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P(t) = P<<sub>0</sub> * K / P<sub>0</sub> + (K-P<<sub>0</sub>) * E^-RT


P(t) is the population size at time t (measured in days)
P0 is the initial population size
K is the carrying capacity of the environment
r is a constant representing population growth or decay

G.

Given all of your observations in the previous problems, and using the logistic model,
P(t)=Po*K/Po+(K-Po)*e^-rt
(1)

without any specific parameter values, explain what happens to the population size when r > 0 and t → ∞.

Do the same thing when r =  0 and r <  0 and t → ∞.

Tutorial

We are interested in knowing what happens to the population size P as t → ∞.

This of course depends on whether r is positive, negative, or zero.

Let's begin with r > 0. The logistic model (1), has only one term that depends on t,

(K-Po)*e^-rt

(2)
Notice that when r > 0, (2) represents exponential decay. Thus, as t → ∞, the term in (2) decays to zero. Substituting zero for the term in (2) into (1) gives,
 
P(t) arrow to the right Po*K/Po = K

 

Thus, we conclude when r > 0 (i.e. birth rate > death rate t → ∞ implies P → K (i.e. the population approaches the carrying capacity.

Now let's consider the special case r = 0 (i.e. birth rate = death rate). If r = 0, the term in (2) reduces to K - P0, and there is no t dependence. Substituting this value into (1) gives,

 
P(t) = Po*K/Po+KL-Po = Po*K/K = Po

 

That is, if r = 0, the population size (at any time t) remains at the initial size, P0. This makes intuitive sense because r= 0 indicates that births are balancing deaths exactly.

Finally, let's consider the case r < 0. I r < 0, the term in (2) represents exponential growth. That is, t → ∞ implies (K - P0) · e-rt . → ∞. Since (2) is in the denominator of (1), we find that as t → ∞ ,
 
P(t) arrow right 0.

 

That is, the denominator in (1) becomes arbitrarily large as t → ∞ , resulting in P(t) → 0.

This makes sense when the death rate is larger than the birth rate.

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