Posted by **Anya** on Wednesday, November 10, 2010 at 1:27am.

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?

- calculus -
**Reiny**, Wednesday, November 10, 2010 at 8:19am
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 = (15-15t) km

AB^2 = (20t)^2 + (15-15t)^2

let's find the minimum distance AB

2AB d(AB)/dt = 800t + 2(15-15t)(-15)

= 0 for a minimum of AD

800t - 30(15-15t) = 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.

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