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Math Help - Conical tank...related rates

  1. #1
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    Exclamation Conical tank...related rates

    Hi,

    I've explained everything here and have written my thoughts for the problem as well:

    http://i44.tinypic.com/241v2m8.jpg

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  2. #2
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    can anyone help wth even parts of my question if it's not all of it?

    -thanks!!
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  3. #3
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    Hi



    Note that \frac{h}{r} = \frac{H}{R}

    At instant t, the height is h
    At instant (t+dt), the height is (h+dh)
    The corresponding volume between t and (t+dt) is

    dV = V(h+dh) - V(h) = \frac{\pi}{3}\:\left(\frac{R^2}{H^2}\:\left(h+dh\r  ight)^3 - \frac{R^2}{H^2}\:h^3\right) = \frac{\pi\:R^2}{3\:H^2}\:3\:h^2\:dh = \frac{\pi\:R^2\:h^2}{H^2}\:dh

    The flowrate is

    Q = \frac{dV}{dt} = \frac{\pi\:R^2\:h^2}{H^2}\:\frac{dh}{dt}

    Therefore \frac{dh}{dt} = \frac{Q\:H^2}{\pi\:R^2\:h^2} = \frac{289}{18\pi}
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  4. #4
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    Hello, janedoe!

    A conical tank is 12 ft across the top and 17 ft deep.
    If water is flowing into the tank at a rate of 8 ft³/min,
    find the rate of change of the depth when the water is 2 ft deep.
    Code:
                  :   6   :
        - *-------+-------*
        :  \      |      /
        :   \     |  r  /
        :    *----+----*
       17     \   |   /
        :      \ h|  /
        :       \ | /
        :        \|/
        -         *

    The volume of the water is: . V \:=\:\frac{\pi}{3}r^2h .[1]

    From the similar right triangles: . \frac{r}{h} \:=\:\frac{6}{17} \quad\Rightarrow\quad r \:=\:\frac{6}{17}\,h

    Substitute into [1]: . V \;=\;\frac{\pi}{3}\left(\frac{6}{17}\,h\right)\!\!  ^2h \quad\Rightarrow\quad V \:=\:\frac{12\pi}{289}\,h^3

    Differentiate with respect to time: . \frac{dV}{dt} \:=\:\frac{36\pi}{289}\,h^2\,\frac{dh}{dt}


    Since \frac{dV}{dt} = 8,\;h = 2. we have: . 8 \:=\:\frac{36\pi}{289}(2^2)\frac{dh}{dt}

    . . Therefore: . \frac{dh}{dt} \:=\:\frac{289}{18\pi} ft/min.



    And thank you, thank you for sharing your work and your thoughts!

    You put in a lot of work before asking for help.
    It is an extremely rare occurence . . .We truly appreciate it!

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  5. #5
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    It's the least I could do!

    Actually I did have that proportion r/h=6/17, but I just plugged in 2 for h... I see that I can't b/c that's a variable right?

    I think I'm getting this now...I'm teaching myself calc so let's hope I do! haha
    Thanks for the help!!!
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