Math Problem Statement

The population of a country dropped from 51.7 million in 1995 to 45.5 million in 2007. Assume that​ P(t), the​ population, in​ millions, t years after​ 1995, is decreasing according to the exponential decay model. ​a) Find the value of​ k, and write the equation. ​b) Estimate the population of the country in 2017. ​c) After how many years will the population of the country be 3 ​million, according to this​ model? Question content area bottom Part 1 ​a) Select the correct answer below and fill in the answer box to complete your choice.

Solution

The population change over time follows the exponential decay model, which has the general form:

P(t)=P0ektP(t) = P_0 e^{-kt}

Where:

  • P(t)P(t) is the population at time tt (in years),
  • P0P_0 is the initial population,
  • kk is the decay constant (rate of decline),
  • tt is the number of years after 1995 (since the population is being measured from 1995).

Step 1: Define the variables

  • In 1995, P0=51.7P_0 = 51.7 million.
  • In 2007, P(12)=45.5P(12) = 45.5 million (since 2007 is 12 years after 1995).
  • We need to find the decay constant kk using the given data.

Step 2: Set up the equation

Using the population at t=12t = 12: P(12)=51.7e12kP(12) = 51.7 e^{-12k} Substitute P(12)=45.5P(12) = 45.5 million: 45.5=51.7e12k45.5 = 51.7 e^{-12k}

Step 3: Solve for kk

First, divide both sides of the equation by 51.7 to isolate the exponential term: 45.551.7=e12k\frac{45.5}{51.7} = e^{-12k}

Now take the natural logarithm (ln) of both sides to remove the exponential: ln(45.551.7)=12k\ln\left(\frac{45.5}{51.7}\right) = -12k

Now solve for kk: k=112ln(45.551.7)k = -\frac{1}{12} \ln\left(\frac{45.5}{51.7}\right)

Let’s compute the value of kk.The decay constant kk is approximately 0.010650.01065.

Thus, the equation for the population model is:

P(t)=51.7e0.01065tP(t) = 51.7 e^{-0.01065t}

Part (b): Estimate the population in 2017

To estimate the population in 2017, we substitute t=22t = 22 (since 2017 is 22 years after 1995) into the population model:

P(22)=51.7e0.01065×22P(22) = 51.7 e^{-0.01065 \times 22}

Let's calculate the population for 2017.The estimated population of the country in 2017 is approximately 40.91 million.

Part (c): Find when the population will be 3 million

We now need to determine when the population will reach 3 million. This means solving for tt when P(t)=3P(t) = 3:

3=51.7e0.01065t3 = 51.7 e^{-0.01065t}

First, divide both sides by 51.7:

351.7=e0.01065t\frac{3}{51.7} = e^{-0.01065t}

Take the natural logarithm of both sides to solve for tt:

ln(351.7)=0.01065t\ln\left(\frac{3}{51.7}\right) = -0.01065t

Now, solve for tt. I'll compute this next.The population will reach 3 million in approximately 267.42 years after 1995, which means around the year 2262.

Summary:

  • (a) The decay constant kk is approximately 0.01065, and the population model is P(t)=51.7e0.01065tP(t) = 51.7 e^{-0.01065t}.
  • (b) The estimated population in 2017 is approximately 40.91 million.
  • (c) The population is expected to reach 3 million about 267.42 years after 1995, which is around the year 2262.

Do you have any questions or would you like more details on a specific part?

Here are 5 related questions to explore:

  1. How does changing the decay constant kk affect the population over time?
  2. What would the population be in 2025 according to this model?
  3. How can we adjust the model to account for immigration or other external factors?
  4. What is the half-life of the population in this decay model?
  5. How would the model differ if the population were growing instead of declining?

Tip: Exponential models are commonly used to describe both growth and decay processes in fields such as biology, physics, and economics!

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Logarithms
Population Modeling

Formulas

P(t) = P_0 e^{-kt}
k = -\frac{1}{12} \ln\left(\frac{45.5}{51.7}\right)
P(22) = 51.7 e^{-0.01065 \times 22}
t = -\frac{1}{0.01065} \ln\left(\frac{3}{51.7}\right)

Theorems

Exponential Growth and Decay Theorem

Suitable Grade Level

Grades 10-12