Contrasting Growth and Decay

In real-life settings, there exists phenomena where quantities either grow or decay exponentially. Typically, these are situations like uninhibited population growth, radioactive decay, or compound interest. Studying these in terms of growth and decay helps understand their behavior in the long run.

Rule

Exponential Growth and Decay

Concept

Exponential Growth

Exponential growth occurs when a quantity increases by the same factor over equal intervals of time. This leads to an exponential function, where the independent variable in the exponent, $t,$ is time. $y = a \cdot b^{t}$ Since the quantity increases over time, the constant multiplier $b$ has to be greater than $1 .$ Thus, the growth factor $b$ can be expressed as $b = 1 + r,$ where $r$ is some positive number. The resulting function is called an exponential growth function.

The constant $r$ can then be interpreted as the rate of growth, in decimal form. A value of $0.06,$ for instance, means that the quantity increases by $6 %$ over every unit of time. As is the case with all exponential functions, $a$ is the $y$ -coordinate of the $y$ -intercept.

Since the growth factor is greater than $1,$ the quantity grows faster and faster, without bound.

Concept

Exponential Decay

The counterpart of exponential growth is exponential decay; when a quantity decreases by the same factor over equal intervals of time, the constant multiplier of the exponential decay function is less than $1 .$ This factor can be expressed as $(1 - r)$ and is known as the decay factor.

The constant $r$ can then be interpreted as the rate of decay, in decimal form. A value of $0.12,$ for instance, would mean that the quantity decreases by $12 %$ over every unit of time.

Since the decay factor is smaller than

1,

the quantity decays toward

0

over time.

Find the rule of the exponential growth function

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Exercise

In an ideal environment, bacteria populations grow exponentially and can be modeled with an exponential growth function. The bacteria Lactobacillus acidophilus duplicates about once every hour. A single bacteria cell is placed in an ideal environment. State and interpret the constants $a$ and $r$ for the growth that will occur. Then, write a function rule describing this growth.

Show Solution

Solution

The constant $a$ is the initial value of the quantity, in this case the number of bacteria. There was only one bacteria placed in the environment, so $a$ is $1 .$ The constant $r$ is the rate of growth. Since the bacteria duplicate every hour, the amount of bacteria doubles every hour. This corresponds to an increase by $100 % .$ Thus, $r$ is $1 .$ Substituting this into the rule of an exponential growth function gives $P (t) = 1 \cdot (1 + 1)^{t},$ which can be simplified as $P (t) = 2^{t} .$ Since this is an exponential growth function, population will grow faster and faster, without bound. In a real environment, this would not happen, since the available space and nutrition would have to be infinite. At some point, the environment would no longer be ideal, so the growth would slow down or stop.

Method

Growth and Decay Factor

For an exponential growth function written as $y = a \cdot (1 + r)^{t},$ the quantity $(1 + r)$ is called the growth factor. Likewise, for an exponential decay function, $y = a \cdot (1 - r)^{t},$ $(1 - r)$ is called the decay factor. In context, it is often the rate of growth or decay, $r,$ that is given. Using $r,$ it is possible to evaluate the corresponding growth or decay factor. As an example, a growth of $10 %$ every unit of time gives the growth factor $(1 + r) = (1 + 0.1) = 1.1 .$ Similarly, it can be necessary to use the growth or decay factor to find the rate of growth or decay. For instance, a factor of $0.85$ indicates a decay. Thus, $r$ can be found by equating $0.85$ with the decay factor $(1 - r),$ and solving the equation.

(1 - r) = 0.85 \Leftrightarrow r = 0.15

Graph the exponential growth or decay function

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Exercise

During a time period, the number of carps in a small lake can be modeled by the function $H (t) = 800 \cdot 0.8 8^{t},$ where $t$ is the time in years. State whether the function shows growth or a decay, and then find the rate of growth or decay, $r .$ Finally, graph the function.

Show Solution

Solution

To begin, let's analyze the given function rule. It's written in the form $y = a \cdot b^{x},$ where $a$ is the initial value and $b$ is constant multiplier/growth factor. The constant multiplier, $0.88,$ is less than $1,$ so it is a decay factor. Therefore, the function shows decay. Since the decay factor is always equal to $1 - r,$ we can write the equation. $0.88 = 1 - r,$ which can be solved for $r .$

0.88 = 1 - r

AddEqn

LHS + r = RHS + r

0.88 + r = 1

SubEqn

LHS - 0.88 = RHS - 0.88

r = 0.12

Thus, the rate of decay is $0.12,$ or $12 %,$ per year. The initial value is $800,$ and the constant multiplier is $0.88 .$ Using this information, we can graph the exponential decay function by plotting some points that lie on $H$ and connecting them with a smooth curve.

Concept

Compound Interest

When money is deposited to a savings account, interest is accrued, often yearly. Different types of interest work in different ways. When the interest earned is then added to the original amount, future interest accrues for a larger amount. This is called compound interest. To calculate the balance on the account at a specific time, an exponential growth function can be used. When the interest is compounded yearly, the balance can be modeled with a function.

$y = P (1 + r)^{t}$

In this context, $P$ stands for the principal, which is the initial amount of money, and $r$ is the interest rate in decimal form. If the interest is not compounded yearly, the function looks a little different.

$y = P (1 + \frac{r}{n})^{n t}$

The constant $n$ is the number of times the interest is compounded per year, while $r$ is still the annual interest rate. For an account with the principal $ $100$ and an annual interest of $15 %$ compounded twice a year, the growth function is:

$B (t) = 100 (1 + \frac{0 . 1 5}{2})^{2 t}$

Notice that this function grows continuously, whereas, in reality, the account balance only increases at the times of compound. Graphing the function together with the actual balance of the account will highlight how it can be used in practice.

Every time the interest is compounded, in this case every half year, the value of

B

is equal to the account balance. However, at all other times, it is not. To find, for instance, the account balance after

1.75

years,

B (1.5)

should then be evaluated, since that was the last time interest compounded.

Find the monthly interest rate

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Exercise

One savings account, with a principal of $ $100,$ offers an annual interest rate of $15 %$ compounded twice a year. Find the balance in the account after $5$ years. Another savings account with the same principal will have the same balance after $5$ years. However, the interest is compounded monthly. Find the interest rate of the second account.

Show Solution

Solution

First, we'll find the function rule describing the growth of the first account. It is given that $P = 100, r = 0.15,$ and $n = 2 .$ Substituting these values in the compound interest formula gives $B (t) = 100 (1 + \frac{0 . 1 5}{2})^{2 t} .$ Let's simplify this function before continuing.

B (t) = 100 (1 + \frac{0 . 1 5}{2})^{2 t}

CalcQuotCalculate quotient

B (t) = 100 (1 + 0.075)^{2 t}

AddTermsAdd terms

B (t) = 100 \cdot 1.07 5^{2 t}

Since the interest will accrute for

5

years,

t = 5 .

Therefore, we can find the account balance by evaluating

B (5) .

B (t) = 100 \cdot 1.07 5^{2 t}

Substitute

t = 5

B (5) = 100 \cdot 1.07 5^{2 \cdot 5}

MultiplyMultiply

B (5) = 100 \cdot 1.07 5^{10}

UseCalcUse a calculator

B (5) = 206.10315 \dots

RoundDecRound to

B (5) \approx 206.10

The account balance is $ $206.10$ after $5$ years. Now we can consider the second account. We know the balance of both accounts is equal, at least when both just had their interest compounded. This means that $B (t)$ also describes the growth in the second account. However, since the interest in the second account accrues monthly, or $12$ times a year, $n = 12 .$ Thus, the exponent in the rule should be $12 t .$ By using the equality $2 t = 1 / 6 \cdot 12 t$ and the power of a power property, we can rewrite $B (t)$ so that it's possible to find the monthly interest rate. Having the exponent $12 t$ means that the base of the power is equal to the monthly growth factor, which we can then use to find the monthly interest rate.

B (t) = 100 \cdot 1.07 5^{2 t}

RewriteRewrite

2 t

as

1 / 6 \cdot 12 t

B (t) = 100 \cdot 1.07 5^{1 / 6 \cdot 12 t}

ProdInExponent

a^{m \cdot n} = (a^{m})^{n}

B (t) = 100 \cdot (1.07 5^{1 / 6})^{12 t}

UseCalcUse a calculator

B (t) = 100 \cdot (1.01212 \dots)^{12 t}

RoundDecRound to

B (t) \approx 100 \cdot 1.01 2^{12 t}

We find an approximate monthly growth factor $1.012,$ which corresponds to a rate of growth that is $0.012 .$ Thus, the monthly interest rate is roughly $1.2 % .$

Contrasting Growth and Decay

Exponential Growth and Decay

Exponential Growth

Exponential Decay

Example

Find the rule of the exponential growth function

Growth and Decay Factor

Example

Graph the exponential growth or decay function

Compound Interest

Example

Find the monthly interest rate