Calculus

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At 2:00PM boat X is heading due west at 15mph towards a dock that is exactly 15 miles away. At the same time boat Y leaves the dock traveling due north at a speed of 20mph. At what time will the boats be closest together?

  • Calculus -

    Let the time passed since the northbound ship left the dock be t hrs
    Make a sketch
    Distance traveled by northbound boat is 20t
    distance from dock of westbound boat is 15-15t.
    let the distance between them be D
    D^2 = (20t)^2 + (15-15t)^2
    = 400t^2 + 225 - 450t + 225t^2
    = 625t^2 - 450t + 225
    2D dD/dt = 1250t - 450
    dD/dt = (625t - 225)/d
    = 0 for a max/min of D
    625t = 225
    t = 225/625 = .36

    so the time is .36 hours past 2:00 pm
    .36 hrs = 21.6 minutes or 21 minutes 36 seconds

    time is 2:21:36 PM

  • Calculus -

    THANK YOU! I get it now!

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