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sveticcg [70]
3 years ago
8

A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re

volutions per minute. How fast is the beam of light moving along the shoreline when it is 1 km from ?

Physics
1 answer:
lbvjy [14]3 years ago
7 0

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

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