Compound Growth Help (page 2)
Introduction to Compound Growth
A quantity (such as a population, amount of money, or radiation level) changes exponentially if the growth or loss is a fixed percentage over a period of time. To see how this works, we will see how the value of an account grows over four years if $100 is deposited and earns 5% interest, compounded annually. Compounded annually means that the interest earned in the previous year earns interest.
After one year, $100 has grown to 100 + 0.05(100) = 100 + 5 = $105. In the second year, the original $100 earns 5% plus the $5 earns 5% interest: 105 + (105)(0.05) = $110.25. Now this amount earns interest in the third year: 110.25 + (110.25)(0.05) = $115.76. Finally, this amount earns interest in the fourth year: 115.76 + (115.76)(0.05) = $121.55. If interest is not compounded, that is, the interest does not earn interest, the account would only be worth $120. The extra $1.55 is interest earned on interest.
Compound growth is not dramatic over the short run but it is over time. If $ 100 is left in an account earning 5% interest, compounded annually, for 20 years instead of four years, the difference between the compound growth and noncompound growth is a little more interesting. After 20 years, the compound amount is $265.33 compared to $200 for simple interest (noncompound growth). A graph of the growth of each type over 40 years is given in Figure 9.1. The line is the growth for simple (noncompounded) interest, and the curve is the growth with compound interest.
We can use a formula to compute the value of an account earning compounded interest. If P dollars is invested for t years, earning r interest rate, then it will grow to A dollars, where A = P (1 + r ) t.
Find the compound amount.
- $5000, after three years, earning 6% interest, compounded annually
We will use the formula A = P (1 + r ) t . P = 5000, r = 0.06, and t = 3. We want to know A , the compound amount.
A = 5000(1 + 0.06) 3 = 5000(1.06) 3 = 5000(1.191016)
The compound amount is $5955.08.
- $10,000 after eight years, interest, compounded annually
A = 10,000(1 + 0.0725) 8 = 10,000(1.0725) 8 ≈ 10,000(1.7505656)
The compound amount is $17,505.66
Many investments pay more often than once a year, some paying interest daily. Instead of using the annual interest rate, we need to use the interest rate per period, and instead of using the number of years, we need to use the number of periods. If there are n compounding periods per year, then the interest rate per period is and the total number of periods is nt . The compound amount formula becomes
Find the compound amount.
- $5000, after three years, earning 6% annual interest
(a) compounded semiannually
(b) compounded monthly
For (a), interest compounded semiannually means that it is compounded twice each year, so n = 2.
The compound amount is $5970.26.
For (b), interest compounded monthly means that it is compounded 12 times each year, so n = 12.
The compound amount is $5983.40.
- $10,000, after eight years, earning annual interest, compounded weekly Interest that is paid weekly is paid 52 times each year, so n = 52.
The compound amount is $17,853.17.
The more often interest is compounded per year, the more interest is earned. $1000 earning 8% annual interest, compounded annually, is worth $1080 after one year. If interest is compounded quarterly, it is worth $1082.43 after one year. And if interest is compounded daily, it is worth $1083.28 after one year. What if interest is compounded each hour? Each second? It turns out that the most this investment could be worth (at 8% interest) is $1083.29, when interest is compounded each and every instant of time. Each instant of time, a tiny amount of interest is earned. This is called continuous compounding. The formula for the compound amount for interest compounded continuously is A = Pe rt , where A , P , r , and t are the same quantities as before. The letter e stands for a constant called Euler’s number. It is approximately 2.718281828. You probably have an e or e x key on your calculator. Although e is irrational, it can be approximated by numbers of the form
where m is a large rational number. The larger m is, the better the approximation for e . If we make the substitution and use some algebra, we can see how is very close to e rt , for large values of n . If interest is compounded every minute, n would be 525,600, a rather large number!
- Find the compound amount of $5000 after eight years, earning 12% annual interest, compounded continuously.
A = 5000 e 0.12(8) = 5000 e 0.96 ≈ 5000(2.611696) ≈ 13,058.48
The compound amount is $13,058.48.
The compound growth formula for continuously compounded interest is used for other growth and decay problems. The general exponential growth model is n ( t ) = n 0 e rt , where n ( t ) replaces A and n 0 replaces P . Their meanings are the same— n ( t ) is still the compound growth, and n 0 is still the beginning amount. The variable t represents time in this formula; although, time will not always be measured in years. The growth rate and t need to have the same unit of measure. If the growth rate is in days, then t needs to be in days. If the growth rate is in hours, then t needs to be in hours, and so on. If the “population” is getting smaller, then the formula is n ( t ) = n 0 e −rt .
- The population of a city is estimated to be growing at the rate of 10% per year. In 2000, its population was 160,000. Estimate its population in the year 2005.
The year 2000 corresponds to t = 0, so the year 2005 corresponds to t = 5; n 0 , the population in year t = 0, is 160,000. The population is growing at the rate of 10% per year, so r = 0.10. The formula n ( t ) = n 0 e rt becomes n ( t ) = 160,000 e 0.10 t . We want to find n ( t ) for t = 5.
n (5) = 160,000 e 0.10(5) ≈ 263,795
The city’s population is expected to be 264,000 in the year 2005 (estimates and projections are normally rounded off).
- A county is losing population at the rate of 0.7% per year. If the population in 2001 is 1,000,000, what is it expected to be in the year 2008?
n 0 = 1,000,000, t = 0 is the year 2001, t = 7 is the year 2008, and r = 0.007. Because the county is losing population, we will use the decay model: n ( t ) = n 0 e −rt . The model for this county’s population is n(t) = 1,000,000e −.007 t . We want to find n ( t ) for t = 7.
n (7) = 1,000,000 e −.007(7) ≈ 952,181
The population is expected to be 952,000 in the year 2008.
- In an experiment, a culture of bacteria grew at the rate of 35% per hour. If 1000 bacteria were present at 10:00, how many were present at 10:45?
n 0 = 1000, r = 0.35, t is the number of hours after 10:00
The growth model becomes n(t) = 1000 e 0.35 t . We want to find n ( t ) for 45 minutes, or t = 0.75 hours.
n (0.75) = 1000 e 0.35(0.75) = 1000 e 0.2625 ≈ 1300
At 10:45, there were approximately 1300 bacteria present in the culture.
Find practice problems and solutions for these concepts at: Exponents and Logarithms Practice Test.
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