At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 18 knots and ship B is sailing north at 23 knots. How fast (in knots) is the distance between the ships changing at 3 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Answer 1

As usual, draw a diagram. The distance between the ships at time t hours is

d^2 = (50+18t)^2 + 23t^2
At 3 pm, t=3, so d = √(104^2+69^2) = 124.8

2d dd/dt = 36(50+18t)+46t
so, dd/dt = 15.6 knots

Answer 2

let the time be t hours past noon.
I see a right-angled triangle , with sides 23t
and 50 + 18t, with D as the distance between them

D^2 = (23t)^2 + (50+18t)^2
D^2 = 529t^2 + 2500 + 1800t + 324t^2
= 853t^2 + 1800t + 2500
2D dD/dt = 1706t + 1800
dD/t = (853t + 900)/D , #1

at 3:00 pm , t = 3
D^2 = (69^2 + 104^2)
D = √15577

dD/dt = (2559 + 900)/√15577 = appr 27.715 knots

check my arithmetic

Answer 3

darn them brackets
23t^2 vs (23t)^2