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Math Help - Tricky Integration word problem...

  1. #1
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    Tricky Integration word problem...

    A fish tank has a triangular cross-section.

    It has a length of 80cm, width of 50cm and a depth of 40cm.

    Initially it is full of water.

    After 4 days, the depth of the water had dropped to 35cm.

    The rate of evaporation is proportional to the surface area of the water.

    Calculate how long the water in the fish tank takes to completely evaporate from when it was full.


    - I've tried what seems like everything I can possibly imagine... I must be missing something so obvious. Please help!
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  2. #2
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    Hello, LLemon!

    This is not a simple problem . . .


    A fish tank has a triangular cross-section.
    It has a length of 80cm, width of 50cm and a depth of 40cm.
    Initially it is full of water.
    After 4 days, the depth of the water had dropped to 35cm.
    The rate of evaporation is proportional to the surface area of the water.
    Calculate how long the water in the fish tank takes to completely evaporatel.
    Code:
                      25
        - *-------*-------*
        :  \      |      /
        :   \     |  x  /
        :    \- - + - -/
       40     \:::|:::/
        :      \::|h:/
        :       \:|:/
        :        \|/
        -         *

    The volume of water is: . V \;=\;\frac{1}{2}(2x)(h)(80) \;=\;80hx .[1]

    From the similar right triangles: . \frac{x}{h} \,=\,\frac{25}{40}\quad\Rightarrow\quad x \:=\:\frac{5}{8}\,h .[2]
    Substitute [2] into [1]: . V \;=\;80h\left(\frac{5}{8}h\right) \;=\;50h^2
    . . Then: . \frac{dV}{dt} \:=\:100h\left(\frac{dh}{dt}\right) .[3]


    We are told that \frac{dV}{dt} is proportional to the surface area: . \frac{dV}{dt} \:=\:k\!\cdot\!A
    . . The surface area is: . A \:=\:(2x)(80) \:=\: 160x
    . . Subtitute [2]: . A \:=\:160\left(\frac{5}{8}h\right) \:=\:100h
    Then: . \frac{dV}{dt} \;=\;k(100h) \;=\;<br />
100kh .[4]


    Equate [3] and [4]: . 100h\left(\frac{dh}{dt}\right)\;=\;100kh \quad\Rightarrow\quad dh \:=\:k\,dt

    Integrate: . h \;=\;kt + C


    When t = 0,\:h = 40\!:\;\;40 \:=\:k(0) + C\quad\Rightarrow\quad C = 40
    . . The height function (so far) is: . h \;=\;kt + 40

    When t = 4,\:h = 35\!:\;\;35\:=\:k(4) + 40\quad\Rightarrow\quad k \,=\,-\frac{5}{4}
    Hence, the height function is: . h \;=\;-\frac{5}{4}\,t + 40


    When is the tank empty (h = 0)?

    . . -\frac{5}{4}\,t + 40 \:=\:0\quad\Rightarrow\quad t \:=\:32


    Therefore, it takes \boxed{{\color{blue}\,32\text{ hours}\,}}.

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  3. #3
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    Nov 2007
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    That was outstanding! Thank you so much. Once you realise about the similar triangles things start falling into place. I was trying to find a relationship using volume of a trapezoid which just got incredibly messy when I plugged it into the related rates of change...
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