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Math Help - Trouble with related rates.

  1. #1
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    Trouble with related rates.

    The problem is as follows:

    A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m^3/min, find the rate at which water level is rising when the water is 3 m deep.

    So the problem tells me that dv/dt=2, h=4, and r=2. After taking my derivative in respect to time, I have to solve dr/dt..but then what do I plug in for dh/dt? Is that where the "3 m deep" part comes into play?

    Please help.
    Thank you.
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  2. #2
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    isn't dr/dt =2m^3??

    so find dh/dt
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  3. #3
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    I thought that because it's cubed, that's the volume.
    Plus, the question asks for the rate, not the height. :/
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  4. #4
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    V = \frac{\pi}{3} r^2 h

    \frac{r}{h} = \frac{2}{4} , so  r = \frac{h}{2} ...

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

    V = \frac{\pi}{12} h^3

    \frac{dV}{dt} = \frac{\pi}{4} h^2 \cdot \frac{dh}{dt}

    plug in your values and determine the value of \frac{dh}{dt}
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  5. #5
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    Quote Originally Posted by skeeter View Post
    V = \frac{\pi}{3} r^2 h

    \frac{r}{h} = \frac{2}{4} , so  r = \frac{h}{2} ...

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

    V = \frac{\pi}{12} h^3

    \frac{dV}{dt} = \frac{\pi}{4} h^2 \cdot \frac{dh}{dt}

    plug in your values and determine the value of \frac{dh}{dt}
    Oh, okay..thank you..
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