Percentage growth

We often express the growth as a percentage. The percent increase or decrease per time step is called the growth rate. For example, assume that in year $2000$ there are $120$ trees in the forest, and each $5$ years the number of trees increases by $2\%$ of the previous number of trees. What type of growth is this? Let's see. After $5$ years, the number of trees increases by $2\%$ of $120$ trees, that is, by

0.02\cdot 120=2.4

trees. So the number of trees in $2005$ is

120+2.4=122.4

After another $5$ years, the number of trees increases by $2\%$ of the $120.4$ trees, that is, by

0.02\cdot 122.4=2.448

So the total number of trees in $2010$ is

122.4+2.448=124.848

and so on. So clearly this is not linear growth, as the increment in not constant from one step to the next. Indeed, it is actually exponential growth, because in each step we multiply by

u=\frac{122.4}{120}=\frac{124.848}{122.4}=1.02

\begin{array}{rllll} \text{trees}:& 120 &\xrightarrow[]{+2\%, \cdot 1.02} & 122.4 &\xrightarrow[]{+2\%, \cdot 1.02} & 124.848 & \xrightarrow[]{+2\%, \cdot 1.02} & ... & \xrightarrow[]{+2\%, \cdot 1.02} & y\\ \text{year}:& 2000 & \xrightarrow[]{+5} & 2005 & \xrightarrow[]{+5} & 2010 & \xrightarrow[]{+5} & ... &\xrightarrow[]{+5} & x\\ \end{array}

We could go on with more examples, but the idea should be clear. In general we have the following:

Theorem 1: Percentage growth

Proof

The general proof is similar to above. Assume the quantity is $y_0$ at the start time $x_0$ . After one time step the quantity increases by $p\%$ of $y_0$ , which is $\frac{p}{100}y_0$ . Thus we have

\begin{array}{lll} y_1 &=& y_0+\frac{p}{100}y_0\\ &=&(1+\frac{p}{100})\cdot y_0 \end{array}

In the next step the quantity increases by $p\%$ of $y_1$ , which is $\frac{p}{100}y_1$ . Thus we have

\begin{array}{lll} y_2 &=& y_1+\frac{p}{100}y_1\\ &=&(1+\frac{p}{100})\cdot y_1 \end{array}

and so on. So we see that $u=1+\frac{p}{100}$ . The proof for the decay works in a similar way.

Consider a quantity that increases after each time step by $p\%$ . Then this is exponential growth with growth factor

u=1+\frac{p}{100}

If the quantity decreases by $p\%$ after each time step, we get exponential decay, and the growth (or decay) factor is

u=1-\frac{p}{100}

Exercise 1

Number of bacteria in a Petri dish culture increase at a rapid pace. At the beginning (at $0 h$ ) there are $100$ bacteria, and every $2 h$ , their number increases by $10\%$ from the previous number. Determine an expression for the number of bacteria at time $x$ .
You bought a boat in the year $2000$ for $50 000$ CHF. Every third month it loses $0.1\%$ of its previous value. Determine an expression for the value of the boat at time $x$ (where $x$ is measured in years).

Solution

Increase by $p=10\%$ , thus, the growth factor is $u=1+\frac{10}{100}=1.1$ . Thus, $f(x)=\underline{100\cdot 1.1^{x/2}}$ .
Decrease by $p=0.1\%$ , thus growth (or decay) factor is $u=1-\frac{0.1}{100}=0.999$ . Every third month is every quarter of a year, so the time step is $0.25$ . Thus $f(x)=\underline{50 000\cdot 0.999^{(x-2000)/0.25}}$ .