Answer:
The beam of light is moving at the peed of:
[tex]\frac{dy}{dt} = \frac{80\pi}{3}[/tex] km/min
Given:
Distance from the isalnd, d = 3 km
No. of revolutions per minute, n = 4
Solution:
Angular velocity, [tex]\omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi [/tex] (1)
Now, in the right angle in the given fig.:
[tex]tan\theta' = \frac{y}{3}[/tex]
Now, differentiating both the sides w.r.t t:
[tex]\frac{dtan\theta'}{dt} = \frac{dy}{3dt}[/tex]
Applying chain rule:
[tex]\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}[/tex]
[tex]sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}[/tex]
Now, using [tex]tan\theta = \frac{1}{m}[/tex] and y = 1 in the above eqn, we get:
[tex](1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}[/tex]
Also, using eqn (1),
[tex]8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}[/tex]
[tex]\frac{dy}{dt} = \frac{80\pi}{3}[/tex]
