Math Problem Statement

The world human population, P billions, is modelled by the equation P = ab^t where a and b are constants and t is the number of years after 2004. Using the estimated population figures for the years from 2004 to 2007, a graph is plotted of log10 P against t. The points lie approximately on a straight line with gradient 0.0054 and intercept 0.81 on the log10 P axis. (a) Estimate, to 3 decimal places, the value of a and the value of b. (b) In the context of the model, (i) interpret the value of the constant a, (ii) interpret the value of the constant b. (c) Use the model to estimate the world human population in 2030. (d) Comment on the reliability of the answer to part (c).

Solution

Let’s break this problem into its components and solve step by step.

Given:

The model is P=abtP = ab^t, where:

  • PP is the population in billions.
  • tt is the number of years after 2004.
  • The log-transformed version is log10P=log10a+tlog10b\log_{10}P = \log_{10}a + t\log_{10}b, showing a linear relationship.

From the question:

  • Gradient of the line (log10b\log_{10}b) = 0.0054
  • Intercept on the log10P\log_{10}P-axis (log10a\log_{10}a) = 0.81

(a) Estimating aa and bb:

1. Determine aa:

The intercept log10a=0.81\log_{10}a = 0.81: a=100.81a = 10^{0.81} Calculating: a6.457a \approx 6.457

2. Determine bb:

The gradient log10b=0.0054\log_{10}b = 0.0054: b=100.0054b = 10^{0.0054} Calculating: b1.012b \approx 1.012

Thus:

  • a6.457a \approx 6.457
  • b1.012b \approx 1.012

(b) Interpretation of the constants:

(i) aa:

The constant aa represents the world population (in billions) in 2004, as t=0t = 0 in that year. Hence, a6.457a \approx 6.457 billion.

(ii) bb:

The constant bb represents the annual growth factor. A value of b1.012b \approx 1.012 indicates a 1.2%1.2\% annual increase in population.


(c) Estimate the population in 2030:

1. Time difference:

2030 corresponds to t=20302004=26t = 2030 - 2004 = 26 years.

2. Model substitution:

P=abt=6.457×(1.012)26P = ab^t = 6.457 \times (1.012)^{26} Calculating: P6.457×1.348858.706billionP \approx 6.457 \times 1.34885 \approx 8.706 \, \text{billion}

Thus, the estimated population in 2030 is approximately 8.706 billion.


(d) Comment on reliability:

The reliability of this estimate depends on:

  1. Assumption of a consistent growth rate (bb) over time, which may not hold due to socioeconomic and environmental factors.
  2. The model is based on data from only 2004 to 2007, making long-term predictions less reliable.
  3. External factors like global crises or policy changes are not accounted for, potentially affecting population growth.

Thus, while the model provides a mathematical estimate, its reliability for 2030 is limited.


Do you have further questions about this solution? Here are five related questions for further exploration:

  1. How does changing the gradient (log10b\log_{10}b) affect the population prediction?
  2. What would the model predict for a shorter or longer time period, such as 2015 or 2100?
  3. How could incorporating additional data points improve the model's reliability?
  4. What alternative growth models could be used to predict population changes?
  5. How does the percentage growth rate relate to bb and tt?

Tip: Always consider external factors (e.g., technology, policy changes) when interpreting long-term predictions from mathematical models.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Logarithmic Functions
Linear Regression
Model Interpretation

Formulas

P = ab^t
log10 P = log10 a + t * log10 b
a = 10^intercept
b = 10^gradient

Theorems

Logarithmic Transformation
Exponential Growth Model

Suitable Grade Level

Grades 10-12