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Math Help - Optimization(rowing across river)

  1. #1
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    Optimization(rowing across river)

    A man can row at 8 km/h, and run at 15 km/h. He needs to get from a point A, on the south bank of a stretch of still water, to point B on the north bank of the water. The direct distance from A to B is 17 km, and the water is 8 km wide. He starts rowing with an angle X between North and the direction in which he rows. Find an expression for the time he will take to get from A to B, in terms of X.

    Any helps would be great
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  2. #2
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    Hello, squeeze101!

    A man can row at 8 km/h, and run at 15 km/h.
    He needs to get from a point A, on the south bank of a stretch of still water,
    to point B on the north bank of the water.
    The direct distance from A to B is 17 km, and the water is 8 km wide.
    He starts rowing with an angle \theta between North and the direction in which he rows.
    Find an expression for the time he will take to get from A to B, in terms of \theta.
    Code:
          : - - - - - - - 15  - - - - - - - - :
          C      8tanθ       P    15-8tanθ    B
          * - - - - - - - - * - - - - - - - - * - -
          |              *              *
          |           *           *
        8 |        *        * 17
          | θ   *     *
          |  *  *
        A * - - - - - - - - - - - - - - - - - - - -

    The distance directly across the water is: AC = 8
    We are told that: AB = 17
    From Pythagorus, we find that: CB = 15

    He will row toward point P at an angle of \theta = \angle CAP

    In right triangle PCA\!:\;\;AP = 8\sec\theta
    . . and: . CP = 8\tan\theta \quad\Rightarrow\quad PB \:=\:15-8\tan\theta


    He will row AP \:=\: 8\sec\theta km at 8 km/hr

    . . This will take him: . \frac{8\sec\theta}{8} \:=\:\sec\theta\,\text{ hours.}


    He will run PB \:=\:15-8\tan\theta km at 15 km/hr.

    . . This will take him: . \frac{15-8\tan\theta}{15} \:=\:1 - \tfrac{8}{15}\tan\theta\,\text{ hours.}


    Therefore, his total time is: . T \;=\;\sec\theta + 1 - \tfrac{8}{15}\tan\theta\,\text{ hours}

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