6.A revolving beacon is situated 380 feet off a straight shore. If the beacon turns at 4.2 radians per minute, how fast does the beam sweep along the shore at (a) its nearest point (b) at a point 400 feet from the nearest point along the shore?
Answers
Given : .A revolving beacon is situated 380 feet off a straight shore.
the beacon turns at 4.2 radians per minute
To Find : how fast does the beam sweep along the shore at
(a) its nearest point
(b) at a point 400 feet from the nearest point along the shore?
Solution:
beacon turns at 4.2 radians per minute
tan α = x / 380
x = 380 tanα
dx/dt = 380 sec²α dα /dt
dα /dt = 4.2 radians per minute
(a) its nearest point
α = 0 => sec²α = 1
dx/dt = 380 (1) 4.2
=> dx/dt = 1596 feet / min
(b) at a point 400 feet from the nearest point along the shore?
tan α =400 / 380 = 20/19
sec²α = 1 + tan² α = 1 + (20/19)² = 761/361
dx/dt = 380 (761/361) 4.2
=> dx/dt = 3364.4 feet / min
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