NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Exponential Growth and Decay

Lesson

Exponential growth or decay occurs when the rate of change of a variable is proportional to the value of the variable itself.

Typical examples of such a relation are:

- compound interest on financial deposits
- radioactive decay
- the cooling of a body that is hot compared with its environment
- the growth of certain microbiological cultures

In all living things, there is a process where new cells are constantly being made to create growth and to replace old cells. One common way these new cells are made is by cell division:

- On day 0, we start with $1$1 cell.
- On day 1, this cell splits into two and forms $2$2 cells.
- On day 2, each of the two cells splits and forms $4$4 cells, and so on.

We have generated the number pattern that is produced by doubling: $1,2,4,8,16,\ldots$1,2,4,8,16,…

We can generalise this elegantly, and say that if $t$`t` represents the number of days that have passed, and $y$`y` represents the number of cells, then $y=2^t$`y`=2`t`.

What we immediately notice is how quickly the number of cells increases. On top of this, the more cells there are the more rapidly the number of cells increases. Another way to describe this is to say that the quantity is **increasing** at a rate that is **proportional to the amount present**. Any time a quantity grows in this way, it is increasing at an exponential rate. This kind of increase is called exponential growth.

In other words, the growth of a quantity $y$`y`, which is written $y'$`y`′ or $\frac{dy}{dt}$`d``y``d``t`, is proportional to $y$`y` itself. If the constant of proportionality is $k$`k`, we can express it mathematically using this equation:

$y'=ky$`y`′=`k``y`

The equation of $y$`y` with initial value $A_0$`A`0 is then given explicitly by this equation:

$y(t)=A_0e^{kt}$`y`(`t`)=`A`0`e``k``t`

We can check this by differentiating both sides of this equation by $t$`t`:

$\frac{dy(t)}{dt}=kA_0e^{kt}$`d``y`(`t`)`d``t`=`k``A`0`e``k``t`

Notice that differentiating this function is the same as multiplying it by $k$`k`, which is the fact represented by the equation $y'=ky$`y`′=`k``y`.

Exponential growth often appears in population models, and we will use a simple model to explore a historical case.

In 1859 a single English settler of Australia released European rabbits onto his property to hunt them for sport. A pair of rabbits is capable of producing two or more offspring every $30$30 days, so in effect the population of rabbits in a particular area doubles every month. If only a single pair of rabbits released by the settler survived and bred like this, then the population would change over the first $6$6 months as shown in the table below.

Months after release | $0$0 | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 | $6$6 |
---|---|---|---|---|---|---|---|

Rabbits | $2$2 | $4$4 | $8$8 | $16$16 | $32$32 | $64$64 | $128$128 |

Under this model, we can see that the number of rabbits isn't just increasing, it is increasing at a faster and faster rate. The more rabbits there are, the faster they grow in number. That is, the number of rabbits is increasing at an exponential rate.

Let's use the mathematical model above, with $y$`y` the population of rabbits and $t$`t` the number of months after their release, and find the correct values of $A_0$`A`0 and $k$`k`. The value of $A_0$`A`0 is the initial value, or the population when they were first released - we can write down $A_0=2$`A`0=2 right away.

To find $k$`k`, we need to use another data point - any one will do. Let's use the fact that after $t=3$`t`=3 months the rabbit population was $y=16$`y`=16. Substitute these values into model equation, together with the value of $A_0$`A`0 we just found:

$16=2e^{k\times3}$16=2`e``k`×3

Dividing both sides by $2$2 and taking the natural logarithm of both sides produces this equation:

$\ln\left(\frac{16}{2}\right)=\ln\left(e^{k\times3}\right)$`l``n`(162)=`l``n`(`e``k`×3)

The logarithm and the exponential cancel each other out on the right-hand side, which allows us to write:

$\ln8=3k$`l``n`8=3`k`

This means that $k=\frac{\ln8}{3}$`k`=`l``n`83, which is $0.6931$0.6931 to four decimal places. Putting it all together we can model the rabbit population using this equation:

$y=2e^{0.6931t}$`y`=2`e`0.6931`t`

Try substituting $t=24$`t`=24 and higher numbers to see how Australia was overwhelmed with millions and millions of rabbits within a decade!

Similarly to exponential growth, any time a quantity is **decreasing** at a rate that is **proportional to the current amount**, the quantity is decreasing at an exponential rate. A rough paraphrase would be "the smaller it gets, the slower it decays". This kind of decrease is called exponential decay. Exponential decay is modelled using the exact same equation as for exponential growth:

$y(t)=A_0e^{kt}$`y`(`t`)=`A`0`e``k``t`

The only change for the decay case is that the value of $k$`k` is **negative**.

Radioactive substances lose mass and release energy over time in a process modelled by exponential decay. The element plutonium-238 decays (via uranium-234) with a half-life of about $88$88 years. This means that every $88$88 years around half of the substance releases its energy and turns into uranium-234 while the other half remains plutonium-238.

If a scientist placed $3$3 kg of plutonium-238 in a vault $60$60 years ago, how much remains today?

Well, we can immediately identify the initial value:

$A_0=3$`A`0=3

To find $k$`k`, we use the fact that half of the plutonium-238 will have decayed $88$88 years after it was placed in the vault - in other words, there would have been $1.5$1.5 kg at this time:

$1.5=3e^{k\times88}$1.5=3`e``k`×88

Just like in the previous example, we divide both sides by $3$3 and take the natural logarithm of both sides:

$\ln\left(\frac{1.5}{3}\right)=\ln\left(e^{k\times88}\right)$`l``n`(1.53)=`l``n`(`e``k`×88)

After cancelling the logarithm and the exponential and rearranging, we find:

$k=\frac{\ln\left(\frac{1}{2}\right)}{88}=-0.0079$`k`=`l``n`(12)88=−0.0079 (4 d. p.)

We can now substitute the values of $A_0$`A`0 and $k$`k` we just found, together with $t=60$`t`=60, into the model equation:

$y=2e^{-0.0079\times60}=1.2450$`y`=2`e`−0.0079×60=1.2450 (4 d.p)

So after $60$60 years there should be about $1245$1245 grams of plutonium-238 remaining in the vault.

Summary of exponential growth and decay

If a quantity $y$`y` undergoes exponential growth or decay, its rate of change $y'$`y`′ is proportional to $y$`y`. This can be expressed as an equation using the constant of proportionality $k$`k`:

$y'=ky$`y`′=`k``y`

The value of $k$`k` is positive for exponential growth, and negative for exponential decay.

The solution to this equation models the quantity directly with respect to the independent variable (usually time $t$`t`), knowing the initial value $A_0$`A`0:

$y=A_0e^{kt}$`y`=`A`0`e``k``t`

Niger's population, $t$`t` years after the year 2000 is modelled by $P=17.7e^{0.0078t}$`P`=17.7`e`0.0078`t`, where $P$`P` is the number of millions of people.

State the population of Niger in the year 2000.

State the continuous growth rate used in this model.

Find Niger's population in 2005.

Give your answer to one decimal place.

Solve for the number of years, $t$

`t`, it will take for Niger's population to reach $22$22 million.Give your answer to one decimal place.

Solve for the number of years, $t$

`t`, it will take for Niger's population to double.Give your answer to one decimal place.

A radioactive isotope decays continuously such that $\frac{dA}{dt}=kA$`d``A``d``t`=`k``A`, where $A$`A` is the amount of isotope remaining after $t$`t` years.

If the rate of decay is $20%$20%, what is the value of $k$

`k`?Initially, $25$25 kg of the isotope were produced in an industrial process.

Write an equation for $A$

`A`, the weight (in kg) of the isotope, at time $t$`t`.Solve for the time, $t$

`t`years, when the instantaneous rate of decay is $1.3$1.3 kg/year.Give your answer to one decimal place.

The population of feral cats in a local council district has been monitored since January 2013.

The population $P$`P` is modelled by $P=6000e^{-0.0018t}$`P`=6000`e`−0.0018`t`, where $t$`t` is the number of years from when the population started being monitored.

State the initial population at the beginning of 2013.

State the continuous decline rate of the number of cats in this district.

Find the population at the beginning of 2018.

Give your answer to the nearest whole number.

Solve for the number of years $t$

`t`it will take for the initial population of cats to halve. Give your answer correct to one decimal place.

Form differential equations and interpret the solutions

Apply differentiation methods in solving problems