# CALCULUS

posted by Gagan

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

1. Steve

well, you know the distance z at time t is

z^2 = (130-30t)^2 + (20t)^2
= 1300t^2 - 7800t + 16900
so, at 4:00, z = 10√145

z dz/dt = 2600t - 7800 = 2600(t-3)

Now plug in t=4

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