Calculus Word Problem

[mhwadester5]

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

 

[galactus]

I assume you're in calc II and touching on that small section that deals with differential equations?.

If so, use the integrating factor.

Rewrite:

\(\displaystyle \frac{dG}{dt}+kG=r\)

The integrating factor comes from the coefficient of the linear term with G.

\(\displaystyle e^{\int kdt}=e^{kt}\)

Multiply through,

\(\displaystyle \frac{d}{dt}[Ge^{kt}]=re^{kt}\)

then integrate and solve for G.

Let me know what you get.

Go to the section in your book dealing with the integrating factor.


In general, if we let the integrating factor be \(\displaystyle \mu=e^{\int p(t) dt}\) and say we have the DE of the form \(\displaystyle \frac{dy}{dt}+p(t)y=q(t)\).........[1]

then \(\displaystyle \frac{d\mu}{dt}=e^{\int p(t)dt}\cdot \frac{d}{dt}\int p(t)dt=\mu p(t)\)

Thus, \(\displaystyle \frac{d}{dt}(\mu y)=\mu\frac{dy}{dt}+\frac{d\mu}{dt}y=\mu\frac{dy}{dt}+\mu p(t)\).....[2]

If [1] is multiplied through by the IF, and then simplified using [2], it becomes:

\(\displaystyle \mu\frac{dy}{dt}+\mu p(t)y=\mu q(t)\)

\(\displaystyle \frac{d}{dt}(\mu y)=\mu q(t)\)

Then, when we integrate we get:

\(\displaystyle \mu y=\int \mu q(t)dt+C\) or \(\displaystyle y=\frac{1}{\mu}\left[\int \mu q(t)dt+C\right]\)

 

[mhwadester5]

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 havn't really covered it yet. Concernign 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.
Sorry in advance.