Math, asked by Anonymous, 3 months ago

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

Answered by amitnrw
0

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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