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:
$$f(t)=\frac{a}{1+be^{-0.9t}}$$
We will need the derivative of this function, so let's compute it now:
$$f'(t)=\frac{0.9abe^{-0.9t}}{\left(1+be^{-0.9t} \right)^2}$$
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:
$$f(0)=\frac{a}{1+be^{-0.9\cdot0}}=\frac{a}{1+b}=20$$
$$f'(0)=\frac{0.9abe^{-0.9\cdot0}}{\left(1+be^{-0.9\cdot0} \right)^2}=\frac{0.9ab}{\left(1+b \right)^2}=16$$
The first equation implies:
$$a=20(b+1)$$
Substituting this into the second equation, we find:
$$\frac{0.9\left(20(b+1) \right)b}{\left(1+b \right)^2}=16$$
$$18b(b+1)=16(b+1)^2$$
We may divide through by $b+1\ne0$ to obtain:
$$18b=16(b+1)$$
$$18b=16b+16$$
$$2b=16$$
$$b=8\implies a=20(8+1)=180$$
Hence, we have found:
$$(a,b)=(180,8)$$
b) To analyze the population in the long run, we may consider:
$$\lim_{t\to\infty}f(t)=\lim_{t\to\infty}\frac{a}{1+be^{-0.9t}}=\frac{a}{1+b\cdot0}=a=180$$
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:
View attachment 1613
