Rate of Change in Distance

Question

At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 23 knots and ship B is sailing north at 20 knots. How fast (in knots) is the distance between the ships changing at 5 PM?

Answer 1

So that you are not carrying around so much extra baggage in solving this problem, first find where the ships are at 5.

Let the origin be Ship B at noon. Using the normal axis orientation, west is negative on the x axis and north is positive on the y axis.

Then, ship A, starting 10 miles west and travelling west at 23 knots, will be at (-10-23*5,0) = (-125, 0) and ship B will be at (0, 20*5) = (0, 100)

Then, since A is traveling west and B is traveling north, the distance apart may be written as

D = sqrt((-125-23t)^2 + (100 + 20t)^2), where t = 0 at 5.

dD/dt =1/2 [46(125+23t)+40(100+20t))/sqrt((-125-23t)^2 + (100 + 20t)^2)

Substituting t = 0, we [get 23(125) + 20(100))/sqrt((-125)^2 + (100)^2) =

125(23+16)/25sqrt(41) =

195/sqrt(41)