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Math Help - Flow rate

  1. #1
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    Flow rate

    a) At a time t=0 a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time t is y. Show that there's a constant b<1 such that y=b^t.

    b) Suppose instead that the tank contains one unit of water at time t=0, but in that addition to the water flowing out as described, water is added at a steady rate a>0. Show that

    dy/dx-ylnb=a and hence find y in terms of a, b, and t.

    Please help and explain to me what I ought to do as I'm not sure about where to start. Thanks in advance.
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  2. #2
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    Quote Originally Posted by free_to_fly View Post
    a) At a time t=0 a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time t is y. Show that there's a constant b<1 such that y=b^t.
    What you are told is that for some k>0:

    dy/dt=-k y

    which has general solution y(t)=e^(-kt) + C, where k and C are constants.
    As y(0)=1, we have C=0, so:

    y=e^(-kt)=(e^-k)^t = b^t

    where b=e^-k.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by free_to_fly View Post
    a) At a time t=0 a tank contains one unit of water. Water flows out of the tank at a rate proportional to the amount of water in the tank. The amount of water in the tank at time t is y. Show that there's a constant b<1 such that y=b^t.

    b) Suppose instead that the tank contains one unit of water at time t=0, but in that addition to the water flowing out as described, water is added at a steady rate a>0. Show that

    dy/dx-ylnb=a and hence find y in terms of a, b, and t.
    In part b you are told that:

    dy/dt = -ky + a

    or:

    dy/dt + ky = a

    but from part a you have that b=e^-k, so k = -ln(b) hence:

    dy/dt - ln(b)y = a

    RonL
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