Oblique tracking? A ship leaves port traveling southwest at a rate of 12 mi/hr. At noon, the ship reaches its closest approach to a radar station, which is on the shore 1.5 mi from the port. If the ship maintains its speed and course, what is the rate of change of the tracking angle ???between the radar station and the ship at 1:30 P.M. (see figure)? (?Hint:? Use the Law of Sines.)
Solution: Step 1 To find what is the rate of change of the tracking angle between the radar station and the ship at 1:30 P.M. Step 2 Given that A ship leaves port traveling southwest at a rate of 12 mi/hr. At noon, the ship reaches its closest approach to a radar station, which is on the shore 1.5 mi from the port. If the ship maintains its speed and course Step3 Step 4 since there is a "closest approach", the radar station is west of the port, and the ship starts out at 45° towards the radar station. the tracking angle will increase from 0° to 90° , and then decrease taking the radar stn. as (0,0), and west as the +ve direction the x-axis, position of the ship at time t is (-1.5 + 6t2 , 6t2) z² = x² + y² z = [(-1.5 + 6t2)² + (6t2)²], the numerator (relevant part) of dz/dt = 12t - (3/4)2, which, on setting to 0 yields z(minimum) = 0.752 at t = 2/16 hr so ship has travelled 2/16 + 1.5 = (24+2)/16 hrs at 1:30 pm tan = y/x = 6t2/(6t2 - 1.5) Differentiate both side we get, sec² d/dt = -0.176777/(0.176777-t)² (1+tan²) d/dt = -0.176777/(0.176777-t)² at t = (24+2)/16, tan = 1.12523, 1+tan² = 2.26614 finally, put values and get , d/dt = -(0.176777/2.66614)/(0.176777-t)² 0.04 rad/hr
