Question

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 25 km/h. Another boat has been heading due east at 10 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together?

minutes

Answer 1

Sounds like parameterization to me. We need to make some functions of T that can describe the position of these.

For the boat leaving, that's easy. Call the dock (0,0), and he's just traveling due south.

X1 = 0
Y1 = -10 (T/60) where T is the number of minutes since 2:00.

For the other boat, we can say:

X2 = -25 + 25(T/60) Because he starts 25 km west and is moving east.
Y2 = 0

Use the distance formula to find the distance between them:

D = sqrt[(X2-X1)^2 + (Y2-Y1)^2]

From there, you can plug in your expressions (2 of which are just 0) and then use derivatives to find the minimum

http://answers.yahoo.com/question/index?qid=20071206160149AA8tAyE

Answer 2

ok @swanny