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Math Help - related rates

  1. #1
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    related rates

    At noon, ship A passes 9 km west of a harbour, heading due south at 8 km/h. At 11 a.m., ship B had left the harbour, sailing due south at 6 km/h. Give an expression for d, the distance between the two ships, in terms of t, the number of hours after noon.


    i really dont even know how to start on this problem,.
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  2. #2
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    Quote Originally Posted by samtheman17 View Post
    At noon, ship A passes 9 km west of a harbour, heading due south at 8 km/h. At 11 a.m., ship B had left the harbour, sailing due south at 6 km/h. Give an expression for d, the distance between the two ships, in terms of t, the number of hours after noon.
    i really dont even know how to start on this problem,.
    start by making a sketch of the situation.

    let the harbor be at the origin and let t = 0 be noon

    at t = 0 , ship A is at position (-9,0) and is heading south at 8 km/hr ...

    its position at any time t is (-9, -8t)

    at t = 0, ship B is at position (0, -6) and is heading south at 6 km/hr ...

    its position at any time t is (0, -6 - 6t)

    use the distance formula to find the distance between the two ships at any time t.
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  3. #3
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    Hello, samtheman17!

    At 11 a.m., ship b leaves the harbor, sailing due south at 6 km/hr,
    At noon, ship a passes 9 km west of the harbor, heading due south at 8 km/h.

    Give an expression for d, the distance between the two ships,
    in terms of t, the number of hours after noon.
    First, make a sketch!
    Code:
                   9
        Q * - - - - - - - - * H
          |                 |
       8t |                 |
          |                 | 6
        A *                 |
          :  *              |
          :     * d         * P
     6-2t :        *        |
          :           *     | 6t
          :              *  | 
        C * - - - - - - - - * B
                   9

    At 11 a.m. ship b starts at H, sails south at 6 km/hr for one hour.
    At noon, ship b is at P\!:\;HP = 6
    In the next t hours, it sails 6t km to point B\!:\;\;PB = 6t

    At noon, ship a starts at Q\!:\;HQ = 9 km,
    . . and sails south at 8 km/hr,
    In the next t hours, it sails 8t km to point A\!:\;QA = 8t

    Note that: . AC \:=\:(6 + 6t) - 8t \:=\:6 - 2t


    In right triangle ACB\!:\;d^2 \;=\;(6-2t)^2 + 9^2

    Therefore: . d \;=\;\sqrt{4t^2 - 24t + 117}

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  4. #4
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    thank you so much, that really helped!
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