The rate of change of a population P of an environment is determined by the logistic formula
dP/dt =0.04P(1-P/20000)
where t is in years since the beginning of 2015. So P (1) is the population at the beginning of 2016. Suppose P (0) = 1000.
(a) Calculate P′(0)
(b) Use the number from the previous part to estimate the population in the middle of 2015. That is, estimate P (0.5)
(c) What assumption is made in the computation in the previous part? Use the formula given for P ′ to see whether or not the assumption is true, to within 1%
(a) We have
$$ P'(t) = 0.04(1-P/20000) \cdot \frac{dP}{dt} = 0.04(1-P/20000) \cdot 0.04P(1-P/20000) = 0.04P(1-P/20000)^2$$
Thus,
$$ P'(0) = 0.04 \cdot 1000 \cdot (1-1000/20000)^2 = 6.4.$$
(b) To estimate $P(0.5)$, we use the fact that $P(0.5)-P(0) \approx P'(0) \cdot (0.5-0)$. Thus,
$$ P(0.5) \approx P(0) + P'(0) \cdot 0.5 = 1000+6.4\cdot 0.5 = 1003.2.$$
(c) The assumption is that the rate of change is approximately constant over the interval $[0,0.5]$. We can check whether this is true by computing $P'(t)$ for $t$ between 0 and 0.5. We have
$$ P'(t) = 0.04P(1-P/20000)^2$$
so
$$ \frac{dP}{dt} = 0.04P(1-P/20000)^2.$$
We can solve this differential equation using separation of variables:
$$ \int \frac{dP}{P(1-P/20000)^2} = \int 0.04 dt.$$
Making the substitution $u=1-P/20000$, we have $du/dP = -1/20000$ and
$$ -\int \frac{1}{u^2} du = -\frac{1}{u} +C = \int 0.04 dt = 0.04t + C',$$
where $C$ and $C'$ are constants of integration. Thus,
$$ \frac{1}{P-20000} = 0.04t + C''$$
where $C''$ is a combination of $C$ and $C'$. Solving for $P$, we have
$$ P = \frac{20000}{1-0.04t+C''}.$$
Using the initial condition $P(0) = 1000$, we can solve for $C''$:
$$ 1000 = \frac{20000}{1+C''} \quad \Rightarrow \quad C'' = 24.$$
Thus,
$$ P(t) = \frac{20000}{1-0.04t+24} = \frac{1000}{1-0.002t+1.2}.$$
We can use this formula to compute $P(0.5)$:
$$ P(0.5) \approx \frac{1000}{1-0.002\cdot 0.5 + 1.2} \approx 1001.2.$$
This is within 1% of the estimate from part (b), so the assumption that the rate of change is constant over the interval $[0,0.5]$ is reasonable.