Q. 4 - Calculus

Question

Theorems: Fill in the blanks to complete each of the following theorem statements.

If a population $P\left(t\right)$ changes over time with natural growth rate $k$ and carrying capacity $L$, then it can be modeled by the differential equation $\frac{dP}{dt}=\_\_\_\_\_$, with solution $P\left(t\right)=\_\_\_\_\_$.

Step-by-Step Solution

Answer

If a population $P\left(t\right)$ changes over time with natural growth rate $k$ and carrying capacity $L$, then it can be modeled by the differential equation $\frac{dP}{dt}=kP\left(1-\frac{P}{L}\right)$, with solution $P\left(t\right)=\frac{L{P}_0}{P_0+\left(L-P_0\right)e^{-kt}}$ .

Step 1. Given information

If a population $P\left(t\right)$ changes over time with natural growth rate $k$ and carrying capacity $L$, then it can be modeled by the differential equation $\frac{dP}{dt}=\_\_\_\_\_$, with solution $P\left(t\right)=\_\_\_\_\_$.

Step 2. Filling the blanks

If a population $P\left(t\right)$ changes over time with natural growth rate $k$ and carrying capacity $L$, then it can be modeled by the differential equation $\frac{dP}{dt}=kP\left(1-\frac{P}{L}\right)$, with solution $P\left(t\right)=\frac{L{P}_0}{P_0+\left(L-P_0\right)e^{-kt}}$.

The differential equation is: $\frac{dP}{dt}=kP\left(1-\frac{P}{L}\right)$

$\Rightarrow \frac{dP}{dt}=\frac{kP\left(L-P\right)}{L}$

By separation of variables and integrating, we get,

$\int \frac{L dP}{P(L-P)}=\int k dt$

By using partial fractions, we get,

$$\begin{gathered} \int \left[\frac{1}{P}+\frac{1}{(L-P)}\right] dP=\int k dt \\ \Rightarrow \ln P-\ln (L-P)=kt+C \\ \Rightarrow \ln \frac{P}{(L-P)}=kt+C \\ \Rightarrow \ln \frac{\left(L-P\right)}{P}=-kt-C \\ \Rightarrow \frac{\left(L-P\right)}{P}=-kt-C \\ \Rightarrow \frac{L}{P}-1=e^{-kt-C} \\ \Rightarrow \frac{L}{P}=1+e^{-kt-C} \\ \Rightarrow \frac{L}{P}=1+A{e}^{-kt} \\ \Rightarrow P=\frac{L}{1+A{e}^{-kt}} \end{gathered}$$

Since, $P\left(0\right)=P_0$ and $P_0=\frac{L}{1+A{e}^{-k\left(0\right)}}$

$$\begin{gathered} \Rightarrow 1+A=\frac{L}{P_0} \\ \Rightarrow A=\frac{L}{P_0}-1 \\ \Rightarrow A=\frac{L-P_0}{P_0} \end{gathered}$$

Therefore, $P\left(t\right)=\frac{L{P}_0}{P_0+(L-P_0)e^{-kt}}$