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Spencer Korman, Homework 2
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We are first given information about a population and its growth rate:
\begin{eqnarray*}
k=0.05\ ,
P(t_0)=100
\end{eqnarray*}
As previously calculated, we can rearrange our initial value problem, \\ $\frac{dP}{dt}=kP(t), P(t_0)=P_0$, to obtain $P(t)=Ce^{kt}$, where $P(t)$ is the population at any given time, {\em C}, which we know is equal to $P_0$, is our initial population at $t=0$, {\em e} is a constant, {\em k} is our annual growth rate, and {\em t} is time elapsed.\par
Checking that {\em C} is indeed 100 can be done by substituting in our given values: $100=Ce^{0.05(0)}$, which simplifies to $100=Ce^{0}$, and once more to obtain $C=100$. So, the solution to our initial value problem is, in fact:
\begin{equation}
P(t)=100e^{0.05t}
\label{IVPsol}
\end{equation}
If we were to assume, however, that 10 animals are killed per year, we would have a new IVP:
\begin{equation}
\frac{dP}{dt}=-kP(t)
\label{newIVP}
\end{equation}
This comes from taking the annual birth rate minus the death rate as a percentage of our initial population: $0.05(100)=5 \Longrightarrow \frac{5-10}{100}=-0.05$.
Pictured below is the solution curve over the direction field of (\ref{newIVP}):
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\includegraphics[scale=0.35]{{"HW2 Graph"}.jpg}
\caption{Solution curve of $\frac{dP}{dt}=-kP(t)$ from $t=0$ to $t=15$ }
\label{NewGraph}
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\end{figure}
According to Figure~\ref{NewGraph} , if 10 animals were to die within the first year, the population would eventually die off. This is because the death rate exceeds the birth rate, as 10 is greater than 5.
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