The population of a pigeons in a city is 100 and is growing exponentially at 20% per year. Write a function to represent the population of pigeons after
t
t years, where the monthly rate of change can be found from a constant in the function. Round all coefficients in the function to four decimal places. Also, determine the percentage rate of change per month, to the nearest hundredth of a percent

Answer:

Step-by-step explanation:

In general, the function will be written ...

  p(t) = (initial value)×(growth factor)^t

where t is in units comparable to those applicable to the growth factor. The growth factor is found from ...

  growth factor = 1 + growth rate

Here, the growth rate is given as 20% per year, so the growth factor per year is ...

  1 +20% = 1.20

The initial value is given as 100, so we can write the exponential function as ...

  p(t) = 100×1.20^t

__

The time period units for t are supposed to be years, but we want to find the growth rate for a month. We can do that by recognizing there are 12 months in a year. In the above equation, we can use (1/12)(12t) in place of t, then figure the growth factor (and growth rate) per month.

  p(t) = 100×(1.20^(1/12))^( 12t)

  p(t) = 100×1.0153^(12t) . . . . population exponential function

This shows the monthly growth factor is 1.0153, so the monthly rate of change (growth rate) is ...

  1.0153 -1 = 0.0153 = 1.53% . . . . monthly rate of change

Answers:

=====================================================

Explanation:

The population starts at 100 pigeons. After a year, the population goes up by 20%, meaning the population goes up by 20. It lands on 120 after a full year (100+20 = 120).

Dividing the "before" and "after" population amounts gets us 120/100 = 1.20

If we didn't worry about the monthly rate, and only cared about the annual rate of growth, then the population function would be

[tex]P(t) = 100(1.20)^t[/tex]

where t is the number of years and P(t) is the population for that year number. Notice how the 1.20 is the base of the exponential to quickly tell the reader the annual rate of growth. We read off the .20 portion and that converts to 20% after moving the decimal 2 spots to the right.

If we were to plug t = 0 into that function, then we'd get P(0) = 100. Inputting t = 1 leads to P(1) = 120. This helps confirm we have the correct annual growth rate function.

---------------------

Unfortunately, your teacher doesn't want the annual growth rate function. Instead, they want you to write a function where the reader can easily pick out the monthly rate of change without having to do any math. This is due to the instructions that "the monthly rate of change can be found from a constant in the function". Let m be this constant.

Since t is the number of years, this means t = 1/12 represents 1/12 of a year, aka 1 month. If you were to compute P(1/12), then you should get roughly 101.530947049973

Divide this amount over the initial population to find m.

m = 101.530947049973/100 = 1.01530947049973

So we get m = 1.0153 after rounding to four decimal places.

The number m is in the form 1+x. Solving 1+x = m will lead to x = 0.0153 to indicate a monthly rate of change is roughly 1.53% since we move the decimal point 2 spots to the right to go from 0.0153 to 1.53%

The pigeon population is increasing by roughly 1.53% each month.

This is why computing this m value leads to the reader able to spot the monthly rate of change without having to resort to a calculator. All they have to do is read the digits to the right of the decimal point and convert that to percentage form.

---------------------

An alternative way to compute this value is to use this formula

m = (1+r)^(1/12)

We'll plug in the annual rate r = 0.20 to get:

m = (1+r)^(1/12)

m = (1+0.20)^(1/12)

m = 1.01530947049973

m = 1.0153

We get the same approximate value as earlier.

---------------------

Whichever method you use, you should get roughly m = 1.0153

We're given that A = 100 is the initial population

So we'll then be able to form this function

[tex]P(t) = A*(m)^{12t}\\\\P(t) = 100(1.0153)^{12t}\\\\[/tex]

This represents the monthly growth of the population where t is in years. Note that after something like t = 3 years, we have 12t = 12*3 = 36 months pass by. So that's why the 12t is in the exponent rather than simply t.

Once again, for comparison we have...

Each function has t represent the number of years. The difference is the base of each exponential to tell us the respective annual or monthly growth rate. Also, the exponent up top must be adjusted depending on which you go for.

---------------------

You may be wondering: "what if the exponent in the monthly growth rate function wasn't 12t?". Let's find out. We'll assume the monthly growth rate function is [tex]P(t) = 100(1.0153)^{t}\\\\[/tex] and show that an error happens.

If we plugged in something like t = 1, we should get P(t) = 120 like mentioned earlier.

[tex]P(t) = 100(1.0153)^{t}\\\\P(1) = 100(1.0153)^{1}\\\\P(1) = 100(1.0153)\\\\P(1) = 101.53\\\\[/tex]

That value of 101.53 is the approximate population number after 1 month, rather than 1 year. Making the exponent 12t will fix the issue.

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