Question Type 6: Formulating a logistic model based on descriptions Exercises

Suppose the intrinsic growth rate doubles to $r=6$ while $K=10$ and $P(0)=12$. Write the new logistic model $P(t)$.

Show that as $t \to \infty$, the logistic solution $P(t) = \frac{10}{1 - \frac{1}{6}e^{-3t}}$ approaches the carrying capacity.

Find the horizontal asymptote of the logistic function $P(t)=\frac{10}{1-\tfrac{1}{6}e^{-3t}}\,.$

Express the solution in the form $P(t)=\frac{10}{1+Ae^{-3t}}$ and determine the constant $A$ given that $P(0)=12$.

A population, $P(t)$, is modelled by the logistic function $P(t)=\frac{10}{1+\frac{1}{6}e^{-3t}}$ where $t \ge 0$ is the time in years.

Determine the time of inflection, $t_{\text{inf}}$, for this population.

Write down the logistic differential equation for a population $P(t)$ with carrying capacity 10 and intrinsic growth rate 3, and solve it to obtain the general explicit solution.

A population model is given by the function $P(t) = \frac{10}{1 - \frac{1}{6} e^{-3t}}$ where $P(t)$ is the population at time $t$ in years, for $t \ge 0$.

Find the time at which the population reaches half the carrying capacity.

Using the model $P(t)=\frac{10}{1-\frac{1}{6}e^{-3t}}$, calculate $P(2)$ correct to three decimal places.

Determine the maximum theoretical growth rate $\left( \max \frac{dP}{dt} \right)$ and the time at which it occurs.

Solve the logistic differential equation $\frac{dP}{dt}=3P\Bigl(1-\frac P{10}\Bigr)$ by separation of variables and verify that $P(t)=\frac{10}{1+Ae^{-3t}}$ is the general solution.

Given the logistic model $P(t)=\frac{10}{1+3e^{-3t}}$, determine the carrying capacity, the intrinsic growth rate, and the initial population.

Determine the time $t$ when the population reaches $P(t)=9$ under the logistic model $P(t)=\frac{10}{1+\frac{1}{6}e^{-3t}}\,.$