Population Change

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.

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.

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.

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