1) Draw a diagram of a right triangle with one vertex at the light source, L, and label this acute angle Θ. The leg adjacent to Θ is 15 ft in length. We are interest in the length of the leg opposite from Θ that we'll call x. We might label the vertex on the wall with the right angle W, and the point on the wall at which the light is pointing P.
Because we are interested in the leg opposite from Θ and the adjacent leg is constant, we will use tanΘ to relate these. We want to calculate dx/dt using related rates.
Lastly, note Θ = π/18 (rads) and dΘ/dt = 2π / 5 secs.
tanΘ = x/15
sec2Θ·dΘ/dt = 1/15·dx/dt
dx/dt ~ 19.4 ft/sec
2) Let P be the location of ship B at noon, A is the location of ship A at 7 pm and B is the location of ship B at 7 pm. We can call the length of PA, the horizontal leg of right triangle APB, x, and call the length of the vertical leg PB, y. h can be the length of the hypotenuse. We are calculating dh/dt.
Note the following: x = 25 , dx/dt = 15 , y = 140, dy/dt = 20 , and h ~ 142.2 (by pythag thm).
x2 + y2 = h2
2x·dx/dt + 2y·dy/dt = 2h·dh/dt
dh/dt ~ 769.7 naut. miles / hr.