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Homework Help: Help with calculus

  1. Mar 16, 2005 #1
    At 3 PM a ship which is sailing due north at 12 knots(nautical miles/hour) is 5 miles west of a westbound ship which is making 16 knots. a) At what rate is the distance between the ships changing at 3 PM

    is it just simply pythagorean theorem of the rates of both ships?
    i got the answer as 20 knots

    i dont see how related rates would be involved on this
    thanks for the help
     
    Last edited: Mar 16, 2005
  2. jcsd
  3. Mar 16, 2005 #2

    xanthym

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    Science Advisor

    Let time t=(0) correspond to 3 PM. Then:
    {North} = (+y Direction)
    {West} = (-x Direction)
    {Northbound Boat Position} = {x=(-5 miles), y=(12*t)}
    {Westbound Boat Position} = {x=(-16*t), y=(0)}

    {Distance Between Boats} = D = sqrt{(Δx)2 + (Δy)2} =
    = sqrt{((-5) - (-16*t))2 + ((12*t) - (0))2} =
    = sqrt{(16*t - 5)2 + (12*t)2} =
    = sqrt{256*t2 - 160*t + 25 + 144*t2} =
    = sqrt{400*t2 - 160*t + 25}

    The time rate of separation distance change can be determined from:
    (dD/dt) = (1/2)*{400*t2 - 160*t + 25}(-1/2)*(800*t - 160)
    Thus, at 3 PM {or t=(0)}:
    (dD/dt) = (1/2)*{0 - 0 + 25}(-1/2)*(0 - 160) = (1/2)*(25)(-1/2)*(-160)
    (dD/dt) = (1/2)*(-160)/5 = (-16 knots)


    ~~
     
    Last edited: Mar 17, 2005
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