a port and a radar station are 4 mi apart on a straight shore

Question

a port and a radar station are 4 mi apart on a straight shore running east to west a ship leaves the port at noon traveling 11 mph. 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 ? (hint use law of sines)

Answer 1

Triangle:

East-West segment from Radar Station to the Port. Side L = 4 miles

Ship leaves the Port at some ‘constant’ Course Angle, usually measured from North. In our Triangle, we have angle "CA" which is ‘constant’ relative to the shore.

Ship speed = v, distance traveled by Ship at speed v = 11 mph ==> Side S

Angle φ : between Ship course track "looking back" toward the Port, and Radar Station

Triangle 3rd side : distance between Radar Station and Ship... Side m

Angle θ : between East-West line (between Radar Station to the Port), and the current position of Ship.

  Radar Station will likely be using angle reference from North, but we convert to Triangle Angle θ relative to East-West shore.

Observations / Hints:

a. Use of Law of Sines is obvious here (below)

b. We are not given our Course Angle "angle CA", but we can derive that from side 'm', using Law of Cosines (later)

c. Without course angle "CA", we will see that there is not much we can do, but we are talking about a Radar Station...

One of the jobs of a Radar Station is to track all vessels, and their courses.

Radar Station is interested in angle θ between E-W Shore and the Ship (but it has it - more correctly, the angle to the Ship from North), but it may be very interested in tracking the rate of change of Angle θ, and it readily has the info on the Ship's track, and more importantly, it "has on hand" the instantaneous distance 'm' to the Ship. (To me, this is a strong Hint)

Radar Station is the only entity that "might" be interested in rate of change of the tracking angle θ, and they have ‘m’ instantaneously, along with the Ship’s actual track.

Wanted:

dθ/dt : "rate of change of the tracking angle theta", at time T = 1/2 hour after Ship departure from Port, at speed v.

Triangle (recap): "Shore" side L, then Angle CA "course angle" inside triangle, Side S distance traveled by Ship, Angle φ looking "back" from Ship, between Port and Radar Station, Side m distance between Ship and Radar Station, Angle θ observed by Radar Station between E-W Coast and position of Ship.

Law of Sines: sin(θ) / S = sin(CA) / m = sin(φ) / L

Observation: φ = π - (θ + CA) ==> sin(φ) = sin(π - (θ + CA)) ==> sin(φ) = sin(θ + CA)

S sin(φ) = L sin(θ)

S sin(θ + CA) = L sin(θ)

S[ sin(θ)cos(CA) + sin(CA)cos(θ) ] = L sin(θ) # Now divide by sin(θ)

S[ cos(CA) + sin(CA)cot(θ) ] = L # Now divide by sin(CA)

S[ cot(CA) + cot(θ) ] = L csc(CA) # Now differentiate with respect to time ’t'

S(-csc²(θ))dθ/dt + [ cot(CA) + cot(θ) ]dS/dt = 0

[ cot(CA) + cot(θ) ]dS/dt = S csc²(θ)dθ/dt

v[ cot(CA) + cot(θ) ] = [S / sin²(θ)] dθ/dt # dS/dt = constant Ship speed ‘v'

dθ/dt = [v*sin²(θ)/S] [ cot(CA) + cot(θ) ]

dθ/dt = [v*sin(θ)/S] [ sin(θ)cot(CA) + cos(θ) ]

dθ/dt = (v/S) (sin(θ) / sin(CA) [ sin(θ)cos(CA) + sin(CA)cos(θ) ]

dθ/dt = (v/S) (sin(θ) / sin(CA) sin(θ + CA)

dθ/dt = (v/S) [sin(θ) / sin(CA)] sin(φ) # sin(θ) / sin(CA) = S/m

dθ/dt = (v/S) (S/m) sin(φ)

dθ/dt = (v/m) sin(φ)         [Eq. A]

dθ/dt = (v/m) (L/m) sin(CA) = (vL/m²)sin(CA) # Observe (from above) that sin(φ) = (L/m) sin(CA)

dθ/dt = (vL/m²)sin(CA)       [Eq. B]

Observation: I was initially surprised by the prominence of side 'm' and angle 'φ' in dθ/dt, but after all, side ‘m’, angle ‘θ’, and angle 'φ' and are the main things changing, along with side S, at its constant rate ‘v’.

Observation 2: The only entity interested in dθ/dt would be the Radar Station, who may have been given the 'intended' Course Angle "CA" of the Ship (initially), but is more interested in what it observes "real time". Radar Station has "live" access to distance 'm', is the only one interested in dθ/dt... so… I suspect we are less interested in knowing "angle CA" in the initial information given, and more on knowing that "distance 'm’ “ is available..

However, it is easy to compute Course angle CA from distance 'm' from Law of Cosines… and vice-versa:

  Given Radar distance 'm': m² = S² + L² -2SL cos(CA) ==> cos(CA) = (S² + L² - m²) / 2SL ==> angle CA

Given distance ‘m’, find angle CA (see above), then:

For Eq. A ==> dθ/dt = (v/m) sin(φ) # with sin(φ) = (L/m)sin(CA)

Given CA, find side ‘m’ (see above)

For Eq. B ==> dθ/dt = (vL/m²)sin(CA)

Answer: Ship must provide CA, or Radar Station can provide instantaneous distance 'm' to the Ship.

   We don't really need to care about Ship's distance made good between Noon - 12:30pm, which help us find *neither* CA nor 'm' (it would seem)