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Math Help - Find the altitude of water inside two cones.

  1. #1
    Dug
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    Find the altitude of water inside two cones.



    I already did ratio and proportion, volume, area, and base formulas.
    But i can't get the answer.
    The answer at the back of the book is 5.1510 ft.
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  2. #2
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    The equilibrium occurs when the water is at the same level in both tanks. (Can you explain why?) Suppose this common level is h. Express the following things through h:

    (1) the diameter of the water surface at height h in each tank;
    (2) the surface area at height h in each tank;
    (3) the volume of water in each tank;
    (4) the total volume of the first tank.

    The equation states that the sum of the volumes of water found in (3) equals (4). From there, h can be found.
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  3. #3
    Dug
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    Quote Originally Posted by emakarov View Post
    The equilibrium occurs when the water is at the same level in both tanks. (Can you explain why?) Suppose this common level is h. Express the following things through h:

    (1) the diameter of the water surface at height h in each tank;
    (2) the surface area at height h in each tank;
    (3) the volume of water in each tank;
    (4) the total volume of the first tank.

    The equation states that the sum of the volumes of water found in (3) equals (4). From there, h can be found.
    sorry i can't understand (1) and (2)
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  4. #4
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    Hello, Dug!

    4. Two inverted conical tanks have their vertices connected by a pipe.
    One tank, initially full of water, has a height of 6 feet and a base radius of 3.5 feet.
    The other tank, initially empty, has a height of 9 feet and a base radius of 4 feet.

    If the water is allowed to flow through the connecting pipe, find the level to which
    the water will rise in the empty tank. (Neglect the water in the pipe.)

    The water will be at the same height h in both tanks.


    Code:
                                         4
                          *---------+---------*
                   3.5    :\        |        /
        *-------+-------* : \       |       /
        :\      |      /  :  \      |      / 
        : \     | r1  /   :   \     | r2  /
        :  \----+----/    9    \----+----/
        6   \   |   /     :     \   |   /
        :    \  |h /      :      \  |h /
        :     \ | /       :       \ | /
        :      \|/        :        \|/
        -       *-------------------*

    The original volume of water is: . V \:=\:\tfrac{\pi}{3}r^2h \:=\:\tfrac{\pi}{3}(\tfrac{7}{2})^2(6) \:=\:\frac{49}{2}\pi\text{ ft}^3


    The volume of water in the small tank is: . V_1 \:=\:\tfrac{\pi}{3}(r_1)^2h

    . . where: . \frac{r_1}{h} \,=\,\frac{\frac{7}{2}}{6} \quad\Rightarrow\quad r_1 \:=\:\tfrac{7h}{12}

    Hence: . V_1 \;=\;\tfrac{\pi}{3}(\tfrac{7h}{12})^2h \;=\;\frac{49\pi}{432}h^3


    The volume of water in the large tank is: . V_2 \;=\;\tfrac{\pi}{3}(r_2)^2h

    . . where: . \frac{r_2}{h} \,=\,\frac{4}{9} \quad\Rightarrow\quad r_2 \:=\:\frac{4h}{9}

    Hence: . V_2 \;=\;\tfrac{\pi}{3}(\tfrac{4h}9})^2h \;=\;\frac{16\pi}{243}h^3


    Since V_1 + V_2 \:=\:V, we have: . \frac{49\pi}{432}h^3 + \frac{16\pi}{243}h^3 \;=\;\frac{49\pi}{2}

    . . \frac{697\pi}{3888}h^3 \:=\:\frac{49\pi}{2} \quad\Rightarrow\quad h^3 \:=\:\frac{95,\!256}{697} \:=\:136.6657102


    \text{Therefore: }\;h \;=\;\sqrt[3]{136.6657102} \;=\;5.150940353

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  5. #5
    Dug
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    thank you very much that was a headache to think about.
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