3.03 Applications of exponential functions
Exponential growth and decay
An exponential function is the appropriate model to use when a quantity is increasing or decreasing at a rate that depends on the quantity present.
For example, in the final rounds of a sports competition, the number of competing teams is halved at every stage. Thus, if $16$16 teams reached the first semifinal, there would be $8$8 in the second semifinal, and so on. The number of teams playing drops from $16$16 to $8$8 and then to $4$4 and finally, $2$2 and we see that the reduction in the number of teams playing at each stage depends on the number in the previous round.
This is an example of exponential decay. The rate of decrease gets progressively smaller. Many processes show the opposite pattern and exhibit exponential growth. In this, the rate of increase increases progressively.
If we are looking at an exponential function of the form $y=ab^x$y=abx, then
- $b$b is tells us whether the function is growing (increasing) or decaying (decreasing)
- If $b>1$b>1, it is growth
- If $0
- $a$a is the initial value, when $x=0$x=0
One of the most common applications is where there is a percentage growth or a percentage decay. For example the amount in a savings account or the remaining balance on a loan respectively. Let's look at the general formula for these scenarios.
Exploration
A particular sports convertible costs $\$48000$$48000. It loses $18%$18% of its value per year. We want to find its value for the next $4$4 years. Let's use a table to do so.
| Year | Amount Lost | Value ($) | Expression for Value |
|---|---|---|---|
| $0$0 | n/a | $48000$48000 | $48000$48000 |
| $1$1 | $0.18\times48000=8640$0.18×48000=8640 | $48000-8640=39360$48000−8640=39360 | $48000-48000\times0.18=48000\left(1-0.18\right)$48000−48000×0.18=48000(1−0.18) |
| $2$2 | $0.18\times39360=7084.80$0.18×39360=7084.80 | $39360-7084.80=32275.20$39360−7084.80=32275.20 | $48000\left(1-0.18\right)^2$48000(1−0.18)2 |
| $3$3 | $0.18\times32275.20=5809.54$0.18×32275.20=5809.54 | $32275.20-5809.54=26465.66$32275.20−5809.54=26465.66 | $48000\left(1-0.18\right)^3$48000(1−0.18)3 |
| $4$4 | $0.18\times26465.66=4763.82$0.18×26465.66=4763.82 | $26465.66-4763.82=21701.84$26465.66−4763.82=21701.84 | $48000\left(1-0.18\right)^4$48000(1−0.18)4 |
Rather than continuing to work through each year in this way, we could approach the problem differently by looking at the column for the Expression for value.
Notice that multiplying by $18%$18% and then subtracting that from the current value is the same as multiplying by $\left(1-0.18\right)=0.82$(1−0.18)=0.82. Since we are doing this repeatedly, we can make a formula.
In fact, for exponential growth and decay by a percentage increase or decrease, we can use the formula below:
$A=P\left(1+r\right)^n$A=P(1+r)n
where:
- $A$A is the resulting amount after the growth/decay
- $P$P is the initial amount
- $r$r is the rate of change as a decimal, and indicates growth if $r>0$r>0, and decay if $r<0$r<0
- $n$n is the number of time periods that has passed from the start
The frequency of $r$r and $n$n must match! So if $r$r is yearly, then $n$n must be in years, but if $r$r is a weekly rate, then $n$n must be in weeks.
Worked examples
Question 1
Justine bought an oil painting for $\$1600$$1600 whose value is given by the formula $A=1600\times0.94^t$A=1600×0.94t, and $t$t is the number of years passed.
a) What's the annual depreciation rate?
Think: What's $r$r in this case?
Do:
$1600\times0.94^t=1600\left(1-0.06\right)^t$1600×0.94t=1600(1−0.06)t
so the annual depreciation rate is $0.06=6%$0.06=6%.
b) How much would the painting be worth in 10 years, round to the nearest dollar.
Think: What parameters do we have, which one are we looking for. It can be helpful to list values before jumping into the formula. For the equation, $A=1600\times0.94^t$A=1600×0.94t, we are given $t=10$t=10 and are looking for $A$A.
Do:
| $A$A | $=$= | $1600\times0.94^t$1600×0.94t |
| $=$= | $1600\times0.94^{10}$1600×0.9410 | |
| $=$= | $861.7841826$861.7841826 |
So the painting would be worth $\$862$$862 in $10$10 years.
Question 2
A certain radioactive isotope decays in such a way that after $175$175 years only half of the original quantity of the isotope remains. Suppose $10$10 kg of the substance existed initially. Create an equation for the amount of the isotope left after $t$t years.
Think: When given a worded problem, it help to list what we were given and what we are looking for.
$A$A: This will be an unknown based on the number of years, $t$t
$P$P: We initially have $10$10 kg, so $P=10$P=10
$r$r: This is the rate of decay, which we can say is $50%$50% if we are looking at the half life
$n$n: This will be a filled in based on the time, we need to remember that the rate is in groups of $175$175 years, so $n$n must be in groups of 175 years
Do:
After $175$175 years, only $5$5 kg will be left and then after a further $175$175 years, only $2.5$2.5 kg will be left, and so on. After $n$n groups of $175$175 years, we have that:
$A=10\times\left(\frac{1}{2}\right)^n$A=10×(12)n
The number of years that have elapsed since the beginning of this experiment must be $t=175n$t=175n. So, we can put $n=\frac{t}{175}$n=t175 into the formula so that we no longer have to convert the time elapsed into groups of $175$175 years but can use just years instead.
Thus, we arrive at a formula for the amount remaining after $t$t years:
$A=10\times\left(\frac{1}{2}\right)^{\frac{t}{175}}$A=10×(12)t175
Reflect: How could we use this equation? How much would be left after $1000$1000 years?
Practice questions
Question 3
The population of a particular mining town increased $160%$160% in $9$9 years, from $5100$5100 in 2004 to $13260$13260 in 2013. Assuming that the population increased at a constant annual rate, answer the following.
Find an expression for $A$A, the size of the population $y$y years after 2004. Write the expression such that it includes the annual rate of growth, correct to four decimal places.
Hence state the annual rate of growth. Give the rate as a percentage correct to two decimal places.
question 4
Consider the function $f\left(t\right)=\frac{8}{7}\left(\frac{3}{8}\right)^t$f(t)=87(38)t, where $t$t represents time.
What is the initial value of the function?
Express the function in the form $f\left(t\right)=\frac{8}{7}\left(1-r\right)^t$f(t)=87(1−r)t, where $r$r is a decimal.
Does the function represent growth or decay of an amount over time?
decay
Agrowth
BWhat is the rate of decay per time period? Give the rate as a percentage.
Compound interest
We can extend the definition of exponential growth to the financial application of compound interest.
Exploration
Suppose a principal amount of money $P$P is invested into an account with an annual interest rate of $r$r, and the interest is compounded or reinvested every year. This means that at the end of each year, the interest earned is added to the account and will be included in the amount of money that earns interest in the future.
| Number of years passed | Account balance after compounding | |||
| 0 | $A_0$A0 | $=$= | $P$P |
The amount in the account is the original investment. |
| 1 | $A_1$A1 | $=$= | $A_0\left(1+r\right)$A0(1+r) |
Interest from the previous year is calculated and added to the account. |
| $=$= | $P\left(1+r\right)$P(1+r) |
$A_0=P$A0=P |
||
| 2 | $A_2$A2 | $=$= | $A_1\left(1+r\right)$A1(1+r) |
Interest from the previous year is calculated and added to the account. |
| $=$= | $P\left(1+r\right)\left(1+r\right)$P(1+r)(1+r) |
$A_1=P\left(1+r\right)$A1=P(1+r) |
||
| $=$= | $P\left(1+r\right)^2$P(1+r)2 |
Simplify
|
||
| 3 | $A_3$A3 | $=$= | $A_2\left(1+r\right)$A2(1+r) |
Interest from the previous year is calculated and added to the account.
|
| $=$= | $P\left(1+r\right)^2\left(1+r\right)$P(1+r)2(1+r) |
$A_2=P\left(1+r\right)^2$A2=P(1+r)2 |
||
| $=$= | $P\left(1+r\right)^3$P(1+r)3 |
Simplify |
The pattern continues so that by $t$t years, the account has a balance of $A\left(t\right)=P\left(1+r\right)^t$A(t)=P(1+r)t.
In financial applications, sometimes the interest is compounded in quarterly, monthly, or even daily. In that case, we can make slight adjustments to the formula. If we let $n$n be the number of times the interest is compounded each year, then we have the following formula:
$A=P\left(1+\frac{r}{n}\right)^{nt}$A=P(1+rn)nt
where:
$A$A is the balance in the account after $t$t years
$P$P is the principal amount
$r$r is the interest rate (this is usually expressed as an annual rate)
$t$t is the total duration of the investment (in years)
$n$n is the number of times the investment is compounded each year
In some instances, interest is compounded continuously. In that case, we can use the base e to denote an infinite number of compounding periods. This leads to the following formula.
If a principal amount $P$P is invested at an annual interest rate r and compounded continuously, then the balance $A$A in the account after t years is given by the formula:
$A=Pe^{rt}$A=Pert
Practice questions
Question 5
Valentina's investment of $\$8000$$8000 earns interest at $2%$2% p.a., compounded semiannually over $2$2 years. Answer the following questions by repeated multiplication.
What is the value of the investment after $2$2 years, to the nearest cent?
What is the amount of interest earned to the nearest cent?
Question 6
Question 7
The price of goods is rising at $0.75%$0.75% per quarter. If $1$1kg of chicken costs $\$12.10$$12.10 today, what is the price expected to be in $2$2 years to the nearest cent?
Population growth
Some questions may not explicitly state a percentage increase or decrease, but do follow an exponential model.
The key idea is that the value we are repeatedly multiplying by will be the base.
Exploration
In a biological experiment, a sample of a microorganism was seen to double in size by cell division every $11$11 minutes. There were initially $100$100 cells. We want to come up with an equation to model this scenario.
We can figure out how many cells there will be after several $11$11-minute periods by a step-by-step calculation if we know how many cells were present initially.
After $11$11 minutes there are $2\times100=200$2×100=200 cells. Then, after a further $11$11 minutes, there are $2\times2n$2×2n cells. After a third period of $11$11 minutes, there are $2\times2\times100=400$2×2×100=400 cells. This pattern continues and we see that after $k$k groups of $11$11 minutes, there will be $2^k\times100$2k×100 cells.
The researcher observes that the increase occurs gradually and not in sudden bursts at the end of each $11$11 minutes. The number of minutes that have elapsed since the beginning of the experiment is $m=11k$m=11k. So, it seems reasonable to put $k=\frac{m}{11}$k=m11 and write down a formula that will give the number of cells $N$N after $m$m minutes, for any number of minutes, not just multiples of $11$11. This is
$N=100\cdot2^{\frac{m}{11}}$N=100·2m11.
Suppose the growth of the microorganism was observed for a hour ($60$60 minutes). How many cells would there be at the end of this time?
| $N$N | $=$= | $100\cdot2^{\frac{m}{11}}$100·2m11 |
| $=$= | $100\cdot2^{\frac{60}{11}}$100·26011 | |
| $=$= | $4385$4385 |
On substituting these numbers into the formula, we calculate $N=4385$N=4385 cells. Since we can't have part of a cell we need to round down to the nearest whole number.
Applications with base e
Exponential functions arise when the rate of change of a variable depends on the current level of the variable. The exponential function then predicts what the level of the variable will be after a given interval of time.
Several typical examples of this phenomenon are often given:
- the rate at which a hot object cools depends on how hot it is in comparison with its surroundings. (This is called Newton's Law of Cooling.) We can predict the temperature of the object after a time interval $\Delta t$Δt.
- The rate at which a mass of radioactive material decays depends on the mass of the material present. Radioactive dating techniques depend on knowing how much radioactive material was originally present and how much is currently present. The rate of decay for the isotope in question is known and therefore, the time interval, the age of the material, can be calculated.
- The number of cells in a biological specimen has a particular doubling time, at least in the initial stages. Thus, the rate of increase depends on the number of cells already present and an exponential function can be constructed that predicts the number of cells that will be present after a given time.
Practice questions
Question 8
Under certain climatic conditions the proportion $P$P of the current blue-green algae population to the initial population satisfies the equation $P=e^{0.007t}$P=e0.007t, where $t$t is measured in days from when measurement began.
Solve for $t$t, the number of days it takes the initial number of algae to double to the nearest two decimal places.
Enter each line of work as an equation.
Question 9
The proportion $Q$Q of radium remaining after $t$t years is given by $Q=e^{-kt}$Q=e−kt, where $k$k is a constant.
After $1679$1679 years, only half the initial amount of radium remains.
Solve for $k$k.
Solve for $t$t, the number of years it takes for only $10%$10% of the initial amount of radium to remain to the nearest two decimal places.
Question 10
The remains of a human body can be dated by measuring the proportion of radiocarbon in tooth enamel.
The proportion of radiocarbon $A$A remaining $t$t years after a human passes away is given by $A=e^{-kt}$A=e−kt, where $k$k is a positive constant.
Solve for the value of $k$k if the amount of radiocarbon present is halved every $5594$5594 years.
Enter each line of work as an equation. Leave your answer in exact form.
For a particular corpse, the amount of radiocarbon present is only $25%$25% of the original amount at death. How many years ago, $t$t, did the person pass away?
Give your answer to two decimal places.