Chapter 7.5, Problem 33E

a.

To determine

To find a formula for the guppy population in terms of $t$ .

Answer to Problem 33E

The formula for the guppy population is

  $P=\frac{150}{1+24e^{-\left(0.225\right)t}}$

Explanation of Solution

Given:

A $2000$ - gallon tank can support no more than $150$ guppies . Six guppies are introduced into the tank . Assume that the rate of growth of the population is

  $\frac{dP}{dt}=0.0015P\left(150-P\right)$

Where time $t$ is in weeks.

Calculation:

The solution of the general logistic differential equation

  $\frac{dP}{dt}=kP\left(M-P\right)$

is given by

  $P=\frac{M}{1+A{e}^{-\left(Mk\right)t}}$

Since, given differential equation is $\frac{dP}{dt}=0.0015P\left(150-P\right)$

on comparing it with general differential equation

  $k=0.0015$

  $M=150$

It is given that Six guppies are introduced into the tank means at $t=0$ , $P=6$

  $\begin{array}{l}P=\frac{M}{1+A{e}^{-\left(Mk\right)t}}\\ 6=\frac{150}{1+A{e}^{-\left(150\times 0.0015\right)\times 0}}\\ 6=\frac{150}{1+A}\\ 1+A=\frac{150}{6}\\ A=24\\ \therefore P=\frac{150}{1+24e^{-\left(0.225\right)t}}\end{array}$

Hence the formula for the guppy population is

  $P=\frac{150}{1+24e^{-\left(0.225\right)t}}$

b.

To determine

To determine the time when guppy population is $100$ and $125$

Answer to Problem 33E

Guppy population is $100$ when $t=17.21\text{ weeks}$

Guppy population is $125$ when $t=21.28\text{ weeks}$ .

Explanation of Solution

Given:

A $2000$ - gallon tank can support no more than $150$ guppies . Six guppies are introduced into the tank . Assume that the rate of growth of the population is

  $\frac{dP}{dt}=0.0015P\left(150-P\right)$

Where time $t$ is in weeks.

Calculation:

From part (a)

The formula for the guppy population is

  $P=\frac{150}{1+24e^{-\left(0.225\right)t}}$

Now, $P=100$

  $\begin{array}{l}P=\frac{150}{1+24e^{-\left(0.225\right)t}}\\ 100=\frac{150}{1+24e^{-\left(0.225\right)t}}\\ 1+24e^{-\left(0.225\right)t}=1.5\\ 24e^{-\left(0.225\right)t}=0.5\\ e^{-\left(0.225\right)t}=0.0208\\ -0.225t=\ln \left(0.0208\right)\\ -0.225t=-3.87\\ t=17.21\text{ weeks}\end{array}$

Now, $P=125$

  $\begin{array}{l}P=\frac{150}{1+24e^{-\left(0.225\right)t}}\\ 125=\frac{150}{1+24e^{-\left(0.225\right)t}}\\ 1+24e^{-\left(0.225\right)t}=1.2\\ 24e^{-\left(0.225\right)t}=0.2\\ e^{-\left(0.225\right)t}=0.00834\\ -0.225t=\ln \left(0.00834\right)\\ -0.225t=-4.786\\ t=21.28\text{ weeks}\end{array}$

Therefore , Guppy population is $100$ when $t=17.21\text{ weeks}$

Guppy population is $125$ when $t=21.28\text{ weeks}$ .