Chapter 7.5, Problem 10E

a.

To determine

To find : an expression for the size of the population after $t$ years

Answer to Problem 10E

The expression for the size of the population after $t$ years $P\left(t\right)=\frac{10000}{1+24{\left(\frac{11}{36}\right)}^t}$

Explanation of Solution

Given information : The question is

Biologist stocked a lake with $400$ fish and estimated the carrying capacity to be $10,000$ . The number of fish tripled in the first year

Calculation : The logistic equation

  $\begin{array}{l}\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\\ P\left(t\right)=\frac{M}{1+A{e}^{-kt}}\\ \text{where }A=\frac{M-P_0}{P_0}\\ M=10000\\ P_0=400\\ A=\frac{10000-400}{400}=\frac{100-4}{4}=\frac{96}{4}=24\\ A=24\\ P\left(t\right)=\frac{10000}{1+24e^{-kt}}\\ P\left(1\right)=3\times P_0=3\times 400=1200\\ P\left(1\right)=1200\\ 1200=\frac{10000}{1+24e^{-k\left(1\right)}}\\ 1+24e^{-k}=\frac{100}{12}\\ 24e^{-k}=\frac{25}{3}-1\\ 24e^{-k}=\frac{22}{3}\\ e^{-k}=\frac{22}{3\times 24}=\frac{11}{36}\\ P\left(t\right)=\frac{10000}{1+24e^{-kt}}=\frac{10000}{1+24{\left(e^{-k}\right)}^t}\\ P\left(t\right)=\frac{10000}{1+24{\left(e^{-k}\right)}^t}=\frac{10000}{1+24{\left(\frac{11}{36}\right)}^t}\\ P\left(t\right)=\frac{10000}{1+24{\left(\frac{11}{36}\right)}^t}\end{array}$

b.

To determine

How long will it take the population will increase to $5000$

Answer to Problem 10E

It takes $2.68years$ the population will increase to $5000$

Explanation of Solution

Given information : The question is

Biologist stocked a lake with $400$ fish and estimated the carrying capacity to be $10,000$ . The number of fish tripled in the first year

Calculation : the logistic equation is

  $\begin{array}{l}P\left(t\right)=\frac{10000}{1+24e^{-1.185623666t}}\\ 5000=\frac{10000}{1+24e^{-1.185623666t}}\\ e^{-1.185623666t}=\left(\frac{10000}{5000}-1\right)\times \frac{1}{24}\\ \text{multiply both sides by }\frac{1+24e^{-1.185623666t}}{5000}\\ t=-\ln \left(\frac{1}{24}\right)\times \frac{1}{1.185623666}\\ t\approx 2.68years\end{array}$