{63335}{4} \approx 15833.75$. $\endgroup$
    – Jared
    Commented Feb 25, 2016 at 5:06
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Put the lighthouse at the origin. Put the shore to be the vertical line $x=7$. Drawing a line segment oriented at angle $\theta$ from the origin to the line, you have ASA so you can solve the triangle. You find the length of the vertical piece is $7 \tan(\theta)$ (assume we're looking at $\theta \geq 0$).

So the speed along the shore is $\frac{d}{dt} 7 \tan(\theta(t)) = 7 \sec(\theta(t))^2 \theta'(t)$. Here $\theta$ changes from $0$ to $2 \pi$ in time $10$ so $\theta'=\frac{\pi}{5}$. The angle with the shore is $\pi/2-\theta$, so we need $\sec(\pi/3)=2$. So the rate in miles per second is $7 \cdot 4 \cdot \frac{\pi}{5} = \frac{28 \pi}{5}$, which is about $18$ miles per second (no calculator, just a rough estimate). Multiplying by 3600 converts to mph. The answer is big, but it's not millions of mph.

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  • $\begingroup$ You made a mistake, you need $sec(\pi/3)$, not $\pi/6$ $\endgroup$ Commented Feb 24, 2016 at 15:51
  • $\begingroup$ @RenéG Ah, yes, the given angle is with the shore. Thanks. $\endgroup$
    – Ian
    Commented Feb 24, 2016 at 15:54
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I used degrees instead of radians, in degrees one doesn't have $\frac{d\tan\theta}{dx} = \sec^2\theta$. When one uses radians one gets 63335 mph.

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