Differential Calculus

At 9am ship B is 65 miles due east of another ship A. Ship B is then sailing due west at 10mi/h and A is sailing due south at 15 mi/hr if they continue in their respective course when will they be nearest to one another? and how near?

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  1. place ship B 65 miles east of ship A and mark them that way.
    Let the time passed be t hrs
    Draw a line BP , so that P is between A and B showing the distance traveled in those t hours.
    Draw a line downwards AQ showing the distance traveled by ship A.
    Join PQ to get the right-angled triangle APQ
    AP = 65-10t
    AQ = 15t
    let D be the distance between P andQ

    D^2 = (65-10t(^2 = (15t)^2
    2D dD/dt = 2(65-10t)(-10) = 2(15t)(15)
    for a minimum of D , dD/dt = 0
    so 2(10)(65-10t) = 2(15t)(15)
    divide by 10
    2(65-10t) = 3(15t)
    130 - 20t = 45t
    65t = 130
    t = 130/65 or 2 hrs

    They will be closest at 9:00 + 2:00 or 11:00 am
    and that distance is ....
    D^2 = (65-20)^2 + 30^2
    = 2925
    D = √2925 = appr 54.1 miles

    check my arithmetic

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    posted by Reiny

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