calculus
posted by Anya .
Boat A leaves the dock at 12pm heading due South at a constant speed of 20 km/h.
Meanwhile, at 12pm Boat B, a pirate ship, is 15 km due West of the dock where boat
A started, and is sailing due East at a constant speed of 15 km/h. If the pirates get
within 10 km of another boat, they will spot it and eventually capture it. Is Boat A
in danger of getting captured by the pirates on Boat B?

Make a diagram showing the ships at t hours after 12:00 pm.
Let D be the position of the dock and A and B the position of the boats.
then DA = 20t km
and BD = (1515t) km
AB^2 = (20t)^2 + (1515t)^2
let's find the minimum distance AB
2AB d(AB)/dt = 800t + 2(1515t)(15)
= 0 for a minimum of AD
800t  30(1515t) = 0
800t  450 + 450t =
t = 450/1250
t = .36 hours
when t = .36
(AB)^2 = 51.84 + 92.16
AB = √144
= 12
Since they have to be within 10 km to be seen, they are ok.