Calculus
posted by Reanna on .
At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 16 knots and ship B is sailing north at 24 knots. How fast (in knots) is the distance between the ships changing at 3 PM?

Make a sketch to get a right angled triangle
let the time passes since noon be t hrs
let the noon position of ship B be the origin O
AO = 10 + 16t miles
OB = 24t miles
AB^2 = AO^2 + OB^2
= (10+16t)^2 + (24t)^2
2 AB d(AB)/dt = 2(10+16t)(16) + 2(24t)((24)
when t = 3 ( at 3:00 pm)
AB^2 = (58)^2 + 72^2 = 8548
AB = √8548
d(AB)/dt = (32(58)+ 48(72) )/(2√8548)
= 28.72
at that moment they are separating at 28.7 knots