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Chapter 6 | Applications of Integration
kt
2 y 0 = y 0 e
2 = e kt ln2 = kt t = ln2 k .
Definition If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is given by Doubling time = ln2 k .
Example 6.44 Using the Doubling Time
Assume a population of fish grows exponentially. A pond is stocked initially with 500 fish. After 6 months, there are 1000 fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches 10,000. When will the owner’s friends be allowed to fish? Solution We know it takes the population of fish 6 months to double in size. So, if t represents time in months, by the doubling-time formula, we have 6= (ln2)/ k . Then, k = (ln2)/6. Thus, the population is given by y =500 e ⎛ ⎝ (ln2)/6 ⎞ ⎠ t . To figure out when the population reaches 10,000 fish, we must solve the following equation: 10,000 = 500 e (ln2/6) t 20 = e (ln2/6) t ln20 = ⎛ ⎝ ln2 6 ⎞ ⎠ t t = 6(ln20) ln2 ≈25.93. The owner’s friends have to wait 25.93 months (a little more than 2 years) to fish in the pond.
6.44 Suppose it takes 9 months for the fish population in Example 6.44 to reach 1000 fish. Under these circumstances, how long do the owner’s friends have to wait?
Exponential Decay Model Exponential functions can also be used to model populations that shrink (from disease, for example), or chemical compounds that break down over time. We say that such systems exhibit exponential decay, rather than exponential growth. The model is nearly the same, except there is a negative sign in the exponent. Thus, for some positive constant k , we have y = y 0 e − kt . As with exponential growth, there is a differential equation associated with exponential decay. We have y ′ =− ky 0 e − kt =− ky .
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