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Math Help - Harder question on connected rates of change

  1. #1
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    Harder question on connected rates of change

    This is the question:

    A hollow right circular cone has height 18 cm and base radius 12 cm.
    It is held vertex downwards beneath a tap leaking at the rate of 2 cm/s.
    Find the rate of rise of water level when the depth is 6 cm.

    I have no idea where to go with this question. This is what i've tried:

    \frac{dh water}{dt}=(\frac{dh water}{dv})(\frac{dv}{dt})

    Then i can find out the volume of the cone easily in respect to either r or h but i don't understand how i can use this in order to find the change in the height of the depth of the water.

    Can someone please explain how i can proceed to complete this question. Thanks in advance.
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  2. #2
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    Quote Originally Posted by Big_Joe View Post
    This is the question:

    A hollow right circular cone has height 18 cm and base radius 12 cm.
    It is held vertex downwards beneath a tap leaking at the rate of 2 cm/s.
    Find the rate of rise of water level when the depth is 6 cm.

    I have no idea where to go with this question. This is what i've tried:

    \frac{dh water}{dt}=(\frac{dh water}{dv})(\frac{dv}{dt})

    Then i can find out the volume of the cone easily in respect to either r or h but i don't understand how i can use this in order to find the change in the height of the depth of the water.

    Can someone please explain how i can proceed to complete this question. Thanks in advance.
    V = \frac{\pi}{3} r^2 h

    \frac{r}{h} = \frac{12}{18} = \frac{2}{3}

    r = \frac{2h}{3}

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

    V = \frac{4\pi}{27}h^3

    find the time derivative, sub in your known values, and determine the value of \frac{dh}{dt}
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