Linear and exponential growth

Science and society, grade 11 (3 units) Lesson 6 of 16Learning

Linear and exponential growth

Linear growth and exponential growth

In linear growth the same number $\displaystyle d$ is added to the amount in each unit of time. In exponential growth it is multiplied by the same number $\displaystyle q$ in each unit of time.

\displaystyle M_t=M_0+d\cdot t

Two towns have 10,000 residents each.
In town A 500 residents are added every year: $\displaystyle 10000+500t$ .
In town B the number of residents grows by 4% every year: $\displaystyle 10000\cdot1.04^{t}$ .

Years	0	1	2	3
Town A	10,000	10,500	11,000	11,500
Town B (rounded)	10,000	10,400	10,816	11,249

In the first years town A grows faster.

Equal differences and equal ratios

In a table of times at equal steps:

in linear growth the difference between each value and the one before is fixed: $\displaystyle {10500-10000}={11000-10500}={500}$ ;
in exponential growth the ratio of each value to the one before is fixed: $\displaystyle {\frac{10400}{10000}}={\frac{10816}{10400}}={1.04}$ .

Exponential growth overtakes

In linear growth the amount added each year is fixed. In exponential growth the amount added itself grows, since 4% of a larger amount is more. So in the end exponential growth overtakes any linear growth.

Years	11	12	20	30
Town A	15,500	16,000	20,000	25,000
Town B (rounded)	15,395	16,010	21,911	32,434

After 12 years town B is larger, and from then on the gap only grows.
After 30 years town B has almost 7,500 more residents than town A.

A straight-line graph and an exponential graph

The graph of linear growth is a straight line, and the graph of exponential growth rises faster and faster.

The graph: $\displaystyle {y}={100+100x}$ – a straight line, and $\displaystyle {y}={100\cdot2^{x}}$ – an exponential graph.

\displaystyle y=100+100x

\displaystyle y=100\cdot2^{x}

\displaystyle (0,100)

\displaystyle (1,200)

\displaystyle (4,500)

\displaystyle (4,1600)

The two graphs meet at two points: $\displaystyle (0,\ 100)$ and $\displaystyle (1,\ 200)$ .
Between $\displaystyle {x}={0}$ and $\displaystyle {x}={1}$ the straight line is a little above the exponential graph.
After $\displaystyle {x}={1}$ the exponential graph is above the line, and the gap grows: at $\displaystyle {x}={4}$ , 1,600 against 500.

Exercise 1

Two amounts change with time:

Time	0	1	2	3
Amount A	20	26	32	38
Amount B	20	30	45	67.5

Which amount grows linearly and which exponentially? Explain with differences and ratios.

What will amount A be at time 4?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

What will amount B be at time 4?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

The graph: $\displaystyle {y}={100+100x}$ and $\displaystyle {y}={100\cdot2^{x}}$ .

Mark the points where the two graphs meet.

\displaystyle y=100+100x

\displaystyle y=100\cdot2^{x}

Exercise 3

Bacteria grow in two dishes in a laboratory.
Dish A has 1,000 bacteria, and 1,000 are added every hour.
Dish B has 100 bacteria, and their number doubles every hour.

How many bacteria will dish A have after 7 hours?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

How many bacteria will dish B have after 7 hours?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Which dish will have more bacteria after 10 hours? Explain without working out every hour.

Exercise 4

The two towns from the explanation: town A has $\displaystyle 10000+500t$ residents, and town B $\displaystyle 10000\cdot1.04^{t}$ .

Write an equation and solve it: after how many years will town A have 16,000 residents?

How many residents will town B have at that time? Round to a whole number.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Which town is larger at that time, and what will happen to the gap between the towns in the years after?

Formulas: science and society, grade 11 (3 units)

Open as a page

Growth and decay

An exponential process – $\displaystyle M_0$ the amount at the start, $\displaystyle q$ the factor the amount is multiplied by in each unit of time, $\displaystyle M_t$ the amount after $\displaystyle t$ units of time
$\displaystyle M_t=M_0\cdot q^{t}$
The growth factor – a growth of $\displaystyle p$ percent in each unit of time
$\displaystyle q=1+\frac{p}{100}$
The decay factor – a decay of $\displaystyle p$ percent in each unit of time
$\displaystyle q=1-\frac{p}{100}$
Decay by half-life – $\displaystyle n$ the number of half-lives gone by
$\displaystyle M_t=M_0\cdot\left(\frac{1}{2}\right)^{n}$
The number of half-lives – $\displaystyle t$ the time gone by, $\displaystyle T$ the half-life
$\displaystyle n=\frac{t}{T}$
Linear growth – the amount grows by $\displaystyle d$ in each unit of time
$\displaystyle M_t=M_0+d\cdot t$

Measures of spread

The mean of $\displaystyle n$ data
$\displaystyle \bar{x}=\frac{x_1+x_2+\dots+x_n}{n}$
The range – the greatest datum minus the least
$\displaystyle x_{\max}-x_{\min}$
The deviation of the datum $\displaystyle x_i$ from the mean
$\displaystyle x_i-\bar{x}$
The standard deviation of $\displaystyle n$ data
$\displaystyle \sigma=\sqrt{\frac{(x_1-\bar{x})^2+(x_2-\bar{x})^2+\dots+(x_n-\bar{x})^2}{n}}$
The standard deviation from a frequency table – $\displaystyle f_1, \dots, f_k$ the frequencies of the values $\displaystyle x_1, \dots, x_k$
$\displaystyle \sigma=\sqrt{\frac{f_1(x_1-\bar{x})^2+f_2(x_2-\bar{x})^2+\dots+f_k(x_k-\bar{x})^2}{f_1+f_2+\dots+f_k}}$

Probability

The complement $\displaystyle \overline{A}$ – $\displaystyle A$ does not happen
$\displaystyle P(\overline{A})$ $\displaystyle =$ $\displaystyle 1-P(A)$
A path of a tree diagram – the product of the probabilities on its branches
$\displaystyle P=p_1\cdot p_2\cdot p_3$
“At least once” in $\displaystyle n$ independent experiments – $\displaystyle p$ the probability that it happens in each
$\displaystyle 1-(1-p)^{n}$
Conditional probability – the probability of $\displaystyle A$ when $\displaystyle B$ is known to have happened; $\displaystyle A\cap B$ – both $\displaystyle A$ and $\displaystyle B$
$\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}$
The probability of “both $\displaystyle B$ and $\displaystyle A$ ”
$\displaystyle P(A\cap B)=P(B)\cdot P(A|B)$
Independent events – knowing that $\displaystyle B$ happened does not change the probability of $\displaystyle A$
$\displaystyle P(A|B)=P(A)$
The expected value – $\displaystyle x_1, \dots, x_k$ the possible values, $\displaystyle p_1, \dots, p_k$ their probabilities
$\displaystyle E=x_1p_1+x_2p_2+\dots+x_kp_k$