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Math Help - Differential Equation Help

  1. #1
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    Differential Equation Help

    Hey everyone! I'm stuck on the following:

    A person is trying to fill a bucket with water. Water is flowing into the bucket from the tap at a constant rate of 0.12 litres/sec. However there is a hole in the bottom of the bucket and water is flowing out of the bucket at a rate proportional to the square of the volume of water present in the bucket. If V(t) is the volume of water (in litres) present in the bucket at time t (in secs) and the bucket initially contains 2 litres of water, write down, but do not solve, the differential equation for V(t) along with its initial condition.

    Okay here is what I have:

    Rate leaving is proportional to (sorry don't know how to write the symbol here) V^2, therefore:

    Rate = kV^2 or should it be ktV^2

    So with that I get a function:

    V(t) = 2 + 0.12t - kV^2

    Okay my question is am I right in putting the rate it's leaving straight into that equation?

    Can I just then sub in dV/dt for V?

    dV/dt = 2 + 0.12t - k(dV/dt)^2

    Any help would be great!
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  2. #2
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    Quote Originally Posted by falconed View Post
    Hey everyone! I'm stuck on the following:

    A person is trying to fill a bucket with water. Water is flowing into the bucket from the tap at a constant rate of 0.12 litres/sec. However there is a hole in the bottom of the bucket and water is flowing out of the bucket at a rate proportional to the square of the volume of water present in the bucket. If V(t) is the volume of water (in litres) present in the bucket at time t (in secs) and the bucket initially contains 2 litres of water, write down, but do not solve, the differential equation for V(t) along with its initial condition.

    Okay here is what I have:

    Rate leaving is proportional to (sorry don't know how to write the symbol here) V^2, therefore:

    Rate = kV^2 or should it be ktV^2

    So with that I get a function:

    V(t) = 2 + 0.12t - kV^2

    Okay my question is am I right in putting the rate it's leaving straight into that equation?

    Can I just then sub in dV/dt for V?

    dV/dt = 2 + 0.12t - k(dV/dt)^2

    Any help would be great!
    dV/dt = rate in - rate out = 0.12 - kV^2 subject to the boundary condition V(0) = 2.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    dV/dt = rate in - rate out = 0.12 - kV^2 subject to the boundary condition V(0) = 2.
    So that is all I needed? I guess I was thinking it would be a bit more complicated.
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  4. #4
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    lol the griffith assignment 3 haha its due thursday
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  5. #5
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    for the rate flowing out why is it kV^2
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  6. #6
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    Quote Originally Posted by mark18 View Post
    for the rate flowing out why is it kV^2
    Read the question more carefully:

    Quote Originally Posted by falconed View Post
    Hey everyone! I'm stuck on the following:

    A person is trying to fill a bucket with water. Water is flowing into the bucket from the tap at a constant rate of 0.12 litres/sec. However there is a hole in the bottom of the bucket and water is flowing out of the bucket at a rate proportional to the square of the volume of water present in the bucket. If V(t) is the volume of water (in litres) present in the bucket at time t (in secs) [snip]
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  7. #7
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    is it possible to find what k is or is more information required
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  8. #8
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    Quote Originally Posted by mark18 View Post
    is it possible to find what k is or is more information required
    More information is required.
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