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Math Help - Parabola

  1. #1
    Member rtblue's Avatar
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    Parabola

    What is the maximum height that an 8-foot-wide truck can be in order to pass through a parabolic tunnel that is 18 feet high and 12 feet wide?

    I drew the parabola, and tried to find similar triangles, but i didn't notice anything. Any help is appreciated.
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  2. #2
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    Well here's a parabola that has x-intercepts 0, 12.

    y = a(x^2-12x) , you should find 'a' given the height is 18, i.e. use the point (6,18)

    What do you get?
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  3. #3
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    Hello, rtblue!

    What is the maximum height that an 8-foot-wide truck can be in order
    to pass through a parabolic tunnel that is 18 feet high and 12 feet wide?
    Code:
            (0,18)|
                  *
               *  |  *
             *    |    *
            *     |     *
                  |
           *      |      *
                  |
                  |
      ----*-------+-------*----
       (-6,0)     |     (6,0)
                  |


    The general equation of a parabola is: . y \:=\:ax^2 + bx + c

    The parabola passes through three points: . (6,0),\:(\text{-}6,0),\:(0,18)


    Substitute these coordinates into the equation:

    . . \begin{array}{ccccccc}<br />
(6,0)\!: & 36a + 6b + c &=& 0 \\<br />
(\text{-}6,0)\!: & 36b - 6b + c &=& 0 \\<br />
(0,18)\!: & 0a + 0b + c &=& 18 \end{array}

    Solve the system of equations: . a = \text{-}\frac{1}{2},\;b = 0,\;c = 18

    Hence, the parabola is: . y \;=\;\text{-}\frac{1}{2}x^2 + 18


    The truck is 8 feet wide.

    How high is the tunnel at x = 4 ?

    Code:
                18|
                  *
               *  |  *
             *- - + - -*
            *|    |    |*
             |    |    |
           * |    |   y| *
             |    |    |
             |    |    |
      ----*--+----+----+--*----
         -6 -4    |    4  6
                  |

    At x = 4\!:\;y \:=\:\text{-}\frac{1}{2}(4^2) + 18 \:=\:10


    Therefore, the truck can be 10 feet high (at most).

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  4. #4
    Super Member bigwave's Avatar
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    with axis of symetry on y axis

    thot I would join the party altho the high cards have already been dealt

    but...

    you could view this as an parabola with axis of symetry on y axis. thus the vertex is (0,18) and we have another point on the graph of (6,0) not using the the (-6,0) point

    then

    vertex = (h,k) or (0,18)

    (x-h)^2 = -4a(y-k)\Rightarrow x^2 = -4a(y-18)

    from the other point (6,0) we plug in to get a
    6^2 = 4a(0-18) so a=-\frac{1}{2}

    so the equation for axis of symmetry over y is

    x^2 = -2(y-18)

    leting x=4 then y=10 or 10ft but this of course actually touches the roof of the tunnel.
    Last edited by bigwave; December 14th 2010 at 11:16 AM. Reason: sign error
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  5. #5
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    The problem was not about "similar triangles", it was about the largest rectangle (the truck) that could be fitted into the parabola.
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  6. #6
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    vertex \displaystyle \frac{-b}{2a} lets assume the parabola crosses at \pm 12 on the x axis and the y intercept is at (0,18). Now, first you need to find the equation of the parabola. Then set the parabola equal to the equation x=8.
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  7. #7
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    Write the equation of your parabola in turning point form...

    \displaystyle y = a(x -h )^2 + k.

    You know the turning point is \displaystyle (6, 18) (since the parabola is symmetrical), and you know that \displaystyle (0,0) and \displaystyle (12,0) are points. Substitute into the equation...

    \displaystyle 0 = a(12 - 6)^2 + 18

    \displaystyle 0 = 36a + 18

    \displaystyle -18 = 36a

    \displaystyle -\frac{1}{2} = a.


    So the equation of the parabola is \displaystyle y = -\frac{1}{2}(x - 6)^2 + 18.


    Your truck is \displaystyle 8\,\textrm{ft} wide, so that means there are \displaystyle 4\,\textrm{ft} on either side. So you want to find the height of your parabola ( \displaystyle y value) when \displaystyle x = 2 or when \displaystyle x = 10.
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