Calculating the Speed of Light in a Revolving Lighthouse

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The problem involves calculating the speed of a light spot from a revolving lighthouse located 1 mile from the shore, with the light making one full revolution per minute. The key equation used is dx/dt = (dx/dθ)(dθ/dt), where the relationship between the angle θ and the distance x is established through trigonometric functions. The solution constructs a geometric representation to find the speed when the light is 2 miles away from the nearest shore point. The calculations yield a speed of 10π for the light spot along the shore. This approach effectively combines geometry and calculus to solve the problem.
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Homework Statement



A light in a revolving lighthouse located 1 mile away from a straight shoreline turns at 1 revolution per minute. Find the speed of the spot of light along the shore when it is 2 miles away from the point of shore nearest the light.

Homework Equations



\frac{dx}{dt} = \frac{dx}{d\theta} \frac{d\theta}{dt}

The Attempt at a Solution



Here is my solution, although I don't know if it (and the answer it produces) is correct or not

Let the lighthouse be point O. Construct a circle with radius 2 around O, and a chord such that the distance from O to the midpoint of the chord is 1. Let the midpoint of the chord be point M, and one of the intersections of the chord with the circle be point N. Now let half the length of the chord be x.

We have \tan{\theta} = x, so \frac{dx}{d\theta} = \sec^2{\theta}. This also equals 1 + \tan^2{\theta}, or 1+x^2. Since x=2, it's basicaaly 5.

Now \frac{dx}{dt} = \frac{dx}{d\theta} \frac{d\theta}{dt}. \frac{dx}{d\theta} is \sec^2{\theta}, and \frac{\text{d} \theta}{\text{dt}} is 2\pi, so the answer is \boxed{10\pi}.
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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