Calculus problem

Question

Two boats are traveling at 30km/h, the first going north and the second going east. The second crosses the path of the first 10 minutes after the first one was there. At what rate is their distance increasing when the second has gone 10km beyond the crossing point.

Answer 1

In 10 minutes, the first boat will have travelled 10 minutes * 1 hr/60 minutes * 30 km/hr = 5km.

Then, as the boats travel at the same rate, when the second boat travels 10 km, the second boat will have also travelled 10 km.

Then, using the standard coordinate system, the first boat's location is (0, 5 + 10) = (0, 15), and the second boat is at (10, 0), so the first boat is (0, 15) - (10, 0) = (-10, 15) from the second and is moving away from it at (-30 km/hr, 30 km/hr)

Thus, the distance is increasing at (-10, 15) . (-30, 30)/|-10,15| =

750/√325 = 150√13 /13

You can also show this result by writing D = sqrt((-10-30t)^2 + (15+30t)^2) =

sqrt(1800t^2 + 1500t + 325), and dD/dt at t = 0 =

1/2(3600t + 1500)/sqrt(1800t^2 + 1500t + 325) = 750/√325 = 150√13 /13, the same answer as above.