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Math Help - finding expression

  1. #1
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    finding expression

    ....S o-----------------------+
    .......|Y *.............(sand)...|
    ..200|.......*.....................| 200
    .......|............*................|
    ....A.+ - - - - - - - o - - - + B
    .......|........a.........X *.......|
    ..200|........................*..Z| 200
    .......|...(mud).................* |
    .......+-----------------------o F
    ........: - - - - 500m - - - - :

    (please ignore the full stops)
    details:
    if the distance from S to F (hypotenuse) , passing through X =200(secY + sec Z)
    and tanY + tanZ=5/2
    and the distance from S to F, passing through X in terms of Y=200(secY + 0.5(4sec^2Y - 20tanY+25)^(1/2)

    if an object can travel 4m/s through sand and 3m/s through mud, determine the best route it should travel through to get from S to F (hypotenuse), passing through X. and the time taken
    Last edited by asian5; August 5th 2006 at 11:53 PM. Reason: bad image
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by asian5
    ....S o-----------------------+
    .......|Y *.............(sand)...|
    ..200|.......*.....................| 200
    .......|............*................|
    ....A.+ - - - - - - - o - - - + B
    .......|........a.........X *.......|
    ..200|........................*..Z| 200
    .......|...(mud).................* |
    .......+-----------------------o F
    ........: - - - - 500m - - - - :

    (please ignore the full stops)
    details:
    if the distance from S to F (hypotenuse) , passing through X =200(secY + sec Z)
    and tanY + tanZ=5/2
    and the distance from S to F, passing through X in terms of Y=200(secY + 0.5(4sec^2Y - 20tanY+25)^(1/2)

    if an object can travel 4m/s through sand and 3m/s through mud, determine the best route it should travel through to get from S to F (hypotenuse), passing through X. and the time taken
    I think you will find everything you need to do this in the answers to your
    earlier question here

    RonL
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  3. #3
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    Hello, asian5!

    You don't need the angles to solve this part of the problem.

    Code:
        S o-----------------------+
          |   *                   |
      200 |       *               | 200
          |           *     500-x |
        A + - - - - - - - o - - - + B
          |       x       P *     |
      200 |                   *   | 200
          |                     * |
          +-----------------------o F
          : - - - -  500m - - - - :

    If an object can travel 4m/s through sand and 3m/s through mud,
    determine the best route it should travel to get from S to F,
    passing through P, and the time taken.

    Let AP = x, then PB = 500-x


    We have: .distance SP\;=\;\sqrt{x^2 + 200^2} meters.

    . . At 4 m/s, it will take: . \frac{\sqrt{x^2+200^2}}{4} seconds.


    We have: .distance PF\;=\;\sqrt{(500-x)^2+200^2} meters.

    . . At 3 m/s, it will take: . \frac{\sqrt{(500-x)^2+200^2}}{3} seconds.


    Hence, the total time is: . T(x)\;=\;\frac{1}{4}\left[x^2+200^2\right]^{\frac{1}{2}} + \frac{1}{3}\left[(500-x)^2+200^2\right]^{\frac{1}{2}}

    . . . and that is the function you must minimize.


    . . . . . . . . . . . Good luck!

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  4. #4
    MHF Contributor Quick's Avatar
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    Soroban, I have already solved this problem (if you see anything wrong with my work in the previous post, please tell me)

    I would show you the work, but something's wrong with this post...

    Last edited by Quick; August 6th 2006 at 07:26 AM.
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  5. #5
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    Hello, Quick!

    I think I've straightened out your format . . .


    Soroban, I have already solved this problem.
    If you see anything wrong with my work in the previous post, please tell me.

    and the answer comes out to:

    t=\frac{\sqrt{40000+x^2}}{4.5} +\frac{\sqrt{290000-1000x+x^2}}{2.5}

    which can then be turned into (solving in terms of Y):

    t=\frac{\sqrt{40000+200^2\tan^2Y}}{4.5} +\frac{\sqrt{290000-1000(200\tan Y + 200^2\tan^2Y}}{2.5}

    t=\frac{\sqrt{40000+40000\tan^2Y}}{4.5} + \frac{\sqrt{290000-200000\tan Y+ 40000\tan^2Y }}{2.5}

    t=\frac{\sqrt{40000(1+\tan^2Y)}}{4.5} +\frac{\sqrt{10000(29-20\tan Y + 4\tan^2Y)}}{2.5}

    t=\frac{200\sqrt{1+\tan^2Y}}{4.5} +\frac{100\sqrt{29-20\tan Y+ 4\tan^2Y }}{2.5}

    I missed your earlier post . . . I will take a look.

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