Math Problem Statement

The population of a town can be modeled by the equation P, left parenthesis, t, right parenthesis, equals, 14, comma, 400, e, start superscript, minus, 0, point, 0, 5, 1, t, end superscriptP(t)=14,400e −0.051t where tt is measured in years since 2010.

a. Is the population increasing or decreasing?

b. What was the population in 2010?

c. Predict the population of the town in 20 years. (Round to the nearest person.)

Solution

Let's analyze the given population model:

The population P(t)P(t) is given by the equation: P(t)=14,400e0.051tP(t) = 14,400 \, e^{-0.051t} where:

  • tt is the time in years since 2010.
  • ee is the base of the natural logarithm (approximately 2.718).
  • The exponent 0.051t-0.051t determines the rate of change.

a. Is the population increasing or decreasing?

To determine whether the population is increasing or decreasing, we need to look at the exponent in the function. The exponent is 0.051t-0.051t. Since the coefficient of tt is negative (-0.051), the exponent is negative, and as time tt increases, e0.051te^{-0.051t} decreases because raising ee to a negative exponent results in a fraction less than 1.

Thus, the population is decreasing over time.

b. What was the population in 2010?

To find the population in 2010, we substitute t=0t = 0 (since 2010 is the starting point).

P(0)=14,400e0.051×0=14,400e0P(0) = 14,400 \, e^{-0.051 \times 0} = 14,400 \, e^0

Since e0=1e^0 = 1:

P(0)=14,400×1=14,400P(0) = 14,400 \times 1 = 14,400

So, the population in 2010 was 14,400.

c. Predict the population of the town in 20 years.

To find the population 20 years after 2010 (i.e., in 2030), we substitute t=20t = 20 into the equation:

P(20)=14,400e0.051×20P(20) = 14,400 \, e^{-0.051 \times 20}

First, calculate the exponent:

0.051×20=1.02-0.051 \times 20 = -1.02

Now, substitute this back into the equation:

P(20)=14,400e1.02P(20) = 14,400 \, e^{-1.02}

Next, let's calculate e1.02e^{-1.02} and find P(20)P(20).The predicted population of the town in 20 years (in 2030) is 5,193 people (rounded to the nearest person).

Summary of Results:

  • a. The population is decreasing.
  • b. The population in 2010 was 14,400.
  • c. The predicted population in 2030 is 5,193.

Would you like more details or have any questions?

Related Questions:

  1. How do exponential growth and decay models differ mathematically?
  2. What is the significance of the base ee in natural exponential functions?
  3. How would the population change if the rate were positive instead of negative?
  4. Can we determine when the population will be halved from its initial value?
  5. How do you compute the rate of change for exponential functions?

Tip: Remember that the sign of the exponent in an exponential model determines whether it represents growth or decay.

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Population Modeling
Algebra

Formulas

P(t) = P_0 * e^(kt)

Theorems

Exponential Growth and Decay Theorem

Suitable Grade Level

Grades 10-12