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Math Help - Boats?

  1. #1
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    Boats?

    A boat leaves a dock at 3:00 P.M. and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 4:00 P.M. How many minutes after 3:00 P.M. were the two boats closest together?
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  2. #2
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    Quote Originally Posted by sheva2291 View Post
    A boat leaves a dock at 3:00 P.M. and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 4:00 P.M. How many minutes after 3:00 P.M. were the two boats closest together?
    Start by setting up an appropriate coordinate system. It will be simplest to set the origin at the dock, the positive x-axis west and the positive y axis south with x and y in km from the dock. Take t in hours with t= 0 at 3:00 P.lM.

    The first boat travels due south so its x coordinate is always 0. Its speed is 20 km/h so its distance from the dock, the y coordinate, after t hours is 20t.

    The second boat travels due east so its y coordinate is always 0. its speed is -15 (because the positive x-axis points west and the boat is going east) so x= -15t+ b. You can find b by using the fact that the boat "reaches the same dock at 4:00 P.M."


    The distance between the two boats is \sqrt{x^2+ y^2}. You can find the minimum distance by differentiating with respect to t and setting the derivative equal to 0 or, for this simple problem, by completing the square.
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  3. #3
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    Hello, sheva2291!

    A boat leaves a dock at 3:00 PM and travels due south at 20 km/hr.
    Another boat has been heading due east at 15 km/h and reaches the same dock at 4:00 PM.
    How many minutes after 3:00 PM were the two boats closest together?
    Code:
          : - - -  15 - - - - :
          P  15t  B   15-15t  D
          o-------o-----------o
                    *         |
                      *       |
                     x  *     | 20t
                          *   |
                            * |
                              o A

    Ship A starts at dock D and sails south at 20 km/hr,
    In t hours, it has sailed 20t km to point A.

    Ship B starts at P (15 km west of D) and sails east at 15 km/hr.
    In t hours, it has sailed 15t km to point B.
    . . Hence: . BD \:=\:15-15t

    Let x \:=\:AB.

    In right triangle BDA\!:\;\;x \;=\;\sqrt{(20t)^2 + (15-15t)^2}

    . . Hence, we have: . x \;=\;\left(625t^2 - 450t + 225\right)^{\frac{1}{2}}

    And that is the function we must minimize.


    I believe the answer is: . t \;=\;\frac{9}{25}\text{ hours} \;=\;21.6\text{ minutes}

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  4. #4
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    ahhh thank you very much, you explained it very clearly. i changed around the numbers and got it right =)
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