CALCULUS

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?

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  1. 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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