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Math Help - CAN ANY1 HELP ME!!! Math C - Rate's of Change

  1. #1
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    Exclamation CAN ANY1 HELP ME!!! Math C - Rate's of Change

    I'm New at math C and this question is bugging me, can any1 solve this equation so i know how to do it?

    The question says:

    The rate of loss of water from a rainwater tank with a leak in the side depends on the pressure at the hole. The pressure depends on the height of water above the leak.
    A cylindrical tank has a leak 20cm from the top, which has not been fixed because it is only a short distance form the top. Five hours after the tank is filled, the level has dropped by 15cm. Using a differential equation, develop a model to describe the height of water above the hole as a function of time. If another hole is made 20cm from the bottom of the tank, which is 2 metres high, how long will it take before half the water has leaked away?
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  2. #2
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    Quote Originally Posted by sterny0 View Post
    I'm New at math C and this question is bugging me, can any1 solve this equation so i know how to do it?

    The question says:

    The rate of loss of water from a rainwater tank with a leak in the side depends on the pressure at the hole. The pressure depends on the height of water above the leak.
    A cylindrical tank has a leak 20cm from the top, which has not been fixed because it is only a short distance form the top. Five hours after the tank is filled, the level has dropped by 15cm. Using a differential equation, develop a model to describe the height of water above the hole as a function of time. If another hole is made 20cm from the bottom of the tank, which is 2 metres high, how long will it take before half the water has leaked away?
    The rate of loss of water is a rate of change of volume, and it is proportional to the head h (the height of the surface above the hole).

    So:

    \frac{dV}{dt}=-kh

    and V=(H+h)A, where H is the height of the hole from the bottom of the tank and A is the crossectional area of the tank, so:

    \frac{dV}{dt}=A\frac{dh}{dt}

    putting there together we have:

    \frac{dh}{dt}=-\frac{k}{A}h

    RonL
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