A town has a population of 6000 people in the year 2009 and is growing at a rate of 3% a year. Let t denote the time in years after 2009, so t = 0 denotes 2009. Let f be the function that, to each year t, assigns the population of the town in year t. In the text box below, enter a formula for f (t). Your entry must begin with f (t) = and to the right of the equals sign you should enter the formula for f (t). Your answer must be mathematically valid. For example, if you believe that

f (t) = t^ - 3 + 5 * t - 3,

then in the text box you should enter f (t) = t^ (-3) + 5*t-3. After entering your answer, click the Save Answer button.

Response to your input section.

Enter The Value Of k Here:

Question

Mathematics

Violet

18 April, 13:48

A town has a population of 6000 people in the year 2009 and is growing at a rate of 3% a year. Let t denote the time in years after 2009, so t = 0 denotes 2009. Let f be the function that, to each year t, assigns the population of the town in year t. In the text box below, enter a formula for f (t). Your entry must begin with f (t) = and to the right of the equals sign you should enter the formula for f (t). Your answer must be mathematically valid. For example, if you believe that

f (t) = t^ - 3 + 5 * t - 3,

then in the text box you should enter f (t) = t^ (-3) + 5*t-3. After entering your answer, click the Save Answer button.

Response to your input section.

Enter The Value Of k Here:

+5

Answers (1)

Know the Answer?

Gilda · Answer 1

f (t) = 6000*1.03^t k = 0.0295588

Step-by-step explanation:

The formula for exponential growth can be written a couple of different ways. One I prefer uses the problem numbers directly:

population = (initial population) * (growth factor) ^t

Here, the initial population is 6000, and the growth factor is 1+3% = 1.03. Then the function can be written as ...

f (t) = 6000·1.03^t

__

Another way to write the function is using the form ...

f (t) = (initial population) * e^ (kt)

where k is the natural logarithm of the growth factor. In this form, we have ...

k = ln (1.03)

k ≈ 0.0295588

so the function would be written ...

f (t) = 6000·e^ (0.0295588t)