In the previous part (Population Growth), I gave you some of the basic background information on exponential growth in an unlimited, constant, and favorable environment. A logarithmic plot of the values indicates a straight line. We call this graph a semi-log plot because the ordinate (vertical axis) is logarithmic and the abscissa (horizontal axis) is arithmetic. The slope of this straight line is an instantaneous rate of change when natural logarithms (base e logs) are used.
What I want to show you now is how to relate some familiar finite rates of change to their instantaneous rate equivalents, and show the usefulness of instantaneous rates in population studies.
If a population doubles each year, then;
R = 2 = finite rate of growth yearly
R = er = 2
r = ln R = 0.693 = instantaneous rate of growth yearly
The instantaneous rate of growth represents the fraction by which a population would grow during a very short period of time. For example, a population of 1000 individuals having an instantaneous growth rate of 0.693 yearly would increase daily as follows:
Number added during first day = 1,000 x (0.693 / 365) = 1.899
Population after first days growth = 1,000 + 1.899 = 1,001.899
or from the basic formula, Nt = N0ert
N1/365 = 1,000 (e0.693 x 1/365)
= 1,000 (e0.001899)
= 1,000 x 1.0019
= 1,001.9
To obtain the population after the second day (by the long-hand method), it would be:
1001.899 + 1001.899 x (0.693 / 365)
= 1001.899 + 1.9 = 1003.8
rather tedious!
I previously said the population would double in a year (R = 2). We should then have 2000 individuals in the population at the end of the year. If we apply the formula, we get:
N1 = 1000 (e0.693 x 365/365)
= 1000 x 1.9937
= 1993.7
This estimate of 1994 is close to the expected 2000 that we would get by applying a finite rate of growth of 2 to the initial population of 1000. The difference here is due to two factors:
mathematical errors due to rounding off decimal places, and
the arbitrary division of time into 365 units (days).
Example Problems:
Problem #1
Given: Population is growing exponentially.
On Jan. 1, there were 1000 animals in the population.
Instantaneous growth yearly is 0.693.
Find: Population size on Feb. 5.
N36/365 = 1000 (e0.693 x 36/365)
= 1000 (e0.07)
= 1000 × 1.0725
= 1072.5 animals
Problem #2
Given: Pheasants start with 1 male + 1 female; and the following graph:
Find: total population at 3rd year (beginning of 4th year)
R = 3
R = er = 3
r = ln 3 = 1.0986
N3 = 2 (e1.0986 x 3)
= 2 (e3.2958)
= 2 (27.11)
= 54.22 pheasants
In the case where a population is declining, let us say 30% each year we have:
R = 0.30 = finite rate of growth yearly
r = -1.20397 = instantaneous rate of growth yearly
Applying this instantaneous rate of growth yearly on a daily basis as before, we have:
N1 = 1000 (e1.20397 × 1)
= 1000 (0.301)
= 301
This compares with an expected population of 300 individuals after one year, when a finite rate of growth of 0.30 yearly is applied to an initial population of 1000.
If we have any group of N0 individuals whose numbers decline because of mortality to Nt individuals at time t, then we might say:
S = Nt / N0 = interval survival rate (for time t)
This is analogous to the situation we considered above where we had a declining population (R < 1.00). Thus, we might have:
Nt / N0 = 0.30 = interval survival rate yearly
Nt / N0 = e-i = 0.3
i = 1.204 = instantaneous mortality rate yearly
You can get i (x) in two ways:
(1) Easiest way is to take ln (3)
(2) Go down e-x column until you come to 0.3, then get x (i)
The instantaneous mortality rate is thus simply the natural logarithm (with sign ignored) of the interval survival rate. |
In field studies, we are often able to estimate survival for some prescribed time interval. We may wish, however, to know just what this survival amounts to for some longer or shorter period of time. Often, for example, we need to compare survival rates, but the time intervals used in our field work may be different. Instantaneous mortality rates provide a quick means of extrapolating, partitioning, or standardizing interval survival rates.
Given: An interval survival rate of 0.60 for 92 days (Oct - Dec)
Find: Interval survival rate for 37 days
S92 = 0.60 = e-i = interval survival rate for 92 days
i = 0.51 = instantaneous mortality rate for 92 days
0.51 (37/92) = 0.205 = instantaneous mortality rate for 37 days
S37 = e-0.205 = 0.81 = instantaneous survival rate for 37 days
It should also be noted that instantaneous mortality rates are additive, whereas interval mortality rates (1 - interval survival rate) are not. |
For example:
Given: A 92-day interval mortality rate of 0.40 followed by a 37-day interval mortality rate of 0.19.
Find: The total mortality rate over the entire 129-day period.
S92 = (1 - 0.40) = 0.60
S37 = (1 - 0.19) = 0.81
S129 = 0.60 × 0.81 = 0.49
i92 = ln (0.60) = 0.51
i37 = ln (0.80) = 0.21
i129 = ln (0.51) + 0.21 = 0.72
e-0.72 = 0.49 = S129
The total mortality rate for the 129-day period would thus be 0.51 (= 1 - 0.49).
Updated 06 August 1996