How do I determine this logistic growth model formula?

The Question

Two yeast cells were placed into a special container to which food was continually added, to keep it at a constant concentration. All other factors were set for optimal yeast growth (for example, temperature, oxygen, and pH). The population was sampled every hour for 21 hours and the results of the estimated population size were recorded in the table below.

Time (hour), Number of yeast cells

(0, 2)

(1, 10)

(2, 15)

(3, 20)

(4, 40)

(5, 60)

(6, 100)

(7, 190)

(8, 260)

(9, 350)

(10, 450)

(11, 530)

(12, 580)

(13, 600)

(14, 600)

(15, 600)

(16, 600)

(17, 600)

(18, 600)

(19, 600)

(20, 600)

What population growth model does the population appear to follow? Write out the mathematical equation it follows.

The Attempt

Looking at the values, I recognize the carrying capacity is reached at hour $$13$$ with $$600$$ cells, signifying a logistic growth model. The formula given to me for logistic growth is:

$$\frac{dN}{dt} = rN[\frac{(K - N)}{K}]$$

From what I have researched:

$$dN$$ = population size change

$$dt$$ = time interval

$$r$$ = (max) growth rate

$$N$$ = starting population size

$$K$$ = carrying capacity

So:

$$dN = 600 - 2 = 598$$?

$$dt = 21$$ hours

$$r =$$?

$$N = 2$$

$$K = 600$$

Plugging the values in gives me:

$$598/21$$ = $$r2[(600 - 2)/600]$$

Update 25/01/19

With some help, I have discovered how to determine the r/rmax value and have the following formula:

$$r = (90+100)/2$$

$$r = 95/hour$$

$$dN/dt = 95N[(K - N)/K]$$

With $$N= 2$$

$$dN/dt=95(2)[(K-2)/K]$$

and if $$K$$ value is carrying capacity, $$K = 600$$

$$dN/dt=95(2)[(600-2)/600]$$

$$dN/dt=190(598/600)$$

$$dN/dt=190(0.997)$$

$$dN/dt=189?$$

However, the question asks me to "Write out the mathematical equation it follows." So am I supposed to leave a value unknown?

• The coefficients of the solution to this differential equation are usually derived by regression. – Roland Jan 25 at 12:50
• @Roland I was able to make some progress with help. Am I on the right track? – Maria Jan 25 at 16:45
• Can you tell how you got the $r$ value, or rather the $r/rmax$ you refer to? – Satwik Pasani Jan 26 at 8:12
• @SatwikPasani From what I understood, the r value refers to the maximum growth rate value, which happens around the 9th hour. So I was directed to average out the growth rate between hours 8 and 9 (90) and 9 and 10 (100) so (90+100)/2 = 95 – Maria Jan 27 at 0:11

I think your problem arises from misunderstanding the $$r$$ coefficient. I've tried to explain it below, let me know if you need more clarification.

Instantaneous vs Average Growth

Differential equations (like the logistic equation you write) talk about the instantaneous drowth, $$dt$$ is infinitesimally small, as opposed to your working where you are subtracting the population at two distinct time points and dividing to find $$r$$, which will only be an average $$r$$ for that time period. If we were to plot population vs time, we can understand $$\frac{dN}{dt}$$ as the slope of the equation, and it is apparent that it is changing with time. So, you cannot use distinct time points to find out R, since that will only give you an average $$r$$. Logistic Curve Fitting

You can solve the differential equation to get $$N$$ as a function of time. The ideal way to do this is to try and fit a logistic curve to the data. This way, we already know $$K$$, $$N_0$$, and out job is to fit a curve to our data by trying different $$r$$ values to see which $$r$$ fits our data the best.

$$N(t) = \frac{KN_0e^{rt}}{K + N_0(e^{rt}-1)}$$

There is a formal way to do this cure fitting. Have a look here, here and here. There will definitely be better resources than these, but I could find these on a cursory google search.