Exponentials

Hong Kong

Stage 4 - Stage 5

Lesson

Exponential growth or decay of a population happens when the *rate of growth* or the *rate of decay* changes in direct proportion to the current size of the population.

So for example, if at a certain time there are $100$100 rabbits on Macquarie Island, and $500$500 rabbits on Flinders Island, assuming exponential growth, we would expect the actual growth in the rabbit population on Flinders Island to be five times that for Macquarie Island.

As another example, two new cars, one valued at $\$100000$$100000 and one valued at $\$20000$$20000, depreciate exponentially. Then the more expensive car will depreciate five times faster than the cheaper car.

In the two examples, as the population changes, the *rate of change* also changes.

For example, suppose that rabbit populations in general tripled every year. Then on Macquarie Island the count each year would look like $100,300,900,2700,\dots$100,300,900,2700,… etc. Thus the average rate of change in the population measured each year would be $200$200 rabbits/year, $600$600 rabbits/year, $1800$1800 rabbits per year etc. The rate of growth is increasing because the population is increasing.

Again, suppose new cars depreciate at $20%$20% per year. Then a $\$100000$$100000 car would depreciate in the first year at an average rate of $\$20000$$20000 per year. In the second year it would depreciate at only $\$16000$$16000 per year ($20%$20% of $\$80000$$80000 is only $\$16000$$16000), and in the third year the rate would be $\$12800$$12800 per year ($20%$20% of $\$64000$$64000). The rate of depreciation is slowing because the price is decreasing.

The equation $y=100\left(3\right)^x$`y`=100(3)`x`, where $x$`x` is the number of years, describes the growth of the rabbit population on Macquarie Island. So, for $x=1$`x`=1, $y=100\left(3\right)^1=300$`y`=100(3)1=300 and for $x=2$`x`=2, $y=100\left(3\right)^2=900$`y`=100(3)2=900 and for $x=3$`x`=3, $y=100\left(3\right)^3=2700$`y`=100(3)3=2700 etc.

The Flinders Island rabbit population would grow according to the function $y=500\left(3\right)^x$`y`=500(3)`x`.

The cars depreciation is modelled by the functions $y=100000\left(0.80\right)^x$`y`=100000(0.80)`x` and $y=20000\left(0.80\right)^x$`y`=20000(0.80)`x`, because after each year only $80%$80% of the previous year's value exists.

So, for the expensive car, the value in successive years becomes,

for $x=1$`x`=1, $y=100000\left(0.80\right)^1=\$80000$`y`=100000(0.80)1=$80000,

for $x=2$`x`=2, $y=100000\left(0.80\right)^2=\$64000$`y`=100000(0.80)2=$64000 and

for $x=3$`x`=3, $y=100000\left(0.80\right)^3=\$51200$`y`=100000(0.80)3=$51200 etc.

Consider the function given by $y=1000x+4000$`y`=1000`x`+4000, which models the running total of all payments made by a purchaser at the end of the $x$`x`th month on a $\$100000$$100000 vintage Rolls-Royce car.

With $x=0$`x`=0, $y=\$4000$`y`=$4000 is the initial deposit. After a year ($x=12$`x`=12) the buyer will have paid a total of $y=1000\left(12\right)+4000=\$16000$`y`=1000(12)+4000=$16000. In fact, for every year that goes by the buyer will pay an additional $\$12000$$12000, and by the end of the $8th$8`t``h` year ($x=96$`x`=96) the entire amount of $\$100000$$100000 will be paid.

The rate of change of $y$`y` is clearly fixed at $\$1000$$1000 per month. The graph of the function is a straight line with a $y$`y`-intercept of $4000$4000 and a constant gradient of $1000$1000.

The rate of change of the running total does not change throughout the entire payment period.

Suppose however the purchaser was given the option of an alternative payment plan. The company offers a new function $y=4000\left(1.49535\right)^{\frac{x}{12}}$`y`=4000(1.49535)`x`12 representing the running total of all payments that must be made.

The payments per month that build the running total in the new model are constantly changing.

After $1$1 year ($x=12$`x`=12), the running total becomes $y=4000\left(1.49535\right)^{\frac{12}{12}}=5981.40$`y`=4000(1.49535)1212=5981.40.

This means that $\$1981.40$$1981.40 would be paid, in addition to the initial $\$4000$$4000 deposit, in the first year.

At the end of the second year ($x=24$`x`=24), a total of $\$8944.29$$8944.29 would be paid comprised of the $\$4000$$4000 deposit, the $\$1981.40$$1981.40 from the first year and $\$2962.89$$2962.89 for the second year.

Note that the payments due each year are rising. The rate of change of the running total is increasing as expected from the exponential model.

Here are the graphs of the two payment plans:

Exponential growth or decay of a particular quantity (like population, savings, radio-active substances etc) is growth where the rate of change varies according to the size of that quantity at any given time.

Exponential growth or decay is also a type of non-linear growth, but there are plenty of other non-linear growth and decay patterns existing in nature that are not exponential. Quadratic decay of gravitational or light intensity, for example, are examples of non-linear phenomena.

The concept of non-constant rates of change is familiar to us in the acceleration of a car or the fall of an elevator. These ideas become very important in the study of the calculus.

Summary

Exponential growth describes a situation where a value is increasing at a rate proportional to its value. Equations of exponential growth will be in the form $y=a^x$`y`=`a``x`.

Exponential decay describes a situation where a value is decreasing at a rate proportional to its value. Equations of exponential growth will be in the form $y=a^{-x}$`y`=`a`−`x`.

Consider the functions $f\left(x\right)=3x$`f`(`x`)=3`x` and $g\left(x\right)=3^x$`g`(`x`)=3`x`.

Complete the table of values for each function.

$x$ `x`$f\left(x\right)$ `f`(`x`)$g\left(x\right)$ `g`(`x`)$0$0 $\editable{}$ $\editable{}$ $1$1 $\editable{}$ $\editable{}$ $2$2 $\editable{}$ $\editable{}$ $3$3 $\editable{}$ $\editable{}$ $4$4 $\editable{}$ $\editable{}$ Graph the functions on the same set of axes.

Loading Graph...Complete the following sentences:

For every $1$1 unit increase in $x$

`x`, the $y$`y`value of $f\left(x\right)$`f`(`x`) increases by $\editable{}$ unit(s).For every $1$1 unit increase in $x$

`x`, the $y$`y`value of $g\left(x\right)$`g`(`x`) increases by a factor of $\editable{}$.For $x\ge2$

`x`≥2, which function increases at a faster rate?The linear function

AThe exponential function

BFind the following function values:

$f$ `f`$\left(8\right)$(8)$=$= $\editable{}$ $g$ `g`$\left(8\right)$(8)$=$= $\editable{}$ Pauline wants to model a quantity that increases at an increasing rate. Which function would be most appropriate to use?

The linear function

AThe exponential function

B

Consider the linear function $f\left(x\right)=5x+2$`f`(`x`)=5`x`+2 and the exponential function $g\left(x\right)=5\left(3\right)^x$`g`(`x`)=5(3)`x`.

Complete the table of values.

$x$ `x`$f\left(x\right)$ `f`(`x`)$g\left(x\right)$ `g`(`x`)$4$4 $\editable{}$ $\editable{}$ $5$5 $\editable{}$ $\editable{}$ $6$6 $\editable{}$ $\editable{}$ Find the value of $f$

`f`$\left(5\right)-f$(5)−`f`$\left(4\right)$(4).Find the value of $\frac{g\left(5\right)}{g\left(4\right)}$

`g`(5)`g`(4).Form an algebraic expression for $f\left(k+1\right)-f\left(k\right)$

`f`(`k`+1)−`f`(`k`).What does the result of the previous part demonstrate?

Over each $1$1 unit interval, the linear function increases by a constant multiplicative factor.

AOver each $1$1 unit interval, the linear function increases by a constant difference.

BForm an algebraic expression for $\frac{g\left(k+1\right)}{g\left(k\right)}$

`g`(`k`+1)`g`(`k`).What does the result of the previous part demonstrate?

Over each $1$1 unit interval, the exponential function increases by a constant multiplicative factor.

AOver each $1$1 unit interval, the exponential function increases by a constant difference.

B