Practice problems for these concepts can be found at: Applications of Definite Integrals Practice Problems for AP Calculus

- , then the rate of change of
*y*is proportional to*y*. - If
*y*is a differentiable function of t with y > 0 , then*y*(*t*)=*y*_{0}e^{kt}; where*y*_{0}is initial value of*y*and*k*is constant. If*k*> 0, then*k*is a growth constant and if*k*< 0, then*k*is the decay constant.

### Example 1 Population Growth

If the amount of bacteria in a culture at any time increases at a rate proportional to the amount of bacteria present and there are 500 bacteria after one day and 800 bacteria after the third day:

- approximately how many bacteria are there initially, and
- approximately how many bacteria are there after 4 days?

#### Solution:

- Since the rate of increase is proportional to the amount of bacteria present, then:

where *y* is the amount of bacteria at any time.

Therefore, this is an exponential growth/decay model: *y* (*t*)= *y*_{0}*e*^{kt}.

Step 1. *y* (1)=500 and *y* (3)=800

500 = y_{0}*e*^{k} and 800= *y*_{0}*e*^{3k}

Step 2. 500= *y*_{0}*e*^{k}

Substitute *y*_{0} = 500^{e–k} into 800= *y*_{0}*e* ^{3k}.

800=(500)(*e*–^{k})(*e*^{3k})

800=500*e*^{2k}

Take the ln of both sides:

STEP 3.

Thus, there are 395 bacteria present initially.

b

Thus there are approximately 1011 bacteria present after 4 days.

### Example 2 Radioactive Decay

Carbon-14 has a half-life of 5750 years. If initially there are 60 grams of carbon-14, how many grams are left after 3000 years?

Step 1. *y*(*t*)= *y*_{0}*e*^{kt}=60*e*^{kt}

Since half-life is 5750 years, 30=60*e*^{k}(5750)

Step 2. *y*(*t*)= *y*_{0}*e*^{kt}

Thus, there will be approximately 41.792 grams of carbon-14 after 3000 years.

Practice problems for these concepts can be found at: Applications of Definite Integrals Practice Problems for AP Calculus

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