Chapter 3 - Section 3.8 - Exponential Growth and Decay - 3.8 Exercises - Page 243: 6

6a) P ( 30 ) = 145 million 6a) Comparison: The model predicts less than the actual data by 5 million. 6b) P ( 40 ) = 225 million 6b) Comparison: The model predicts more than the actual data by 11 million. 6c) P ( 30 ) = 256 million 6c) Comparison: The model predicts more than the actual data by 13 million. 6d) P ( 40 ) = 305.3 million 6d) Comparison: The model predicts more than the actual data by 31.8 million.

a) Since we are assuming that the “rate of population growth is proportionate to the population size” ( alternately “the relative growth rate is constant”) , we use the $e^{kt}$ formula from page 237 ( Formula 2 ). P ( t ) = P ( 0 ) $e^{kt}$ P ( 0 ) = year 1950 = 83 P ( 0 ) = 83 P ( t ) = 83 $e^{kt}$ 1960 – 1950 = 10 = t ← year 1960 = 100 million P ( 10 ) = 83 $e^{10k}$ = 100 83 $e^{10k}$ = 100 $e^{10k}$ = $\frac{100}{83}$ ln $e^{10k}$ = ln $\frac{100}{83}$ 10k = ln $\frac{100}{83}$ k = $\frac{ln \frac{100}{83}}{10}$ ≈ 0.01863 P ( t ) = 83 $e^{kt}$ P ( t ) = 83 $e^{0.01863t}$ 1980 – 1950 = 30 = t ← year 1980 P ( 30 ) = 83 $e^{0.01863( 30 )}$← year 1980 6a) P ( 30 ) = 145 million ← year 1980 prediction with model Actual data for year 1980 = 150 million 6a) Comparison: The model predicts less than the actual data by 5 million. b) Use figures for 1960 and 1980 to predict year 2000. P ( t ) = P ( 0 ) $e^{kt}$ P ( 0 ) = year 1960 = 100 P ( 0 ) = 100 P ( t ) = 100 $e^{kt}$ 1980 – 1960 = 20 = t ← year 1980 = 150 million P ( 20 ) = 100 $e^{20k}$ = 150 100 $e^{20k}$ = 150 $e^{20k}$ = $\frac{150}{100}$ ln $e^{20k}$ = ln $\frac{150}{100}$ 20k = ln $\frac{150}{100}$ k = $\frac{ln \frac{150}{100}}{20}$ ≈ 0.02027 P ( t ) = 100 $e^{kt}$ P ( t ) = 100 $e^{0.02027t}$ 2000 – 1960 = 40 = t ← year 2000 P ( 40 ) = 100 $e^{0.02027( 40 )}$ 6b) P ( 40 ) = 225 million ← year 2000 Actual data for year 2000 = 214 million 6b) Comparison: The model predicts more than the actual data by 11 million. c) Use figures for years 1980 and 2000 to predict 2010. Actual data for 2010 = 243 million P ( t ) = P ( 0 ) $e^{kt}$ P ( 0 ) = year 1980 = 150 P ( 0 ) = 150 P ( t ) = 150 $e^{kt}$ 2000 – 1980 = 20 = t ← year 2000 = 214 million P ( 20 ) = 150 $e^{20k}$ = 214 150 $e^{20k}$ = 214 $e^{20k}$ = $\frac{214}{150}$ ln $e^{20k}$ = ln $\frac{214}{150}$ 20k = ln $\frac{214}{150}$ k = $\frac{ln \frac{214}{150}}{20}$ ≈ 0.01777 P ( t ) = 150 $e^{kt}$ P ( t ) = 150 $e^{0.01777t}$ 2010 – 1980 = 30 = t ← year 2010 P ( 30 ) = 150 $e^{0.01777( 30 )}$ 6c) P ( 30 ) = 256 million ← year 2010 Actual data for year 2010 = 243 million 6c) Comparison: The model predicts more than the actual data by 13 million. d) d) Use the model from 6c: ( P ( t ) = 150 $e^{0.01777t}$ Predict year 2020 – I think the model will continue to be greater than the actual data. 2020 – 1980 = 40 = t ← year 2020 P ( t ) = 150 $e^{0.01777t}$ P ( 40 ) = 150 $e^{0.01777( 40 )}$ 6d) P ( 40 ) = 305.3 million ← year 2020 Actual data for year 2020 = 273.5 million ( Wikipedia, retrieved on May 18, 2022) 6d) Comparison: The model predicts more than the actual data by 31.8 million.