Calculus problem

Question

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 10 km/h. Another boat has been heading due east at 5 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?

Answer 1

Assuming the origin is the dock and that 2 PM is t = 0...

The equation of position for boat 1 is y = -10t where 10 is km/hr, t is in hrs, and y is in km

Boat 2 goes 5 km/hr east and reaches the dock at 3PM or t = 1. You can back out that the boat was at x = -5 km at t = 0 and the position of boat 2 is x = -5 + 5t

The distance between the boats is the distance between (0, y₁) and (x₂, 0)

d = sqrt( (x₂ - 0)² + (0 - y₁)² ) = ((-5 + 5t)² + (10t)²)^1/2

To minimize this, we'll take the derivate with time of d² to get rid of square root sign and set it to 0:

2(-5 + 5t) (5) + 200t = 0 and solve for t. we know it's a minimum from the 2nd derivative or change of sign of the 1st derivative.

remember that you need to express t in minutes and add to 2 to answer the question.

Good luck and consider hiring a tutor.