# 6.02 Applications of exponential functions

Lesson

## 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.

Remember!

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 $00<b<1, it is decay •$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$$48000. It loses 18%18% of its value per year. We want to find its value for the next 44 years. Let's use a table to do so. Year Amount Lost Value () Expression for Value 00 n/a 4800048000 4800048000 11 0.18\times48000=86400.18×48000=8640 48000-8640=39360480008640=39360 48000-48000\times0.18=48000\left(1-0.18\right)4800048000×0.18=48000(10.18) 22 0.18\times39360=7084.800.18×39360=7084.80 39360-7084.80=32275.20393607084.80=32275.20 48000\left(1-0.18\right)^248000(10.18)2 33 0.18\times32275.20=5809.540.18×32275.20=5809.54 32275.20-5809.54=26465.6632275.205809.54=26465.66 48000\left(1-0.18\right)^348000(10.18)3 44 0.18\times26465.66=4763.820.18×26465.66=4763.82 26465.66-4763.82=21701.8426465.664763.82=21701.84 48000\left(1-0.18\right)^448000(10.18)4 Rather than continuing to work through each year in this way, we could approach the problem differently by looking at the column for the Expression for value. Notice that multiplying by 18%18% and then subtracting that from the current value is the same as multiplying by \left(1-0.18\right)=0.82(10.18)=0.82. Since we are doing this repeatedly, we can make a formula. In fact, for exponential growth and decay by a percentage increase or decrease, we can use the formula below: Exponential growth and decay formula A=P\left(1+r\right)^nA=P(1+r)n where: • AA is the resulting amount after the growth/decay • PP is the initial amount • rr is the rate of change as a decimal, and indicates growth if r>0r>0, and decay if r<0r<0 • nn is the number of time periods that has passed from the start The frequency of rr and nn must match! So if rr is yearly, then nn must be in years, but if rr is a weekly rate, then nn must be in weeks. #### Worked examples ##### Question 1 Justine bought an oil painting for \1600$$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(10.06)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 1010 years. ##### Question 2 A certain radioactive isotope decays in such a way that after 175175 years only half of the original quantity of the isotope remains. Suppose 1010 kg of the substance existed initially. Create an equation for the amount of the isotope left after tt years. Think: When given a worded problem, it help to list what we were given and what we are looking for. AA: This will be an unknown based on the number of years, tt PP: We initially have 1010 kg, so P=10P=10 rr: This is the rate of decay, which we can say is 50%50% if we are looking at the half life nn: This will be a filled in based on the time, we need to remember that the rate is in groups of 175175 years, so nn must be in groups of 175 years Do After 175175 years, only 55 kg will be left and then after a further 175175 years, only 2.52.5 kg will be left, and so on. After nn groups of 175175 years, we have that: A=10\times\left(\frac{1}{2}\right)^nA=10×(12)n The number of years that have elapsed since the beginning of this experiment must be t=175nt=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 175175 years but can use just years instead. Thus, we arrive at a formula for the amount remaining after tt 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 10001000 years? #### Practice questions ##### Question 3 The population of a particular mining town increased 160%160% in 99 years, from 51005100 in 2004 to 1326013260 in 2013. Assuming that the population increased at a constant annual rate, answer the following. 1. Find an expression for AA, the size of the population yy years after 2004. Write the expression such that it includes the annual rate of growth, correct to four decimal places. 2. 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)^tf(t)=87(38)t, where tt represents time. 1. What is the initial value of the function? 2. Express the function in the form f\left(t\right)=\frac{8}{7}\left(1-r\right)^tf(t)=87(1r)t, where rr is a decimal. 3. Does the function represent growth or decay of an amount over time? decay A growth B decay A growth B 4. What 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 PP is invested into an account with an annual interest rate of rr, 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_0A0​ == PP The amount in the account is the original investment. 1 A_1A1​ == 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=PA0​=P 2 A_2A2​ == 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)^2P(1+r)2 Simplify 3 A_3A3​ == 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)^2A2​=P(1+r)2 == P\left(1+r\right)^3P(1+r)3 Simplify The pattern continues so that by tt years, the account has a balance of A\left(t\right)=P\left(1+r\right)^tA(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 nn be the number of times the interest is compounded each year, then we have the following formula: Compound interest formula A=P\left(1+\frac{r}{n}\right)^{nt}A=P(1+rn)nt where: AA is the balance in the account after tt years PP is the principal amount rr is the interest rate (this is usually expressed as an annual rate) tt is the total duration of the investment (in years) nn 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. Continuous compound interest If a principal amount PP is invested at an annual interest rate r and compounded continuously, then the balance AA 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. 1. What is the value of the investment after$2$2 years, to the nearest cent? 2. What is the amount of interest earned to the nearest cent? ##### Question 6 A$\$70000$$70000 investment earns interest at 4%4% p.a., compounded semiannually over 2020 years. Using the compound interest table, calculate: Total amount from a \1000$$1000 investment using compound interest. Periods Interest rate per period$35$35$40$40$45$45$50$50$55$55$1%$1%$1417.83$1417.83$1490.34$1490.34$1566.55$1566.55$1646.67$1646.67$1730.88$1730.88$2%$2%$2006.76$2006.76$2216.72$2216.72$2448.63$2448.63$2704.81$2704.81$2987.80$2987.80$3%$3%$2835.46$2835.46$3290.66$3290.66$3818.95$3818.95$4432.05$4432.05$5143.57$5143.57$4%$4%$3999.56$3999.56$4875.44$4875.44$5943.13$5943.13$7244.65$7244.65$8831.18$8831.18$5%$5%$5632.10$5632.10$7209.57$7209.57$9228.86$9228.86$11813.72$11813.72$15122.56$15122.56 1. the value of this investment, correct to the nearest cent. 2. the amount of interest earned, correct to the nearest cent. ##### 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$N=4385 cells. Since we can't have part of a cell we need to round down to the nearest whole number. ### Applications with base e Exponential functions arise when the rate of change of a variable depends on the current level of the variable. The exponential function then predicts what the level of the variable will be after a given interval of time. Several typical examples of this phenomenon are often given: 1. the rate at which a hot object cools depends on how hot it is in comparison with its surroundings. (This is called Newton's Law of Cooling.) We can predict the temperature of the object after a time interval$\Delta t$Δt. 2. The rate at which a mass of radioactive material decays depends on the mass of the material present. Radioactive dating techniques depend on knowing how much radioactive material was originally present and how much is currently present. The rate of decay for the isotope in question is known and therefore, the time interval, the age of the material, can be calculated. 3. The number of cells in a biological specimen has a particular doubling time, at least in the initial stages. Thus, the rate of increase depends on the number of cells already present and an exponential function can be constructed that predicts the number of cells that will be present after a given time. #### Practice questions ##### Question 8 Under certain climatic conditions the proportion$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$t, the number of days it takes the initial number of algae to double to the nearest two decimal places. ##### Question 9 The proportion$Q$Q of radium remaining after$t$t years is given by$Q=e^{-kt}$Q=ekt, where$k$k is a constant. 1. After$1679$1679 years, only half the initial amount of radium remains. Solve for$k$k. 2. Solve for$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=Ce^{-kt}$A=Cekt, where$k$k is a positive constant. 1. Solve for the value of$k$k if the amount of radiocarbon present is halved every$5594$5594 years. Leave your answer in exact form. 2. 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.

### Outcomes

#### A.CED.A.2^

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. ^Equations using all available types of expressions, including simple root functions

#### F.IF.B.4'''

For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity. '''Include rational, square root and cube root; emphasize selection of appropriate models.

#### F.IF.C.8'''

Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. '''Include rational and radical; Focus on using key features to guide selection of appropriate type of model function