Our world is constantly changing. Thus, rates of change have applications in a variety of fields, one of which is ecology.
The rate of change of a population depends on the rate at which births and deaths occur, the resources available, and potential competition for these resources. You can use these parameters to formulate an equation that models population. Then, of course, the derivative or rate of change of that equation is the rate at which the population changes.
Population growth rates help scientists in making predictions on future population sizes and the resources necessary to maintain it. You can apply population growth rates to the spread of viral infections. Calculating the rate at which a virus infects humans or animals can give an indication of how widespread an infection is (or will be), the resources necessary to help those who are infected, and how to best combat visual infections.
Definition of Population Change
Populations refer to more than just people. Every living thing is part of a population. And populations of every kind are rarely a static value. Populations change constantly due to a variety of factors (environment, food sources, migration, predation, disease, etc.).
Population change is the natural change in the number of a particular organism in an environment.
Oftentimes, the change in the population will follow a trend, usually either increasing or decreasing over time.
Measures of Population Change
If you know a how a population is trending, you can find out how much that population has changed over a certain period of time.
The measure of population change is the difference in a population between the beginning of a certain interval of time and the end of that interval.
Because populations change over time, the measure of a population can often be represented as a function of time called \(P(t)\). To find the measure of the population change of that function \(P(t)\) between times \(t_{1}\) and \(t_{2}\), you simply find the difference \(P(t_{2})-P(t_{1})\).
The population of a colony of bacteria can be measured in thousands by the equation \(P(t)=t^{2}+4t-1\) where \(t\) is measure in hours. Find the measure of population change between hour 3 and hour 5.
Answer:
To find the measure of the population change over the time period, you plug in \(5\) for \(t_{2}\) and \(3\) for \(t_{1}\).
\[\begin{align}\text{population change } &=P(t_{2})-P(t_{1})\\&=P(5)-P(3)\\&=(5^{2}+4(5)-1)-(3^{2}+4(3)-1)\\&=44-20\\&=24.\\\end{align}\]
The bacteria population of the colony increased by \(24{,}000\) in that particular two hour span.
What if you want to know how fast a population is changing instead of how much it has changed?
Rate of Population Change
Sometimes populations change slowly if the environment and other factors are relatively stable. Other times the population increases or decreases quickly over a short period of time. It can be very informative to find out how fast those changes are happening.
If \(P(t)\) is the number of organisms in a population at time \(t\), the rate of population change at time \(t\) is \(P'(t)\).
The rate of population change tells you how fast the population is increasing or decreasing over a given time period or at a particular instant. And, as rate of change usually does in calculus, this means you can use the derivative of the population function \(P(t)\). Calculus provides us with two formulas for computing population change: average population change and instantaneous population change.
Population Change Formulas
Average Rate of Population Change
The average rate of population change measures how much the population changes over a specific time period. It is an application of the Amount of Change Formula.
Let \(P(t)\) be a population equation that represents the number of people, animals, or organisms in a population at time \(t\). The average rate of population change of \(P(t)\) between times \(t_{1}\) and \(t_{2}\) is
\[\begin{align} \text{average rate of change }&=\frac{\Delta P(t)}{\Delta t}\\ &=\frac{P(t_{2})-P(t_{1})}{t_{2}-t_{1}} . \end{align}\]
How is the average rate of population change formula derived?
The change in the population size between \(t_{1}\) and \(t_{2}\) can be explained as difference between the population size at time \(t_{2}\) and the population size at \(t_{1}\). In mathematical terms, you can write this as
\[\Delta P(t)=P(t_{2})-P(t_{1}).\]
If you want to find the amount the population changed between \(t_{1}\)and \(t_{2}\), then you simply divide the change in the population size \(\Delta P(t)\) by the amount of time that passes, \(\Delta t=t_{2}-t_{1}\).
Let's look at an example.
The population of a city of \(100{,}000\) people triples every \(10\) years. What will be the population of the city in \(4\) years?
Answer:
If \(P(t)\) represents the population of the city at time \(t\), then \(P(0)=100{,}000\). You can expect that \(P(10)=300{,}000\).
You can find the average population growth rate per year over the ten year periods using the formula:
\[\begin{align} \text{average rate of change } & = \frac{P(t_{2})-P(t_{1})}{t_{2}-t_{1}} \\&=\frac{300{,}000-100{,}000}{10-0}\\&=\frac{200{,}000}{10}\\&=20{,}000.\end{align}\]
So, on average, the population of the city increases by 20,000 people per year.
To estimate the population of the city after four years
\[\begin{align}P(4)&=P(0)+4(\text{average rate of change} )\\&=100{,}000+4(20{,}000)\\&=180{,}000.\end{align}\]
After four years, the total population of the city is about \(180{,}000\) people.
Instantaneous Rate of Population Change
To find the exact rate of change of a population at a certain moment of time, you find the instantaneous rate of population change. Notice that the instantaneous rate of population change is synonymous with the derivative of the population.
For a reminder on things like rates of change and how they relate to derivatives, see the article Rates of Change
Let \(P(t)\) be a population equation that represents the number of people, animals, or organisms in a population at time \(t\). The instantaneous rate of population change at time \(t\) is
\[\begin{align}\text{instantaneous rate of change } &=\lim_{\Delta t\to 0} \frac{\Delta n}{\Delta t} \\ &=P'(t). \end{align}\]
Notice that this means the derivative of the population equation is the instantaneous rate of change!
The only difference between the average rate of population change formula and the instantaneous rate of population change formula is the limit in front. Because you want to find the rate of population change formula at a single moment in time, \(\Delta t\) should be really close to \(0\). So, you find the limit of the average rate of population change formula as the change in time approaches \(0\). Let's look at an example of this.
The population of a colony of bacteria can be measured in thousands by the equation \(P(t)=t^{2}+4t-1\) where \(t\) is measure in hours. Find the rate of change of the population growth at \(2\) hours. Then, find the average rate of change of the population growth over \(2\) hours.
Answer:
To find the instantaneous rate of change of the population growth, you plug in \(P(t)\) into the instantaneous rate of change formula with \(t_{1}=2\) hours and \(t_{2}=t\) to get
\[\begin{align} P'(2) &=\lim_{t \to 2}\frac{P(t)-P(2)}{t-2}\\&=\lim_{t\to2}\frac{(t^{2}+4t-1)-(2^{2}+4(2)-1)}{t-2}\\&=\lim_{t\to2}\frac{t^{2}+4t-12}{t-2}\\&=\lim_{t\to2}\frac{(t-2)(t+6)}{t-2}\\&=\lim_{t\to2} (t+6)\\&=8.\\\end{align}\]
At \(2\) hours, the population of bacteria is growing at a rate of \(8{,}000\) bacteria per hour.
Let's find the average rate of population growth over two hours to see how the rates compare.
\[\begin{align}\text{average rate of change } &=\frac{P(2)-P(0)}{2-0}\\&=\frac{(2^{2}+4(2)-1)-(0^{2}+4(0)-1)}{2}\\&=\frac{11-(-1)}{2}\\&=6.\\\end{align}\]
On average, over the first two hour period, the population of the bacteria colony grows at a rate of about \(6{,}000\) bacteria per hour.
Population Change Example
Now let's find the average rate of population change and the instantaneous rate of population change for a population's function.
The frog population of a certain pond was observed during the spring months and was modeled using the function \(P(t)=1.25^t+2\) where \(t\) is time measured in weeks and \(P\) is the population in hundreds of frogs.
a) Find the average rate of population change between week \(3\) and week \(7\).
b) Find the instantaneous rate of population change at week \(5\).
First, it may help to know what the graph of the function looks like. The graph below shows the function \(P(t)\) along with a table of values for several weeks:
Fig. 1 - Graph of our population exponential growth function with a table of values
Solution:
For both parts of the question, it may also help to remember that another term for rate of change is slope. You should be able to draw a line on the graph to represent the rate of population change.
Part a) To find the average rate of population change between two times, \(t_1\) and \(t_2\), you can draw a line between those points, here week \(3\) and week \(7\), on the graph as shown below. This is called a secant line. The slope of the secant line between two points is the average rate of change between those points.
Fig. 2 - Our exponential function with a secant line.
To find the slope of this secant line, you need the formula for the average rate of population change (which is very similar to the formula for the slope of a line between two points):
\[\begin{align} \text{average rate of change }&=\frac{\Delta P(t)}{\Delta t}\\ &=\frac{P(t_{2})-P(t_{1})}{t_{2}-t_{1}} . \end{align}\]
You can use the table of values to substitute:
\[\begin{align} \text{average rate of change}&=\frac{P(7)-P(3)}{7-3}\\ &=\frac{6.77-3.95}{7-3}\\ &=\frac{2.82}{4}\\ &\approx 0.71 . \end{align}\]
Recall that the population is being measured in hundreds, so an average rate of change of \(0.71\) means that the population is increasing by an average of \(71\) frogs per week from week \(3\) to week \(7\).
Part b) To find the instantaneous rate of population change at \(5\) weeks, you use the formula above which says that \[\text{instantaneous rate of change}=P'(t).\]
Let's first remember that the derivative of a function at a single point tells you the slope of the tangent line at that point. On the graph, that would look like this:
Fig. 3 - Our exponential function with a tangent line
To find the slope of that tangent line, first you need to find the derivative of your function \(P(t)\). Since \(P(t)\) is an exponential function, you will need the rule for the derivative of the exponential function. And recall that the derivative of a constant is \(0\):
\[\begin{align} P'(t)&=\frac{d}{dt}(1.25^{t}+2)\\ &=1.25^{t}ln(1.25).\\ \end{align}\]
Now you can substitute in \(5\) for \(t\) to find the slope at \(5\) weeks:
\[\begin{align} P'(t)&=1.25^{t}ln(1.25)\\ P'(5)&=1.25^{5}ln(1.25)\\ &\approx0.68.\\ \end{align}\]
So the frog population is growing at a rate of about \(68\) frogs per week at \(5\) weeks.
Note that week \(5\) is halfway between week \(3\) and week \(7\). If you look at both the secant line and the tangent line on the graph at the same time, shown in the image below, you will see how similar the lines are. This shows that the average rate of population change is a pretty good approximation of the instantaneous rate of population change.
Fig. 4 - Our exponential function with both secant line and tangent line
Population Change - Key takeaways
- Rate of change, or derivatives, has useful applications in the field of ecology - more specifically measuring population changes
- The measure of population change between times \(t_{1}\) and \(t_{2}\) is given by the equation \[\text{population change} =P(t_{2})-P(t_{1}).\]
- Average rate of population change between time \(t_{2}\) and \(t_{1}\) is measured with the formula\[\frac{P(t_{2})-P(t_{1})}{t_{2}-t_1}.\]
- Instantaneous rate of population change at time \(t\) is measured with the formula \[\lim_\limits{\Delta t \to 0}\frac{\Delta P(t)}{\Delta t}=P'(t).\]
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