Math 2000 - Solutions 7

Spring 2001

Unit 7A

2a. Because the absolute growth rate (505 people per year) is constant, the population of Danbury is growing linearly.

2b. Because the relative or percentage growth rate (30% per year) is constant, the price of food is growing exponentially.

10a. Table 7-3 shows that one minute before 12:00 the bottle is 1/2 full; two minutes before 12:00 the bottle is 1/4 = 1/(2²) full; three minutes before 12:00 the bottle is 1/8 = 1/(2³) full; and so on. Therefore, 10 minutes before 12:00 at 11:50, the bottle is 1/(2¹⁰) full or about 1/1000 full.

10b. Table 7-3 shows that 50 minutes before 12:00 at 11:10, the bottle is 1/2⁵⁰ full or 8.9 ´ 10^-16 full.

13a.

Year	Population	Year	Population
2000	6 ´ 10^9	2550	1.229 ´ 10^13
2050	1.2 ´ 10^10	2600	2.458 ´ 10^13
2100	2.4 ´ 10^10	2650	4.915 ´ 10^13
2150	4.8 ´ 10^10	2700	9.830 ´ 10^13
2200	9.6 ´ 10^10	2750	1.966 ´ 10^14
2250	1.92 ´ 10^11	2800	3.932 ´ 10^14
2300	3.84 ´ 10^11	2850	7.864 ´ 10^14
2350	7.68 ´ 10^11	2900	1.573 ´ 10^15
2400	1.536 ´ 10^12	2950	3.146 ´ 10^15
2450	3.072 ´ 10^12	3000	6.291 ´ 10^15
2500	6.144 ´ 10^12

b. We see that there will be 5.1 ´ 10¹⁴ people on the Earth occupying 5.1 ´ 10¹⁴ m² sometime between 2800 and 2850.

c. The number of people that could be supported assuming each person needs 10⁴ m² is

This limit would be reached shortly after the year 2150.

Unit 7B (In these solutions, T_double refers to the doubling time.)

2b. There are three 4-month intervals in 12 months, so three doublings will occur in 12 months. This means that the population increases by a factor of 2³ = 8 in 12 months.

There are five 4-month intervals in 20 months, so five doublings will occur in 20 months. This means that the population increases by a factor of 2⁵ = 32 in 20 months.

2c. An eightfold increase is three doublings, so it will take three doubling times or 15 years for the population to increase by a factor of eight.

4b. We set initial value = 15,600 and T_double = 30 years. After t = 12 years, the population will be

new value = 15,600 ´ 2^12/30 = 15,600 ´ 2^0.4

= 15,600 ´ 1.32 = 20,584.

After t = 40 years, the population will be

new value = 15,600 ´ 2^40/30 = 15,600 ´ 2^1.33333

= 15,600 ´ 2.52 = 39,310.

5. Let’s take the initial value to be the 1990 population of 5.2 billion and let T_double = 40 years.

After t = 20 years (for the year 2010 which is half of a doubling time in the future), the population will be

new value = 5.2 billion ´ 2^20/40 = 5.2 billion ´ 1.4142

= 7.35 billion.

After t = 70 years (for the year 2060 which is about one and a half of a doubling times in the future), the population will be

new value = 5.2 billion ´ 2^70/40 = 5.2 billion ´ 3.3636

= 17.49 billion.

After t = 110 years (for the year 2100 which is almost three doubling times in the future), the population will be

new value = 5.2 billion ´ 2^110/40 = 5.2 billion ´ 6.7271

= 34.98 billion.

6a. We are told that the annual growth rate is 7% per year. Therefore, the doubling time is approximately 70/P = 70/7 = 10 years. Setting t = 3, the CPI increases by a factor of 2^3/10 = 1.23.