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Math Help - Solving dG/dt = r - kG

  1. #1
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    Solving dG/dt = r - kG

    Hey first time user and thread starter,
    I have a calculus word problem that I need to do for assignment but I absolutely don't know where to start.

    Problem:
    A glucose solution is administered intravenously at a constant rate r. The glucose
    is consumed in the bloodstrem at a rate proportional to its concentration at any time.
    Then if G is the glucose concentration at any time,

    dG/dt = r - kG

    where k is the proportionality constant. Determine the glucose concentration as a
    function of time if the initial concentration is G0. What is the expect concentration
    after a long period of time?

    I have read the conditions for posting and in no way want a complete solution (Against rules, and doesn't help me learn) I just want to know where to start or what to look at first, then I can hopefully figure it out myself.

    As I already know how to do the majority of the calculus that is in my course, just not good with interpreting word problems.

    Thanks, Any help be appreciated
    Last edited by mr fantastic; September 28th 2011 at 05:44 PM. Reason: Re-titled.
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  2. #2
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    Re: Calculus Word Problem

    what do you know about separable differential equations?
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  3. #3
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    Re: Calculus Word Problem

    Hovering 'above' that differential equation is an equation with no differentials in it, which is what you want.

    You can think of the given equation as the result of doing 'implicit differentiation' on the 'higher' equation. (Higher and lower will make sense, at least, if like me you were taught to think of differentiation as 'down' and integration 'up'.)

    And you can see that the differentiation was with respect to t.

    As things stand, though, you can't integrate both sides with respect to t.

    But dG/dt + kG = r has potential. You just need to multiply through by an 'integrating factor'.

    Edit: D'oh! let's not do all that (now in spoiler).

    Spoiler:
    The idea is to turn the left hand side into the result of a product-rule differentiation with respect to t.

    I have a wacky diagram for this rule... (key in spoiler)



    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    But this is wrapped inside the legs-uncrossed version of...



    ... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.

    The general drift...





    So I would put the equation into this form...



    ... and then choose the integrating factor to be e to the power of the anti-derivative (wrt t) of k, as indicated in the key in the spoiler above. Here's the next stage...

    Spoiler:



    I hope that gives you an overview of the process.



    Let's 'separate' as Skeeter says. And here's a pic just in case it helps.




    __________________________________________________ __________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; September 28th 2011 at 12:32 PM.
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    Re: Solving dG/dt = r - kG

    Ok what it seems like from my conclusion is mostly to do with integration as you somewhat said which is kinda weird considering that we have not really covered it yet. Concerning the course I am doing I am not doing a full time maths degree or anything doing another course with one maths subject in it that seems to cover the majority of I guess the basics. So if I see somewhat confused by some of the stuff you explained we may not have covered or touched on it yet. So Skeeter to tell the truth I don't know much at all about "Separable Differential Equations", which I guess could make this harder to help me so.
    Sorry in advance.

    PS: That weird balloon diagram sorta confused me more
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  5. #5
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    Re: Solving dG/dt = r - kG

    Quote Originally Posted by MHwadester5 View Post
    Ok what it seems like from my conclusion is mostly to do with integration as you somewhat said which is kinda weird considering that we have not really covered it yet. Concerning the course I am doing I am not doing a full time maths degree or anything doing another course with one maths subject in it that seems to cover the majority of I guess the basics. So if I see somewhat confused by some of the stuff you explained we may not have covered or touched on it yet. So Skeeter to tell the truth I don't know much at all about "Separable Differential Equations", which I guess could make this harder to help me so.
    Sorry in advance.

    PS: That weird balloon diagram sorta confused me more
    dG/dt = r - kG

    => dt/dG = 1/(r - kG)

    => t = \int \frac{1}{r - kG} \, dG

    You are now expected to integrate, use the given initial condition, and solve for G. Then consider the limit of this solution as t --> +oo.

    Note that the answer to "What is the expect concentration after a long period of time?" can be calculated by solving dG/dt = 0 ....
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  6. #6
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    Re: Solving dG/dt = r - kG

    Quote Originally Posted by mr fantastic View Post
    dG/dt = r - kG

    => dt/dG = 1/(r - kG)

    => t = \int \frac{1}{r - kG} \, dG

    You are now expected to integrate, use the given initial condition, and solve for G. Then consider the limit of this solution as t --> +oo.

    Note that the answer to "What is the expect concentration after a long period of time?" can be calculated by solving dG/dt = 0 ....
    Actually you do not need the solution of the ODE to obtain the solution after a "long" period of time, since it will be independent of time (that is a steady state solution) you have dG/dt=0 and so r-kG=0

    CB
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