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Math Help - HELP with function

  1. #1
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    HELP with function

    Hey can some one pls explain this problem to me! tnx in advance

    The swimming pool shown in Figure 1.54 is 3 ft. deep at the shallow end, 8 ft. deep at the deepest end, 40 ft. long, 30 ft. wide and the bottom is an incline plane. Express the volume V of water in the pool as a function of the height h of the water above the deep end.(hint: V will be a piecewise define function)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by geeko View Post
    Hey can some one pls explain this problem to me! tnx in advance

    The swimming pool shown in Figure 1.54 is 3 ft. deep at the shallow end, 8 ft. deep at the deepest end, 40 ft. long, 30 ft. wide and the bottom is an incline plane. Express the volume V of water in the pool as a function of the height h of the water above the deep end.(hint: V will be a piecewise define function)
    from your diagram, are you aware that you can split the volume of the figure up into the volume of a box and a triangular prism? can you find the volume of each of those figures?
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  3. #3
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    Hello, geeko!

    A swimming pool is 3 ft. deep at the shallow end, 8 ft. deep at the deep end,
    40 ft. long, 30 ft. wide, and the bottom is an incline plane.
    Express the volume V of water in the pool as a function of the height h
    of the water at the deep end. (Hint: V will be a piecewise-defined function)
    Code:
                           40
      -   * - - - - - - - - - - - - - - - - - *
      :   |                                   |
      3   |                                   | 3
      :   |                   40              |
      -   * - - - - - - - - - - - - - - - - - *
      :   |          x                  *
      :   + - - - - - - - - - - - *
      6   |:::::::::::::::::*
      :  h|:::::::::::*
      :   |:::::*
      -   *

    For the first six feet, the volume of water is a triangular prism.
    . . Its volume is: . \text{Area of triangle} \times \text{30-foot width}

    From the similar right triangles: . \frac{x}{h} = \frac{40}{6} \quad\Rightarrow\quad x \:=\:\frac{20}{3}h .[1]


    The area of the triangle is: . A \:=\:\frac{1}{2}hx

    Substitute [1]: . A \:=\:\frac{1}{2}h\left(\frac{20}{3}h\right) \:=\;\frac{10}{3}h^2

    Then the volume is: . V \:=\:\frac{10}{3}h^2 \times 30 \:=\:100h^2

    . . \boxed{\text{For }0 \leq h \leq 6\!:\;\;V \;=\;100h^2\text{ ft}^3}



    For the last three feet of water, the bottom (triangular) portion is already filled.
    . . It contains: . 100(6^2) \:=\:3600 ft³ of water.

    The upper portion is a rectangular slab of water:
    . . length 40, width 30, and height h-6
    Its volume is: . 40 \times 30 \times (h-6) \:=\:1200(h-6) ft³.

    The total volume is: . V \:=\:1200(h-6) + 3600

    . . \boxed{\text{For }6 \leq h \leq 9\!:\;\;V \;=\;1200(h-3)\text{ ft}^3}



    Therefore: . V \;=\;\bigg\{ \begin{array}{ccc}100h^2 & & 0 \leq h \leq 6 \\ 1200(h-3) && 6 \leq h \leq 9 \end{array}

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