Welcome to our community

Be a part of something great, join today!

αℓєєиα♥'s question at Yahoo! Answers regarding the logistic population growth model

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
αℓєєиα♥'s question at Yahoo! Answers regarding the logistic population growth model

Here is the question:

Calculus Exponential Growth Problem?

Really hating calculus right now...

The number of yeast cells in a laboratory culture increases rapidly initially but levels off eventually. The population is modeled by the function:

n = f(t) = a / (1+be^(-0.9t))

Where t is measured in hours. At time t = 0 the population is 20 cells and is increasing at a rate of 16 cells/hour.

a) Find the values of a and b.

b) According to this model, what happens to the yeast population in the long run?

If you could explain how to do this problem that'd be great. It's just that this equation of the growth is unfamiliar.
I have posted a link there to this topic so the OP can see my work.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: αℓєєиα♥'s question at Yahoo! Answers regarding the logistic population growth model

Hello αℓєєиα♥,

We are given the following function which models the population of yeast cells:

\(\displaystyle f(t)=\frac{a}{1+be^{-0.9t}}\)

We will need the derivative of this function, so let's compute it now:

\(\displaystyle f'(t)=\frac{0.9abe^{-0.9t}}{\left(1+be^{-0.9t} \right)^2}\)

Incidentally, this population model is the so-called Logistic function - Wikipedia, the free encyclopedia where resources are limited and competition for these resources limits the population growth.

a) To find the values of the parameters $a$ and $b$, we may take the information provided about the initial values to get two equations in two unknowns:

\(\displaystyle f(0)=\frac{a}{1+be^{-0.9\cdot0}}=\frac{a}{1+b}=20\)

\(\displaystyle f'(0)=\frac{0.9abe^{-0.9\cdot0}}{\left(1+be^{-0.9\cdot0} \right)^2}=\frac{0.9ab}{\left(1+b \right)^2}=16\)

The first equation implies:

\(\displaystyle a=20(b+1)\)

Substituting this into the second equation, we find:

\(\displaystyle \frac{0.9\left(20(b+1) \right)b}{\left(1+b \right)^2}=16\)

\(\displaystyle 18b(b+1)=16(b+1)^2\)

We may divide through by $b+1\ne0$ to obtain:

\(\displaystyle 18b=16(b+1)\)

\(\displaystyle 18b=16b+16\)

\(\displaystyle 2b=16\)

\(\displaystyle b=8\implies a=20(8+1)=180\)

Hence, we have found:

\(\displaystyle (a,b)=(180,8)\)

b) To analyze the population in the long run, we may consider:

\(\displaystyle \lim_{t\to\infty}f(t)=\lim_{t\to\infty}\frac{a}{1+be^{-0.9t}}=\frac{a}{1+b\cdot0}=a=180\)

Thus, we have found the limiting number of yeast cells is 180.

Here is a plot of the population function for the first 24 hours:

yeastpop.jpg