Suppose that a population grows according to a logistic model with carrying capacity 7,000 and $k = 0.05$ per year. Choose the logistic differential equation for these data.
A) $\frac { d P ( t ) } { d t } = 7,000 P \left( 1 + \frac { p } { 0.05 } \right)$
B) $\frac { d P ( t ) } { d t } = P \left( 1 + \frac { p } { 7,000 } \right)$
C) $\frac { d P ( t ) } { d t } = 0.05 P \left( 1 + \frac { p } { 7,000 } \right)$
D) $\frac { d P ( t ) } { d t } = 0.05 P \left( 1 + \frac { p } { 5 } \right)$
E) $\frac { d P ( t ) } { d t } = 0.05 P \left( 1 - \frac { p } { 7,000 } \right)$

Question

Quizplus · Accepted Answer

The logistic differential equation for population growth is given by $\frac { d P ( t ) } { d t } = k P \left( 1 - \frac { P } { M } \right)$, where P is the population, t is time, k is the growth rate coefficient and M is the carrying capacity.

Substituting k = 0.05 and M = 7,000, we get the logistic differential equation as:
$\frac { d P ( t ) } { d t } = 0.05 P \left( 1 - \frac { P } { 7,000 } \right)$

Therefore, option E is the correct choice.

Suppose That a Population Grows According to a Logistic Model k=0.05k = 0.05k=0.05

Suppose That a Population Grows According to a Logistic Model $k = 0.05$