A port and a radar station are 4 mi apart on a straight shore running east and w

Question

A port and a radar station are 4 mi apart on a straight shore running east and west. A ship leaves the port at noon traveling at a rate of 12 mi/hr. If the ship maintains its speed and course, what is the rate of change of the tracking angle theta between the shore and the line between the radar station and the ship at 12:30 PM? The rate of change of the tracking angle theta between the shore and the line between the radar station and the ship at 12:30 PM is rad/hr. (Round to four decimal places as needed.)

Explanation / Answer

Given that, S is the distance the ship has traveled and is the angle that the radar station measures between the port and the ship.

Consider the locations of the ship, the radar station, and the port as points on a triangle. The angle opposite the shoreline would be = - /4 - = 3/4 -

So by law of sines,
sin(3/4 - )/4 = sin/s

Now we will use trig identities (and algebra) to change the expression and then differentiate implicitly (and do slightly more algebra) to find d/dt.

s[sin(3/4)cos - sin cos(3/4)]/sin = 4
s[sin(3/4)cot - cos(3/4)] = 4
s[1/sqrt(2)][cot + 1] = 4
s[cot + 1] = 4sqrt(2) ----------- (1)

(ds/dt)[cot + 1] + s[-cosec²](d/dt) = 0
d/dt = (ds/dt)[cot + 1] / [s{cosec²}]
d/dt = (ds/dt)(sin)[cos + sin] / s

Find s and at 12:30 PM.
s = (12 mi/hr)(1/2 hr) = 6 mi

from rquation (1)

(6 mi)[cot + 1] = 4sqrt(2)
cot = [4sqrt(2) - 6]/6
tan = 6/[4sqrt(2) - 6]

-1.5136675998 1.6279 rad

d/dt = 12[sin(1.6279)][cos(1.6279) + sin(1.6279)] / 6 = 1.879 radian/hour

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