Math 2000 - Solutions 6

Fall 2000

Unit 7A

2c. Because the relative or percentage rate of change (-2.5% per year) is constant, the number of births is declining exponentially.

2d. Because the absolute rate of change (-$150 per year) is constant, the value of the equipment is declining linearly.

Unit 7B

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.

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.

6b. Because the percentage growth rate is 5.5% per year, the doubling time is approximately 70/P = 70/5.5 = 12.7 years. After five years the value of the certificate will increase by a factor of 2^5/12.7 = 1.3 giving the certificate a value of $100 ´ 1.3 = $130. After 50 years the value of the certificate will increase by a factor of 2^50/12.7 = 15.3 giving the certificate a value of $100 ´ 15.3 = $1530. We could also use the compound interest formula: after 5 years, the balance is $100 × 1.055⁵ = $130.70 and after 50 years, the balance is $100 × 1.055⁵⁰ = $1454.20. The latter figures are more accurate.

6c. With a percentage growth rate of 2% per year, the doubling time is approximately 70/P = 70/2 = 35 years. After 75 years the population will increase by a factor of 2^75/35 = 4.4.

Unit 5A

4a. Setting P = $3000, APR = 0.04, and Y = 12, we find that the accumulated balance is

6a. Setting P = $1000, APR = 0.07, Y = 15, and n = 12 (for monthly compounding), the accumulated balance is