How Do You Find The Rate At Which A Population Grows?

Problems that ask you to solve for the rate r in exponential growth require the use of roots or creative use of exponents. Let’s look at an example.

Problem Suppose a town has a population of 5000 in 2010. If there are 7000 people in the town in 2012, what is the annual rate at which the population is growing?

Solution The easiest way to approach this problem is to think in terms of the exponential growth (compound interest formula),

This formula applies when population grows at an annual rate.

Let’s look at the quantities in the problem statement:

  • 5000 people originally > P = 5000
  • There is 7000 in the population after 2 years > A = 7000 and n = 2

Putting these values into the formula above gives us

We need to find the annual rate r. Since the r is hidden in the parentheses, we start by isolating the parentheses.

To get at the r, we need to remove the square on the parentheses.

Using a calculator to do the square root, we get r ≈ 0.183 or 18.3%.

Although most calculators have a square root key, when removing powers it is often useful to raise both sides to a power. For instance, we could remove the square by raising both sides to the ½ power:

When you raise a power to another power, you multiply the exponents 2 ∙ ½ = 1. The right side simply becomes 1 + r. Now we can solve for r:

Using the power key on your calculator gives the same answer as before. Make sure the 1/2 power is entirely in the power. You can make sure this happens using parentheses: (7000/5000)^(1/2)-1.

Now what if the population grows over six years instead of two years? Instead of a square on the parentheses we now have a sixth power.

To solve for r in this equation, we follow similar steps.

The root can be computed on a graphing calculator using the MATH button or put into WolframAlpha:

Either method gives r ≈ 0.577 or 5.77%. Notice that the annual rate is lower when it is earned over a longer period of time.

If we use a 1/6 power to solve for r, we would carry out the steps below:

Using a 1/6 power on your calculator gives the same answer as above.

How Long Would It Take to Double the Population?

In a previous FAQ, I looked at an equation for modeling the growth of the population on Gilligan’s Island,

A = 7 (1+0.02)n

This equation assumed that the initial was 7 in 1964 and grew by 2% in each year after that. That means that if you put in n = 14, you will have multiplied the initial population by 1.02 a total of 14 times (1978).

How many years would you need to go to double the original population?

Continue reading “How Long Would It Take to Double the Population?”

What Should Have Happened on Gilligan’s Island?

Those of you who are fans of classic TV will recognize this as introduction to the sitcom “Gilligan’s Island”. This series ran from September 26, 1964 to September 4, 1967. The series revolved around 7 castaways (4 men and 3 women) marooned on an island somewhere near Hawaii.

Over the course of three seasons, the series followed attempts to leave the island, visitors to the island and general incompetence in getting rescued. When the series was cancelled in 1967, the castaways were never rescued.

In 1978, a made for TV movie called “Rescue from Gilligan’s Island” aired in which the castaways were rescued. However, at the end of this movie they decided to go on a reunion cruise and became stranded on the same island again after another freak storm. In 1979, another made for TV movie called “The Castaways on Gilligan’s Island” aired in which they were rescued once again. This time they decide to convert the island to a resort. It was hoped that this premise would generate a new series, but this never happened.

In 1981, a second sequel was created, “The Harlem Globetrotters on Gilligan’s Island” in which nefarious forces plotted to take over the island and the castaways are saved by the Harlem Globetrotter.

Although this series began before I was born, I watched years and years of reruns throughout my childhood. I must have watched each of the 98 episodes several times each. But there was one question that nagged at me. If there were two women on the island of child bearing age (Ginger and Mary Ann), why didn’t the population grow? Why didn’t nature take its course and lead to new castaways?

Let’s find out how many castaways there should have been in 1978, 14 years after they were originally shipwrecked.

To answer this question, let’s assume that the population on the island is an example of exponential growth. This may or may not be a good choice due to the small size of the population. However, with this assumption the amount of peoplec A on the island n years after 1964 will be given by the equation

A = P (1+r)n

In this situation, P is the initial amount of people on the island and r is the annual growth rate in percent per year.

Since the population of Gilligan’s Island was initially 7, we’ll set P = 7. For the growth rate r, we’ll use a fairly conservative rate of 2% per year. This is about what the world birth rate is. Around the world birth rates vary from a little less than 1% (China) to around 5% (African countries). With these values, we model the population by

A = 7 (1+0.02)n

To predict the population in 1978, I’ll find the value at n = 14.

A = 7 (1+0.02)14 = 7 (1.02)14 = 9.24

According to this growth model, there should have been about 9 people on the island in 1978.

 

This equation assumed that the initial was 7 in 1964 and grew by 2% in each year after that. That means that if you put in n = 14, you will have multiplied the initial population by 1.02 a total of 14 times (1978).

How many years would you need to go to double the original population?

Since the original population was 7, double that number is 14. So if I put that number into A, I need to solve

14 = 7 (1.02)n

for the number of years n after 1964.

Start by dividing both sides by 7 to isolate the piece with the exponent in it,

2 = (1.02)n

Now take the logarithm of both sides,

log(2) = log((1.02)n)

The reason we need to use logarithms is that powers in front of logarithms may be moved to the front,

log(2) = n log(1.02)

Now divide both sides by log(1.02),

log(2)/log(1.02) = n

Using a calculator to compute the logs gives a value of n of about 35. So in 1964 + 35 or 1995, the population would have increased from 7 to 14 according to the equation.