Linear and exponential growth

A common constant in our life is that things keep changing. Often, these changes can be described by some rule. Two such simple rules are linear and exponential growth, which we will discuss now.

We start by measuring how something changes over time by first deciding on a fixed time step $\tau$ (in minutes, day, years, ...), and we then make measurements every $\tau$ , starting at some point in time $x_0$ . Thus we get a sequence of measurements

\begin{array}{rllll} \text{measurements}:& y_0 &\xrightarrow[]{} & y_1 &\xrightarrow[]{} & ... & \xrightarrow[]{} & y\\ \text{time}:& x_0 & \xrightarrow[]{+\tau} & x_1 &\xrightarrow[]{+\tau} & ... &\xrightarrow[]{+\tau} & x\\ \end{array}

where $y_0$ is the first measurement at time $x_0$ , $y_1$ is the second measurement at time $x_1$ (the time $\tau$ later), and so on. Time $x$ denotes an arbitrary point in time, and $y$ is the corresponding measurement. Generally we want to know is there is a rule to determine $y$ for any time $x$ .

Example 1

A study looks at the growth of a forest. In the year $x_0=2010$ the forest contains $y_0=500$ trees. Every $\tau=10$ years the number of trees in the forest are counted. In year $x_1=2020$ there are $y_1=600$ trees, and another ten years later, in $x_2=2030$ , there are $y_3=700$ trees, and so on.

\begin{array}{rllll} \text{trees}:& 500 &\xrightarrow[]{+100} & 600 &\xrightarrow[]{+100}& 700 &\xrightarrow[]{+100} & ... & \xrightarrow[]{+100} & y\\ \text{year}:& 2010 & \xrightarrow[]{+10} & 2020 &\xrightarrow[]{+10} & 2030 &\xrightarrow[]{+10} & ... &\xrightarrow[]{+10} & x\\ \end{array}

Is there a rule according to which the number of trees in a forest grow, that is, is the a rule which let's us determine the number of trees $y$ for any time $x$ ? For example, how many trees will there be in the year $x=4040$ ? Of course we need more data, but just based on the three data points we have, we might guess that the rule is that every $10$ years $100$ new trees are added to the forest. This is an example of a quantity that grows linearly.

Let us define linear growth more formally:

Definition 1

A quantity grows linearly, if it increases (or decreases) after every time step $\tau$ by the same amount $u$ :

\begin{array}{rllll} \text{measurements}:& y_0 &\xrightarrow[]{+u} & y_1 &\xrightarrow[]{+u} & ... & \xrightarrow[]{+u} & y\\ \text{time}:& x_0 & \xrightarrow[]{+\tau} & x_1 &\xrightarrow[]{+\tau} & ... &\xrightarrow[]{+\tau} & x\\ \end{array}

If $u>0$ then the quantity is increasing, and $u$ is called the increment. If $u<0$ then the quantity decreases (or decays), and $u$ is called the decrement. The term linear growth can be used in both situations, but in the latter case we can also be more specific and say linear decay.

So what is exponential growth? Let us return to our forest example.

Example 2

A study of another forest finds the following: in the year $x_0=2010$ the forest contains $y_0=500$ trees. Then years later, in the year $x_1=2020$ there are $y_1=1000$ trees, and another ten years later, in year $x_2=2030$ , there are $y_2=2000$ trees, and so on.

\begin{array}{rllll} \text{trees}:& 500 &\xrightarrow[]{\cdot 2} & 1000 &\xrightarrow[]{\cdot 2}& 2000 &\xrightarrow[]{\cdot 2} & ... & \xrightarrow[]{\cdot 2} & y\\ \text{year}:& 2010 & \xrightarrow[]{+10} & 2020 &\xrightarrow[]{+10} & 2030 &\xrightarrow[]{+10} & ... &\xrightarrow[]{+10} & x\\ \end{array}

So every $10$ years the number of trees double. This is an example of a quantity that grows exponentially.

Here is the formal definition:

Definition 2

A quantity grows exponentially, if it increases (or decreases) after every time step $\tau$ by the same factor $u$ :

\begin{array}{rllll} \text{measurements}:& y_0 &\xrightarrow[]{\cdot u} & y_1 &\xrightarrow[]{\cdot u} & ... & \xrightarrow[]{\cdot u} & y\\ \text{time}:& x_0 & \xrightarrow[]{+\tau} & x_1 &\xrightarrow[]{+\tau} & ... &\xrightarrow[]{+\tau} & x\\ \end{array}

$u$ is called the growth factor. Note that if $u>1$ then the quantity is increasing. If $0<u<1$ , then the quantity decreases (or decays). The term exponential growth can be used in both situations, but in the latter case we can also be more specific and say exponential decay ( $u$ is then the decay factor).

Formulas for linear and exponential growth

Let us derive a formula for calculating the quantity $y$ at any given time $x$ for linear and exponential growth. In both cases we first find the number of steps of size $\tau$ we need to get from the time of the first measurement $x_0$ to time $x$ . If we denote this number by $n$ , we have

n=\frac{x-x_0}{\tau}

Indeed, $x-x_0$ is the distance from $x_0$ to $x$ , and from $n\cdot \tau=x-x_0$ we get the formula above. Now, at $x_0$ the quantity is $y_0$ , and each step in time the quantity grows by the amount $u$ . In case of linear growth, each step adds the amount $u$ :

\begin{array}{lll} y&=&y_0+\underbrace{u+u+...+u}_{n \text{ steps}}\\ &=&y_0+n\cdot u\\ &=&y_0+u\cdot \frac{x-x_0}{\tau} \end{array}

and for exponential growth, each step multiplies by $u$ :

\begin{array}{lll} y&=&y_0\cdot\underbrace{u\cdot u\cdot ... \cdot u}_{n \text{ steps}}\\ &=&y_0\cdot u^n\\ &=&y_0\cdot u^{(x-x_0)/\tau} \end{array}

Thus, we have the following:

Theorem 1: Linear and exponential growth

A quantity is $y_0$ at time $x_0$ , and each time step $\tau$ the quantity is measured. If the quantity grows linearly by the amount $u$ , then the quantity at time $x$ is given by the linear function

f(x)=y_0+u\cdot \frac{x-x_0}{\tau}

If the quantity grows exponentially by the factor $u$ , then the quantity at time $x$ is given by the exponential function

f(x)= y_0\cdot u^{(x-x_0)/\tau}

Here are two examples which show again how we can derive the formulas.

Example 3

Try to solve the problems below in the same way as was shown above, that is, determine first the number of steps $n$ you need, and then by how much you have to add or multiply after each step.

Each second week, you get $20$ CHF from your parents. You have $15$ CHF in week 1. How much money do you have
1. in week $27$ ?
2. in week $x$ ?
Starting from $6$ cells (at $0$ hours), each cell in a growing human divides on average every $20$ hours. How many cells are there
1. after $360$ hours?
2. after $x$ hours?

Solution

It is $\tau=2$ weeks, $u=20$ CHF, $x_0=1$ week, and $y_0=15$ CHF.
$\begin{array}{rllll} \text{money}:& 15 &\xrightarrow[]{+20} & 35 &\xrightarrow[]{+20} & ... & \xrightarrow[]{+20} & y\\ \text{week}:& 1 & \xrightarrow[]{+2} & 3 &\xrightarrow[]{+2} & ... &\xrightarrow[]{+2} & x\\ \end{array}$
1. The number of steps to get from week $x_0=1$ to week $x=27$ is $n=(27-1)/2=13$ steps. In each step we add $20$ CHF, the amount of CHF in week $27$ is
  $y=15+\underbrace{20 + 20 + ... + 20}_{n=13 \text{ times}} =15 + 20 \cdot 13 = \underline{275}$
2. The number of steps to get from week $1$ to week $x$ is $n=(x-1)/2$ , so the amount of money in week $x$ is
  $y=15+\underbrace{20 + 20 + ... + 20}_{n=(x-1)/2 \text{ times}} = 15 + 20\cdot \frac{x-1}{2}=\underline{10x+5}$
It is $\tau=20$ hours, $u=2$ , $x_0=0$ hour, and $y_0=6$ cells.
$\begin{array}{rllll} \text{cells}:& 6 &\xrightarrow[]{\cdot 2} & 12 &\xrightarrow[]{\cdot 2} & ... & \xrightarrow[]{\cdot 2} & y\\ \text{hour}:& 0 & \xrightarrow[]{+20} & 20 &\xrightarrow[]{+20} & ... &\xrightarrow[]{+20} & x\\ \end{array}$
1. The number of steps to get from hour $x_0=0$ to hour $x=360$ is $n=(360-0)/20=18$ steps. In each step we multiply by factor $2$ , the amount of cells in hour $360$ is
  $f(360)=6\cdot \underbrace{2\cdot 2\cdot ... \cdot 2}_{n=18 \text{ times}} = 6\cdot 2^{18}$
2. The number of steps to get from hour $x_0=0$ to hour $x$ is $n=(x-0)/20=x/20$ , so the amount of money in week $x$ is
  $y=6\cdot \underbrace{2\cdot 2\cdot \cdot ... \cdot 2}_{n=(x-1)/2 \text{ times}} = \underline{6\cdot 2^{x/20}}$

Exercise 1

In the year $1998$ the number of inhabitants in a village is $5000$ . The number of inhabitants is counted every $12$ years. Find the function equation of the function which gives then number of inhabitants at time $x$ if
1. the number of inhabitants in the year $2010$ is $5150$ and growth is linear.
2. the number of inhabitants in the year $2010$ is $5150$ and growth is exponential.
3. the number of inhabitants triples every $12$ years.
4. the number of inhabitants is divided by three every $12$ years.
The growth of a population is described by the function $f(x)=2x+1$ , where $x$ is years. Determine the increment for every times step $\tau=5$ years.
The growth of a population is described by the function $f(x)=3 \cdot 4^{0.25x}$ , where $x$ is years. Determine the growth factor for every times step $\tau=2$ years.

Solution

$x_0=1998$ , $y_0=5000$ , $\tau=12$ years.
1. Increment $u=5150-5000=150$ , thus $f(x)=\underline{5000+150\cdot \frac{x-1998}{12}}$
2. Growth factor $u=\frac{5150}{5000}=1.03$ , thus $f(x)=\underline{5000\cdot 1.03^{(x-1998)/12}}$
3. Growth factor $u=3$ , thus $f(x)=\underline{5000\cdot 3^{(x-1998)/12}}$
4. Growth factor $u=\frac{1}{3}$ , thus $f(x)=\underline{5000\cdot \left(\frac{1}{3}\right)^{(x-1998)/12}}$
We need to find out by how much the quantity increases in a time step of $\tau=5$ , e.g. from $x=0$ to $x=5$ :
$u=f(5)-f(0)=11-1=\underline{10}$
We need to find out by how what factor we need to multiply the quantity in a time step of $\tau=2$ , e.g. from $x=0$ to $x=2$ :
$u=\frac{f(2)}{f(0)}=\frac{3\cdot 4^{0.5}}{3\cdot 4^0}=\underline{2}$

A fundamental difference between linear and exponential growth is the speed by which a quantity grows. Exponential growth is much faster. Consider the following exercise:

Exercise 2

In 1805, an oak tree has height $5.25 m$ , in 1827 the height is $11.15 m$ . Find the height of the tree in year 2017 under the assumption that

growth is linear, and
growth is exponential.

Which growth model is more realistic?

Solution

It is $\tau=22$ years, $x_0=1805$ , $y_0=5.25 m$ . The diagram is

\begin{array}{rllll} \text{height}:& 5.25 &\xrightarrow[]{} & 11.15 &\xrightarrow[]{} & ... & \xrightarrow[]{} & y\\ \text{year}:& 1805 & \xrightarrow[]{+22} & 1827 &\xrightarrow[]{+22} & ... &\xrightarrow[]{+22} & 2017\\ \end{array}

Linear growth: $u=11.15-5.25=5.9$ , thus
$f(x)=5.25+5.9\cdot \frac{x-1805}{22}$
and thus
$f(2017)=5.25+5.9\cdot \frac{2017-1805}{22}=\underline{62.1 m}$
Exponential growth: $u=\frac{11.15}{5.25}=2.123$ , thus
$f(x)=5.25\cdot 2.123^{(x-1805)/22}$
and thus
$f(2017)=5.25\cdot 2.123^{(2017-1805)/22}=\underline{7453.5m}$
Clearly, linear growth is more realistic.