Goals: introduce the logistic model of population growth; relate it to intraspecific competition and density dependence. Define the carrying capacity. Show that maximum population growth rate is at half the carrying capacity. Plot population growth rate and per capita growth rate versus population size. Compare plots for logistic growth to those for exponential growth.
The Lecture:
For exponential growth, we assumed that the per capit growth rate, r, was constant. This is not realistic for many situations. As populations grow, we expect intraspecific (within a species) competition for resources to increase the deathrate and decrease the birthrate. r should thus decrease as the population size increases. Logistic growth is a model of population growth that is density dependent: the per capita growth rate (r, dN/Ndt) declines as population size (density) increases. This will reflect the effects of intraspecific competition.
To develop the model of logistic growth, we assume that the per capita growth rate r (or dN/Ndt) decreases linearly (in a straight line) as population size increases. This is shown in the following graph:
The highest possible value of per capita growth is called rm, where the "m" stands for "maximum." This is determined by the reproductive abilities of whatever species we're modelling -- grasshoppers would have a much higher rm, for example, than do elephants, because grasshoppers can reproduce so much faster than can elephants.
When r is positive, birthrates must be greater than deathrates (since r=b-d), and the population is increasing in size. When r is negative, deathrates must be greater than birthrates, and the population is decreasing in size. The population is staying the same size (growth is zero) when r=0. On this graph, note that r is positive for low populations so small populations increase until the point where r is zero. r is negative for large populations so large populations decline in size until the point where r is zero. The point where r is zero is an equilibrium point; populations do not increase or decrease at this point. Further, it is a stable equilibrium: populations that are different from the equilibrium move to the equilibrium. This equilibrium point where the population size does not increase or decrease is called the carrying capacity: it represents the maximum number of individuals that the resources in the environment can maintain over time. The symbol given for the carrying capacity is K.
Development of the Logistic Growth Equation: From this graph, we can develop the equation for logistic growth. The graph shows a straight line, and the formula for a straight line, as you know, is Y=mX+b where b is the Y intercept and m is the slope. In this case, the Y variable is dN/Ndt (the per capita growth rate.) The X variable is N (population size.) The Y intercept is rm. The slope is - rm/K since the line has a negative rise (a decrease, a negative slope) of rm over the run of K. So if we plug these values into the formula for a straight line, we get:
dN/Ndt = (- rm/K )N + rm
Simplifying this, we get:
dN/Ndt = rm- rmN/K
dN/Ndt = rm(1-(N/K))
then multiply both sides by N to get an equation for population growth:
dN/dt = rmN(1-(N/K)) which is one of the standard ways the logistic growth equation is written. The other is:
dN/dt=rmN(K-N)/K
Note that the first part of the equation, dN/dt=rmN, is the same as exponential growth. The part that is mulitplied by this, (1-(N/K)), must be what makes the model density dependent. To see how it has this effect, consider first a very small population, then a population at the carrying capacity.
In a very small population, N is much much less than K. If N is much much less than K, then N/K is a very very small number, and 1-(N/K) is almost equal to 1. When 1-(N/K) is almost equal to 1, rmN(1-(N/K)) is almost equal to just rmN. Since dN/dt=rmN is the formula for exponential growth, this tells us that for very small populations, population growth is virtually exponential.
Now consider a population at the carrying capacity. When a population is at the carrying capacity, that means that N=K. When N=K, N/K=K/K=1, and 1-(N/K) = 1-1=0. Since this is multiplied by rmN to give population growth, this means that at carrying capacity, population growth is zero.
So you should be able to see how the (1-(N/K)) factor in the logistic growth equation is what is causing growth to depend on the population size -- that is, it is what makes the equation show density dependent growth.
Now let's see what this looks like. It should start out like exponential growth, but then the growth rate should decline to zero at the carrying capacity. Here's the graph of population size (N) versus time (t) for logistic growth:
Determining the population size at which population growth is maximum: Ecologists have been interested in identify where populations will grow at the fastest rate for various reasons. One reason is that it would be interesting to know whether species that depend on a population, such as predators, keep that population at a size where it will grow fastest -- that size will result in the highest production of new individuals, so it might also maintain the highest predator population. There are theoretical reasons to think that this will NOT be the case, since predators should evolve through natural selection to take the number of prey that benefits each individual predator the most, not what will maintain the best predator population (these aren't the same thing.) Another reason we're interested in knowing the population size where growth rate is maximum is that we may be interested in managing a population to produce most new individuals, either it we're looking at a game species for which we want to maximize individuals produced for hunting or fishing, or of we're looking at an endangered species and want to manage conditions so that most individuals of the species get produced.
It turns out that the maximum rate of population growth is at a population size of half the carrying capacity (K/2) We will consider two explanations for the point where population growth is highest -- one verbal, one mathematical (to show you yet again that ecologists should know some calculus.) YOU WILL BE EXPECTED TO KNOW one of these arguments on the next exam -- you can pick whether you want to do the verbal argument or the mathematical argument.
Verbal argument to show that maximum growth rate is at a population size of half the carrying capacity:
When populations are very small, the per capita population growth rate is close to the maximum -- this means that per individual, population growth is very fast. There are not, however, many individuals (the population is small.) A small number of individuals only has the potential to produce as much as those few individuals can reproduce. So while per capita growth is high, the growth of the population is slow. In contrast, when populations are approaching the carrying capacity, there are many individuals who could reproduce. Per capita growth is, however, low because of intraspecific competition for resources. Since the individuals present are not reproducing many offspring because of competition, population growth is slow. In a population halfway to the carrying capacity, there are enough individuals so that a significant amount of reproduction is possible, and there isn't too much negative impact of intraspecific competition. As a result, at this population size, population growth is highest.
Calculus argument to show that maximum growth rate is at a population size of half the carrying capacity:
To find where a function has a maximum, take the derivative of that function and set it equal to zero. If we do this for the growth rate function, it means we take the derivative with respect to N, the population size, of:
dN/dt = rmN(1-(N/K))
Step 1: Multiply this out; it will be easier to take the derivative if we do.
dN/dt=rmN - rmN2/K
Step 2: Take the derivative with respect to N:
Derivative with respect to N = rm- 2rmN/K
Step 3: Set equal to zero:
0 = rm- 2rmN/K
Step 4: Factor:
0 = rm(1-2N/K) Now we have two terms multiplied by each other. For the equation to be zero, one must be zero, so either rm is zero or 1-2N/K is zero. If rm were zero, it would mean that the maximum possible growth rate for a population is zero. Such a population would be dead before it could get started so this situation is not ecologically relevant. We'll just consider the case for which 1-2N/K is zero.
Step 5: Set 1-2N/K equal to zero and solve for N;
1-2N/K = 0
1 = 2N/K
K = 2N
N= K/2
So the maximum growth rate is at half the carrying capacity.
[Notes for the mathematically inclined: really you'd need to check to be sure this is a maximum and not a minimum, but if you look at the graph of population size versus time you can see the slope is steepest at about half the carrying capacity so we can use common sense and say it must be the steepest slope, not the shallowest. Also, there is a way to find the maximum of this function using algebra, not calculus; if you would like you can figure that out and use that on the exam.]
Now that we've seen how the population growth rate depends on population size, let's make a graph of it -- it should start low, increase to a maximum at K/2, and then decrease to 0 at the carrying capacity. So here it is:
At this point, you should compare plots of N versus t, dN/dt versus N, and dN/Ndt versus N for exponential and logistic growth. Why do we see the differences that we see?
Logistic growth turns out to be a good general model of population growth. It fits, at least roughly, the pattern of population growth seen in many different species. Some species, however, show different patterns. In the next lecture we will consider how we can make the logistic growth model a little more accurate for specific species, realizing that as we do, we make it less general.