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Chapter 6 | Applications of Integration
growth models, we must always be careful to interpret the function values in the context of the phenomenon we are modeling. Example 6.42 Population Growth Consider the population of bacteria described earlier. This population grows according to the function f ( t ) =200 e 0.02 t , where t is measured in minutes. How many bacteria are present in the population after 5 hours (300 minutes)? When does the population reach 100,000 bacteria?
Solution We have f ( t ) =200 e 0.02 t . Then
f (300) =200 e 0.02(300) ≈80,686.
There are 80,686 bacteria in the population after 5 hours. To find when the population reaches 100,000 bacteria, we solve the equation 100,000 = 200 e 0.02 t 500 = e 0.02 t ln 500 = 0.02 t t = ln500 0.02 ≈ 310.73. The population reaches 100,000 bacteria after 310.73 minutes.
6.42 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?
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
⎛ ⎝ 1+ 0.02 3
⎞ ⎠
3
1000 = $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
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