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# Exponential Growth Decay Problems for AP Calculus

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

1. , then the rate of change of y is proportional to y.
2. If y is a differentiable function of t with y > 0 , then y (t)= y0ekt; where y0 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:

1. approximately how many bacteria are there initially, and
2. approximately how many bacteria are there after 4 days?

#### Solution:

1. 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)= y0ekt.

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

500 = y0ek and 800= y0e3k

Step 2. 500= y0ek

Substitute y0 = 500ek into 800= y0e 3k.

800=(500)(ek)(e3k)

800=500e2k

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.

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)= y0ekt=60ekt

Since half-life is 5750 years, 30=60ek(5750)

Step 2. y(t)= y0ekt

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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