Calculating the Speed of Light in a Revolving Lighthouse

  • Thread starter Thread starter professordad
  • Start date Start date
  • Tags Tags
    Light
Click For Summary
SUMMARY

The discussion focuses on calculating the speed of the spot of light along a shoreline from a lighthouse located 1 mile away, revolving at 1 revolution per minute. The solution involves using the equation \(\frac{dx}{dt} = \frac{dx}{d\theta} \frac{d\theta}{dt}\) to derive that the speed of the light spot is \(10\pi\) miles per minute when it is 2 miles away from the nearest point on the shore. The calculations utilize trigonometric identities, specifically \(\tan{\theta}\) and \(\sec^2{\theta}\), to determine the relationship between the angle and the distance from the lighthouse.

PREREQUISITES
  • Understanding of trigonometric functions and identities
  • Familiarity with calculus concepts, particularly derivatives
  • Knowledge of angular motion and its relation to linear motion
  • Basic geometry involving circles and chords
NEXT STEPS
  • Study the application of derivatives in related rates problems
  • Explore trigonometric identities and their applications in calculus
  • Learn about angular velocity and its impact on linear velocity
  • Investigate real-world applications of revolving light sources in navigation
USEFUL FOR

Students studying calculus, particularly those focusing on related rates, as well as educators looking for practical examples of trigonometric applications in physics and engineering.

professordad
Messages
18
Reaction score
0

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}.
 
Last edited by a moderator:
Physics news on Phys.org
Mod note: I removed all of the \text stuff in your LaTeX. Now it renders correctly.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
1K
Replies
28
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
9
Views
4K
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K