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Math Help - Related Rates

  1. #1
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    Related Rates

    A swimming pool is 25 ft wide, 60 ft long, 3 ft deep at the shallow end, and 15
    ft at its deepest point. If the pool is being filled at a rate of 800 ft^3/min,
    at what rate is the water level rising when the depth at the deepest point is
    5ft.

    I'm stuck in this exercise. I just need a tip for the exercise. Thanks for you
    help.

    I did this:
    Let x feet be the wide, and y the long, t for time measured in minutes.

    dx/dt= 25 ft and I need to find dy/dt when y = 60ft.

    I know that I have to write an equation that relates the various quantities of
    the problem. But I don't know what equation I should put.
    Attached Thumbnails Attached Thumbnails Related Rates-relrates.jpg  
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, Ron!

    A swimming pool is 25 ft wide, 60 ft long, 3 ft deep at the shallow end, and 15 ft at its deepest point.
    If the pool is being filled at a rate of 800 ft³/min, at what rate is the water level rising
    when the depth at the deepest point is 5ft?
    Side view of the pool . . . Ignore the top 3 feet.
    Code:
            : - - - -  60 - - - - - :
      -     *-----------------------*
      :     |       x           *
      :   - +---------------*
      12  : |:::::::::::*
      :   y |:::::::*
      :   : |:::*
      -   - *
    The area of the water is: . \,A\,=\,\frac{1}{2}xy
    From similar right triangles, we have: . \,\frac{x}{y}\,=\,\frac{60}{12}\;\;\Rightarrow\;\;  x\,=\,5y
    The area of the water is: . \,A\:=\;\frac{1}{2}(5y)(y)\:=\:\frac{5}{2}y^2
    The volume of the water is: . V\:=\:\frac{5}{2}y^2 \times 25\:=\:\frac{125}{2}y^2

    Differentiate with respect to time: . \frac{dV}{dt}\:=\:125y\cdot\frac{dy}{dt}


    We are given: \frac{dV}{dt} = 800 ft³/min and y = 5

    So we have: . 800 \:=\:125(5)\cdot\frac{dy}{dt}

    Therefore: . \frac{dy}{dt}\:=\:\frac{800}{625}\:=\:1.28 ft/min.
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  3. #3
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    I have that idea but I was confussed. But THANKS A LOT!!!
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