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Math Help - Related rates problem.. :)

  1. #1
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    Related rates problem.. :)

    Related rates problem.. :)-dumb.pngHello So first this is the problem:

    A Swimming pool is 20 ft. wide, 40 ft. long, 3 ft. deep at the shallow end and 9 ft. deep at it's deepest point. A cross section is shown in the figure. If the pool is being filled at a rate of 0.8 cu ft. per min., how fast is the water level rising when the depth at the deepest end is 5ft.

    Thank you.. mind if you please show me the step by step procedure thank you...
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  2. #2
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    Re: Related rates problem.. :)

    Was that picture part of the given problem or did you draw it yourself? It is not what I would get from your description- nor is it my experience with swimming pools. You have a "lowered" section on the left with slanted sides and then up to a height of 3 feet beow the top, then flat for the rest of the forty foot length. Where did you get that "6- 12- 6- 16" along the bottom (which I assume are distances). I would, rather, have drawn the side view of the swimming pool as a single trapezoid with the two parallel ends of lengths 3 and 9 feet and the "height" 40 feet. I don't know if you were given that rather strange picture or drew it yourself.
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    Re: Related rates problem.. :)

    No sir, I didn't... It was given to our midterms exam just a while ago, me and my other colleague discussed the solutions and answered all of our questions.. excepts one... that One exactly... yep, i tried to use the area, which then substituted it to the volume then differentiated it.. but I'm pretty sure I'm still lost..
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  4. #4
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    Re: Related rates problem.. :)

    Hello, kspkido!

    A swimming pool is 20 ft. wide, 40 ft. long, 3 ft. deep at the shallow end and 9 ft. deep at it's deepest point.
    A cross-section is shown in the figure.
    If the pool is being filled at a rate of 0.8 cu ft. per min.,
    . . how fast is the water level rising when the depth at the deepest end is 5 ft?
    Code:
          : - - - - - - - - - -  40 - - - - - - - - - :
        - *-------------------------------------------*
        3 |                                           |
        - *  -  -  -  -  -  -  -  *-------------------*
          :*~~~~~~~~~~~~~~~~~~~~~*: - - -  16 - - - - :
        6 : *:::::::::|:::::::::* :
          : -*::::::::|h:::::::*  :
        - + - *-------------- * - :
          : 6 : - -  12 - - - : 6 :

    Since h \le 6, we are concerned with the trapezoid only.
    The cross-section looks like this:
    Code:
            h :   12  : h
          *-------+-------*
           *::::::|::::::*
            *:::::|h::::*
             *::::|::::*
              *---*---*
                  12
    The area is: . A \:=\:\tfrac{h}{2}(12 + 12 + 2h) \:=\:12h+h^2

    The volume is: . V \:=\:20A \:=\:20(12h+h^2)

    Then: . \frac{dV}{dt} \:=\:20(12 + 2h)\frac{dh}{dt}

    When \frac{dV}{dt} = 0.8,\;h - 5, we have: . 0.8 \:=\:20(22)\frac{dh}{dt}

    . . \frac{dh}{dt} \:=\:\frac{0.8}{440} \:=\:\frac{1}{55}


    The water is rising at the rate of \tfrac{1}{55} ft/min.
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    Re: Related rates problem.. :)

    Wow... I never thought you could do that... h=6, but you used it as an integer just so that you could derive ayt?... but one question sir... h-5? is equals to 5? how is that? but thank you sir...
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  6. #6
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    Re: Related rates problem.. :)

    you've posted this problem before ... and it was answered.

    Rate Problem Involving Changing Height..
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