Calculus

At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 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. At time t hours, we have the distance z between the ships as

    z^2 = (130-25t)^2 + (20t)^2
    at 4:00, z^2 = 30^2+80^2=7300, so z=10√73

    Now,
    2z dz/dt = 2(130-25t)(-25)+2(20t)(20)
    z dz/dt = 1025t-3250
    so, at t=4,
    dz/dt = 85/√73 km/hr

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