Question

Limited Growth Equation Another differential equation that models limited growth of a population \(P\) in an environment with carrying capacity \(M\) is \(d P / d t=k(M-P)\) (where \(k>0\) and \(M>0\) ). (a) Show that \(P=M-A e^{-k t},\) where \(A\) is a constant determined by an appropriate initial condition. (b) What is lim \(P(t) ? ~ M\) (c) For what time \(t \geqslant 0\) is the population growing the fastest? (d) Writing to Learn How does the growth curve in this model differ from the growth curve in the logistic model? See answ

Short answer

(a) P = M - A e^{-kt}, where A is determined by the initial conditons. (b) lim P(t) = M as t approaches infinity. (c) The population is growing the fastest at time t = 0. (d) This model's growth increment is maximum at the start and approaches carrying capacity slower compared with the logistic model.

Step-by-step solution

Separate and Integrate

Separate the variables in the limited growth equation dP/dt = k(M-P), to get M-P = dP, and kdt = dp. Then integrate both sides to get -k t = P - M + C, where C is the integration constant.

Find the constant A

Rearrange the equation to solve for P, yielding P = M - A e^{-kt}, where A = C e^{kt}. This constant A will be determined by specific initial conditions given.

Evaluate Limit

As time approaches infinity, the exponent part of the equation e^{-kt} approaches 0, leaving us with P = M, meaning the population will approach the carrying capacity of the environment.

Fastest Growth

The growth rate is fastest when d^2P/dt^2 = 0. Solving the equation, we find that it happens at t = 0.

Comparison with Logistic Model

In comparison with the logistic model, this limited growth curve approaches the carrying capacity M more slowly, and the population growth increment is maximum at the start, t = 0, whereas in logistic model it's maximum when population is half of the carrying capacity