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Math Help - Relative rates, cone filling with water

  1. #1
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    Relative rates, cone filling with water

    I think there is a problem in the d/dt part
    Problem:
    A cone 4 m high, base diameter of 6 m, water is entering at 3 m^3 per minute
    when water is at a height of 3 m, what is the rate of rise?

    1/3pi(r^2)h = V
    1/3pi(3^2)(3) = 28.274 m^3 at level of 3 m
    dh/dt = ?
    dv/dt= 2m^3/min
    f'ghi + fg'hi + fgh'i + fghi' = dv/dt
    0 + 0 + 1/3pi(2r)(dh/dt) + 1/3pi(r^2)(1)
    working this out gets me a rate of
    1.18 m^3/min

    but I feel there is a problem in the fghi part
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  2. #2
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    Re: Relative rates, cone filling with water

    Hello, togo!

    A cone 4 m high, base diameter of 6 m.
    Water is entering at 3 m^3 per minute
    When water is at a height of 3 m, what is the rate of rise?

    I assume it is an inverted cone.
    Code:
                      3
        - *-------+-------*
        :  \      |      /
        :   \-----+-----/
        :    \    |  r /
        4     \   |   /
        :      \  |h /
        :       \ | /
        :        \|/
        -         *
    The volume of a cone is: . V \:=\:\frac{\pi}{3}r^2h .[1]

    From similar triangles we have: . \frac{r}{h} = \frac{3}{4} \quad\Rightarrow\quad r = \frac{3}{4}h

    Substitute into [1]: . V \:=\:\frac{\pi}{3}\left(\frac{3}{4}h\right)h \quad\Rightarrow\quad V \:=\:\frac{3\pi}{16}h^3


    Differentiate with respect to time: . \frac{dV}{dt} \:=\:\frac{9\pi}{16}h^2\,\frac{dh}{dt}


    We are given: . \frac{dV}{dt} = 3,\;h = 3.

    Therefore: . 3 \:=\:\frac{9\pi}{16}(3^2)\frac{dh}{dt} \quad\Rightarrow\quad \frac{dh}{dt} \:=\:\frac{16}{27\pi}\text{ m/min}
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  3. #3
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    Re: Relative rates, cone filling with water

    thanks, however I was really hoping someone could point out where my method went wrong, because I am most comfortable with that one
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    Re: Relative rates, cone filling with water

    I'm confused with pi. Google says the derivative of pi is zero. However in some of the answers in this book, the derivative of pi ends up being pi.
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    Re: Relative rates, cone filling with water

    Quote Originally Posted by togo View Post
    I'm confused with pi. Google says the derivative of pi is zero. However in some of the answers in this book, the derivative of pi ends up being pi.
    \displaystyle \begin{align*} \pi \end{align*} is a constant, so its derivative is 0. HOWEVER, in this case you have \displaystyle \begin{align*} \pi \end{align*} times a FUNCTION and the derivative of a function times a constant is the same as the constant times the function. I.e. \displaystyle \begin{align*} \frac{d}{dx} \left[ k\, f(x) \right] = k \, \frac{d}{dx} \left[ f(x) \right] \end{align*}.
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    Re: Relative rates, cone filling with water

    however can someone explain how to find derivative of this:
    V = pi(r^2)h

    f = pi
    g = r^2
    h = h

    f' = pi
    g' = 2r
    h' = 1

    so f'gh + fg'h + fgh' =
    pi(r^2)h + pi(2r)h + pi(r^2)(1)

    so is this correct?? thanks!!

    also edit: say related rates question includes:
    Volume remains constant
    Radius decreases 2cm/min
    How fast does surface area change when radius 10 cm and height 8 cm

    so then:
    pi(r^2)h + pi(2r)h + pi(r^2)(1) =
    pi(r^2)h + pi(2r)dr/dt(h) + pi(r^2)dh/dt = V

    correct?
    Last edited by togo; February 9th 2013 at 10:37 AM.
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    Re: Relative rates, cone filling with water

    Quote Originally Posted by togo View Post
    thanks, however I was really hoping someone could point out where my method went wrong, because I am most comfortable with that one
    You went wrong by treating r as if it were a constant. As soroban told you that is not true- as h increases, r must also increase
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