The logistic solution

Three quick questions from earlier lessons.

The logistic solution is $P(t) = \dfrac{K}{1 + Ae^{-rt}}$ . What does $P(t)$ approach as $t \to \infty$ ?

The explicit solution, fitted to the start.

\dfrac{dP}{dt} = rP\left(1 - \dfrac{P}{K}\right)

,

P(0) = P_0

has the solution

P(t) = \dfrac{K}{1 + Ae^{-rt}}

with

A = \dfrac{K - P_0}{P_0}

.

$K = 1000$ , $P_0 = 200$ , $r = 0.5$ .

A = 4

(1000 − 200)/200.

P(t) = \dfrac{1000}{1 + 4e^{-0.5t}}

Substitute.

P(0) = 200

Check t = 0: 1000/5.

Your turn: $K = 600$ , $P_0 = 100$ . Find $A$ .

A = 5

(600 − 100)/100.

The logistic solution $P(t) = \dfrac{ 203 }{1 + Ae^{-rt} }$ has $P(0) = 58$ . Find $A$ .
The logistic solution $P(t) = \dfrac{ 755 }{1 + Ae^{-rt} }$ has $P(0) = 48$ . Find $A$ .

$\dfrac{dP}{dt} = \dfrac{ 2 }{10}P\left(1 - \dfrac{P}{ 253 }\right)$ . How fast is the population growing when $P = \dfrac{ 253 }{2}$ ?
$\dfrac{dP}{dt} = \dfrac{ 7 }{10}P\left(1 - \dfrac{P}{ 556 }\right)$ . How fast is the population growing when $P = \dfrac{ 556 }{2}$ ?

Three questions to check you have got it.

The logistic solution $P(t) = \dfrac{ 203 }{1 + Ae^{-rt} }$ has $P(0) = 58$ . Find $A$ .
The logistic solution $P(t) = \dfrac{ 755 }{1 + Ae^{-rt} }$ has $P(0) = 48$ . Find $A$ .
The logistic solution $P(t) = \dfrac{ 770 }{1 + Ae^{-rt} }$ has $P(0) = 24$ . Find $A$ .

The logistic solution

$P(t) = \dfrac{K}{1 + Ae^{-rt}}$ , $A = \dfrac{K - P_0}{P_0}$ , $P \to K$ .

Watch out: check $A$ by computing $P(0)$ .

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