Question

A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the gulcose is added, it is converted into other substances and removed from the bloodstream at a rate that is proportional to the concentration at the time. Thus a model for the concentration C=C(t) of the glucose solution in the bloodstream is
dC/dt=r-kC
Where r an dk are positive constants.
1. Suppose that the concentration at time t=0 is C0. Determine the concentration at any time t by solving the differential equation.
2. Assuming that C0

Answer 1

Answer:

$C(t) =\dfrac{ r}{k} - \left (\dfrac{r-kC_{0}}{k} \right )e^{ -kt}$

$C(t) =\dfrac{ r}{k}- e^{ -kt}$   ,thus, the  function is said to be an increasing function.

Step-by-step explanation:

Given that:

$\dfrac{dC}{dt}= r-kC$

$\dfrac{dC}{r-kC}= dt$

Taking integration on both sides ;

$\int\limits\dfrac{dC}{r-kC}= \int\limits \ dt$

$- \dfrac{1}{k}In (r-kC)= t +D$

$In(r-kC) = -kt - kD \\ \\ r- kC = e^{-kt - kD} \\ \\ r- kC = e^{-kt} e^{ - kD} \\ \\r- kC = Ae^{-kt} \\ \\ kC = r - Ae^{-kt} \\ \\ C = \dfrac{r}{k} - \dfrac{A}{k}e ^{-kt}$

$C(t) =\frac{ r}{k} - \frac{A}{k}e^{ -kt}$

here;

A is an integration constant

In order to determine A, we have $C(0) = C0$

$C(0) =\frac{ r}{k} - \frac{A}{k}e^{0}$

$C_0 =\frac{r}{k}- \frac{A}{k}$

$C_{0} =\frac{ r-A}{k}$

$kC_{0} =r-A$

$A =r-kC_{0}$

Thus:

$C(t) =\dfrac{ r}{k} - \left (\dfrac{r-kC_{0}}{k} \right )e^{ -kt}$

2. Assuming that C0 < r/k, find lim t→[infinity] C(t) and interpret your answer

$C_{0} < \lim_{t \to \infty }C(t) \\ \\C_0 < \dfrac{r}{k} \\ \\kC_0$

The equation for C(t) can  be rewritten as :

$C(t) =\dfrac{ r}{k} - \left (\dfrac{r-kC_{0}}{k} \right )e^{ -kt}C(t) =\dfrac{ r}{k} - \left (+ve \right )e^{ -kt} \\ \\C(t) =\dfrac{ r}{k}- e^{ -kt}$

Thus;  the  function is said to be an increasing function.