Modeling with First Order DE’s — Question 8

Question 8 – Solution

Strategy. Convert total growth observations to per-capita rates before estimating the crowding coefficient.

Step 1: Calibrate the rate law. The observations give $$a-100b=0.20,\qquad a-200b=0.10.$$ Subtracting yields $$\boxed{a=0.30\text{ day}^{-1},\qquad b=0.001\text{ individual}^{-1}\text{day}^{-1}}.$$ The model becomes $P'=0.30P(1-P/300)$. Both specified populations give total rate $20$ individuals/day.

Step 2: Solve the IVP. Separation for the branch $0<P<300$ gives $$\ln\frac{P}{300-P}=0.30t+C.$$ Using $P(0)=100$ gives $C=-\ln 2$ and $$\boxed{P(t)=\frac{300}{1+2e^{-0.30t}}},\qquad t\ge 0.$$ If $z=2e^{-0.30t}$, then $P'=90z/(1+z)^2=0.30P(1-P/300)$, and the initial value is $100$. Applying the calibration assumes the same fixed resources and crowding law, no migration, and a population scale for which the continuous approximation is appropriate.

Step 3: Locate the target. Setting $P=250$ yields $e^{-0.30t}=1/10$, so $$\boxed{t=\frac{\ln 10}{0.30}\approx 7.675\text{ days}}.$$ The formula increases strictly from $100$ toward $300$ for $t\ge 0$, so $250$ is attained exactly once. Its denominator remains positive throughout this physical time range.

Step 4: Test the competing exponential model. For $P'=rP$, the initial observation forces $r=20/100=0.20$ day$^{-1}$. At $P=200$, that model predicts $40$ individuals/day, double the observed rate. Matching one population and one instantaneous slope does not validate a constant per-capita growth law.

The second observation distinguishes the two proposed laws. It supports the calibrated crowding model over this exponential alternative under the stated exact-data assumptions, but two rate observations alone do not establish that the model remains accurate at every population size.