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Consider a population of bacteria that grows according to the function f ( t ) = 500 e 0.05 t , where t is measured in minutes. How many bacteria are present in the population after 4 hours? When does the population reach 100 million bacteria?

There are 81,377,396 bacteria in the population after 4 hours. The population reaches 100 million bacteria after 244.12 minutes.

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Let’s now turn our attention to a financial application: compound interest . Interest that is not compounded is called simple interest . Simple interest is paid once, at the end of the specified time period (usually 1 year). So, if we put $ 1000 in a savings account earning 2 % simple interest per year, then at the end of the year we have

1000 ( 1 + 0.02 ) = $ 1020 .

Compound interest is paid multiple times per year, depending on the compounding period. Therefore, if the bank compounds the interest every 6 months, it credits half of the year’s interest to the account after 6 months. During the second half of the year, the account earns interest not only on the initial $ 1000 , but also on the interest earned during the first half of the year. Mathematically speaking, at the end of the year, we have

1000 ( 1 + 0.02 2 ) 2 = $ 1020.10 .

Similarly, if the interest is compounded every 4 months, we have

1000 ( 1 + 0.02 3 ) 3 = $ 1020.13 ,

and if the interest is compounded daily ( 365 times per year), we have $ 1020.20 . If we extend this concept, so that the interest is compounded continuously, after t years we have

1000 lim n ( 1 + 0.02 n ) n t .

Now let’s manipulate this expression so that we have an exponential growth function. Recall that the number e can be expressed as a limit:

e = lim m ( 1 + 1 m ) m .

Based on this, we want the expression inside the parentheses to have the form ( 1 + 1 / m ) . Let n = 0.02 m . Note that as n , m as well. Then we get

1000 lim n ( 1 + 0.02 n ) n t = 1000 lim m ( 1 + 0.02 0.02 m ) 0.02 m t = 1000 [ lim m ( 1 + 1 m ) m ] 0.02 t .

We recognize the limit inside the brackets as the number e . So, the balance in our bank account after t years is given by 1000 e 0.02 t . Generalizing this concept, we see that if a bank account with an initial balance of $ P earns interest at a rate of r % , compounded continuously, then the balance of the account after t years is

Balance = P e r t .

Compound interest

A 25-year-old student is offered an opportunity to invest some money in a retirement account that pays 5 % annual interest compounded continuously. How much does the student need to invest today to have $ 1 million when she retires at age 65 ? What if she could earn 6 % annual interest compounded continuously instead?

We have

1,000,000 = P e 0.05 ( 40 ) P = 135,335.28.

She must invest $ 135,335.28 at 5 % interest.

If, instead, she is able to earn 6 % , then the equation becomes

1,000,000 = P e 0.06 ( 40 ) P = 90,717.95.

In this case, she needs to invest only $ 90,717.95 . This is roughly two-thirds the amount she needs to invest at 5 % . The fact that the interest is compounded continuously greatly magnifies the effect of the 1 % increase in interest rate.

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Suppose instead of investing at age 25 b 2 4 a c , the student waits until age 35 . How much would she have to invest at 5 % ? At 6 % ?

At 5 % interest, she must invest $ 223,130.16 . At 6 % interest, she must invest $ 165,298.89 .

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Practice Key Terms 4

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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