12) Scientists are interested in the population of a particular species. They attempt to model the population $P$ at time $t$ days using a differential equation. Initially the population is observed to be 50 and after 10 days the population is 100 .

The first model the scientists use assumes that the rate of change of the population is proportional to the population.

(a) Write down a differential equation for this model and solve it for $P$ in terms of $t$.

[5]

To allow for constraints on population growth, the model is refined to

$$

\frac{\mathrm{d} P}{\mathrm{~d} t}=\lambda P(500-P)

$$

where $\lambda$ is a constant.

(b) Solve this differential equation to find $P$ in terms of $t$.

[6]

(c) Using the refined model, state the population of this species in the long term. Comment on how this value suggests the refined model is an improvement on the first model.

[2]

(a) $P=50\left(2^{0.1 t}\right) $

(b) $P=\frac{500\left(2.25^{0.1 t}\right)}{9+2.25^{0.1 t}} ; 500$

(a)

$\frac{d P}{d t}=k P$, where $k$ is a positive constant.

$\int \frac{1}{P} d P=\int k d t$

$P=e^{k t+c}$

$P=A e^{k t}$, where $A=e^c$ is an arbitrary constant.

When $t=0, P=50$, then $A=50$

When $t=10, P=100$, then $e^k=2^{0.1}$

Thus, $P=50\left(2^{0.1 t}\right)$

(b)

$\frac{d P}{d t}=\lambda P(500-P)$

$\int \frac{1}{P(500-P)} d P=\int \lambda d t$

$\frac{1}{500} \int \frac{1}{P}+\frac{1}{500-P} d P=\int \lambda d t$

$\ln P-\ln |500-P|=500 \lambda t+c$

$\frac{P}{500-P}=B e^{500 \lambda t}$, where $B=\pm e^c$ is an arbitrary constant.

When $t=0, P=50$, then $B=\frac{1}{9}$

When $t=10, P=100$, then $e^{5000 \lambda}=2.25 \Rightarrow e^{500 \lambda=2.25^{0.1}}$

$\frac{P}{500-P}=\frac{1}{9} 2.25^{0.1 t}$

$P=\frac{500\left(2.25^{0.1 t}\right)}{9+2.25^{0.1 t}}$

(c) Using $P=\frac{500\left(2.25^{0.1 t}\right)}{9+2.25^{0.1 t}}=\frac{500}{9\left(2.25^{-0.1 t}\right)+1}$, we we that as $t \longrightarrow \infty, e^{-0.1 t} \longrightarrow 0$, then

$P \longrightarrow 500$. Thus, the population will increase and tend to 500 .

Refined model is more realistic as it suggests a limiting value for the population while the first model suggests the population will go to indefinitely.