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A train is heading due west from St. Louis. At noon, a plane flying horizontally due north at a fixed altitude of 4 miles passes directly over the train. When the train has
traveled another mile, it is going 80 mph, and the plane has traveled another 5 miles and is going 500 mph. At that moment, how fast is the distance between the train and the plane increasing?

Is the answer 105 mph.

  • Calculus -

    Measuring time (t) in hours from that moment, the vector distance R between train and plane at later times is
    R = (1+80t) i + (5+ 500t) j + 4 k
    where i, j and k are unit vectors west, north and vertical.

    dR/dt = 80 i + 500 j

    The magnitude of that rate of change is
    sqrt[(80)^2 + (500)^2] = 506 mph.

    This could also have been solved simply by taking the magnitude of the relative velocity vector, 80 i + 500 j mph

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