Exponential Growth
A model for growth of a quantity for which the rate of growth is directly proportional to the amount present. The equation for the model is A = A0bt (where b > 1 ) or A = A0ekt (where k is a positive number representing the rate of growth). In both formulas A0 is the original amount present at time t = 0.
This model is used for such phenomena as inflation or population growth. For example, A = 7000e0.05t is a model for the exponential growth of $7000 invested at 5% per year compounded continuously.

Exponential Decay
A model for decay of a quantity for which the rate of decay is directly proportional to the amount present. The equation for the model is A = A0bt (where 0 < b < 1 ) or A = A0ekt (where k is a negative number representing the rate of decay). In both formulas A0 is the original amount present at time t = 0.
This model is used for phenomena such as radioactivity or depreciation. For example, A = 50e–0.01t is a model for exponential decay of 50 grams of a radioactive element that decays at a rate of 1% per year.

Summary

A quantity is said to be subject to exponential decay/growth if it decreases/increases at a rate proportional to its value.

Problem It is often assumed that 1/3 acre of land is needed to provide food for one person. It is also estimated that there are 10 billion acres of arable land in the world, and therefore a maximum population of 30 billion people can be sustained if no other sources of food are known. The total world population at the beginning of 1970 was 3.6 billion.
(a) Assuming that the population continues to increase at the rate of 2 percent per year, what will the population be in the year 2000? When will the maximum population be reached?
(b) What allows us to use the exponential growth and decay model y=Cekt in his problem? Solution(a) In the exponential growth and decay model y=Cekt, the number C represents initial amount, or 3.6 billion=3,600,000,000 (initial because t=0, which in this case is 1970). We know that a 2% growth occurs annually, so 1971 results in a population of N*1.02, or 3,672,000,000. Plugging that number in for y and putting our C where it belongs, we can solve for k. In this example, k=.019802627296. Barring a discussion of significant digits, we know that we like our answer to be as accurate as possible for the AP graders; round-off done incorrectly will likely cause inaccuracy. Now, to figure out population in 2000, we say that t = 30, since 1970 + 30 = 2000 and plug in our values for C (always the initial amount in this example) and k (our constant of proportionality). We easily find that the population in the year 2000 will be approximately 6,520,901,703 (quite a bit higher than our population in 1970).
In order to figure out when our population maxes out, we plug in 30,000,000,000 for y (which represents the result variable--what happens in our model once we figure in the rest), and our C and k as before. Solving for t gives us a startling answer of t = 107. If t = 0 corresponds to 1970, this answer corresponds to 2077. Don't forget units! Since this will likely happen at night when I am sleeping and will wake to find my refrigerator empty, the correct units are p.m. (just kidding).
(b) We are allowed to use the exponential growth and decay model because our variable in question (population) is proportional to the rate of change of itself. In simpler words, the more people there are, the more people will be produced.

## Table of Contents

## Definition

Exponential GrowthA model for growth of a quantity for which the rate of growth is directly proportional to the amount present. The equation for the model is A = A0b

t(whereb> 1 ) or A = A0ekt(wherekis a positive number representing the rate of growth). In both formulas A0 is the original amount present at timet= 0.This model is used for such phenomena as inflation or population growth. For example, A = 7000

e0.05tis a model for the exponential growth of $7000 invested at 5% per year compounded continuously.Exponential DecayA model for decay of a quantity for which the rate of decay is directly proportional to the amount present. The equation for the model is A = A0b

t(where 0 <b< 1 ) or A = A0ekt(wherekis a negative number representing the rate of decay). In both formulas A0 is the original amount present at timet= 0.This model is used for phenomena such as radioactivity or depreciation. For example, A = 50

e–0.01tis a model for exponential decay of 50 grams of a radioactive element that decays at a rate of 1% per year.## Summary

A quantity is said to be subject toexponential decay/growthif it decreases/increases at a rate proportional to its value.## Formulas

## Examples

ProblemIt is often assumed that 1/3 acre of land is needed to provide food for one person. It is also estimated that there are 10 billion acres of arable land in the world, and therefore a maximum population of 30 billion people can be sustained if no other sources of food are known. The total world population at the beginning of 1970 was 3.6 billion.(a) Assuming that the population continues to increase at the rate of 2 percent per year, what will the population be in the year 2000? When will the maximum population be reached?

(b) What allows us to use the exponential growth and decay model

y=Cektin his problem?Solution(a) In the exponential growth and decay modely=Cekt, the numberCrepresents initial amount, or 3.6 billion=3,600,000,000 (initial becauset=0, which in this case is 1970). We know that a 2% growth occurs annually, so 1971 results in a population ofN*1.02, or 3,672,000,000. Plugging that number in foryand putting ourCwhere it belongs, we can solve fork. In this example,k=.019802627296. Barring a discussion of significant digits, we know that we like our answer to be as accurate as possible for the AP graders; round-off done incorrectly will likely cause inaccuracy. Now, to figure out population in 2000, we say thatt= 30, since 1970 + 30 = 2000 and plug in our values forC(always the initial amount in this example) andk(our constant of proportionality). We easily find that the population in the year 2000 will be approximately 6,520,901,703 (quite a bit higher than our population in 1970).In order to figure out when our population maxes out, we plug in 30,000,000,000 for

y(which represents the result variable--what happens in our model once we figure in the rest), and ourCandkas before. Solving for t gives us a startling answer oft= 107. Ift= 0 corresponds to 1970, this answer corresponds to 2077. Don't forget units! Since this will likely happen at night when I am sleeping and will wake to find my refrigerator empty, the correct units are p.m. (just kidding).(b) We are allowed to use the exponential growth and decay model because our variable in question (population) is proportional to

the rate of changeof itself. In simpler words, the more people there are, the more people will be produced.## Practice

Problems with Hints and Answers!!Problems 1 <---4 exponential growth questions

Problems 2 <---4 exponential decay questions

Problems 3 <---4 continually compounded interest questions

Problems 4 <---4 compound interest questions

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