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Math Help - water is flowing from a shallow concrete conical reservoir...

  1. #1
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    Post water is flowing from a shallow concrete conical reservoir...

    Water is flowing at a rate of 50 m^3/min from a shallow concrete conical reservoir, vertex down, of base radius 40m and height 6m. How fast in cm per minutes, is the water level falling when the water is 3ft deep?

    I know the volume of a cone is V = (pi/3)rh
    and I know to use the chain rule: dV/dt = (dh/dt)(dV/dh)

    but I don't know how to find dh/dt or dv/dh
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  2. #2
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    Quote Originally Posted by genlovesmusic09 View Post
    Water is flowing at a rate of 50 m^3/min from a shallow concrete conical reservoir, vertex down, of base radius 40m and height 6m. How fast in cm per minutes, is the water level falling when the water is 3ft deep?

    I know the volume of a cone is V = (pi/3)rh
    and I know to use the chain rule: dV/dt = (dh/dt)(dV/dh)

    but I don't know how to find dh/dt or dv/dh
    \frac{r}{h} = \frac{40}{6} = \frac{20}{3}

    r = \frac{20h}{3}

    V = \frac{\pi}{3} r^2 h

    V = \frac{\pi}{3} \left(\frac{20h}{3}\right)^2 h

    V = \frac{400\pi}{27} h^3

    take the time derivative of the last equation to get the relationship between \frac{dV}{dt} and what you are looking for ... \frac{dh}{dt}
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    \frac{{dv}}<br />
{{dt}} = \frac{{400\pi }}<br />
{9}{h^2}\frac{{dh}}<br />
{{dt}}<br />

    i know you plug 50 into dv/dt but I'm not sure what to plug in for h... 6m or 3ft?

    and i know after that you solve for dh/dt
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    Quote Originally Posted by genlovesmusic09 View Post
    \frac{{dv}}<br />
{{dt}} = \frac{{400\pi }}<br />
{9}{h^2}\frac{{dh}}<br />
{{dt}}<br />

    i know you plug 50 into dv/dt but I'm not sure what to plug in for h... 6m or 3ft?

    and i know after that you solve for dh/dt
    the question asks for dh/dt in cm/min when h = 3 ft(?) ... (sure it's in feet? if so, then you'll have to convert to meters to get it in units of m/min)

    I wouldn't convert m to cm until the very end.
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    yes the question says "How fast in cm per minutes, is the water level falling when the water is 3ft deep?"
    so I don't need to incorporate the 3ft into the equation?
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    Quote Originally Posted by genlovesmusic09 View Post
    yes the question says "How fast in cm per minutes, is the water level falling when the water is 3ft deep?"
    so I don't need to incorporate the 3ft into the equation?
    yes you do ... h = 3ft in the derivative equation. you'll need to convert to meters like I told you previously.
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  7. #7
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    so i plugged 3ft or 0.9144meters into h^2
    and found \frac{{dh}}<br />
{{dt}} = \frac{{450}}<br />
{{365.76\pi }}<br />
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