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

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

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

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

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

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

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

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

Find the time of inflection $t_{\text{inf}}$ for the logistic curve $P(t)=\frac{10}{1-\tfrac{1}{6}e^{-3t}}\,.$

Find the time at which the population reaches half the carrying capacity under $P(t)=\frac{10}{1-\tfrac{1}{6}e^{-3t}}\,.$

Determine the maximum growth rate $\bigl(\max\frac{dP}{dt}\bigr)$ and the time at which it occurs for the logistic model with $K=10$ and $r=3$.

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

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.