Answer:
[tex]C(t) =\dfrac{ r}{k} - \left (\dfrac{r-kC_{0}}{k} \right )e^{ -kt}[/tex]
[tex]C(t) =\dfrac{ r}{k}- e^{ -kt}[/tex] ,thus, the function is said to be an increasing function.
Step-by-step explanation:
Given that:
[tex]\dfrac{dC}{dt}= r-kC[/tex]
[tex]\dfrac{dC}{r-kC}= dt[/tex]
Taking integration on both sides ;
[tex]\int\limits\dfrac{dC}{r-kC}= \int\limits \ dt[/tex]
[tex]- \dfrac{1}{k}In (r-kC)= t +D[/tex]
[tex]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} \\ \\[/tex]
[tex]C(t) =\frac{ r}{k} - \frac{A}{k}e^{ -kt}[/tex]
here;
A is an integration constant
In order to determine A, we have [tex]C(0) = C0[/tex]
[tex]C(0) =\frac{ r}{k} - \frac{A}{k}e^{0}[/tex]
[tex]C_0 =\frac{r}{k}- \frac{A}{k}[/tex]
[tex]C_{0} =\frac{ r-A}{k}[/tex]
[tex]kC_{0} =r-A[/tex]
[tex]A =r-kC_{0}[/tex]
Thus:
[tex]C(t) =\dfrac{ r}{k} - \left (\dfrac{r-kC_{0}}{k} \right )e^{ -kt}[/tex]
2. Assuming that C0 < r/k, find lim t→[infinity] C(t) and interpret your answer
[tex]C_{0} < \lim_{t \to \infty }C(t) \\ \\C_0 < \dfrac{r}{k} \\ \\kC_0 <r[/tex]
The equation for C(t) can be rewritten as :
[tex]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}[/tex]
Thus; the function is said to be an increasing function.
